SyllabusMATH4210 Financial Mathematics

Lecturer: Prof. KAZUFUMI ITOOffice: Room 227, LSB Email:kito@math.cuhk.edu.hk

Lecture Notes: By Prof. Raymond Chan

Description The course will cover the following topics

— Basic Option Theory: Financial Markets; What is an Option? What are Options For?Types of Options; Put & Call Options; Trading of Options; American & European Options;Interest Rates.

— Stochastic Tools: Normal Distributions; Wiener Processes (Brownian motion); PoissonProcess, Stochastic Differential Equations; Ito’s Lemma; etc.

— Black-Scholes Model: Black-Scholes Analysis; Black-Scholes Equations; Black-ScholesFormulas; Boundary and Final Conditions; Hedging in Practice; Implied Volatility; Risk-neutral Valuation.

— Numerical Methods for Option Pricing

Assessment Scheme

— Six homework assignments 15%— One programming assignment (in any language you like) 5%— Two quizzes 20%— Final examination 60%

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Forward contract— A binding agreement where one party promises to buy an asset from

another party at some specified time in the future and at some specified price.— No money changes hands until the delivery date or maturity of the

contract. The terms of the contract make it an obligation to buy the assetat the delivery date,

— No negotiation in the matter. The asset could be a stock, a commodityor a currency.

Example 2.On May 31, you receive an acceptance letter from Oxford University for

the coming September. The tuition fee, payable on September 1, is 1,000GBP. This is equal to HK$15,400 on May 31 (the current exchange rate 1.00GBP = HK$15.40). You have found a summer job that will pay HK$5,200each month for the months of June to August, i.e. by August 31, you willhave HK$15,600. What should you do then?

1.Try your luck: If the exchange rate is less than HK$15.60 by August 31,you earn some extra cash besides the registration fee. If the exchange rate ishigher than HK$15.60, you apply for Chinese University instead.

2.Buy a forward on British pounds: Look for forward contracts on Britishpounds that entitles you to use HK$15.600 to buy 1,000 GBP on August 31.Then you have eliminated, or hedged, your risk. Forward contracts eliminateyour risk. But why would someone want to buy HK$15,600 with 1,000 GBPon August 31? Consider the following example.

Example 3.Suppose it is May 31, and the Chief Financial Officer (CFO) of a corpora-

tion knows that the corporation will receive 1 million GBP in three months(on August 31) and wants to hedge against exchange rate moves. The CFOcan contact a bank, find out that the exchange rate for a three-month forwardcontract on pounds sterling is, say 15.600, and agree to sell 1 million GBP.The corporation then has a short forward contract on sterling, i.e. it hasagreed that on August 31, it will sell 1 million GBP to the bank for HK$15.6million. The bank has a long forward contract on sterling, i.e. it has agreedthat on August 31 it will buy 1 million GBP for HK$15.6 million. Both sideshave made a binding commitment.

We see from the two examples that forward contracts eliminate the risk of

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the parties involved. Note that neither party has to pay any money upfront.So the forward contract is worth $0 at the beginning. But between theinitial time and the expiration date, the value of the contract will change.For example if on June 30, the exchange rate is HK$20 =1GBP, then yourcontract in Example 2 may worth around HK$4,400. You can sell the contractfor this amount to anyone who is thinking the exchange rate of pounds will goup further. This is in general true for any derivatives (forwards and options):their values will fluctuate between now and the expiration date, so that onecan earn money by selling the derivatives itself without waiting to exercisethem on the expiration date.

European call option— At a prescribed time in the future, known as the expiration date T .— Holder of the option may purchase a underlying asset for a prescribed

amount, known as the strike price E.— Up-front the holder pay the premium c.— Payoff c(T ) = max(0, S(T )− E).

Example

The current price of ABC Corp stock is $45 per share, and investor ’Greg’expects it will go up significantly. Greg buys a call contract for 100 sharesof ABC Corp from ’Terence,’ who is the call writer/seller. The strike pricefor the contract is $50 per share, and Greg pays a premium up front of $5per share, or $500 total. If ABC Corp does not go up, and Greg does notexercise the contract, then Greg has lost $500.

ABC Corp stock subsequently goes up to $60 per share before the contractexpires. Greg exercises the call option by buying 100 shares of ABC fromTerence for a total of $5,000. Greg then sells the stock on the market atmarket price for a total of $6,000. Greg has paid a $500 contract premiumplus a stock cost of $5,000, for a total of $5,500. He has earned back $6,000,yielding a net profit of $500.

If, however, the ABC stock price drops to $40 per share by the time thecontract expires, Greg will not exercise the option (i.e., Greg will not buy astock at $50 per share from Terence when he can buy it on the open marketat $40 per share). Greg loses his premium, a total of $500. Terence, however,keeps the premium with no other out-of-pocket expenses, making a profit of$500.

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The break-even stock price for Greg is $55 per share, i.e., the $50 pershare for the call option price plus the $5 per share premium he paid for theoption. If the stock reaches $55 per share when the option expires, Greg canrecover his investment by exercising the option and buying 100 shares of ABCCorp stock from Terence at $50 per share, and then immediately selling thoseshares at the market price of $55. His total costs are then the $5 per sharepremium for the call option, plus $50 per share to buy the shares from Greg,for a total of $5,500. His total earnings are $55 per share sold, or $5,500 for100 shares, yielding him a net $0. (Note that this does not take into accountbroker fees or other transaction costs). Note, however, that the $5 per sharepremium is a sunk cost that he has already paid regardless of whether hechooses to exercise the call. Thus, while he breaks even at $55 per share,actually, he makes a $5 per share profit that covers the earlier expense of $5per share. Thus it is in his interest to exercise the call if the price goes above$50, even if it does not reach $55, because the profit he makes will reduce hisnet loss.

Factors to determine the value (price) c(t) of a call option

1. S(t), today’s share value. The higher the share price is now then thehigher the price is likely to be in the future.

2. E, the strike price. The lower the exercise price is, the less the investorhas to pay on exercise, and hence the higher the option value should be.

3. T , the expiry date. If T is large, we have more chance of seeing theshare value rises above E. Therefore the option price should be higher.

4. r, the interest rate. The option price is usually paid for up-front at theopening of the contract, whereas the payoff, if any, does not come until later.The option price should reflect the income that would otherwise have beenearned by investing the premium c(t) in the bank.

5. D, the dividend. It will be seen in Chapter 3 that the stock price willdecrease right after dividend are issued.

6. σ, the volatility, which measures randomness of the asset price.

Thus, the value of a call option is in fact a function of several variables

c(t) = c(S(t), E, T − t, r,D, σ).

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An arbitrage opportunity— is a situation where one can have the possibility of earning a positive

return without any risk. An arbitrage opportunity is a trading opportunitythat either (1) takes a negative amount of cash to enter (i.e. cash flow toyour pocket is positive) and promises a non-negative payoff to leave; or (2)takes a non-positive amount of cash to enter (cash flow to you is either zero orpositive) and promises a non-negative payoff with the possibility of a positivepayoff when leaving. If an arbitrage opportunity exists, an investor can havea positive probability of a risk less return.

Lemma 1 The option price cannot be negative, i.e. c(t) ≥ 0 for al t ≤ T .

Proof: We prove it by contradiction. If c(t) < 0, then that means some onewill pay us −c(t) > 0 amount of money to own the option (positive cash flowto enter the deal). At the expiry of the option, we are certain that the payoffis nonnegative because as owner of the option, we can choose to exercise it ornot. Hence it is precisely case 1b arbitrage opportunity. Since by assumption,there should not exist such an opportunity, we have a contradiction. Hencethe only conclusion is that c(t) ≥ 0.

Suppose with equal probabilities the ABC share price takes the values $40or $60 The expected payoff is given by:

Expected payoff =1

2× 0 +

1

2× 10 = 5

Forget about the interest rate for the moment. It seems reasonable that thevalue of the option (premium) should be around $5.

Lemma 2. If it is possible for the stock price to exceed the strike price at theexpiry date, then the call option price c(t) > 0, i.e.

Prob(S(T ) > E) > 0⇒ c(t) > 0.

Lemma 3 If the stock price is zero at some time t0, then the stock price hasto be zero for all time t ≥ t0.

Lemma 4 If the stock price is zero, then the value of a European call optionmust be zero too, i.e.,

S(t) = 0⇒ c(t) = 0.

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The mathematics in this course will emphasize two financial concepts thathave had a startling impact over the last three decades on the way the finan-cial industry views derivative trading: We will emphasize investments thatreplicates equities, and we will explore mathematical models of how equitiesbehave in the absence of arbitrage opportunities. The combination of thesetwo concepts furnishes a powerful tool for finding prices.

From this simple example (E = 250, c = 10) we can see why someonewould prefer to buy an option rather than the underlying asset. Supposethat the holder does pay $10 premium for this option, and if the share pricerises to $270 at T , then he has made a net profit of $10 ($20 payoff onexercising less $10 the cost of the option). Observe that this net profit of $10is 100% of the up-front premium. However, the downside of this speculationis that if the share price is less than $250 at expiry date T , the loss is also$10, which is 100% of what he invested in the option. Say, if the investorpurchases instead the share for $250 on January 11, the corresponding profitor loss of $20 will be only 20/250 = 8% of the original investment

Option prices thus respond in an exaggerated way to changes in the un-derlying asset price. This effect is called gearing or leverage.

Put Options The right to sell an asset is known as a put option with thepayoff

p(T ) = max(0, E − S(T ))

and its payoff properties are opposite to those of a call.A put option can serve as a protective insurance. For illustration, see the

following example.

Example 3 Suppose that an investor owns a substantial amount of telecom-munication stocks of XYZ company, which is now trading at $50 per share.The investor believes that the stock price is likely to vary widely in the monthsahead, and hence he wishes to begin selling this stock soon. But since he hasa substantial amount of the stocks, by selling them, the stock price is likely toplumbet. What can the investor do? Well, he can begin a program of buyingputs of approximately three months to expiration, say, with strike price $45.He has to pay an option premium, which, for such a case, costs $2.80 for eachput. This strategy protects the investor in the following way: a future lowstock price will cause the puts to be exercised, and hence it will allow the

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investor to obtain at least $45 for each share called away by a put. If thestock price stays above the $45 threshold, the put expires worthless. So, inany case, the investor is guaranteed a minimum price for shares of the stockhe sells, as long as he owns puts for these shares. The $2.80 expense for eachput may be regarded as an ”insurance premium” that guarantees the abilityto sell some of the investor’s stock at or above the level $45.

Long and Short

The term short sell means that we borrow shares and sell them out to themarket immediately, and in the future we buy back the same amount ofshares and return them. This is a trading strategy that yields a profit whenthe price of a security goes down and a loss when it goes up.

Example 4 Consider the position of an investor who shorts 500 IBM sharesin July when the price per share is $120 and closes out his or her positionby buying them back in October when the price per share is $100. Supposethat a dividend of $4 per share is paid in August. The investor receives500×$120 = $60000 in July when the short position is initiated. The dividendleads to a payment by the investor of 500 × $4 = $2000 in August. Theinvestor also pays 500 × $100 = $50000 when the position is closed out inOctober. The net gain is, therefore,

$60000− $2000− $50000 = $8000.

Essentially, when speaking of stocks, long positions are those that areowned and short positions are those that are owed. An investor whoowns 100 shares of XYZ stock is said to be long 100 shares. This investor haspaid in full the cost of owning the shares. An investor who has sold 100 sharesof XYZ stock without currently owning those shares is said to be short 100shares. The short investor owes 100 shares at settlement and must fulfill theobligation by purchasing the shares in the market to deliver. Oftentimes, theshort investor borrows the shares from a brokerage firm in a margin accountto make the delivery. Then, with hopes the stock price will fall, the investorbuys the shares at a lower price to pay back the dealer who loaned them.

When an investor uses option contracts in an account, long and shortpositions have slightly different meanings. Buying or holding a call or putoption is a long position because the investor owns the right to buy or sellthe security to the writing investor at a specified price. Selling or writing

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a call or put option is just the opposite and is a short position because theinvestor owes the holder the right to buy the shares from or sell the sharesto him at the holder’s discretion.

——————————————————Regardless of the direction of a stock, when the price changes, some will

make money while others will lose. The reason for this is the differencebetween buying long and selling short. Our most common conception ofinvesting in stocks is to buy while the price is low and sell when the priceis high. However, there are many investors (typically with a great deal ofmoney at work in the market) that do the reverse - sell high and buy low.

Buying Long Unless you’re an eternal pessimist, buying a stock in the hopesthat its share price will appreciate is a good way to invest in the market.After all, stocks have the greatest return of all major asset classes, averaginganywhere from 8% to 17% depending on the time frame surveyed. Since theodds are generally in the investor’s favor over time, buying today and waitingfor the share price to increase down the road seems to be a good bet. As aresult, the overwhelming majority of investors buy long positions in stocksand hope to receive a good return over a long period of time.

Selling Short For pessimists and opportunists, selling short is a valuableway to play the market. As noted above, selling short has the goal of sellingshares at a high price and buying them back at a low price. This begs thequestion: How can you sell shares that you don’t own? The answer is thatyou borrow them.

When you open a brokerage account (commission, fee) with a company likeFidelity, E-Trade, or Schwab, buried in the fine print is usually a provisionthat allows the company to ’lend’ your shares to other account holders. Whileyou don’t really lose possession of the shares, your ownership stake allows ashort seller to borrow the shares and sell them in the open market.

During this period where the shares are on ’loan’, you won’t see anychanges on your statement, but the short seller will have a short positionin the borrowed stock. During this period of time, the short seller will payinterest to the brokerage company for use of the borrowed stock. These ratescan vary greatly, but usually aren’t too far from the prime rate (for largeaccounts, anyway).

If the stock drops as the short seller hopes, he or she will then ’buy to cover’

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which will close out the position. The proceeds from the sale are then takenagainst the cost of the purchase plus the interest charges. If the proceeds aregreater than the costs, the short seller will make a profit.

Application: Buying Long As an example, let’s suppose we purchase 100shares of XYZ stock for $20 per share. This means that the shares have atotal cost basis of $2,000. A year later, the shares are worth $40 per share.This results in a profit as calculated here:

$4, 000 sale proceeds − $2, 000 cost basis = $2, 000 profit.

Application: Selling Short Taking the same process in reverse, we sell 100shares of XYZ stock for $40 per share and receive $4,000 in sale proceeds.One year later, the stock has fallen to $20 per share and 100 shares are boughtto cover for a total of $2,000 in cost basis. During the holding period, interestwas charged in the amount of $180. This results in a profit shown below:

$4, 000 sale proceeds − $2, 000 cost basis − $180 interest = $1, 820 profit.

As you can see from these two calculations, the only difference in the mathis the added cost of interest for short selling, but it’s the same concept forboth - sale proceeds minus costs equals profit.

Wrapping Up From this article, you should remember that buying long isthe most prevalent method of investment in the stock market. This has manyreasons, but the best one is that the market pays quite well over time. Shortselling is simply the reverse of buying long, but with an important difference.Short sellers must pay interest on borrowed shares.

Critical Thinking If you plan to invest your money for retirement, whichinvestment strategy is best - buying long or selling short? Are short sellersmore likely to be long-term investors or short-term traders? Why do shortsellers exist in the marketplace? Do they serve an important function? Often,when a stock on the decline has been sold short in large quantities, shortsellers must eventually buy to cover so they can cash in on their profits andstop paying interest. What will happen to the share price if all of these shortsellers buy to cover at the same time? At least in theory, stock prices can goup forever, but can only drop to zero. What is the short seller’s maximumprofit opportunity?

4. Why Would Anyone Buy Options? Options have two primary uses:(i) speculation, and (ii) hedging.

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In a more general setting, hedging is used to guarantee that the lost in onefinancial product is compensated by the gain of another financial product.As an example, observe that the value of a put option p(t) rises when theunderlying asset price S(t) falls. What happens to the value of a portfolio(i.e. an investment plan) containing both assets and puts. Buy x units ofXYZ shares at S(t) and buy y units of put on the XYZ shares at p(t). Theportfolio value, denoted by Π(t) = xS(t) + y p(t). When S(t) increases, p(t)decreases; when S(t) decreases, p(t) increases. A natural question one wouldlike to ask is for which (x, y) the risk of the above portfolio is minimized. Theactivity of finding such (x, y) is called hedging.

5, Why Would Anyone Write an Option?

Instead, their profit comes from selling at slightly above the ”true value”and buying at slightly below, the less risk associated with this policy, thebetter. If a market maker can sell an option for more than it is worth andthen hedge away all the risk for the rest of the option’s life, he has locked ina guaranteed, risk-free profit. This idea of reducing or eliminating risk bringsus back to the subject of hedging.

Example 5 Suppose a market maker has $100 to invest for the next threemonths. He decides to use the money to buy a stock at S(t) = 100 Thenhe will lose money if the stock price falls, see Figure 2 (left). To protectedagainst the possible lost, he can buy a put as in Example 3. But it is morefun to write a call option say with exercise price $105 and expiry date threemonths from now. The option is out-of-money now, but it still has the timevalue. Say the fair price for the option is $3 and the market maker can sellit above the fair price at $4. Then the payoff at expiry for the market makeris given in Figure 2 (right). Notice that by writing a call, the market maker

1. is insured from the downside with $4 to spare; i.e., the breakeven pointis shifted from $100 down to $96, and

2. earns more than he should if S(T ) ≤ 109.Of course, the market maker has forfeited part of the upside?when S(T ) >

109, he would earn less than he should have if he had not written the call.What if S(T ) < 96. The market maker will lose money eventually. One wayto avoid that is to sell the stock whenever S(t) drops below $96 and buyit back when it gets back to $96. If there is no transaction cost and thereis no sudden jump in the stock price, then the market maker is completely

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protected against the downside.

Portfolios A portfolio is a combination of buying and selling certain fi-nancial products. That is, a portfolio is an investment plan. The value of aportfolio, denoted by Π(t) is the sum of the values of all financial productswithin the portfolio. This will also be the amount of money a third party paidto the holder of the portfolio so that the holder can transfer the ownershipof the portfolio to the third party.

Example 6 Let us look at the following simple portfolios:

(i) Consider a portfolio consisting of just 1 share at price S(t). The value ofthis portfolio is Π(t) = S(t), which is positive. That means you need moneyto set up (i.e. own) such a portfolio, and earn money when selling it.

(ii) You have just borrowed E amount of money from the bank, then Π(t) =−E. The portfolio in this case is the mortgage payment agreement. Its valueis negative as you earn money when setting it up, but you need to pay moneyto disown it. The value of the portfolio will become more negative as timegoes by because of the mortgage interest. The negative in value means thatyou owe some obligation by having or setting up this portfolio.

(iii) You have just short sold 1 share at price S(t). If you hold this portfolioand you want to transfer the ownership of it to a third party. Then you haveto pay the third party S(t). Namely, the third party has to pay me −S(t) inorder to transfer the ownership. So in this case, the value for this portfolio isΠ(t) = −S(t). Again the negative portfolio price means that the holder owessome obligation. If I am getting this portfolio, I am getting your obligation.Hence I should get money from you when I get this portfolio.

(iv) You have just bought 1 share at price S(t) and 1 call option c(t). Thevalue for this portfolio at t is Π(t) = S(t) + c(t). The value of this portfoliois positive. That means you need money to set this up.

(v) As in Example 5, you have just bought 1 share at price S(t) and sold(or written) 1 call option c(t). The value for this portfolio at t is Π(t) =S(t)− c(t). One important point to remember is if Π(t) > 0, like Example 7(i), then there is a positive cash flow to the holder if he wants to sell it, andthere is a negative cash flow to the buyer or anyone wanting to set up sucha portfolio, i.e. one has to pay to own or set up the portfolio. On the otherhand, if Π(t) < 0, like Example 7 (ii), then there is a negative cash flow for

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the holder if he wants to sell it, i.e. he has to pay money to the buyer inorder to sell it. However, there will be a positive cash flow to the buyer ofsuch portfolio, or more precisely, one earns money in order to buy or set upsuch a portfolio.

Example 7 Consider this more useful example. Suppose that a local XYZcompany is negotiating a joint venture project with PCCW in January, ofwhich the outcome and the details of the contract will be made known toinvestors in late September. It is predicted that the XYZ share price willeither rise or fall by more than 10% after the announcement is made. Say,today?s XYZ share price is $10, and European calls and puts are availablewith expiry dates in April, July, October, December and with strike pricesof $10 plus or minus $0.5, respectively. Under such a circumstance, whatportfolio of options will do well? Consider buying a European call c withstrike price at $9.50, maturity in October; and a European put p with strikeat $10.50 at the same expiry date. Such a portfolio (with a call and a put onthe same expiry date) is called a strangle. The value of this portfolio is

Π(t) = c(S(t), 9.5, T − t, r) + p(S(t), 10.5, T − t, r))

At T (in October), the terminal payoff is given by

Π(T ) = max(0, S(T )− 9.5) + max(0, 10.5− S(T )).

It is plotted in Figure 3, where again we do not consider the premium ofbuying this portfolio upfront, which should be something more than $1 if theinterest rate is zero (why?). We see that this portfolio speculates a big swingin the stock price far above or below its current price of $10. What if webuy one call with exercise price at $10.5 and one put at $9.5 instead? Canyou draw the payoff diagram as in Figure 3? What can you say about thepremium in this case? We will study strangles and other combinations ofoptions in Chapter 5.

Proposition 1 The current stock price S(t): S(t) ≥ c(t), the call optionprice c(t).Proof: Assume

Π(t) = S(t)− c(t) < 0

But,Π(T ) = S(T )− c(T ) = S(T )−max(0, S(T )− E) ≥ 0

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which implies arbitrage.

Proposition 2 For 0 ≤ E1 ≤ E2

0 ≤ c(S(t), E1, τ, r)− c(S(t), E2, τ, r) ≤ E2 − E1

0 ≤ p(S(t), E2, τ, r)− p(S(t), E1, τ, r) ≤ E2 − E1

Proposition 3 c(t) ≥ S(t)− E.Proof: Suppose there exists a time t such that c(t) < S(t) − E. Then attime t, we construct a portfolio of buying one stock and selling one call, sothe value of the portfolio at time t is Π(t) = c(t)− S(t) < −E. At maturity,we have Π(T ) = max(S(T )− E, 0)− S(T ) = −min(E, S(T )) ≥ −E. HenceΠ(T ) − Π(t) > 0 and an arbitrary opportunity exists. According to theprevious part, S(t) = 82, E = 80 and c(t) = 1.5 violates c(t) ≥ S(t) − E.Hence an arbitrage opportunity exists and one can lock in a riskless profit atmaturity by buying one put and one stock at time t.

American Options European options can only be exercised at expiry. Be-sides European options, many options nowadays are what is called American.An American option is one that may be exercised at any time prior to expiryby its holder. Mathematically speaking, American options are more inter-esting since they can be interpreted as free boundary problems. A questionthat the holder of an American option must determine is when it is best timeto exercise the option. We will see that the ”best” time to exercise is notsubjective, but that it can be determined in a natural and systematic way(Optimal Stopping Time).

European and American call and put options form a small section of theavailable derivative products. They are called vanilla options because theyare ubiquitous and simple. Nowadays the trade in the vanilla options is sogreat that it can, in some markets, have a value in excess of that of the tradein the underlying assets. There are however other more complicated options,including the so-called exotic or path-dependent options. These options havevalues which depend on the history of an asset price, not just on its valueon exercise. An example is an option to purchase an asset for the arithmeticaverage value of that asset over the month before expiry. An investor mightwant such an option in order to hedge sales of a commodity, say, which occurcontinually throughout this month. There are also options which depend on

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the geometric average of the asset price, the maximum or the minimum ofthe asset price. For instances,

Digital option An option that at maturity pays a fixed amount if in-the-money, zero otherwise. A digital option is also called a binary option.

Barrier options An option which can either come into existence or becomeworth-less if the underlying asset reaches some prescribed value before expiry.Asian options the price S(T ) depends on some form of average.

— S(T ) replaced by the average = 1T

∫ T0 S(t) dt, it is called Asian price

option— If E is replaced by the average, it is called Asian strike option

Look back options The price depends on the asset price maximum or mini-mum.

Example 8 Warrants in the HK Stock Exchange are like European options.They can only be exercised at the expiry date. Of course, one can sell thembefore expiry if the price of the warrant is already high enough. One questionis how one computes the stock price at expiry. The use of the closing priceof the stock on the expiry date may subject the stock price and hence thewarrant price to big fluctuation. Thus HKEx uses the average of the closingprices in the final 5 days as the price of the stock at T when computing thepayoff of the warrant. In this sense, warrants are like Asian price options.The term ”gearing” also refers to the ratio between a company’s stock priceand the price of its warrants.

Chapter 3 Interest Rates and Forward

Let M(t) be the current balance at time t. Recall the compounded m

times with annual interest rate r is given by

M(t)(1 + r/m)m →M(t)er.

after the one year. Let us consider a small subsequent time interval (t, t+∆t)during which the balance M(t) becomes M(t+ ∆t). The return now can beexpressed as

M(t+ ∆t)−M(t)

M(t)∼ r(t)∆t.

As ∆t→ 0dM/M = r(t) dt.

It is clear that r(t) gives a measure of the average rate of growth of the bank

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deposit. Equivalently, (3) can also be expressed as dMdt = rM(t) and

M(T ) = M(t)exp(

∫ T

t

r(s) ds)

Note that a promise of a payment of E at time T in the future, no matterhow sincerely made, is not worth E today. A lesser amount V today couldbe invested safely and would grow to reach E at time T . Then V is calledthe present value of E. In fact, from (5), we have

V = Eexp(

∫ T

t

r(s) ds), V = Ee−r(T−t).

Proposition 1 Assume that no dividend payment is made over the lifetimeof options. Then c(S(t), E, τ, r) ≥ S(t)− Ee−rτ .

Proof. Again we use a no arbitrage argument. Suppose by contradictionthat Proposition were false, i.e., c(S(t), E, τ, r) − S(t) < −Ee−rτ . Form aportfolio: short sell one share, buy a call, and put Ee−rτ in bank. Thus,

Π(t) = c(t)− S(t) + E−rτ < 0

the terminal payoff is given by

Π(T ) = E − S(T ) + max(0, S(T )− E) = max(0, E − S(T )) ≥ 0.

Suppose a five-year zero rate with continuous compounding is quoted as5% per annum. This means that $100, if invested for five years, grow to$100 × e5×.05 = 128.4. Conversely the n-year zero rate rn can be computedfrom the current price V of a zero-coupon bond to be mature exactly n yearsfrom now. In fact, at maturity, since we get back the face value, say $1, wehave V enrn = 1,

rn = − logV

n.

Example 2 Consider the five bonds in Table 1, where for the coupon bonds(Bonds 4 and 5), the coupons are paid every six months. Because the firstthree bonds are zero-coupon bonds, the zero rates corresponding to the ma-turity of these bonds can be computed via (12). For example, the 0.25-year

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zero rate is given by −ln(0.975)/.25 = 10.127%. This gives the first threerows of Table 2. Table 1. Data for Bootstrap Method

The 1.5-year zero rate R is computed from the fourth bond. Note thatthe bond has $4 payments after 6 months, $4 after 1 year, and $104 after1.5 year. Using the zero rates in the first three rows of Table 2, we add thepresent values of all these payments and equate the sum to the present valueof the bond:

4e0.10469×.5 + 4e0.10536×1. + 104e−R×1.5 = 96.

This gives R = 0.10681 = 10.681% This is the only zero rate that is consistentwith the six-month rate, one-month rate, and the data in Table 1.

Once we have the 1.5-year zero rate, we can repeat the same technique tocompute the 2-year zero rate R from the fifth bond, which is given by theequation:

6e0.10469×.5 + 6e0.10536×1.0 + +6e0.10681×1.5 + 106e−R×2.0 = 101

The answer is R = 10.808%

Bond Yield and Forward Rates When we buy a bond at a market price differ-ent from the bond?s face value, three numbers are commonly used to measurethe annual rate of return we get on our investment:

1. Coupon rate: the periodic payment measured annually, as a percentageof the bond’s face value;

2.Current yield: the annual payout as a percentage of the current marketprice of the bond;

3. Yield-to-maturity: the percentage rate of return paid on the bond if itis bought and held to its maturity date.

The yield-to-maturity, also called the bond yield, is perhaps the best measureof the return rate. It is obtained by discounting. For a bond, the yield-to-maturity is the rate R that, when used to discount all future payouts fromthe bond to their present value, will yield the present price of the bond. Oncewe know the present market value V (t) for the bond at the present time t, Rcan be determined from

V (t) =∑

Π(Ti)e−R(Ti−t),

where Π(Ti) is the payoff size to occur at some future time Ti > t.

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Example 3 Consider the fifth bond in Example 2. Its face value is $100 andcurrent value is $101.6. Since it pays $6 every six months, its coupon rate is6% semi-annually, or 12% annually. Its current yieldblue is 6/101.6=5.91%semi-annually or 11.82% annually. Its yield-to-maturity R can be solved fromthe equation:

6e−R×.5 + 6e−R×1.0 + 6e−R×1.5 + 106e−R×2.0 = 101.6

This is a nonlinear equation in R. We can solve it using the bisection methodor Newton’s method, and R is found to be 0.1079 or 10.79%. Thus if we holdthe bond for 2 years till it matures, it will give us an annual return rate of10.79%. Comparing it with the 2-year zero rate in Table 2, we see that thisbond is under-performed.

As we have seen in the example, it is not easy to calculate the yield-to-maturity of a coupon bond. However, for zero-coupon bonds maturing attime T in the future, since there are no payouts in between t to T , R(t, T ) iseasily computable. In fact, if the bonds have face value 1, i.e. Π(T ) = 1 andare trading at V (t, T ) now, then

R(t, T ) = −lnV (t, T )/(T − t)(Note the similarity of this equation with (12)). The yield-to-maturity on the30-year Treasury bond is widely regarded as the bell-wether (i.e. the leader)rate of the bond market. It mirrors the direction of interest rates. The finaltype of rates that we are going to discuss are called forward interest rates.Zero rates are rates between now and some times in the future. Forward ratesare the rates of interest implied by zero rates for periods of time beginningin the future. For illustration, consider the following example:

Example 4 Suppose the one-year interest rate (i.e. zero rate) is 8% andthe two-year zero rate is 8.5%. The forward rate for the second year is theinterest rate that, when combined with 8% for the first year, gives 8.5% forthe two-year rate. Assume that the interest is continuously compounded, howto calculate this forward rate? Denote the forward rate for the second yearby r(1, 2) (in this notation, the n-year zero rate can be denoted by r(1,n)).Say, we invest $100. It grows to $100e0.08 at the end of the first year. If weterminate this investment but reinvest the money for the second year, thefinal amount would be

er(1,2)100e0.08 = 100e0.08+r(1,2).

17

But the two-year rate is 8.5%. That is, the same $100 grows to 100e2×.0085.To rule out arbitrage opportunities, these amounts must match. We equatethem 100e2×0.085 = 100e0.08+r(1,2), so that 0.17 = 0.08 + r(1, 2). We see thatthe forward rate, r(1, 2), is 9%.

Example 5 Consider Example 2 again. The 1.5-year zero rate is 10.681%while the 2-year zero rate is 10.808%. Hence the forward rate r(1.5, 2) is givenby 2× .10808 = 1.5× .10681 + 0.5r(1.5, 2) ,which gives r(1.5, 2) = 11.189%.

In general, if r1 is the T1-year zero rate of interest and r2 is the T2-yearrate for a longer maturity, then the forward interest rate between T1 andT2,is given by

r(T1, T2) =r2T2 − r1T1

T2 − T1= r2 + (r2 − r1)

T1

T2 − T1.

If we let T1 approach to T2, so that r1 approaches r2, we see that the forwardrate, for a very short period of time, beginning at any time T

r(T, T ) = r + T∂r

∂T.

This is known as the instantaneous forward rate for maturity T .

4 Forwards and Futures A forward contract is an agreement between twoparties whereby one contracts to buy a specified asset, e.g. oil, cotton orstock, from the other for a specified price, K, known as delivery price on amaturity date T in the future. Roughly speaking, a forward contract is anagreement to buy (long position) or sell (short position) an asset at a certainfuture time for a certain price. Recall that futures contracts are just forwardcontracts that sold through a clearinghouse. Often people enter into forwardcontracts without knowing that is what they are called.

What should be the fair delivery price K of a forward contract and what isits fair price at any time before expiry? To answer that, we need the followingtheorem.Proposition 3 Let Π1(t) and Π2(t) be two portfolios such that the combinationof products within the portfolios cannot be changed between now and theexpiry time T . If Π1(T ) = Π2(T ), then we have Π1(t) = Π2(t).

Proof: If by contradiction. Π1(t) < Π2(t), at a time t < T , then we forma portfolio Π(t) = Π1(t)−Π2(t) by shorting one Π2(t) and longing one Π1(t).

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Then initially, Π(t) < 0 but Π(T ) = 0, a contradiction to no arbittrageassumption.

As an application, if a portfolio Π1(t) gives a fixed amount of money Mat time T , then Π1(t) = Me−r(T−t). One can easily prove that by usingΠ2(t) = Π1(t) as a deposit in a bank that gives M at time T . Using theproposition, we can find the value of any forward contract easily.

Proposition 4 Consider a forward contract that expires at T with deliveryprice K. If the current price of the underlying commodity is S(t), then thevalue f(t) of the forward contract is

f(t) = S(t)−Ke−r(T−t).

Proof: Consider two portfolios at time t. (i) one long forward contractf(t) plus an amount of cash of Ke−r(T−t) in a bank, and (ii) one S(t). In thefirst portfolio, the cash will grow to an amount K at time T . It then can(and have to) be used to pay for one S(T ) at the maturity of the forwardcontract. Thus both portfolios will therefore consist of one S(T ) at time T .It follows from Proposition 3 that they must be equally valuable at all earliertime t, i.e. f(t) +Ke−r(T−t) = S(t) for all t ≤ T .

As in Example 6, if the air-ticket S(t) becomes very high, then f(t) willbe very high too. Since the value of the forward contract should be 0 (nomoney down) at the time the contract is written, i.e. f(t0) = 0 where t0 isthe time the contract is written, we can use that to get the fair delivery priceK of the contract. In fact, substitute t0 into (15), we have

K = S(t0)er(T−t0).

Hence

f(t) = S(t)− S(t0)er(T−t0)e−r(T−t) = S(t)− S(t0)e

r(t−t0).

5 Stock Splitting and Dividends

It is customary for shares in the US to have prices between $10 and $100,perhaps because then typical daily changes are of the same sort of size asthe last digit or two, and perhaps so that average purchase sizes for retailinvestors are a sensible number of shares. A company whose share price rises

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above this range will usually issue new shares to bring it back. This is calleda stock split. That is, the existing shares are ”split” into more shares. Forexample,in a 3-for-1 stock split, three new shares are issued to replace eachexisting share. Equivalently, one may describe it as two additional, newly-created, new shares for each old one are issued. Note that, basically, a stocksplit does not change the assets or the earning ability of a company, and oneshould not expect it to have any effect on the wealth of the company’s shareholders. With this understanding, it is easy to see the effect of a 3-for-1 issueon the share price and the option price. Let V denote the total value of thecompany and let N be the total number of the shares. Then the value of ashare S is equal to V/N . In a 3-for-1 stock split, it results in 3N total shares,yet V is unchanged. So, the share value becomes now V/(3N) per unit.In general, an n-for-m stock split (which means n new shares are issued toreplace m existing shares) should cause the stock price to go down to m/n ofits previous value. The terms of option contracts are also adjusted to reflectexpected changes in a stock price arising from a stock split. After an n-for-mstock split, the exercise price is reduced to m/n of its previous value and thenumber of shares covered by one contract is increased to n/m of its previousvalue.

Example 7. Consider a call option to buy 100 shares of a company for $30per share (i.e., the strike price). Suppose that the company makes a 2-for-1stock split. The terms of the option contract are then changed so that itgives the holder the right to purchase 200 shares for $15 per share.

A stock dividend involves a company issuing more shares to its existingshareholders. For example, a 20% stock dividend means that investors receiveone new share for each five already owned. Like a stock split, a stock dividendhas no effect on either the assets or the earning power of a company. Thestock price can be expected to go down as a result of a stock dividend. The20% stock dividend referred to is essentially the same as a 6-for-5 stock split.All else being equal, it should cause the stock price to decline to five-sixths ofits previous value. The terms of an option are adjusted to reflect the expectedprice decline arising from a stock dividend in the same way as they are forthat arising from a stock split.

Example 8 Consider a put option to sell 100 shares of a company for $15per share. Suppose that the company declares a 25% stock dividend. This

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is equivalent to a 5-for-4 stock split. The terms of the option contract arechanged so that it gives the holder the right to sell 125 shares for $12 pershare.

Many assets, such as equities, pay out cash dividend. These are paymentsto share- holders out of the profits made by the company concerned, andthe likely future dividend stream of a company is reflected in today’s shareprice. To an average investor the value of holding the stock comes from thedividends and any growth in the stock’s value. When a dividend is declared,an ex-dividend date is specified. Investors who own shares of the stock up tothe ex-dividend date receive the dividend. Usually, when we model dividendpayments, we need to consider two issues: (i) When, and how often, aredividend payments made? (ii) How large are the payments? There are severalpossible different structures for dividend payments. Individual companiesusually make two or four payments per year, which may need to be treateddiscretely, but the large number of dividend payments on an index such asS&P500. are so frequently that it may be best to regard them as a continuouspayment rather than as a succession of discrete payments.

Suppose that our asset pays a dividend of amount D at time t = td. Letus consider only the case in which the dividend yield , which is given bydy = D/S is a known constant. Obviously, 0 ≤ dy ≤ 1 (usually it is a fewpercent at most). Thus, at time td, holders of the asset receive payment dyS,where S is the asset price just before the dividend is paid. Let us see theeffect of the dividend payment on the asset price. Its value just before thedividend date, at time t cannot equal its value just after, at time td. If itdid, the strategy of buying the asset immediately before td, collecting thedividend, and selling straight away, would yield a risk-free profit. In fact, itis clear that, in the absence of other factors such as taxes, the asset pricemust fall by exactly the amount of the dividend payment. That is,

S(t+d ) = S(t−d )− dy S(t−d ) = (1− dy)S(t−d ).

An easy and intuitive explanation can be given as follows: Say, a stock priceis S just before a (discrete) dividend D = dyS is paid. What is the shareprice immediately after the dividend payment? Well, let V be the total valueof the company and let N be the total number of the shares. Then S = V/N .As the company pays out a dividend D per share, the value of the companyreduces to V − ND. Thus, the share price immediately after the payment

21

will be given by

V −NDN

=V

N−D = S − dyS = (1− dy)S.

Obviously, the price of an option on an underlying asset that pays dividendsis affected by the payments. We have just seen that dividends have the effectof reducing the stock price on the ex-dividend date. This is bad news forthe value of call options and good news for the put options. The value ofa call option is, therefore, negatively related to the size of any anticipateddividend, and the value of a put option is positively related to the size of anyanticipated dividend. We will see the impact of dividends on options later inthe course.

Chapter 4 Put-Call Parity1. Bull and Bear

Financial analysts use words such as ”bull” and ”bear” to describe thetrend in stock markets. Generally speaking, a bull market is characterizedby rising prices. Indeed, an investor is sometimes called a bull when hebuys commodities or securities in anticipation of a rise in prices, or tries byspeculative purchases to effect such a rise. These bulls are potential writersof put options and buyers of call options. The word ”bear” is used just in theopposite way to ”bull”. In short, a bearish market is usually characterizedby falling stock-market prices. The ”bears” are also potential customers forput options and writers of call options on the underlying, as these investorsexpect (or speculate) the asset price to fall.

In the sequel, we say that we are bullish about the market if we think thatthe price of the market will rise; and we say that we are bearish about themarket if we think that the price of the market will fall. A portfolio is saidto be good for a bullish market if it will bring a profit when the market rises;and we say a portfolio is good for a bearish market if the portfolio will bringa profit when the market falls. By combining calls and puts with variousexercise prices one can construct portfolios that suit one?s market view.Example 1 Consider the following two portfolios:

1. (Bull market) Buy one c(S(t), 10, τ, r) and and sell one c(S(t), 20, τ, r):

Π1(t) = c(S(t), 10, τ, r)− c(S(t), 20, τ, r).

2. (Bear market) Buy one c(S(t), 20, τ, r) and and sell one c(S(t), 10, τ, r)

Π2(t) = c(S(t), 20, τ, r)− c(S(t), 10, τ, r).

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2. Properties of Stock OptionsIntuitively, one should exercise an American call option if it is deep in the

money. But we prove in the following that this is not the case.Proposition 1 Suppose a share pays no dividend between t and T . Then,

C(S(t), E, τ, r) = c(S(t), E, τ, r).

Proof: It has been shown in Proposition 3.2 that C(t) ≥ c(t). We need to

show that the strict inequality C(t) > c(t) is not valid. Suppose, however,C(t) > c(t). Then we can form the portfolio: write one American call andbuy a European call:

Π(t) = c(t)− C(t)

Observe that, there is an initial cash inflow in forming such a portfolio. Aswe are the writer of an American call, we have to monitor the action takenby the buyer of this option. There are two possibilities:

(i)The buyer of this American call does not exercise the option early. Inother words, the American call option survives till the end of the contract.The terminal value of this portfolio is then given by Π(T ) = 0 and Π(T ) −Π(t) > 0 and thus, an arbitrage opportunity exists, which is a contradiction.

(ii) The buyer exercises the option early, say, at time t1 ≤ T . Let τ1 =T − t1 < τ At this time, S(t1)−E > 0 otherwise the buyer would not exerciseit (why?). Now that the buyer exercises it at t1, we, as the writer, have tosell him the underlying asset at the price E. What can we do? We can shortsell the asset S(t1) to the buyer, from whom we receive E., Alternatively, wecan short sell the asset S(t1) in the market, and give the buyer S(t1) − E

from S(t1). Note that at this stage, we still use no money out of our ownfund.) Next, let us deposit E in a bank account so that it earns the risk-freeinterest rate r After having done that, the value of the ”adjusted” portfolioat time t1 is given by

Π(t1) = c(S(t1), E, τ1, r)− S(t1) + E

At T

Π(T ) = max(0, S(T )− E)− S(T ) + Eerτ1 ≥ E(erτ1 − 1) > 0

As the original Π(t1) < 0 and thus Π(T )−Π(t1) > 0, we have made a profit,and hence found an arbitrage opportunity. A contradiction.

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Proposition 2 Given interest rate r > 0, it is never optimal to exercise anAmerican call option on a non-dividend-paying stock before the expiry date.

Note that we impose in this proposition the assumption that no dividendpayment is made over the lifetime of options . A natural question one wouldlike to ask is that: What would happen if dividend payments are made? Weshall return to the question later in the chapter. Nevertheless, it can beoptimal to exercise an American put option on a non- dividend-paying stockearly. Indeed, at any given time during its life, a put option should alwaysbe exercised early if it is sufficiently deep in the money. To illustrate, weconsider an extreme situation.

Example 2 Suppose that the strike price is $10 and the stock price is vir-tually zero. By exercising immediately, a trader makes an immediate gainof $10. If the trader waits, the gain from exercising might be less than $10but cannot be more than $10, because negative stock prices are impossible.Furthermore, receiving $10 now is preferable to receiving $10 in the future.It follows that in this case, the put option should be exercised immediately.

3. Put-Call Parity Relation Although call and put options are superfi-cially different, in fact they can be combined in such a way that they areperfectly correlated. This is demonstrated by the following identity.

Proposition 6 Assume that no dividend payment is made over the lifetimeof options. The following identity holds:

S(t) + p(S(t), E, τ, r)− c(S(t), E, τ, r) = Ee−rτ .

This relationship between the underlying asset and its options is called theput-call parity

Proof: Suppose that S(t) + p(t)− c(t) < Ee−rτ . Form the following port-folio: long one share S(t), long one put p(t), short one call c(t) and borrowEe−rτ from a bank. Note that both the call and the put have the same expirydate T and the same exercise price E. The value of this portfolio is

Π(t) = S(t) + p(t)− c(t)− Ee−rτ < 0

24

At T

Π(T ) = S(T ) + max(0, E − S(T ))−max(0, S(T )− E)− E

= max(0, S(T )− E)−max(0, S(T )− E) = 0.

Thus, the profit is Π(T )−Π(t) > 0, which implies that an arbitrage opportu-nity is found. On the other hand, suppose S(t) +p(t)− c(t) > Ee−rτ . We canalso present a corresponding portfolio Π(t) = −S(t)−p(t) + c(t) +Ee−rτ < 0to show that an arbitrage opportunity exists and hence a contradiction.

The parity identity shows that there is a close relationship between thedifferent financial instruments. For example, a long position in a stock com-bined with a short option in a call is equivalent to a short put option plusa certain amount of cash. The put-call parity (9) can be yet derived byconsidering the following two portfolios:

(A) long one European call option plus an amount of cash equal to Ee−rτ .Π1(t) = c(t) + Ee−rτ .

(B) long one European put option plus one share. Π2(t) = p(t) + S(t).Both the call and put options have the same strike price and the expiration

date. One can check easily that both portfolios are worth Π1(T ) = Π2(T ) =max(S(T ), E) at expiration of the options. Because the options are Euro-pean, they cannot be exercised prior to the expiration date. By Proposition3.3, the portfolios must, therefore, have identical values today. Thus, (9) isestablished.

We illustrate in the next two examples that if the values of portfolios (A)and (B) are not the same, then there will be an arbitrage opportunity opento a trader.

Example 3 Suppose European call and put options on a non-dividend-payingstock each have a strike price $30 and an expiration date in three months.Say, the stock price is $31 today, the risk-free interest rate is 10% per annum,the price of a three- month European call option is $3, and the price of a three-month European put option is $2.25. Identify the arbitrage opportunity opento a trader. Then the value of Portfolio (A) is c(t)+Ee−rτ = $3+$30e−.1×.25 ∼$32.26, while the value of Portfolio (B) is given by p(t)+S(t) = $2.25+$31 =$33.25. In such a case, Portfolio (B) is overpriced compared to Portfolio (A).An arbitrage strategy is to buy the securities in Portfolio (A) and short thesecurities in Portfolio (B). This involves buying the call and shorting both

25

the put and the stock. The strategy generates a positive cash flow of

M = −$3 + $2.25 + $31 = $30.25

upfront. We put this money M in the bank. Our portfolio in essence is:Π(t) = c(t) − p(t) − S(t) + M = 0. When invested at the risk-free interestrate, M grows to $30.25. in three months. Three months later, if the stockprice at expiration of the option is greater than $30, the call will be exercised.If it is less than $30, the put will be exercised. In either case, the investorends up buying one share for $30. This share can be used to close outthe short position. Hence, Π(T ) = −E + Mer×.25 and the net profit isΠ(T )− Π(t) = $31.02− $30.00 = $1.02.

Example 4 As in Example 3, but assume that the call price is $3 and theput price is $1. In such case, one can easily check that Portfolio (A) is nowoverpriced to Portfolio (B). Indeed,

c(t) + Ee−r×.25 = $3 + $30e−.1×.25 = $32.26

p(t) + S(t) = $1.00 + $31.00 = $32.00.

An arbitrage can short the securities in Portfolio (A) and buy the securities inPortfolio (B) to lock in a profit. Take note that this strategy involves an initialinvestment of $31+$1−$3 = $29. When financed at the risk-free interest rate,a repayment of $29e.1×.25 ∼ $29.73 is required at the end of the three months.As in Example 3, either the call or the put will be exercised at expiration,and the net profit is $30.00− $29.73 = $0.27. To write it mathematically, weset up a portfolio Π(t) = p(t)+S(t)− c(t)−M , where M = $29 is the moneywe get by selling a put and a stock and buying a call. We put M in the bank.Thus initially Π(t) = 0. At T , Π(T ) = E −Me.1×.25 = 30− 29.97 = 0.27.

Suppose we have special information that the stock price will increasesignificantly over the next month. How do we take advantage of this infor-mation? Do we buy calls, sell puts, buy the stock, or buy the stock withborrowing? The put-call parity gives us the answer. If you believe the stockprice cannot fall, you should buy stock with borrowing. According to (9),S(t)−Ee−r(T−t) < c(t), i.e. you need less money than buying a call. Do notjust buy a call because by put-call parity this involves purchasing an insur-ance policy (the put option) that you do not need. On the other hand, you

26

could simultaneously buy calls and write puts, but again by put-call paritythis is the same as a levered stock position. Note that the put-call parityin Proposition 6 holds only for European options. However, it is possible toderive some relationships for American option prices. We will here mentiononly one of such relationships.

Proposition 7 When the stock pays no dividends,

S(t)− E ≤ C(s(t), E, τ, r)− P (s(t), E, τ, r) ≤ S(t)− Ee−rτ .

Proof: Observe that C(t) = c(t) for non-dividend-paying stocks, and andP (t) ≥ p(t), always. Thus, it is trivial to see

C(t)− P (t) ≤ c(t)− p(t) = S(t)− Ee−rτ .

by the put-call parity for European options. It remains to derive the lowerbound for C(t)−P (t). For this it suffices to show that c(t)+E ≥ S(t)+P (t)since C(t) = c(t). If by contradiction, c(t) + E < S(t) + P (t) , then weconsider the portfolio Π(t) = c(t) +E −P (t)− S(t) < 0, where E is put in abank account earning an interest rate r. The value of the portfolio is negative(positive cash flow) at the initial time t. First consider the case where theput option is never exercise in [t, T ] then

Π(T ) = max(0, S(T )− E)− S(T ) + Eerτ > 0.

If the put option is exercised at t0, then

Π(t0) = c(t0) + Eer(t0−t) − (E − S(t0))− S(t0) = c(t0) + E(er(t−t0) − 1) ≥ 0.

Thus, for the both case arbitrage exits.

Example 5 An American call option on a non-dividend-paying stock withexercise price $20.00 and maturity in five months is worth $1.50. This mustalso be the value of a European call option on the same stock with the sameexercise price and maturity. Suppose that the current stock price is $19.00and the risk-free interest rate is 10% per annum. By the put-call parityrelationship for European options, we get the price of a European put withexercise price $20 and maturity in five months

1.50 + 20e−.1×5/12 − 19 = 1.68

27

From (10),19− 20 ≤ C − P ≤ 19− 20e−.1×5/12 ∼ .18

or equivalently .18 ≤ P − C ≤ 1. This shows that P lies between $1.68 and$2.50. In other words, upper and lower bounds for the price of an Americanput with the same strike price and expiration date as the American call are$2.50 and $1.68.

4 The Effect of Dividends The results produced so far have assumed thatwe are dealing with options on a non- dividend-paying stock. In the follow-ing we will examine the impact of dividends. In the US, exchange-tradedstock options generally have less than a year to maturity. The dividends Dpayable during the life of the option can usually be predicted with reasonableaccuracy. Recall from (3.1) that if td is the ex-dividend date, recall

S(t+d ) = S(t−d )−D.

Note that since the holder of an option does not receive any dividend, thevalue of the option should be the same just before and after the ex-dividenddate

Proposition 8 For any options V (S(t), t) = V (S(t), E, τ, r) on an asset S(t)with dividend payment at td ∈ (t, T ), we have V (S(t+d ), t+d ) = V (S(t−d ), t−d ).

Proof: Let us prove this for American put, and leave the proofs for theother options as exercises. Since S drops across td, we know that P (S(t−d ), t−d ) ≤P (S(t+d ), t+d ). Now, if P (S(t−d ), t−d ) < P (S(t+d ), t+d ), then one can buy the op-tion at t−d and sell it back at t+d to earn an immediate riskless profit.

The proposition seems to be counter-intuitive in that dividends have theeffect of reducing the stock price S(t) across td, and hence it should be badnews for the value of call options and good news for the put options. Thefact that the option price is continuous across td, even though the asset valueis not, does not mean that the option value is unaffected by the dividendpayments. The effect of (11) is felt throughout the life of the option, and ispropagated by the underlying equation (the Black-Scholes partial differentialequation) that governs its value. More precisely, a call option should be lessvaluable if one foresees that the underlying stock will pay cash dividendswithin the lifetime of the option. Next we derive the put-call parity equalitywhen we have dividends.

Proposition 9 Suppose the dividend payment D is only made once at td ∈

28

(t, T ) and the asset does not pay dividend at any other time over the lifetimeof options. Then, for European options, the following identity holds :

S(t) + p(S(t), E, τ, r)− c(S(t), E, τ, r)−De−r(td−t) = Ee−rτ .

Proof: We consider two portfolios: (A) one European call option plus anamount of cash equal to De−r(td−t)−Ee−rτ and (B) one European put optionplus one share. It is easy to show that both give Der(T−td) + max(S(T ), E)at T . Thus by Proposition 3.3, they must have the same value at t < td.

Notice that if t > td, then portfolio (B) in the proof above is S(t) + p(t) +De−r(t−td) while portfolio (A) is c(t) + De−(t−td + Eerτ Hence by equatingthe two portfolios and canceling De−(t−td on both sides, we get S(t) + p(t) =c(t) +Eerτ , which is precisely the put-call parity we have in (9). We get thisput-call parity because t > td and the stock will not pay dividend between t

and T .By (12), we get the following inequality:

c(S(t), E, τ, r)) ≥ S(t)−De−r(t−td) − Ee−rτ (p(t) ≥ 0)

p(S(t), E, τ, r)) ≥ Ee−rτ +De−r(t−td) − S(t) (c(t) ≥ 0).

for t < td. For American options, we have the following result.

Proposition 10 Suppose the dividend payment D is only made once at td ∈(t, T ) and the asset does not pay dividend at any other time over the lifetimeof options. Then, the put-call parity becomes

S(t)−De−r(td−t) − E ≤ C(t)− P (t) ≤ S(t)− Ee−rτ .

The proof of this equation is similar to that in Proposition 7. We leave it asan exercise.

By Propositions 2.3 and 4.1, we know that for a non-dividend paying stock,

C(t) = c(t) ≥ S(t)− E for all t ≤ T.

However the same inequality is true for dividend-paying stocks.

Proposition 11 For any stock, whether it is dividend-paying or not,

C(t, S(t), τ, r) ≥ S(t)− E, and P (t, S(t), τ, r) ≥ E − S(t) for all t ≤ T.

29

Proof: If say at any time t ≤ T C(t) < S(t) − E, then the holder of theoption can exercise his call option to get a profit of S(t) − E and then hecan use C(t) amount of money (which is straightly less than S(t) − E tobuy back the option. By doing so, he reaps an instantaneous riskless profitof S(t) − E − C(t) > 0, a contradiction. The same can be said about theAmerican puts.

With no dividend payment to be made during the life of an American calloption, it has been explained that it is never optimal to exercise it beforethe expiry date. When dividends are expected, however, we can no longerassert that an American call option will not be exercised early. Sometime itis optimal to exercise an American call immediately prior to an ex-dividenddate. This is because the dividend causes the stock price to jump down,making the option less attractive. It turns out that it is never optimal toexercise a call at other times.

Proposition 12 Suppose the dividend payment D is only made once at td ∈(t, T ) and the asset does not pay dividend at any other time over the lifetimeof options. Then, an American call option will only be exercised at td or T .

Proof: If the option is exercised at T , then we are done. Suppose it isnot. First consider exercising at (td, T ). Since there are no dividends inthis time interval, the American call behaves like the European call. Henceaccording to Proposition 2, one would not exercise the call option in thisperiod. One would not exercise at t+d also. For if one exercise, one getsS(t+d )− E = S(t−d )−D − E.. This is clearly less than S(t−d )− E which onewould get if one exercises at t−d .

It remains to show that the option will not be exercised in any time in theinterval (t, td) By Proposition 11, we have at t−d

(14) S(t−d )− E ≤ C(t−d )

Thus, we claim that S(t) − E ≤ C(t) for all t < td and hence it is neveroptimal to exercise the option before td. The proof of this is similar to theproof in Proposition 2, but we repeat it here for clarity. By contradiction, ifat some t ∈ [t, td) S(t)−E > C(t), then we buy a C(t), short a S(t) and putE in the bank. The portfolio at t Π(t) = C(t)− S(t) + E ≤ 0. Since we arethe holder of the American call C(t) we can choose to keep it at least until

30

td Then we have

Π(t−d ) = C(t−d )− S(t−d ) + Eer(td−t) ≥ C(t−d )− S(t−d ) + E,

which is nonnegative by (14). Thus Π(t−d )− Π(t) > 0, a contradiction.

From the proof, we see that the only time that C(t) can be (and notnecessarily must be) equal to S(t) − E, the exercise price, is either at td orat T . At all other time, C(t) > S(t)−E and therefore one must not exercisethe option.

The following proposition for American options, first shown in Proposition2.2 for European options, holds regardless of whether dividends are paid outor not.

Proposition 13 For 0 ≤ E1 ≤ E2

0 ≤ C(S(t), E2, τ, r)− C(S(t), E1, τ, r) ≤ E2 − E1

0 ≤ P (S(t), E1, τ, r)− P (S(t), E2, τ, r) ≤ E2 − E1.

5 Put-Call Parity for Digital Options The original and still the most com-mon contracts are the vanilla calls and puts. Increasingly important are thebinary or digital options. These contracts have a payoff at expiry that is dis-continuous in the underlying asset price. Say, for a simple example of a binarycall, it pays a fixed amount Q at expiry time T , if the asset price is greaterthan or equal to the exercise price E and it pays nothing at expiry if the assetprice ends up below the strike price. This is a kind of cash-or-nothing call.

Why would you invest in a binary call? If you think that the asset pricewill rise by expiry, to finish above the strike price, then you might chooseto buy either a vanilla call or a binary call. The vanilla call has the bestupside potential, growing linearly with S beyond the strike. The binarycall, however, can never pay off more than Q, If you expect the underlyingto rise dramatically, then it may be best to buy the vanilla call. If youbelieve that the asset price rise will be less dramatic, then buy the binarycall. The gearing of the vanilla call is greater than that for a binary callif the move in the underlying is large. A cash-or-nothing binary put can bedefined analogously to a cash-or-nothing binary call. The holder of such a putreceives Q if the asset is straightly below E at expiry. The binary put wouldbe bought by someone expecting a modest fall in the asset price. There is a

31

particularly simple binary put-call parity relationship. What do you get atexpiry if you hold both a binary call and a binary put with the same strikesand expiries? The answer is that you will always get Q regardless of the levelof the underlying at expiry. Thus, according to Proposition 3.3,

Binary call + Binary put = Qe−r(T−t).

by Π1(t) = Qe−r(T−t) −Bc(t) and Π2(t) = Bp(t) with Π1(T ) = Π2(T ).

Chapter 5 Trading Strategies— protective measure— easy to set up the portfolio— smaller initial investment— call-put partiality— bull and bear market

2 Spreads A spread trading strategy involves taking a position in two ormore options of the same type (i.e., two or more calls or two or more puts).2.1 Bull Spreads One of the most popular types of spreads is a bull spread.It can be formed by buying a call option on a stock with a strike price E1

and selling a call option on the same stock with a strike price E2 whereE2 > E1 Both options have the same expiration date. Since we usually usethe x-axis to denote time and y-axis to denote money, suchspreads are alsocalled vertical spreads Since a call price always decreases as the strike priceincreases (see Proposition 2.2)

Π(t) = c1(S(t), E1, τ, r)− c1(S(t), E2, τ, r) > 0

i.e., when created from calls, a bull spread requires an initial investment. AtT , the terminal payoff is equal to

Π(T ) = max(0, S(T )−E1)−max(0, S(T )−E2) =

0 for S(T ) ≤ E1

S(t)− E1 for E1 ≤ S(T ) ≤ E2

E2 − E1 for S(T ) ≥ E2

Proposition 1 Suppose that European options expiring at time T are avail-able with every single possible strike price. Then it is possible to combinethem together to get any payoff functions that we prefer at time T.

Proof: Consider the butterfly spread ΠE,δS(t) of buying two call optionswith strike prices E − δS and E + δS and selling two call options with strike

32

price E. Note that ΠE,δS(T, x) (x = S(T )) is a piecewise linear function withΠE,δS(T )(E) = δS and ΠE,δS(T )(x) = 0 for |x−E| ≥ δS. Let y = f(x) be thepayoff function at T . Given pairs (Ei, yi) on the x-y plane, i.e., f(Ei) = yi,the linear interpolation f of f is given by

f(x) =∑i

yiδSi

ΠEi,δSi(T, x)

and it approximates f(x) as accurate as possible, i.e., f(Ei) = f(Ei). Here,f is the linear spline interpolation of f and 1

δSiΠEi,δSi(T, x) is the linear

spline functions.

Chapter 6 Geometric Brownian Motion Recall a random variable X :Ω → R is a function from the sample space Ω to R, i.e., X(ω) is a sampledoutcome of X. The probability P assigns the probability 0 ≤ P (X(ω) ∈A) ≤ 1 for an event X(ω) ∈ A, where A is an arbitrary interval (a, b) onR.

1 Normal Distributions and Gaussian random variables We begin byrecalling the normal distribution briefly. Let Z be a random variable dis-tributed as standard normal, i.e., Z ∼ N(0, 1) The probability density func-tion (PDF) of Z is given by

pZ(x) =1√2πe−

x2

2 , x ∈ (∞,∞)

This is the famous bell-curve of normal distributions. That is,

Prob(Z(ω) ∈ (a, b)) =

∫ b

a

pZ(x) dx

Thus, ∫ ∞−∞

pZ(x) dx = 1

with

mean = E[Z] =

∫ ∞−∞

xpZ(x) dx = 0

and

variance = E[|Z|2] =

∫ ∞−∞

x2 pZ(x) dx = 0

33

In general, the expectation of f(Z) is given by

E[f(Z)] =

∫ ∞−∞

f(x) pZ(x) dx.

For example the moment generating function of Z is defined by

E[etZ ] =

∫ ∞−∞

etx pZ(x) dx

=

∫ ∞−∞

etx−|x|2

2 dx =

∫ ∞−∞

e−|x−t|2

2 et2

2 dx = et2

2 .

Since

E[etZ ] =∞∑k=1

tk

k!E[|Z|k],

we havedk

dtkE[etZ ]|t=0 = E[|Z|k]

and thus

E[|Z|n] =

0 n is odd

(2k)!2kk!

n = 2k is even

Let X be a Gaussian random variable with mean µ and standard deviationσ, i.e., X ∼ N(µ, σ2) Its probability density function is given by

pX =1√

2πσ2e−|x−µ|2

2σ2 , x ∈ (∞,∞)

i.e.,

Z =X − µσ

∼ N(0, 1)

and

E[|X − µ|n] =

0 n is odd

σn (2k)!2kk!

n = 2k is even.

Example Recall that for α ∈ (0, 12) zα is the 100α percentage point of the

standard normal distribution is defined as the number if

α = Prob(Z > zα) =

∫ ∞zα

pZ(x) dx.

34

Referring to a statistical table, we have for examples, z0.05 = 1.6449 andz0.025 = 1.960. More precisely, if Z ∼ N(0, 1), then within 95% confidence,

−1.96 ≤ Z ≤ 1.96

If x ∼ N(µ, σ)µ− 1.96σ ≤ X ≤ µ+ 1.96σ.

X signifies the daily return of a stock, then we say the stock will have a 2.5%value at risk (VAR) of µ − 1.96σ. That implies that on average the stockholder expects a daily return of no more than µ − 1.96σ in 1 day out of 40trading days.

A Model for Asset PricesThe asset price and stock price were previously denoted byS(t) Here we

are going to take the ”volatility” into account. The standard, generic symbolfor such a ”random-ness” is ω(sample) Thus we denote the asset price attime t by S(t, ω) or simply by St(ω). It is a stochastic process, i.e. for eachfixed t St(ω) is a random variable depending on ω (OUTCOME depends onsample ω). For those who are not familiar with stochastic processes, you mayconsider each ω, t→ S(t, ω) is a realization of the process.

We will build our model of the stock prices based on the above observation.We first recall the return is defined to be the change in the price divided bythe original value, i.e.

return = change in price/original value.

Consider a small subsequent time interval (t, t+ ∆t)

return =S(t+ ∆t)− S(t)

S(t)= µ∆t+ σ∆X,

where∆X = X(t+ ∆t)−X(t) ∼ N(0,∆t)

i.e., a random sample drawn from a normal distribution, and σ > 0 is anumber called the volatility, which measures the standard deviation of thereturn, i.e.,

∆S

S∼ µ∆t+ σ

√∆tε, ε ∼ N(0, 1).

35

Let us adopt the differential notation used in calculus. Namely, we usethe notation dt for the small change in any quantity over this time intervalwhen we intend to consider it as an infinitesimal change.

(10) dS(t)/S(t) = µ dt+ σdXt,

which is the mathematical representation of our simple recipe for the priceof a non- dividend-paying stock. Equivalently, (10) can also be expressed as

dSt = µSt dt+ σSt dX(t)

and integrating this in time

St = St0 +

∫ t

t0

µ(ω)Ss ds+

∫ t

t0

σ(ω)Ss(ω) dXs(ω).

When σ = 0, we recall

St = St0e∫ tt0µ(s) ds

We may intituitively think

St = e∫ tt0µt+σ(X(t)−X(t0))

St0.

But, dlnSt 6= dSt/St and there are difference between stochastic differentialequations and ordinary differential equations. The difference can fortunatelybe handled easily by Ito’s Lemma discussed in the next chapter. We willshow (S(t) is the geometric Brownian motion ) is

St = e(µ−σ2

2 )(t−t0)+σ(X(t)−X(t0))St0.

3. Wiener Process and its Generalizations The term dXt in (10) which givesthe randomness to St is certainly the main feature of the geometric Brownianmotion. In fact, Xt(ω) is a Wiener process and is known to follow Brow-nian motions. Historically speaking, such a random process was observedby Robert Brown, an English botanist, in the summer of 1827, that ”pollengrains suspended in water performed a continual swarming motion. ” Henceit was named after Robert Brown, called Brownian motion. Brownian mo-tion is a process of tremendous practical and theoretical significance. It wasused as an model to explain the ceaseless irregular motions of tiny particles

36

suspended in a fluid. It had also been used as a model of the stock marketin Louis Bachelier’s (1900) work. His paper was at first largely ignored byacademics for many decades, but now his work stands as the innovative firststep in a mathematical theory of stock markets that has greatly altered the fi-nancial world today. In 1905, Albert Einstein gave a satisfactory explanationand asserted that the Brownian motion originates in the continued bombard-ment of the pollen grains by the molecules of the surrounding water, withsuccessive molecular impacts coming from different directions and contribut-ing different impulses to the particles. (Incidentally, 1905 is the same year inwhich Einstein set forth his theory of relativity and his quantum explanationfor the photoelectric effect.) But, Brownian motion is complicated and it isnot surprising that it took more than another decade to get a clear picture ofthe Brownian motion stochastic process.A rigorous mathematical foundationupon which Brownian motion could be built had to wait until 1920?s. In1923, Norbert Wiener (1894–1964) laid a rigorous mathematical foundationand gave a proof of its existence. Hence, it explains why it is now also called aWiener process. In the sequel, we will use both Brownian motion and Wienerprocess interchangeably.Definition 1 We say a stochastic process Xt(ω) is a standard Wiener processif it satisfies the following conditions:

(a) X0(ω) = 0(b) for all ω, the map t→ Xt(ω) is a continuous function.(c) for every t and h ≥ 0 the increment Xt+h − Xt ∼ N(0, h) (Gaussian

process) and Xu −Xv and Xt −Xs are independent for all 0 ≤ u ≤ v ≤ t ≤s(Independent increments)

(Events Ω1 and Ω2 are independent if P (Ω1∩Ω2) = P (Ω1)P (Ω2). P (Xu−Xv ∈ A ∪ Xt −Xs ∈ B) = P (Xu −Xv ∈ A)P (Xt −Xs ∈ B).

Condition (a) says that the starting point of a standard Wiener processis at the origin. Condition (b) follows from (c) (Kolmogorov’s theorem).For each fixed ω the function t → Xt(ω) is called sample path (realization,trajectory) of the Wiener process associated with ω. For any set

Prob(Xt+h −Xt ∈ A) =1√2πh

∫A

e−|x|22h dx.

From (c) and (a), one can deduce that, for 0 ≤ t0 < t1 < · · · ≤ tn

Xt0, Xt1 −Xt0, · · · , Xtn −Xtn−1

37

are independent, i.e., has independent, normally distributed, increments. Inparticular, we have Xt+h−Xt is independent with Xt. Hence a Wiener processhas the so-called Markov property, which means that only the present valueof the process is relevant for predicting the future, while the past historyof the process and the way that the present has emerged from the past areirrelevant. A stochastic process which satisfies the Markov property is calleda Markov process. Generally speaking,the following expression says that aWiener process is Markov: for t ≥ 0 , i.e.,

Prob(Xt+h −Xt ∈ A|Xs, 0 ≤ s ≤ t) = P ((Xt+h −Xt ∈ A|Xt).

Take note that stock prices are usually assumed to follow a Markov process.Sup- pose that the price of XYZ stock is $50 now. If the stock price followsa Markov process, our prediction for the future should be unaffected by theprice one week ago, one month ago, or one year ago. The only relevantpiece of information is the price now which is $50. Obviously, predictionsfor the future are uncertain and must be expressed in terms of probabilitydistributions. The Markov property implies that the probability distributionof the price at any particular future time is not dependent on the particularpath followed by the price in the past.

Definition 2 The process

Yt = σXt + µt+ ξ

is called a generalized Wiener process, starting at ξ with a drift parameter µand variance rate σ2.

Properties of Brownian motion First, we will show that the quadraticvariation

∑nk=1 |∆Bk|2, ∆Bk = Btk − Btk−1

converges in mean square tothe length of the interval as the length of the subdivision tends to zero;

E[(∑n

k=1 |∆Bk|2 − t)2] =∑

k,`(|∆Bk|2 −∆tk)(|∆B`|2 −∆t`)

=n∑k=1

E[(|∆Bk|2 −∆tk)2] =

∑k

3(∆tk)2 − 2(∆tk)

2 + (∆tk)2

=n∑k

2(∆tk)2 ≤ 2tmax

k|∆tk| → 0.

38

where we used if k 6= `

E[(|∆Bk|2 −∆tk)(|∆B`|2 −∆t`)] = E[(|∆Bk|2 −∆tk)]E[(|∆B`|2 −∆t`)] = 0.

E[(|∆Bk|2 −∆tk)2] = E[|∆Bk|4 − 2E[|∆Bk|2]∆tk + (∆tk)

2.

Proposition 6. (Chebyshev’s inequality) For any random variable R andany positive δ > 0 we have

Prob(|R| > δ) ≤ 1

δ2E[|R|2].

In fact,E[|R|2] ≥ E|R≥δ[|R|2] ≥ δ2Prob(|R| ≥ δ).

Thus, E[|∑n

k=1 |∆Bk|2− t|2]→ 0 as ∆t→ 0 implies that∑n

k=1 |∆Bk|2 → t inprobability (Theorem 7):

Prob(|n∑k=1

|∆Bk|2 − t| > δ)→ 0

for all δ > 0.On the other hand, the total variation, defined by V = sup

∑nk=1 |∆Bk|

over all partition 0 = t0 < t1 < · · · < tn = t, is infinite with probability one.In fact, using the continuity of the trajectories of the Brownian motion, wehave

n∑k=1

|∆Bk|2 ≤ supk|∆Bk|n∑k=1

|∆Bk| ≤ V supk|∆Bk| → 0

if V < ∞, which contradicts the fact that∑n

k=1 |∆Bk|2 converges in meansquare to t.

Ito’s Rule |dBt|2 = dt, dt dBt = 0, |dt|2 = 0.

E = [|∆Bt|2] = E|Bt+h −Bt|2 = h

and

|∆Yt|2 = |σ∆Bt+µ∆t|2 = |µ|2|∆t|2+|σ|2|∆Bt|2+2σµ∆t∆Bt ∼ |σ|2∆t+o(∆t)

Letting ∆t→ 0,(dYt)

2 = σ2 dt.

39

Chapter 7 Ito’s stochastic calculusWe define the Ito’s stochastic integral by∫ t

0

b(Ys, s) dBs = lim∆t→0+

In, In =n∑k=1

b(Ytk−1, tk−1)(Btk −B(tk−1))

One can prove the limit exists and

E[|∫ t

0

b(Ys, s) dBs|2] = E[

∫ t

0

|b(Ys, s)|2 ds].

Ito’s stochastic calculus Let Yt be an Ito’s process, i.e.,

dYt = a(Yt, t) dt+ b(Yt, t) dBt

Equivalently,

Yt(ω) = Y0(ω) +

∫ t

0

a(Ys(ω), s) ds+

∫ t

0

b(Ys(ω), s) dBs(ω).

Then, f ∈ C2,1([0, T ]×R), f(Yt, t) is an Ito’s process satisfying

df(Yt, t) = (a(Yt, t)∂f

∂Y+

1

2b(Yt, t)

2 ∂2f

∂Y 2+∂f

∂t) dt+ b(Yt, t)

∂f

∂YdBt.

equivalently

df(Yt, t) =∂f

∂YdYt +

1

2

∂2f

∂Y 2(dYt)

2 +∂f

∂tdt

where(dYt)

2 = (a dt+ b dBt)2 = b2 dt

Proof: By the Taylor’s series for t = tn

f(Yt)− f(Y0) =n∑k=1

[∂f

∂t(Ytk−1

, tk−1)(tk − tk−1) +n∑k=0

∂f

∂Y(Ytk−1

, tk−1)(Ytk − Ytk−1)

+1

2

n∑k=1

∂2f

∂Y 2(Ytk−1

, tk−1)(Ytk − Ytk−1)2 +O((∆t)2, |∆Y |3).

40

Here,

|∆Yk|2 = |b∆Bt+a∆t|2 = |a|2|∆t|2 + |b|2|∆Bt|2 +2ab∆t∆Bt ∼ |b|2∆t+o(∆t)

Thus,n∑k=1

∂2f

∂Y 2(Ytk−1

, tk−1)(Ytk − Ytk−1)2 →

∫ t

0

b(Ys, s)2 ds.

Exampled|Bt|2 = 2BtdBt + dt

Equivalently, ∫ t

0

2Bs dBs = |Bt|2 − |B0|2 − t

In general

df(Bt) = f ′(Bt)dBt +1

2f ′′(Bt) dt

Equivalently, ∫ t

0

f ′(Bs) dBs = f(Bt)− f(B0)−∫ t

0

1

2f ′′(Bs) ds

d(g(t)Bt) = g′(t)Bt + g dBt

Equivalently,∫ t

0

g(s) dBs = g(t)Bt − g(0)f(B0)−∫ t

0

g′(s)Bs ds

Geometric Brownian motion

dlnSt =dStSt− 1

2

(dSt)2

S2t

wheredSt = µSt dt+ σSt dBt

and(dSt)

2 = σ2S2t dt

Thus, we obtain

(15) dlnSt = (µ− 1

2σ2) dt+ σdBt

41

lnSt = lnS0 + (µ− 1

2σ2)t+ σBt

and thusSt = S0e

µ− 12σ

2)t+σBt

That is,

(17) lnStS0∼ N((µ− σ2

2)t, σ2 t).

Modern mathematical economists usually prefer the geometric Brownianmotion over the Brownian motion as a model for prices of assets, say sharesof stock, that are traded in a perfect market. By (7), such prices are nonneg-ative and exhibit random fluctuations about a long-term exponential decayor growth curve. Both of these properties are possessed by the geometricBrownian motion, but not by the Brownian motion.

Example 5. Consider a stock with an initial price of $40, an expected returnof 16% per annum, and a volatility of 20% per annum. Let us compute itsprice distribution in six months time. Take t = 0.5 year. Clearly, µ = .16and σ = .20. By (16)

lnS.5 ∼ N((ln 40 + .16− .202

2) .5, .0202 .5). ∼ N(3.759, 0.1412)

Recall from (6.8) that there is a 95% probability that a standard normalrandom variable has a value within 1.96 standard deviations of its mean.Hence, within 95% confidence,

3.759− 1.96× 0.141 < lnS.5 < 3.759 + 1.96× 0.141.

Equivalently,e3.759−1.96×0.141 < S.5 < e3.759+1.96×0.141.

Thus, there is a 95% probability that the stock price in six months will liebetween $32.55 and $56.56.

Theorem 2 The probability density function of St is given by

pSt(s) =1

σs√

2πte|ln s

S0−(µ−σ

2

2 )t|2

2σ2t

42

Proof: The probability density function of ln St is given by

pln St(x) =1

σ√

2πte|x−ln S0−(µ−σ

2

2 )t|2

2σ2t .

By the definition∫ w

−∞plnS

dw = Prob(lnS ≤ w) = Prob(S ≤ ew) =

∫ ew

−∞pS ds.

Differentiate the both sides with respect to w, we obtain

plnS(w) = pS(ew)ew.

Letting w = ln s we obtain the claim.

4 Expected Returns of Geometric Brownian Motions From (9), we knowthat

(22) µ dt = E[dS

S] > E[dlnS] = (µ− σ2

2) dt

The difference comes from measuring the returns differently. It will givedifferent rates of return, and is closely related to an issue in the reporting ofmutual fund returns. We illustrate that with an example first.

Example 7 A mutual fund, starting at $100 five years ago, performed asfollows in the last five years:

115, 138, 179.4, 143.52, 179.4.

Then the returns per annum measured using annual compounding will be:

15%, 20%, 30%, −20%, 25%.

i.e.(S(2)− S(1))/S(1) = .15, (S(3)− S(2))/S(2) = .20, etc.

The arithmetic mean of the returns, calculated by taking the sum of thereturns and dividing by 5, is 14%. However, an investor would actually earnless than 14% per annum by leaving the money invested in the fund for fiveyears. In fact, for the $100 he invested, at the end of the five years, he getsonly $179.40. If we measure that using continuous compound interest, thatwill be 11.69% as

100× e0.1169×5 = 179.40.

43

Note that the average return 11.69% can also be obtained by computing thereturns annually by compounding continuously and they are:

14%, 18.2%, 26.2%, −22.3%, 22.3%,

where

S(1) = S(0)e0.14(i.e. ln S(1)− lnS(0)=.14).S(2) = S(1)e.182,

etc. Then the average return 11.69% is obtained by averaging these five re-turns. The fact that 14% > 11.69% is just a validation of (22). So whataverage return should the fund manager report? It is tempting for the man-ager to make a statement such as: ”The average of the returns per year thatwe have realized in the last five years is 14%”. Although true, this is mis-leading. It is much less misleading to say ”The average return realized bysomeone who invested with us for the last five years is 11.69% per year.” Insome jurisdictions regulatory standards require fund managers to report re-turns the second way. In the following, to distinguish the two different waysof computing average return, we will call the one computed by compoundingdiscretely (like the 14% in Example 7) the ?short-term? return rate, whilethe one computed by compounding continuously (like the 11.69% in Example7) the ”long-term return” rate. We call the latter long- term because it canbe computed using the beginning price and the ending price only, (after 5years in Example (7), see (23)). Thus (22) states that the long-term returnrate µ− σ2

2 is always less than the short-term return rate µ. In the following,we verify that from their original definitions and show how exactly the termσ2

2 comes in. We start with the long-term return rate.

Proposition 4 Assume that the asset price follows the geometric Brownianmotion. Let η be the continuously compounded rate of return per annumrealized by the stock between times t and t+ ∆t. Then

E[η] = µ− σ2

2.

Proof: By the definition S(t+ ∆t) = eη∆tS(t), i.e.

(24) η =1

∆tlnS(t+ ∆t)

S(t)∼ N(µ− σ2

2,σ2

∆t).

Thus η, the continuously compounded rate of return per annum for the stockis normally distributed with mean µ− σ2

2 and standard deviation σ√∆t

.

44

We see that the rate of return on the stock, when measured in the con-tinuously compounded way, will tend to µ− σ2

2 as ∆t increases (because thevariance decreases with ∆t). That is, if we measure the return rate in a longertime frame, we are more certain that it will be closer to µ− σ2

2 .

Example 8 Consider a stock with an expected return of 17% per annum anda volatility of 20% per annum. Then, by (25), the probability distributionfor the actual rate of return (continuously compounded) realized over threeyears is normal with mean

.17− 0.22

2= 0.15,

or 15% per annum and standard deviation

0.2√3

= 0.1155

or 11.55% per annum. Because there is a 95% chance that a normal randomvariable will lie within 1.96 standard deviations of its mean, we can be 95%confident that the actual return realized over three years will be between

7.6%(= 0.15−1.96×0.1155) and +37.6%(= 0.15+1.96×0.1155) per annum.

Next we compute the ”short-term” expected return rate of the stock.Proposition 5. Assume that the asset price follows the geometric Brownianmotion Let λ be the rate of return of the stock over the short time interval

S(t+ ∆t) = S(t)eλt

Then E[λ] = µ.Proof. By (17), we know that

S(t+ ∆t)− S(t)

S(t)= e(µ−σ

2

2 )∆t−σ∆B − 1

and

E[e(µ−σ2

2 )∆t−σ∆B] =

∫ ∞−∞

1√2π∆t

e−|x|22∆te(µ−σ

2

2 )∆t−σx dx = eµ∆t

Thus,

E[S(t+ ∆t)− S(t)

S(t)] = eµ∆t − 1 ∼ µ∆t.

45

Thus we have shown that the expected continuously compounded returnof a stock following geometric Brownian motion is µ− σ2

2 , which is less thanµ, the expected return on the stock computed discretely. In fact this is truefor any asset models. (Indeed, in Example 7, we did not assume any modelon the fund prices.) The phenomenon is well known: the geometric mean ofa set of numbers (not all the same) is always less than the arithmetic mean.

5 Calibrating Geometric Brownian Motions None of the analysis thatwe have presented so far is of much use unless we can estimate the param-eters in our model. In the next chapter, we will see that only the volatilityparameter σ appears in the value of an option and the drift µ does not. Buthow do we obtain σ, the volatility, of the asset S(t). One way is to estimateit from historic data. From Definition 6.5, the volatility σ of a stock can bedefined as the standard deviation of the return provided by the stock in oneyear when the return is expressed using continuous compounding. Equation(16) shows that σis also the standard deviation of the natural logarithm ofthe stock price at the end of one year. Thus a simple approach for estimatingσ from past data is as follows. Suppose we have the values of asset price S atn + 1 equal time-steps ∆t. Call these values S0, S1, · · · , Sn in chronologicalorder with length ∆t of time interval in years. We first form the data series

Ui = lnSi+1 − lnSi = lnSi+1

Si.

SinceSi+1 − Si

Si= e(µ−σ

2

2 )∆t+σ∆Bi − 1,

where ∆Bi are independent, the successive ratios

S1

S0, · · · Sn

Sn−1

are independent random variables too. Next we find the mean U and andvariance σ2 by

U =1

n

n∑i=1

Ui, σ2 =1

n− 1

n∑i=1

|Ui − U |2.

These statistics are estimates for the theoretical mean and variance of thepopulation Ui. In fact, by (17),

Ui ∈ N((µ− σ2

2)∆t, σ2∆t).

46

Hence the theoretical mean and variance of Ui are (µ − σ2

2 )∆t and σ2∆t,respectively. Thus we can solve the equations

U = (µ− σ2

2)∆t, σ2 = σ2∆t

for µ and σ. The algebra is easy and the answers are

µ =U + σ2/2

∆t, σ =

σ√∆t.

Chapter 8 Black-Scholes EquationsUp to now, we only consider hedgings that are done upfront. For example,

if we write a naked call (see Example 5.2), we are exposed to unlimited riskif the stock price rises steeply. We can hedge it by buying a share of theunderlying asset. This is done at the initial time when the call is sold. Weare then protected against any steep rise in the asset price. However, if wehold the asset until expiry, we are not protected against any steep dive inthe asset price. So is there a hedging that is really riskless? The answer wasgiven by Black and Scholes, and also by Merton in their seminal papers onthe theory of option pricing published in 1973. The idea is that a writer ofa naked call can protect his short position of the option by buying a certainamount of the stock so that the loss in the short call can be exactly offsetby the long position in the stock. This is standard in hedging. The ques-tion is how many stocks should he buy to minimize the risk? By adjustingthe proportion of the stock and option continuously in the portfolio duringthe life of the option, Black and Scholes demonstrated that investors cancreate a riskless hedging portfolio where all market risks are eliminated. Inan efficient market with no riskless arbitrage opportunity, any portfolio witha zero market risk must have an expected rate of return equal to the risk-less interest rate. The Black-Scholes formulation establishes the equilibriumcondition between the expected return on the option, the expected returnon the stock, and the riskless interest rate. We will derive the formula inthis chapter. Since the publication of Black-Scholes’ and Merton’s papers,the growth of the field of derivative securities has been phenomenal. TheBlack-Scholes equilibrium formulation of the option pricing theory is attrac-tive since the final valuation of the option prices from their model depends on

47

a few observable variables except one, the volatility parameters. Thereforethe accuracy of the model can be ascertained by direct empirical tests withmarket data. When judged by its ability to explain the empirical data, theoption pricing theory is widely acclaimed to be the most successful theorynot only in finance, but in all areas of economics. In recognition of their pi-oneering and fundamental contributions to the pricing theory of derivatives,Scholes and Mer- ton received the 1997 Nobel Prize in Economics. Unfortu-nately, Black was unable to receive the award since he had already passedaway then. To begin with the Black-Scholes model, let us state the list ofassumptions underlying the Black-Scholes model.

i) The asset price follows the geometric Brownian motion discussed inChapter 6. That is,

(1) dSt = µSt dt+ σSt dBt

ii) The risk-free interest rate µ and the asset volatility σ are known func-tions.

iii) There are no transaction costs.iv)The asset pays no dividends during the life of the option.v) There are no arbitrage possibilities.vi) Trading of the asset can take place continuously.vii) Short selling is permitted.viii) We can buy or sell any number (not necessarily an integer) of the

asset. We note that the Black-Scholes model can be applied to asset modelsother than (1), such as jump-diffusion models, but it may be difficult to de-rive explicit formulas then, as we do have for geometric Brownian motions.However, this should not discourage their use, since an accurate numericalsolution is usually quite straightforward. We will relax assumptions (ii)–(iv)in the next chapter. For example, (iv) can be dropped if the dividends areknown beforehand. They can be paid either at discrete intervals or continu-ously over the life of the option. We will discuss them in the next chapter.

2 Derivation of the Black-Scholes Differential EquationSuppose that we have an option whose value V (S, t) depends only on S

and t. It is not necessary at this stage to specify whether V is a call ora put; indeed, V can be the value of a whole portfolio of different options

48

although for simplicity we can think of a simple call or put. Using Ito’slemma (Theorem 7.1) and noting that St follows (1), we can write

(2) dV = (µS∂V

∂S+

1

2σ2S2∂

2V

∂S2+∂V

∂t) dt+ σS

∂V

∂SdBt

This gives the stochastic process followed by V . Note that by (2) werequire V to have at least one t derivative and two S derivatives. Next weconstruct a portfolio consisting of longing one option and shorting a number∆ of the underlying asset. Here if ∆ < 0, we are in fact buying ∆ amountof the underlying asset. For example, if we have bought one put option, wemay want to buy certain amount of stock to minimize the risk. In that case,∆ < 0. The Black-Scholes idea is first to find this proportion ∆ so that theportfolio becomes deterministic. Note that the value of this portfolio is

(3) Π(t) = V (t)−∆S

The change in the value of this portfolio in one time-step dt is

(4) dΠ(t) = dVt −∆ dSt

where we assume ∆ is held fixed during the time-step. Substituting (1) and(2) into (4), we find

dΠt = (µS∂V

∂S+

1

2σ2S2 ∂

2V

∂S2+∂V

∂t− µ∆S) dt+ (σS(

∂V

∂S−∆) dBt)

since dSt = µS dt+ σS dBt. Note that there are two terms in the right handside. The first term is deterministic while the second term is stochastic asit involves the Brownian motion Bt. But if we choose ∆ = ∂V

∂S , then thestochastic term is zero, and it becomes

(6) dΠt = (1

2σ2S2∂

2V

∂S2+∂V

∂t) dt

Thus, it reduces the stochastic expression into a deterministic expression.We now appeal to the concepts of arbitrage and the assumption of no

transaction costs. The return on an amount Π invested in riskless assetswould see a growth of rΠ dt in a time dt. In fact, if dΠ were greater thanthis amount, rΠ dt, an arbitrager could make a guaranteed riskless profit byborrowing an amount Π to invest in the portfolio. The return for this risk-free strategy would be greater than the cost of borrowing. Conversely, dΠ

49

were less than rΠ dt, then the arbitrager would short the portfolio and investΠin the bank. Either way the arbitrager would make a riskless, no cost,instantaneous profit. The existence of such arbitrageurs with the ability totrade at low cost ensures that the return on the portfolio and on the risklessaccount are more or less equal. Thus, we should have

dΠ = rΠ dt,

and hence by (6),

(8) rΠ dt = (1

2σ2S2∂

2V

∂S2+∂V

∂t) dt

Now replace Π in (8) by V −∆S and letting ∆ = ∂V∂S and then divide both

sides by dt, We arrive at

(9)∂V

∂t+ rS

∂V

∂S+

1

2σ2S2 ∂

2V

∂S2− r V = 0.

This is the Black-Scholes partial differential equation. It is hard to over-emphasize the fact that, under the assumptions stated earlier, any derivativesecurity whose price depends only on the current value of S and on t, andwhich is paid for up-front, must satisfy the Black-Scholes equation. Manyseemingly complicated option valuation problems, such as exotic options,become simple when looked at in this way. Before moving on, we make threeremarks about the derivation we have just seen.

(i) By definition, the ”delta” ∆ = ∂V∂S is the amount of assets that we need

to hold to get a riskless hedge. For example, at expiry T , V (S, T ) = c(S, T ) =max(S − E, 0). Hence ∆ = 1 if S > E and ∆ = 0 if S < E. That means weneed to hold 1 stock if S > E as the buyer will come to exercise the option;and if S < E there is no need to hold any stock. The value of ∆ is thereforeof fundamental importance in both theory and practice, and we will returnto it repeatedly. It is a measure of the correlation between the movements ofthe option or other derivative products and those of the underlying asset.

(ii) Second, the linear differential operator given by

(10) LBS =∂

∂t+ rS

∂

∂S+

1

2σ2S2 ∂2

∂S2− r

has a financial interpretation as a measure of the difference between thereturn on a hedged option portfolio (the first two terms) and the return on

50

a bank deposit (the last two terms)–see (8). Although this difference mustbe identically zero for a European option in order to avoid arbitrage, we seelater that this need not be so for an American option.

(iii) Third, the Black-Scholes equation (9) does not contain the drift pa-rameter µ of the underlying asset. Hence the price of the options will beindependent of how rapidly or slowly an asset grows. The price will dependon the volatility σ however. A consequence of this is that two people mayhave quite different views on µ, yet still agree on the value of an option. Wewill return to this in Section 5.

3 Boundary and Final Conditions for European Options Equation (9)is the first partial differential equation (PDE) that we have derived in thiscourse. We now introduce a few basic points in the theory of PDE so that weare aware of what we are trying to achieve. By deriving the partial differentialequation (9) for a quantity such as an option price, we have made an enormousstep towards finding its value. we just need to solve the equation. Sometimesthis involves solution by numerical means if exact formula cannot be found.

The final condition (backward in time) is given by

c(S, T ) = max(S − E, T )

The boundary condition is given by

(14) c(0, T ) = 0( at S = 0), c(S, T ) = S − Ee−r(T−t)( as S →∞).

Theorem 1 The value of the vanilla European call is given by

(18) c(S, t) = c(S, T − t, E, T ) = S N(d1)− E−r(T−t)N(d2)

where

(19) N(d) = erf(d) =1√2π

∫ d

−∞e−

x2

2 dx,

the cumulative distribution function for the standard normal distribution,

(20) d1 =ln (SE ) + (r + σ2

2 )(T − t)σ√T − t)

, d2 =ln (SE ) + (r − σ2

2 )(T − t)σ√T − t)

51

Proof. We first check that c(S, t) really satisfies the Black-Scholes equation(9). We first note that for ω = t or S, we have

∂N(d)

∂ω=∂N(d)

∂d

∂d

∂ω= e−

d2

2∂d

∂ω

where

∂d1

∂t=

d1

2(T − t)− 1√

T − t(r

σ+σ

2),

∂d2

∂t=

d2

2(T − t)− 1√

T − t(r

σ− σ

2).

Hence we have(21)∂c∂t = S ∂N(d1)

∂d1

∂d1

∂t − re−r(T−t)N(d2)− e−r(T−t) ∂N(d2)

∂d2

∂d2

∂t

= Se−d212√

2π( d1

2(T−t) −1√T−t(

rσ + σ

2 ))− re−r(T−t)N(d2)− Ee−r(T−t)−d212√

2π( d1

2(T−t) −1√T−t(

rσ −

σ2 ))

Also, since∂d1

∂S=

1

Sσ√T − t

,

(22)∂c

∂S= N(d1) +

Se−d212

√2π

1

Sσ√T − t

− Ee−r(T−t)−d212

√2π

1

Sσ√T − t

×rS

Differentiating it once more, we get

(23)

∂2c∂S2 = Se−

d212√

2πSσ√T − t)− d1e

−d212√

2πSσ√T−t + Ee−r(T−t)−

d212√

2πS2σ√T−t + Ed2e

−r(T−t)−d212√

2πS2σ2(T−t)

= 2S2σ2 (S e−

d212√

2π( σ

σ√T−t)− d1

2(T−t))−Ee−r(T−t)−

d212√

2π( σσ√T−t + d2

2(T−t)). ×σ2S2

2

By substituting (18), (21)–(23) into the left hand side of the Black-Scholesequation (9), we see that it is indeed identically equal to zero. For theboundary condition at S = 0, we first note that by (19)–(20), d1 = d2 = −∞,N(−∞) = 0 and thus c(0, t) = 0. For the boundary condition at S =∞ wenote again that d1 = d2 = ∞, N(∞) = 1 and thus c(∞, t) = S − Ee−r(T−t).Finally, we consider the final condition (12). At t = T if S > E, then d1, d2 →∞ and if S < E, then d1, d2 → −∞. Hence c(S, T ) = max(S − E, 0).

Next we give the formula for European put options.

52

Theorem 2. The value of the vanilla European put is given by

(24) p(S, t) = e−r(T−t)N(−d2)− SN(−d1)

Proof. One can of course verify that the formula (24) does satisfy theBlack-Scholes equation and the boundary and final conditions for Europeanputs as we did in the proof of Theorem 1. However, there is a better wayto verify that. We can derive (24) immediately by using the put-call parityformula (see (4.7))

c(S, t)− p(S, t) = Ee−r(T−t).

Theorem 1, and the identity

N(d) +N(−d) = 1

We remark that although (18) and (24) seem to be closed-form solutions forthe vanilla options, one still has to compute the integral N(d1) and N(d2),numerically by quadrature rules such as Simpson’s rule or Gaussian rule.Next we compute the ”deltas”, the amount of the underlying asset that oneshould hold at any time t if one has short sell the option. Recall from (3)that the riskless portfolio is Π(t) = V (t) − ∆S with ∆ = σ ∂V∂S That is,whenever we buy (or sell) one option, we have to short sell (or buy) ∆ unitsof the underlying stock in order that the portfolio is riskless. Note that ∆ ischanging with time, and that means we have to do the hedging continuously.If one cannot compute one should not buy or sell the option.

Theorem 3 The deltas of vanilla call and put options are

∆c(S, t) = N(d1), ∆p(S, t) = N(d1)− 1

By using the ”deltas” to do the hedging, we see in (5) that the largest randomcomponent of the portfolio is eliminated. This hedging process is called deltahedging. Once can in fact hedge away higher order effect by knowing thefollowing quantities.

Definition 4 The gamma, theta, vega and rho of a portfolio Π are definedrespectively,

5 Pricing Options Using Risk Neutrality An important observation of the Black-Scholes equation (9) is that it does not contain the drift parameter µ of theunderlying asset. Hence we see in Theorems 1 and 2 that the price of vanilla

53

European options is independent of how rapidly or slowly an asset grows.The only parameter from the stochastic differential equation (1) for the assetprice that affects the option price is the volatility σ. A consequence of thisis that two people may differ in their estimates for µ yet still agree on thevalue of an option. Moreover, the risk preferences of investors are irrelevant:because the risk inherent in an option can all be hedged away, there is noreturn to be made over and above the risk-free return. The same conclusionis true for vanilla options as well as other derivative products. It is gener-ally the case that if a portfolio can be constructed with a derivative productand the underlying asset in such a way that the random component can beeliminated?as was the case in our derivation of the Black-Scholes equation(9)–then the derivative product may be valued as if all the random walks in-volved are risk- neutral. This means that the drift parameter in the stochasticdifferential equation for the asset can be replaced by r wherever it appears.The option is then valued by calculating the present value of its expectedreturn at expiry with this modification to the random walk. To apply thisoption pricing idea to our geometric Brownian motion model, we can firstreplace (1) by

dSt = rSt dt+ σS dBt

It is a risk-neutral world: we pretend that the random walk for the returnon St has drift r instead of µ. From this, we can calculate the probabilitydensity function of the future values of St. This is given by Theorem 7.2 withµ there replaced by r:

pSt(s) =1

σs√

2πte−|ln ( s

S0−(r−σ

2

2 )t|2/2σ2t

To evaluate the option price, we first calculate the expected payoff of theoption at expiry. Suppose its payoff function at T for any St is given byV (ST ). Then the expected payoff of the option at time T is

E[V (ST , T )] =1

σ√

2πT

∫ ∞0

V (s, T )

se−|ln ( s

S0−(r−σ

2

2 )T |2/2σ2T ds

The value of the option at present time t is then obtained by discounting this

54

amount of money at expiry back to current time:

(26)

V (S, t) = e−r(T−t)E[V (ST , T )|St = S]

= e−r(T−t)1

σ√

2π(T − t)

∫ ∞0

V (s, T )

se−|ln ( s

S0−(r−σ

2

2 )(T−t)|2/2σ2(T−t) ds.

One can verify that this solution indeed satisfies the Black-Scholes equation(9). If the payoff function V (S, T ) is simple, such as in the case of binaryoptions or vanilla options, one can integrate the integral to get the optionprice. If it is complicated, then one can use numerical quadrature rules orMonte Carlo methods to compute the integration.

Note that by replacing µ by r in our geometric Brownian motion model forSt in (1), we do not mean that µ = r. If it were correct, then all assets wouldhave the same expected return as a bank deposit and no one would investin the stock markets. It is just a trick to obtain the option price becausewe know that the value of the options does not depend on µ, and in a risk-neutral world, everything grows at a rate of r. We finally note that in therisk-neutral world, the asset price grows like:

St = S0e(r−σ

2

2 )t+σ Bt

cf. (7.7) where we have µ instead of r. It should be noted that from thederivation of the Black-Scholes equation, in particular from (8), in the risk-neutral world, everything grows with risk-free return rate r.

The call option price in (18) can be rewritten as

c(S0, 0) = e−rT (S0e−rTN(d1)− EN(d2)) (t = 0).

Note that the term N(d2) is the probability that the option will be exercisedin a risk-neutral world, i.e.,

Prob(ST ≥ E) =1

σ√

2πT

∫ ∞E

1

se−|ln ( s

S0−(r−σ

2

2 )T |2/2σ2T ds

=1

σ√

2πT

∫ ∞lnE

e−|w−ln S0−(r−σ2

2 )T |2/2σ2T dw = N(d2) (s = ew).

Therefore,

E[E 1ST≥E] = E

∫ ∞E

pST ds = EN(d2)

55

The term S0erTN(d1) is the expected value of a variable that is equal to ST

if ST ≥ E and to zero otherwise in a risk-neutral world, i.e.,

E[ST 1ST≥E] =

∫ ∞E

spST ds =

∫ ∞E

1

σ√

2πTe−|ln ( s

S0−(r−σ

2

2 )T |2/2σ2T ds

=1

σ√

2πT

∫ ∞lnE

ewe−|w−ln S0−(r−σ2

2 )T |2/2σ2T dw = S0erT N(d1).

Hence, we have Fynman-Kac formula

c(S, t) = e−r(T−t)E[(ST − E)1St≥E] = E[e−r(T−t)(ST − E)+]

which is just (26) with V (S, t) = c(S, t).

Chapter 9 Numerical Methods for Option PricingEquation (8.26) provides a way to evaluate option prices. For some simple

options,such as the European call and put options, one can integrate (8.26)directly and obtain a closed-form solution as in (8.18) or (8.24). However,for more complicated options, it may not be easy to do the integration andone has to resort to numerical means finding the option prices. There arethree common methods for evaluating option prices numerically: the binomialmethod, the Monte Carlo method,and the finite difference method. We beginwith the binomial method.

1. One-step Binomial ModelConsider a very simple situation where a stock price is currently $20 and it

is known that at the end of three months the stock price will be either $22 or$18. We suppose that the stock pays no dividends and that we are interestedin evaluating a European call option with exercise price $21 expiring in threemonths. This option will have one of the two values at the end of the threemonths. If the stock price turns out to be $22, the value of the option willbe $1; if the stock price turns out to be $18, the value of the option will bezero, see Figure1.

Consider a portfolio consisting of a long position in shares of the stockand a short position in one call option, i.e. Π(t) = −V + ∆S, where V isthe option price. If the stock price moves up from 20 to 22, the value of theshares is 22∆ and the value of the option is 1, so that the total value of theportfolio is 22∆− 1. If the stock price moves down from 20 to 18, the value

56

of the shares is 18∆ and the value of the option is 0. So that the total valueof the portfolio is 18∆. The portfolio is riskless if the value of is chosen sothat the final value of the portfolio is the same for both of the alternativestock prices. This means that 22∆− 1 = 18∆, or ∆ = .25.

So the value of the portfolio at expiry is

Π(T ) = 22∆− 1 = 4.5 = 18∆.

Riskless portfolio must, in the absence of arbitrage opportunities, earn therisk-free rate of interest. Suppose that in this case the risk-free rate is 12%per annum. It follows that the value of the portfolio to day (when t = 0)must be the present value 4.5, i.e.,

Π(0) = 4.5e−.12∗.25 = 4.367

The value of the stock price to day is known to be 20. If the option price isdenoted by V , the value of the portfolio to day is also given by

Π(0) = 20∆− V = 5− V.

It follows that 5− V = 4.367, or V = 0, 633In general, if the stock price S moves up to uS or down to dS with u > 1and

d < 1, and the corresponding payoffs for the option are Vu and Vd respectively,then we have

Π(T ) = uS∆− Vu = dS∆− Vdor

(1) ∆ =Vu − VdSu− Sd

.

Denoting the risk-free interest rate by r, the present value of the portfoliomust be

Π(0) = Π(T )e−rT ,

Since the cost of setting up the portfolio at t = 0 is Π(0) = S∆−V where Vis the option price now, it follows that

S∆− V = Π(0) = (Su∆− Vu)e−rT = (Sd∆− Vd)e−rT .

Thus,

V = (S − Sue−rT )∆ + Vue−rT = Vu−Vd

u−d (1− ue−rT ) + Vue−rT

= e−rT (erT−uu−d (Vu − Vd) + Vu) = e−rT (e

rT−du−d −

erT−uu−d )

57

Equivalently,

(2) V = e−rT (pVu + (1− p)Vd),

where

(3) p =erT − du− d

.

Notice that from (2), we have

pVu + (1− p)Vd = V erT

which is the expected option price at time T when there is no arbitrage. SinceVu is what the option will be worth when the stock price goes to uS, and Vdis what the option will be worth when the stock price goes to dS, the variablep here can be interpreted as the probability of an up movement in the stockprice, where as the quantity i−p is then the probability of a down movementin the stock price. More precisely, if the probability of S going up to uS isassumed to be p and there are no arbitrage, then the expected option priceat time T is precisely given by the fair value pVu + (1 − p)Vd. Moreover, inthis case, the expected stock price at time T is also given by

E[ST ] = pSu+ (1− p)Sd = pS(u− d) + Sd.

Using (3), it simplifies to

(4) E[ST ] = SerT .

which means that the stock price grows on average at the risk-free rate (andhence is the option). Setting the probability of the up movement equal to pis therefore equivalent to assuming that the return on the stock equals therisk-free rate. This is the same idea as the risk-neutrality we mentioned inSection 8.6, cf. (4) and (8.28). This risk-neutrality interpretation can beused to find the value of the option faster. In the previous example, p mustsatisfy the equation or

22p+ 18(1− p) = E[ST ] = 20e0.12×0.25,

and

(5) p =1

4(20e0.12×0.25 − 18) = 0.6523.

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Thus, the expected payoff for the option is

0.6523× 1 + 0.3477× 0 = 0.6523.

Discounting back to today at the risk-free rate, the value of the option todayis

0.6523e−0.12×0.25 = 0.633 = V.

2 Two-step Binomial Model We can extend the analysis above one stepfurther to a two-step binomial model. Consider a call option with a strikeprice $21 and expiring half year from now. Assume that the current stockprice is $20, and the risk-free interest rate is 12% per annum. We break thetime interval of half a year into two time-steps, each of length 3 months. Ineach of the two time-steps, we assume the stock price can only go up 10% ordown 10%. In Figure 3, the upper number of each node shows the stock price,which we can construct easily from the given information. For example, ifthe stock price is $22 at the 3-month time, then it can only be $24.2 or $19.8at the 6-month time. Note that because $22 x d = $19.8 = $18 x u, we onlyhave three nodes at the second time-step, and not four.

Next we compute the probability of p for each time-step according to(5). Using the data, u = 1.1, d = 0.9, r = 0.12, time-step= 0.25, we getp = 0.6523. Since we have the stock prices for the nodes D, E and F at theexpiry date, we can obtain the payoffs of the option at these nodes. Theyare given by the lower numbers at the nodes. With the payoffs of the optiongiven at nodes D and E, we can use (2) to compute the value of the optionat node B:

e−0.120.25(0.6523× 3.2 + 0.3477× 0) = 2.0257.

Similarly, using the value of the option at nodes E and F, we can find thevalues of the option at the node C (which obviously is equal to 0). Now werepeat the computation once again, using the value of the options at nodesB and C, we get the option price at node A:

e−0.12×0.25(0.6523× 2.0257 + 0.3477× 0) = 1.2823.

In general, if the stock prices move up or down by a factor u or d respectivelyin each time-step δt, then we have the binomial tree of Figure 4. As in theabove example, we find

Vu = e−rδt (p Vuu + (1− p)Vud), Vd = e−rδt (p Vud + (1− p)Vdd),

59

where δt is the length of one time-step. Then

V = e−rδt (p Vu + (1− p)Vd).

3 The Binomial Tree Model The binomial models described above can begeneralized to an M -step binomial model. We first construct a binomial treeof possible asset prices, and then evaluate the option prices backward onestep at a time as we did in the 2-step binomial method. Assume that at timet = 0 we know the asset price S0

0 = S. Then at the next time-step δt thereare two possible asset prices, S1

0 = dS00 and S1

1 = uS00 . Given u, d, r, δt we

can compute p as in (3):

p =erδt − du− d

At the following time, 2δt there are three possible asset prices, S20 = d2S0

0 , S21 =

udS00 and S2

2 = u2S00 . In general, at the m-th time mδt there are m+ 1 pos-

sible asset prices, Smn = dm−nun S00 , 0 ≤ n ≤ m. Thus, at the final time Mδt

there are M + 1 possible asset prices.

Example 1 If we set u = 1/d, then the possible stock prices can be simplifiedfurther

SMn = dM−2nS0 − 0 = u2n−MS00 .

Next we apply the payoff function for the option on SnM . For a put wehave,

V Mn = max(0, E − SMn ), 0 ≤ n ≤M,

where E is the exercise price and V Mn denotes the n-th possible value of

the put at time-step M where the asset value is SMn . For a call we haveV Mn = max(0, SMn − E). Starting with V M

n , using the risk-neutral argumentwe can calculate the value of the option at each possible asset price for time-step m by

V mn = e−δt(p V m+1

n+1 + (1− p)V m+1n )

4 Other Options by Binomial Methods To evaluate the option prices ofother options, say American options, we use the same idea. First, similar tothe European options, we construct a binomial tree of possible asset prices.Consider the situation at time-step m and at asset price Sm+1. The option

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n can be exercised prior to expiry to yield a profit determined by the payofffunction. For a put V m

n = max(0, E − Smn ) and V mn = max(0, Smn − E) for a

call. If the option is retained, its value is, as in the European case,

V mn = e−δt(p V m+1

n+1 + (1− p)V m+1n ), 0 ≤ n ≤ m.

The value of the American option is the maximum of two choices: the choiceof exercising the option and the choice of keeping the option, i.e.

V mn = max(max(E − Smn , 0), e−δt(p V m+1

n+1 + (1− p)V m+1n ))

for a put and

V mn = max(max(Smn − E, 0), e−δt(p V m+1

n+1 + (1− p)V m+1n ))

for a call.

5 Determination of the Parameters In Sections 1–4, we assume we knowu and d, and then we calculate p using (6) and then V accordingly. In general,the parameters p, u and d are unknown, and have to be chosen so that theygive the correct values for the mean and variance of stock price changes duringa time interval of length δt.

Under the risk-neutral assumption, the expected return from a stock is therisk-free interest rate, r Hence at the end of the time interval δt, the expectedvalue of the stock. Hence at the end of the time interval δt, the expectedvalue of the stock E[Sδt] is Seδt, see (4) or Corollary 8.5. It follows that

Seδt = E[Sδt] = p Su+ (1− p)Sd,

or

(8) eδt = p u+ (1− p) d.

By Corollary 8.5 again,

V ar(Sδt) = S2e2δt(eσ2δt − 1).

Since V ar(Sδt) = E[S2δt]− E[Sδt]

2 we and

E[S2δt] = p(Su)2 + (1− p)(Sd)2,

we have

S2e2δt(eσ2δt − 1) = p(Su)2 + (1− p)(Sd)2 − (p Su+ (1− p)Sd)2

61

or

(9) e2rδt+σ2δt = p u2 + (1− p) d2.

Equations (8) and (9) impose two conditions on p, u and d. We still needone more equation to determine them. One can equate the third moment ofSδt, but it will lead to extremely difficult expressions. A popular choice forthe third condition is u = 1/d and we obtain

(11) d = A−√A2 − 1, u = A+

√A2 − 1

where

A =1

2(e−rδt + e(r+σ2)δt).

Another popular choice is to set p = 12 and

d = eδt(1−√eσ2δt − 1), u = eδt(1 +

√eσ2δt − 1)

These assumptions may give different pricing of the options, but one canshow that if δt→ 0, then the two pricings should be the same.

The binomial method is extremely memory efficient for options with onlyone underlying asset: although there are O(M 2) nodes, the memory requiredgrows only linearly with the number of time-steps, i.e. it is of order O(M)only. The reason is that we can re-use the memory in the (m + 1)-th time-step for the m-th time-step. Thus the largest memory requirement is at thelast time-step M , where we require 2(M + 1) memory for the stock pricesand the option prices. Since there are O(M 2) nodes, and each node requiresO(1) calculations, the execution time grows quadratically with the numberof time-steps, i.e. of order O(M 2). For an option whose price depends ond stock prices, the binomial method will require O(Md) memory locationsand O(Md+1) calculations. Thus if d is large, it is not an efficient method orone needs a very fast computer to compute such options. It is interesting tonote that about 10% of the world top 500 fastest computers are installed infinancial institutions across the world to compute option prices.6 Monte Carlo Method Monte Carlo method is another name for simula-tion method. Here we try to simulate the stock price at the expiry date Taccording to the log-normal process (7.7). If we know the stock price S(T )at T , then we know the option price c(S, T ) at T . By discounting that backto the current time t, we know the option price c(S, t) at t.

62

However, in a risk-neutral world (or when we are only interested in com-puting the option prices), µ is taken as the interest rate r, i.e.,

ST = Ste(r−σ

2

2 )(T−t)+σε√T−t, ε ∼ N(0, 1).

In general, we fix a time-step δt = T−tM and the number of paths N which

usually goes from 10,000 to 1,000,000, depending on the accuracy we want.Then we generate the j-th path from t to T by

(15) Sji = Sji−1e(r−σ

2

2 )δt+σεji√δt, 1 ≤ i ≤M, 1 ≤ j ≤ N

where εji are random numbers distributed as N(0, 1).Note that if we are only interested in evaluating European options, there

is no need to know what the stock prices are in between t and T ; we onlyneed to know the stock price at T . We can modify (15) to get it directly:

SjT = Se(r−σ2

2 )(T−t)+σεj√T−t

i.e., for each εj sampled from N(0, 1), we can generate one sample of SjT .From each of them, a payoff for the option can be calculated at expiry. Forexample, if it is a European call option, we can computeV j = max(SjT −E, 0), 0 ≤ j ≤ N . Then the expected value of the payoff can be estimatedas the arithmetic average of these payoffs, i.e.

E[VT ] =1

N

N∑j=1

V j

The current value of the option can be calculated by discounting the expectedpayoff value, i.e.,

V = e−r(T−t)E[VT ].

The number of simulation is carried out depends on the accuracy required.The Central Limit Theorem states that suppose X1, X2, · · · is a sequence ofi.i.d. (independently identically distributed) random variables with E[Xi] =µ and V ar[Xi] = σ2 Then as n (number of samples) approaches infinity, therandom variables

√n(Sn−µ) converge in distribution to a normal distribution

N(0, σ2), i.e.√n

((1

n

n∑i=1

Xi

)− µ

)d−→ N(0, σ2).

63

This shows that our uncertainty about the value of the option is inverselyproportional to the square root of the number of trials N . That would meanthat to double the accuracy of the simulation, we must quadruple the numberof trials; and to increase the accuracy by a factor of 10, the number of trialsmust be increased by a factor of 100.

Monte Carlo simulation tends to be numerically more efficient than otherprocedures when there are three or more underlying assets that the optionprice is dependent on. This is because the time taken to carry out a MonteCarlo simulation increases approximately linearly with the number of un-known variables, i.e. it is of order Q(MN). In contrast, the time taken formost other procedures increases exponentially with the number of unknownvariables. For example, the cost of the binomial method and the finite dif-ference method both increase like O(Md+1) where d is number of underlyingassets. Monte Carlo simulation also has the advantage that it provides astandard error for the estimates that are made. It is an approach that canaccommodate complex payoffs and complex stochastic processes. It can beused when the payoff depends on some function of the whole path followed bya variable, not just its terminal value. For example, we can use it to computethe following exotic options easily.

Barrier options are exotic options where the payoff depends on whether theunderlying asset’s price reaches a certain level during a certain period of time.A down-and-out put option is a regular European put option with strike priceE and maturity T but ceases to exist if the asset price drops below or equalto a given barrier level B, i.e.,

payoff at maturity =

0 if St′ ≤ B for some t′ ≤ T

max(E − ST , 0), otherwise.

Lookback options are exotic options where the payoff depends on the maxi-mum or minimum asset price reached during the life of the option. A Euro-pean fixed lookback put option with strike price E and maturity T has thefollowing payoff:

payoff at maturity = max(E − Smin, 0),

where Smin is the minimum asset price achieved during the life of the option.

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Asian options are exotic options where the payoff depends on the averageprice of the underlying asset during the life of the option. An average priceput option with strike price E and maturity T has the following payoff:

payoff at maturity = max(E − Save, 0),

where Save is the average value of the underlying asset calculated during thelife of the option.The limitations of the Monte Carlo simulation are that it converges slowly,and it is used normally for European options, for it is very difficult to use forAmerican options.7 Finite Difference Approximation Besides the Black-Scholes equation,in practice, there are many other options that can be modeled by partialdifferential equations too. Unfortunately not many of them have closed-formsolutions as in (8.18) or (8.24). In this section, we discuss how to solvepartial differential equations by finite difference approach. The main ideais to approximate the differential operators in the differential equation bydifference operators.

Given a function u(x), using Taylor’s expansion, we have

u(x+ δx) = u(x) + δx u′(x) + δx2

2 u′′(x) + δx3

6 u′′′(x) +O(δx4)

u(x− δx) = u(x)− δx u′(x) + δx2

2 u′′(x)− δx3

6 u′′′(x) +O(δx4)

If δx is small, we can approximate the derivative u′(x) at x by the backwarddifference or more accurately, by the ”central difference”;

(18)du

dx=u(x)− u(x− δx)

δx+O(δx),

(19)du

dx=u(x+ δx)− u(x− δx)

2δx+O(δx2).

From (16) and (17), we can also approximate the second order derivativeu′′(x) at x by the central difference;

(20)d2u

dx2=u(x+ δx)− 2u(x) + u(x− δx)

δx2+O(δx2).

Suppose we are to solve the Black-Scholes equation for a European optionV (S, t),

Vt +σ2S2

2

∂2V

∂S2+ rS

∂V

∂S− r V (S, t) = 0

65

with given terminal condition V (S, T ) (for 0 < S < ∞) and boundary con-ditions V (S, t) and V (∞, t) (for 0 < t < T ). Since it is impossible to workwith S =∞ numerically, we first replace the boundary S =∞ by S = Smaxwhere Smax is a sufficiently large stock price. Then we replace the conditionV (∞, t) by V (Smax, t), see Figure 9.

Next we partition the solution domain [0, Smax]× [0, T ] by grid lines. Moreprecisely, we partition [0, [0, Smax] into N equal sub-intervals, each of lengthδS and [0, T ] into M equal sub-intervals, each of length δt see Figure 9. LetSj = jδSand ti = iδt for 0 ≤ j ≤ N and 0 ≤ i ≤M . Note that for 0 ≤ i ≤Mand 0 ≤ j ≤ N , V (S0, ti), V (SN , ti) and V (Sj, tM) are given boundary andfinal conditions. They are known numbers. Our problem is to find V (Sj, ti)for 0 ≤ i ≤M and 0 ≤ j ≤ N , which are the option prices inside the solutiondomain. Our idea of finding them is to replace the derivatives in (21) by thefinite differences in (18)–(20). This reduces the partial differential equationto a difference equation. We then solve the difference equation in a time-marching manner, marching back from time T to 0, at one δt a time. Let usillustrate that by computing Vj,i ∼ V (Sj, ti) for 0 ≤ j ≤ N, 0 ≤ i < M . Letus consider (21) at the point (Sj, ti) and apply a backward difference (18) atti and central differences (19) and (20) at Sj;(23)Vj,i − Vj,i−1

δt+σ2

2SjVj+1,i − 2Vj,i + Vj−1.i

δS2+ rSj

Vj+1,i − Vj−1,i

2δS− r Vj,i = 0,

0 < j < N , i.e., we update Vj,i−1, 1 < j < N from Vj,i, 1 < j < N.Solution domain of European options and the grids. Once we have ob-

tained all the option prices at the time step ti, we can march one δt backwardin time to obtain Vj,i−1 for all 0 ≤ j ≤ N from Vj,i starting from i = M .

Obviously, the accuracy of the solution depends on the size of δt and δS.From (18)–(20), we see that the error in approximating the partial differentialequation (21) by the finite difference equation (23) is O(δt) and O(δS2). Inorder to balance the two errors, one would choose δt = O(δS2). That wouldimply that δt will be a very small number and hence M will be very big(of O(N 2)). That would mean that we will have to do many time-marchingsteps. Unfortunately, there is no way to overcome this. In fact, one canshow that a = δt/(δS)2 has to be less than a certain constant a0 in orderthat Vj,i of (23) converges to the true solution of (21) when δt and δS go to0. If a ≥ a0, then Vj,i actually diverges. For some problems, the condition

66

a < a0 will restrict δt to a very small number, and hence the method may betoo slow to find the solution. There are better finite difference schemes thatcan circumvent this problem but they will require more computational costs.Example 8. Consider a European put option with the exercise price E = 10and T = 1/2 year. Assume r = 0.05 and σ = 0.2. Using the method, we canobtain the approximate option prices for different asset values S0, which areshown in Table 7. Here we choose Smax = 25, and the number of intervals for[0, T ] is M = 60, i.e., every three days. The step-size δS is determined by a.The exact solution is obtained by the closed-form formula (8.24). We see thatthe method diverges when a > 0.1, and the solution becomes oscillatory, seeFigure 10. The finite difference method is extremely efficient for options withonly one underlying asset: its computational cost is of O(MN). However,it also has the curse of dimension in that the cost is O(MNd) for an optionwhose price depends on d stocks.

The time-marching scheme (23) for computing European option prices canbe adapted easily to compute the American option prices. The main idea isthat once Vj,i is computed by (23), we have to compare it with the exerciseprice at (Sj, ti) to determine the true value of the option at (Sj, ti). Forexample, if we are computing the call option price, and ch(Sj, ti), is the pricecomputed by (23), then c(Sj, ti) = max(ch(Sj, ti), Sj − E).

67