Stochastic Modelling in Finance - Personalpersonal.strath.ac.uk/x.mao/talks/sm_fin10.pdf ·...

Option and ProbabilityStochastic ModellingWell-known Models

Monte Carlo Simulations

Stochastic Modelling in Finance

Xuerong Mao FRSE

Department of Mathematics and StatisticsUniversity of Strathclyde

Glasgow, G1 1XH

April 2010

Xuerong Mao FRSE SM in Finance



Outline1 Option and Probability

OptionOption value

2 Stochastic ModellingLinear modellingNonlinear modelling

3 Well-known ModelsThe Black–Scholes worldNon-linear SDE models

4 Monte Carlo SimulationsWhy Monte Carlo simulationsTest problemEM method for nonlinear SDEsEM method for financial quantities




OptionOption value


OptionOption value







OptionOption value

Problem

Assume that on 10 April 2010, Mr King has $100K to invest for1 year and he has two choices:

(a) invest the money in a bank saving account to receive arisk-free interest.

(b) buy a $100K house and then sell it on 10 April 2011.

Which choice should Mr King take?




OptionOption value

Assume the annual interest rate r = 1% and let X denote theprice of the house on 10 April 2011. Consider cases:

(i) P(X = $110K ) = 0.5 and P(X = $90K ) = 0.5. That is, theprice will increase or decrease by 10% equally likely.In this case, EX = $100K so it is better to deposit themoney in the bank which gives Mr King $101K by 10 April2011.

(ii) P(X = $110K ) = 0.6 and P(X = $90K ) = 0.4.In this case, EX = 0.6× $110K + 0.4× $90K = $102K ,which is $1K better than the return of the saving account.




OptionOption value

Assume that you trust the housing market will obey Case (ii).Should you have $100K available, you would have invested itinto the house to obtain the expected profit of $2K.

The problem is that you do NOT have the capital of $100K andyou just feel unfair to give the opportunity to rich people like MrKing.

However, Professor Mao would like to help. On 10 April 2010,Professor Mao (the writer) writes a European call option thatgives you (the holder) the right to buy 1 house for $100K on 10April 2011 from Prof Mao if you wish.




OptionOption value

European call option

DefinitionA European call option gives its holder the right (but not theobligation) to purchase from the writer a prescribed asset for aprescribed price at a prescribed time in the future.

The prescribed purchase price is know as the exercise price orstrike price, and the prescribed time in the future is known asthe expiry date.




OptionOption value

On 10 April 2011 you would then take one of two actions:

(a) if the actual value of a house turns out to be $110K youwould exercise your right to buy 1 house from ProfessorMao at the cost $100 and immediately sell it for $110Kgiving you a profit of $10K.

(b) if the actual value of a house turns out to be $90K youwould not exercise your right to buy the house fromProfessor Mao—the deal is not worthwhile.




OptionOption value

Note that because you are not obliged to purchase the house,you do not lose money. Indeed, in case (a) you gain $10K whilein case (b) you neither gain nor lose.

Professor Mao on the other hand will not gain any money on 10April 2011 and may lose an unlimited amount.

To compensate for this imbalance, when the option is agreedon 10 April 2010 you would be expected to pay Professor Maoan amount of money to buy the "right". (The fair amount isknown as the value of the option.)

Question: Should Professor Mao charge you $2K, do you wantto sign the option?




OptionOption value

Let C denote the payoff of the option on 10/04/2011. Then

C =

{$10K if X = $110K ;$0 if X = $90K .

Recalling the probability distribution of X

P(X = $110K ) = 0.6, P(X = $90K ) = 0.4.

we obtain the expected payoff

EC = 0.6× $10K + 0.4× $0 = $6K .

But $1K saved in a bank for a year will only grow to

(1 + 1%)× $1K = $1.01K .

Therefore, the option produces the expected profit

$6K − $1.01 = $4.99K .




OptionOption value

It is significant to compare your profit with Mr King’s.

Mr King invests his $100K in the house and expects to make$1K more profit than saving his money in a bank.

You pay only $2K for the option and expect to make $4.99Kmore profit than saving your $2K in a bank.

It is even more significant to observe that you only need $2K,rather than $100K, in order to get into the market.




OptionOption value

However, should Professor Mao charge you $5.99K, doyou want to sign for the option?If you save your $5.99K in a bank, you will have

(1 + 1%)× $5.99K = $6.05K

which is $50 better off than EC = $6K , the expected payoffof the option. You should therefore not sign the option.




OptionOption value


OptionOption value







OptionOption value

Key question

How much should the holder pay for the privilege of holding theoption? In other words, how do we compute a fair price for thevalue of the option?




OptionOption value

In the simple problem discussed above, the fair price of theoption is

EC1 + r

=$6K

1 + 1%= $5940.59

However, the idea can be developed to cope with morecomplicated distribution.




OptionOption value

Example

Assume that the house price will increase by 5% per half a yearwith probability 60% but decrease by 4% per half a year withprobability 40%. Then the house price X on 2011 will have theprobability distribution:

X (in K$) | 92.16 100.80 110.25P | 0.16 0.48 0.36

Hence

EC = 0.48× 0.80K + 0.36× $10.25K = $4.074K

and the option value is

EC1 + r

=$4.074K1 + 1%

= $4.034K




OptionOption value

Example

Assume that the house price will increase by 3% per quarterwith probability 60% but decrease by 2% per quarter withprobability 40%. Then the house price X on 10/04/2011 willhave the probability distribution:

X (in K$) |92.237 96.943 101.889 107.087 112.551P |0.0256 0.1536 0.3456 0.3456 0.1296

Hence

EC = 0.3456×1.889+0.3456×7.087+0.1296×12.551 = 4.729

and the option value is

EC1 + r

=4.729

1 + 1%= 4.682




OptionOption value

The model discussed before is the well-knownCox–Ross–Rubinstein (CRR) binomial model. This model canbe simulated easily by R.




OptionOption value

> n=4 # number of periods > up=0.03 # increase percentage > dw=0.02 # decrease percentage > upno <- 0:n > p0=100 # initial house price > pT <-p0*(1+up)^upno*(1-dw)^(n-upno) #prices at expiry date > pT [1] 92.23682 96.94278 101.88884 107.08725 112.55088 > upprob <- 0.6 # prob of increase > prob <- dbinom(0:n,n,upprob) \#binomial distribution > prob [1] 0.0256 0.1536 0.3456 0.3456 0.1296 > K=100 #strike price > payoff <-numeric() > {for (i in 1:(n+1)) + if (pT[i]>K) payoff[i]=pT[i]-K else payoff[i]=0} > payoff [1] 0.000000 0.000000 1.888836 7.087246 12.550881 > meanpayoff <- sum(payoff*prob) > meanpayoff [1] 4.728728 > r =0.01 # riskfree interest rate > optionvalue <- meanpayoff/(1+r) > optionvalue [1] 4.681909

x

x




OptionOption value

R-simulation for the12-month CRR binomial model > n=12 > up=0.01 > dw=0.009 > upno <-0:n > p0=100 > pT <- p0*(1+up)^upno*(1-dw)^(n-upno) > upprob= 0.6 > prob<-dbinom(0:n,n,upprob) > K=100 > payoff <- numeric() > {for (i in 1:(n+1)) + if (pT[i]>K) payoff[i]=pT[i]-K else payoff[i]=0} > meanpayoff <-sum(payoff*prob) > r=0.01 > optionvalue <- meanpayoff/(1+r) > optionvalue [1] 3.238385

x

x




OptionOption value

R-simulation for 365-day CRR model:

CRR(365, 0.00028, 0.00028, 100, 0.6, 100, 0.01)

produces the option value $2.044K




OptionOption value

However, the housing price, or more generally, an asset price ismuch more complicated than the binomial distributionsassumed above.

How might we model an asset price?




Linear modellingNonlinear modelling

Now suppose that at time t the underlying asset price is x(t).Let us consider a small subsequent time interval dt , duringwhich x(t) changes to x(t) + dx(t). (We use the notation d · forthe small change in any quantity over this time interval when weintend to consider it as an infinitesimal change.) By definition,the intrinsic growth rate at t is dx(t)/x(t). How might we modelthis rate?






OptionOption value








If, given x(t) at time t , the rate of change is deterministic, say r ,then

dx(t)x(t)

= rdt .

This gives the ordinary differential equation (ODE)

dx(t)dt

= rx(t).

Thenx(t) = x(0)ert

For example, if you invest x(0) into a bond with the risk-freeinterest rate r , then your return (capital plus interest) by time tis x(t).





However the rate of change is in general not deterministic as itis often subjective to many factors and uncertainties e.g.system uncertainty, environmental disturbances. To model theuncertainty, we may decompose

dx(t)x(t)

= deterministic change + random change.





The deterministic change may be modeled by

µdt

where µ is the average rate of change. So

dx(t)x(t)

= µdt + random change.

How may we model the random change?





In general, the random change is affected by many factorsindependently. By the well-known central limit theorem thischange can be represented by a normal distribution with meanzero and and variance σ2dt , namely

random change = N(0, σ2dt) = σ N(0,dt),

where σ is the standard deviation of the rate of change , andN(0,dt) is a normal distribution with mean zero and andvariance dt . Hence

dx(t)x(t)

= µdt + σN(0,dt).





A convenient way to model N(0,dt) as a process is to use theBrownian motion B(t) (t ≥ 0) which has the followingproperties:

B(0) = 0,dB(t) = B(t + dt)− B(t) is independent of B(t),dB(t) follows N(0,dt).





The stochastic model can therefore be written as

dx(t)x(t)

= µdt + σdB(t),

or

dx(t) = µx(t)dt + σx(t)dB(t)

which is a linear stochastic differential equation (SDE)—theNobel prize winning Black–Scholes model.






OptionOption value








If the rate of change and the standard deviation depend on x(t)at time t , the the model become nonlinear. In this case, thedeterministic change may be modeled by

R̄dt = R̄(x(t), t)dt

where R̄ = r̄(x(t), t) is the average rate of change given x(t) attime t , while

random change = N(0,V 2dt) = V N(0,dt) = VdB(t),

where V = V (x(t), t) is the standard deviation of the rate ofchange given x(t) at time t .





The stochastic model can therefore be written as

dx(t)x(t)

= R̄(x(t), t)dt + V (x(t), t)dB(t),

or

dx(t) = R̄(x(t), t)x(t)dt + V (x(t), t)x(t)dB(t)

which is a nonlinear stochastic differential equation (SDE).




The Black–Scholes worldNon-linear SDE models


OptionOption value








The Nobel prize winning Black–Scholes model

dx(t) = µx(t)dt + σx(t)dB(t)

is also known as the geometric Brownian motion.






Given the asset price S(t) = S at time t , a European call optionis signed with the exercise price K and the expiry date T . Thevalue of the option is denoted by C(S, t).

The payoff of the option at the expiry date is

C(S,T ) = (S − K )+ := max(S − K , 0).

The Black–Scholes PDF

∂V (S, t)∂t

+ 12σ

2S2∂2V (S, t)∂S2 + rS

∂V (S, t)∂S

− rV (S, t) = 0.





Regardless whatever the growth rate µ the individual holdermay think, the fair option value should be priced based on thefollowing SDE

dx(u) = rx(u)du + σx(u)dB(u), t ≤ u ≤ T , x(t) = S,

where r is the risk-free interest rate, rather than the individualSDE

dy(u) = µy(u)du + σy(u)dB(u), t ≤ u ≤ T , y(t) = S

that the holder may think. Hence the expected payoff at theexpiry date T is

E(x(T )− K )+

Discounting this expected value in future gives

C(S, t) = e−r(T−t)E[max(x(T )− K ,0)].





The solution

x(T ) = S exp[(r − 1

2σ2)(T − t) + σ(B(T )− B(t))

]gives

log(x(T )) = log(S)+(

r−12σ2)

(T−t)+σ(B(T )−B(t)) ∼ N(µ̂, σ̂2),

where

µ̂ = log(S) +(

r − 12σ2)

(T − t), σ̂ = σ√

T − t .

HenceZ :=

log(x(T ))− µ̂σ̂

∼ N(0,1)

which givesx(T ) = eµ̂+σ̂Z .





TheoremThe explicit BS formula for the value of the European call optionis

C(S, t) = SN(d1)− Ke−r(T−t)N(d2),

where N(x) is the c.p.d. of the standard normal distribution,namely

N(x) =1√2π

∫ x

−∞e−

12 z2

dz,

while

d1 =log(S/K ) + (r + 1

2σ2)(T − t)

σ√

T − tand

d2 =log(S/K ) + (r − 1

2σ2)(T − t)

σ√

T − t.






OptionOption value








Square root process

IfR̄(x(t), t) = µ, V (x(t), t) =

σ√x(t)

,

then the SDE becomes the well-known square root process

dx(t) = µx(t)dt + σ√

x(t)dB(t).





Mean-reverting square root process

IfR̄(x(t), t) =

α(µ− x(t))

x(t), V (x(t), t) =

σ√x(t)

,

then the SDE becomes

dx(t) = α(µ− x(t))dt + σ√

x(t)dB(t).

This is the mean-reverting square root process.





Theta process

IfR̄(x(t), t) = µ, V (x(t), t) = σ(x(t))θ−1,

then the SDE becomes

dx(t) = µx(t)dt + σ(x(t))θdB(t),

which is known as the theta process.




Why Monte Carlo simulationsTest problemEM method for nonlinear SDEsEM method for financial quantities


OptionOption value








Most of SDEs used in practice do not have explicit solutions.Monte Carlo simulations have been widely used to simulate thesolutions of nonlinear SDEs. There are two main motivationsfor such simulations:

using a Monte Carlo approach to compute the expectedvalue of a function of the underlying underlying quantity, forexample to value a bond or to find the expected payoff ofan option;generating time series in order to test parameterestimation algorithms.

Question: Can we trust the Monte Carlo simulations?






OptionOption value








Solution of a linear SDEThe linear SDE

dX (t) = 2X (t)dt + X (t)dB(t), X (0) = 1

has the explicit solution

x(t) = exp(1.5t + B(t)).

The Monte Carlo simulation can be carried out based on theEuler-Maruyama (EM) method

x(0) = 1, x(i + 1) = x(i)[1 + 2∆ + ∆Bi ], i ≥ 0,

where ∆Bi = B((i + 1)∆)− B(i∆) ∼ N(0,∆).





0.0 0.5 1.0 1.5 2.0

24

68

1012

t

X(t)

or x

(t)

true solnEM soln

0.0 0.5 1.0 1.5 2.0

05

1015

2025

t

X(t)

or x

(t)

true solnEM soln

0.0 0.5 1.0 1.5 2.0

020

6010

014

0

t

X(t)

or x

(t)

true solnEM soln

0.0 0.5 1.0 1.5 2.0

24

68

10

t

X(t)

or x

(t)

true solnEM soln

x

x





Example - the Black-Scholes model

Consider a BS model

dS(t) = 0.05S(t)dt + 0.03S(t)dB(t), S(0) = 10

and a European call option with the exercise price K = 10.05 atexpiry time T = 1, where 0.05 is the risk-free interest rate and0.03 is the volatility. By the well-known Black-Scholes formulaon the option, we can compute the value of a European calloption at time zero is

C = 0.4487318.

On the other hand, we can let ∆ = 0.001, simulate 1000 pathsof the SDE, compute the mean payoff at T = 1, discounting itby e−0.05, we get the estimated option value

C∆ = 0.4454196





To be more reliable, we can carry out such simulation, say 10times, to get 10 estimated values:

0.4454196, 0.4611569, 0.4512847, 0.4490462, 0.4294038,

0.4618921, 0.4556195, 0.4559547, 0.4399189, 0.4489594.

Their mean valueC̄∆ = 0.4498656

gives a better estimation for C.






OptionOption value








Typically, let us consider the square root process

dS(t) = rS(t)dt + σ√

S(t)dB(t), 0 ≤ t ≤ T .

A numerical method, e.g. the Euler–Maruyama (EM) methodapplied to it may break down due to negative values beingsupplied to the square root function. A natural fix is to replacethe SDE by the equivalent, but computationally safer, problem

dS(t) = rS(t)dt + σ√|S(t)|dB(t), 0 ≤ t ≤ T .





Discrete EM approximation

Given a stepsize ∆ > 0, the EM method applied to the SDEsets s0 = S(0) and computes approximations sn ≈ S(tn), wheretn = n∆, according to

sn+1 = sn(1 + r∆) + σ√|sn|∆Bn,

where ∆Bn = B(tn+1)− B(tn).





Continuous-time EM approximation

s(t) := s0 + r∫ t

0s̄(u))du + σ

∫ t

0

√|s̄(u)|dB(u),

where the “step function” s̄(t) is defined by

s̄(t) := sn, for t ∈ [tn, tn+1).

Note that s(t) and s̄(t) coincide with the discrete solution at thegridpoints; s̄(tn) = s(tn) = sn.





The ability of the EM method to approximate the true solution isguaranteed by the ability of either s(t) or s̄(t) to approximateS(t) which is described by:

Theorem

lim∆→0

E(

sup0≤t≤T

|s(t)−S(t)|2)

= lim∆→0

E(

sup0≤t≤T

|s̄(t)−S(t)|2)

= 0.






OptionOption value








Bond

If S(t) models short-term interest rate dynamics, it is pertinentto consider the expected payoff

β := E exp

(−∫ T

0S(t)dt

)from a bond. A natural approximation based on the EM methodis

β∆ := E exp

(−∆

N−1∑n=0

|sn|

), where N = T/∆.

Theorem

lim∆→0|β − β∆| = 0.






A European call option with the exercise price K at expiry timeT pays S(T )− K if S(T ) > K otherwise 0.

TheoremLet r be the risk-free interest rate and define

C = e−rT E[(S(T )− K )+

],

C∆ = e−rT E[(s̄(T )− K )+

].

Thenlim

∆→0|C − C∆| = 0.





Up-and-out call option

An up-and-out call option at expiry time T pays the Europeanvalue with the exercise price K if S(t) never exceeded the fixedbarrier, c, and pays zero otherwise.

TheoremDefine

V = E[(S(T )− K )+I{0≤S(t)≤c, 0≤t≤T}

],

V∆ = E[(s̄(T )− K )+I{0≤s̄(t)≤c, 0≤t≤T}

].

Thenlim

∆→0|V − V∆| = 0.


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