MATH 360 (Mathematics of Investment)

MATH 360: Theory of Investment and Credit

Albert Cohen

Actuarial Sciences ProgramDepartment of Mathematics

Department of Statistics and ProbabilityC336 Wells Hall

Michigan State UniversityEast Lansing MI

[email protected]@stt.msu.edu

Albert Cohen (MSU) MATH 360: Theory of Investment and Credit MSU Spring 2014 1 / 223

Logistics

All course materials available on class page (www.math.msu.edu)

Syllabus too

Course textbook is S.A. Broverman’s ”Mathematics of Investmentand Credit”, 5th edition or later

Some questions on these slides, and on in class exam preparationslides, are taken from the third edition of Broverman’s book. Pleasenote that Actex owns the copyright for that material. No portion ofthe ACTEX textbook material may be reproduced in any part or byany means without the permission of the publisher. We are verythankful to the publisher for allowing posting of these notes on ourclass website.

Supplementary book is Finan, available online


Introduction

The value of money is a function of the time that passes while it is stuffedunder a mattress, deposited in a bank account, or invested in an asset.Just what that function is depends on many circumstances, and we willspend our time investigating many real-world examples


Main Tools

Calculus, and the study of Geometric Series, provides

the machinery necessary for solving for the final value of aninvestment

the optimal time to switch accounts

the interest rate needed to plan fixed income investments

among other wonderful things


Main Tools

Calculus (1-d only!)

Can you take a derivative?

Can you find the maximum of a function?

Sequences and Series

Do you know what a geometric series is? What happens as n→∞ ?

n∑k=0

λk =1− λn+1

1− λ(1)


Interest Accumulation and Effective Rates of Interest (Ex1.1)

Begin by investing 1000 at 9% per annum. Compound annually for 3years. End up with...


Interest Compounded Monthly

Sometimes, interest rate is quoted per annum, but compounded monthly.

For previous example, assume again 9% but compounded monthly. Then

FV = 1000 ·

[(1 +

0.09

12

)12]3

= 1000 · (1.0938)3

(2)

In other words, the equivalent or effective annual rate is 9.38%


Fluctuating Interest Rates

It may be that over a longer period of time, the rate of return on an initialinvestment A(0) may fluctuate. At time n, the value of the investment isnow A(n).


Fluctuating Interest Rates

Guess: A(n) = A(0) · (1 + r)n

Reality: A(n) = A(0) · (1 + r1) · ... · (1 + rn)

In a bank account or bond, the interest rate, ri is guaranteed to be smallbut positive. For a general investment, there is no guarantee that aninvestment will not shrink in value over a period of time.


Arithmetic vs Geometric Means

One other possible guess for an average rate of return is r = r1+...rnn . The

question of ordering, ie which average rate is bigger, reduces to thequestion:

Is(

1 +r1 + ...rn

n

)n> (1 + r1) · ... · (1 + rn)? (3)

To answer this, define for k ∈ 1, 2, .., n,

xk = 1 + rk . (4)

Then, our question is reframed as:

Is(x1 + ...xn

n

)n> (x1) · ... · (xn)? (5)



The answer is yes, and there are many proofs. One of them is byinduction. Another is to apply Jensen’s Inequality:

For

a concave function f , for example if f ′′(x) ≤ 0 for all x in our domain

real numbers ak such that∑n

k=1 ak 6= 0,

real numbers xk ,

it follows that

f

(∑nk=1 akxk∑nk=1 ak

)≥∑n

k=1 ak f (xk)∑nk=1 ak

. (6)



By Jensen’s Inequality, if we define f (x) = ln (x) and each ak = 1n , then

we retain

ln(x1 + ...+ xn

n

)≥ 1

n

n∑k=1

ln (xk) = ln(

n√

x1 · ... · xn)

(7)

and so by taking exponentials of both sides we are done.

The inequality can be shown to be strict if the xknk=1 are not all thesame. In other words, if even one interest rate is different between periods,the the arithmetic average rate is strictly greater than the geometic rate.


Example 1.2

Table: Average vs. Annual Rate of Return

Year 2002 2001 2000 1999 1998

Annual Rate 6.9% 6.4% 9.4% −3.0% 7.8%Average Rate 6.9% 6.65% 7.56% 4.82% 5.41%

We can calculate the average interest rate iav via

(1 + iav )n = (1 + r1) · (1 + r2) · ... · (1 + rn) (8)


Example 1.3

A very nice example is the following: On Jan 1, 2000, Smith deposits 1000into an account with 5% annual interest. The interest is paid on every Dec31. Smith withdraws 200 on Jan 1, 2002, deposits 100 on Jan 1 2003 andagain withdraws from the account on Jan 1 2005, this time 250. What isthe balance in the account just after the interest is paid on Dec 31, 2006?

Hint: Think of a deposit as a loan to the bank, a withdrawal as a loan toSmith, and at the end of the term (7 years,) both bank and Smith settleup accounts.


Example 1.3

We can replicate the cash flows of this example by stating that Smith hasinvested both positive and negative amounts of money over the period oftime:

Table: Example 1.3

Year Period

1000 7 years

−200 5 years

100 4 years

−250 2 years

Hence,

A(7) = 1000 · (1.05)7 + (−200) · (1.05)5

+ 100 · (1.05)4 + (−250) · (1.05)2 = 997.77(9)


Example 1.3

Another way of computing A(7) is to do so recursively:

A(0) = 1000

A(1) = A(0) · 1.05 = 1000 · (1.05)

A(2) = A(1) · (1.05)− 200

= 1000 · (1.05)2 − 200

A(3) = A(2) · (1.05) + 100

= 1000 · (1.05)3 − 200 · (1.05) + 100

A(4) = A(3) · (1.05)

A(5) = A(4) · (1.05)− 250

A(6) = A(5) · (1.05)

A(7) = A(6) · (1.05)

= 1000 · (1.05)7 + (−200) · (1.05)5

+ 100 · (1.05)4 + (−250) · (1.05)2 = 997.77

(10)


Linear Interest

Another way interest can be designed to accrue is linearly Symbolically, ifA(0) is the initial value of our investment, then the final value is

Alinear (t) = A(0) · (1 + i · t) (11)

Notice that at any time k ≤ t = n,

Acompound(n) = A(k) · (1 + i)n−k

= A(0) · (1 + i)k · (1 + i)n−k

Alinear (n) = A(0) · (1 + i · n)

6= A(k) · (1 + i · (n − k))

= A(0) · (1 + i · k) · (1 + i · (n − k))

(12)


Bonus Question

Assume you can select

from two interest rates j , i where 0 < i < j < 1 and

time t ∈ [0, 1]

where you can decide to switch from a bank paying j for a time of length tto one paying i for the remaining time 1− t .

At what time t would this be optimal to do, if at all?


Bonus Question

In general, if we switch at, then it must be that

1 + j < (1 + jt)(1 + (i(1− t))

⇒ 1 + j < 1 + jt + i(1− t) + ijt(1− t)

⇒ (j − i)(1− t) < ijt(1− t)

⇒ j − i

ij< t.

(13)

It follows that if 0 < j−iij < 1, we are able to justify such a switch!

The inequality linking i to j is now

i < j <i

1− i. (14)


Bonus Question

The optimal time to switch is when f (t) := (1 + jt)(1 + i(1− t)) ismaximized:

0 = f ′(t) =d

dt

[(1 + jt)(1 + i(1− t))

]= (j − i) + ij(1− 2t).

⇒ toptimal =1

2+

1

2· j − i

ij.

(15)

Of course, toptimal < 1 as i < j < i1−i and so the optimal value of the

bank account is

f (toptimal) =[1 + j ·

(1

2+

1

2· j − i

ij

)][1 + i ·

(1

2− 1

2· j − i

ij

)]. (16)


Accumulated Amount Function

A(t) is also known as the A.A.F., and

A (t2) = A (t1) (1 + it2)

it2 :=A (t2)− A (t1)

A (t1)

(17)

it2 is known as the effective rate for our investment from t1 to t2.


Present Value

A dollar today is certainly more valuable than a dollar tomorrow, and evenmore valuable than a dollar next year. To reflect this idea, we say that thepresent value of a unit of currency one year from now is ν, where

ν ≡ 1

1 + i. (18)

ν is also known as the discount factor. This factor works as the inverse ofthe interest gained on an investment of a unit of currency for one year.For example, the present value of 25, 000 at a rate of 5% per annum, 25years from now, is 25000

(1.05)25


Equation of Value

In calculating the present value of an investment, we are workingbackwards from a fixed outcome to calculate it’s value to us today.

But, there may be more than one payment expected in the future, such aswhen planning for retirement. They may also be balanced by furtherinvestment made in the future.


Example 1.5

Loan Shark pays you 1000 every week, starting today. There are 4payments in total, at 8% weekly interest. Starting the week after the last1000, you are to repay the loan in 3 consecutive weekly installments of1100 at 8% weekly interest, plus a fourth payment X at 8% weeklyinterest. What is X ?


Example 1.5

Table: Example 1.5: Payments In

Week 0 1 2 3

Payment in 1000 1000 1000 1000

Table: Example 1.5: Payments Out

Week 4 5 6 7

Payment out 1100 1100 1100 X


Example 1.5

PV0(in) = PV0(out)

PV0(in) = 1000 +1000

1.08+

1000

1.082+

1000

1.083

PV0(out) =1100

1.084+

1100

1.085+

1100

1.086+

X

1.087

X = 2273.79

(19)


Nominal Interest Rates

When a credit card has a 24% nominal annual interest rate, you do notonly pay 24% on the balance. Rather, that 24% is broken down into 12monthly interest charges on the average balance over a 30 day billingcycle. So, one pays an effective annual rate of(

1 +0.24

12

)12

− 1 = 0.2682. (20)


Example 1.9

Which is better:

(A) 15.25% compounded semi-annually, or

(B) 15% compounded monthly ?

iA(eff ) =

(1 +

0.1525

2

)2

− 1 = 0.1583

iB(eff ) =

(1 +

0.15

12

)12

− 1 = 0.1608

(21)


Notation

Actuaries reserve i for effective annual rate and i (m) for the nominal ratecompounded m times annually. The link between the two is

i =

(1 +

i (m)

m

)m

− 1

i (m) = m ·(

(1 + i)1m − 1

)i (∞) = lim

m→∞i (m) = ln (1 + i)

< i (m) ∀m

(22)


Notation


i =

(1 +

i (m)

m

)m

− 1

i (m) = m ·(

(1 + i)1m − 1

)

i (∞) = limm→∞

i (m) = ln (1 + i)

< i (m) ∀m

(22)


Notation


i =

(1 +

i (m)

m

)m

− 1

i (m) = m ·(

(1 + i)1m − 1

)i (∞) = lim

m→∞i (m) = ln (1 + i)

< i (m) ∀m

(22)


Example 1.10

Assume i = 0.12. Then

Table: Example 1.10

m m ·(

(1 + i)1m − 1

)1 0.12

2 0.1166

6 0.1144

52 0.1135

365 0.113346

∞ 0.113329


Comparison of continuous to compound interest earned

Assume that we are promised 12% annual interest, and we wish tocalculate the daily interest gained on 10, 000, 000:

interestovernight = 10, 000, 000 · 0.12 · 1

365= 3287.67

interestcontinuous = 10, 000, 000 ·(

e1

365 − 1)

= 3288.21,(23)

a difference of 0.54 on a principle of 10 million


Discount Rates

Sometimes, interest paid up front: Receive A(0) up front, pay backA(1) > A(0).Consider the example

A(0) = 900

A(1) = 1000

d =A(1)− A(0)

A(1)

1

1 + i= ν = 1− d

(24)

Essentially, the discount is the forward price of a dollar (or any other unitof currency) in the interest market with no carrying cost or dividends(coupons) paid out.


Example 1.11

A T-bill represents the forward value today of 100 delivered at time T .The price is calculated via simple discount:

P(0) = P(T ) · (1− d · t) = 100 ·(

1− d · T

365

)

100 = P(T ) = P(0) · (1 + i · t) = P(0) ·(

1 + i · T

365

) (25)

Table: Example 1.11

Term Discount (%) Investment (%) Price per 100

12−day 0.965 0.974 99.680

28−day 0.9940 0.952 99.927

91−day 1.130 1.150 99.174

128−day 1.400 1.430 99.292


Example 1.11

A T-bill represents the forward value today of 100 delivered at time T .The price is calculated via simple discount:

P(0) = P(T ) · (1− d · t) = 100 ·(

1− d · T

365

)100 = P(T ) = P(0) · (1 + i · t) = P(0) ·

(1 + i · T

365

) (25)

Table: Example 1.11

Term Discount (%) Investment (%) Price per 100

12−day 0.965 0.974 99.680

28−day 0.9940 0.952 99.927

91−day 1.130 1.150 99.174

128−day 1.400 1.430 99.292


Nominal Annual Discount Rate

d (m) is the quoted annual discount rate that is applied m times over the

year, with the effective discount rate as d (m)

m for the period of 1m years.

Also, d is the effective discount rate :

1− d =

(1− d (m)

m

)m

d (m) = m ·(

1− (1− d)1m

)d (∞) = lim

m→∞d (m) = ln

(1

1− d

)= i (∞)

(26)


Force of Interest

From time t1 to t2, an investment grows by a rate of

it1→t2 :=A(t2)− A(t1)

A(t1)(27)

Now, fix a time t1 = t and correspondingly, let t2 = t + 1m . It follows that

the nominal rate that gives the same annual growth from t to t + 1m is

i (m) = m ·A(t + 1

m

)− A(t)

A(t)=

1

A(t)·

A(t + 1

m

)− A(t)

1m

limm→∞

i (m) =1

A(t)· limm→∞

A(t + 1

m

)− A(t)

1m

=1

A(t)

dA

dt=: δ(t)

(28)


Force of Interest


it1→t2 :=A(t2)− A(t1)

A(t1)(27)



i (m) = m ·A(t + 1

m

)− A(t)

A(t)=

1

A(t)·

A(t + 1

m

)− A(t)

1m

limm→∞

i (m) =1

A(t)· limm→∞

A(t + 1

m

)− A(t)

1m

=1

A(t)

dA

dt

=: δ(t)

(28)


Force of Interest


it1→t2 :=A(t2)− A(t1)

A(t1)(27)



i (m) = m ·A(t + 1

m

)− A(t)

A(t)=

1

A(t)·

A(t + 1

m

)− A(t)

1m

limm→∞

i (m) =1

A(t)· limm→∞

A(t + 1

m

)− A(t)

1m

=1

A(t)

dA

dt=: δ(t)

(28)


Force of Interest

Since δ(t) = 1A(t)

dAdt = d

dt [ln (A(t))], it follows that∫ t

0δ(s)ds =

∫ t

0

d

ds[ln (A(s))] ds = ln (A(t))− ln (A(0)) (29)

and so

A(t) = A(0)e∫ t0 δ(s)ds (30)


Some Examples...

Calculate δ(t) for

A(t) = e−t2

A(t) = t · e−t

A(t) = (t − 1)2

A(t) = A(0) · (1 + i · t)

A(t) = A(0) · (1 + i)t


Example 1.14

Assume δ(t) = 0.08 + 0.005t and your asset’s initial value is A(0) = 1000.Then

A(t) = A(0)e∫ t0 (0.08+0.005s)ds = 1000e0.08t+0.0025t2

A(5) = 1000e0.08·5+0.0025·52 = 1588.04(31)

But, if we were give A(2) = 1000, then

A(7) = A(2)e∫ 72 δ(s)ds = 1000e0.08·5+0.0025·(72−22) = 1669.46 (32)


Some thoughts...

This limiting term, known as the force of interest represents thetheoretical upper limit for interest gained on this asset. However, In thederivation above, there is no randomness involved in obtaining the ordinarydifferential equation

dA(t) = δ(t)A(t)dt (33)

To handle this lack of stochasticity, one common asset model (GeometricBrownian Motion) is defined by the stochastic differential equation

dA(t) = δ(t)A(t)dt + σ (A(t)) dW (t), (34)

where W (.) is the Wiener process.


Introduction and Geometric Series

An annuity is a series of periodic payments.

In this class, they are notevent-contingent, but rather only depend on the passage of time. Tocompute the value of the most foundational annuity, we must recall thefollowing fact: If x ∈ R : x 6= 1, then

Xn := 1 + x + ...+ xn

(1− x) · Xn = 1− xn+1

⇒ Xn =1− xn+1

1− x

(35)



An annuity is a series of periodic payments. In this class, they are notevent-contingent, but rather only depend on the passage of time.

Tocompute the value of the most foundational annuity, we must recall thefollowing fact: If x ∈ R : x 6= 1, then

Xn := 1 + x + ...+ xn

(1− x) · Xn = 1− xn+1

⇒ Xn =1− xn+1

1− x

(35)



An annuity is a series of periodic payments. In this class, they are notevent-contingent, but rather only depend on the passage of time. Tocompute the value of the most foundational annuity, we must recall thefollowing fact:

If x ∈ R : x 6= 1, then

Xn := 1 + x + ...+ xn

(1− x) · Xn = 1− xn+1

⇒ Xn =1− xn+1

1− x

(35)



An annuity is a series of periodic payments. In this class, they are notevent-contingent, but rather only depend on the passage of time. Tocompute the value of the most foundational annuity, we must recall thefollowing fact: If x ∈ R : x 6= 1, then

Xn := 1 + x + ...+ xn

(1− x) · Xn = 1− xn+1

⇒ Xn =1− xn+1

1− x

(35)


Example 2.1

The federal government sends Smith a family allowance of 30 every monthfor Smith’s child. Smith deposits the payments in bank account on thelast day of each month. The account earns interest at the annual rate of9% compounded monthly and payable on the last day of each month, onthe minimum monthly balance. If the first payment is deposited on May31, 1998, what is the account balance on December 31,2009, including thepayment just made?


Example 2.1

There are 140 total payments

imonthly = i12 = 0.0075

TotalValue = 30 + 30 · (1.0075)1 + ...+ 30 · (1.0075)139

= 30 ·139∑k=0

1.0075k = 30 · 1− 1.0075140

1− 1.0075

= 7385.91

(36)


Level Payment Annuities

Number of payments in a series of payments is called the term of theannuity

Time between the successive payments is called the payment period,or frequency

A series of payments whose value is found at the time of the finalpayment is known as accumulated annuity immediate

sn i :=n−1∑k=0

(1 + i)k =(1 + i)n − 1

i(37)

Equivalently, (1 + i)n = 1 + i · sn i .


Ex 2.2

What level amount must be deposited on May 1 and Nov 1 each year from1998 to 2005, inclusive, to accumulate to 7000 on November 1, 2005 ifthe nominal annual interest rate, compounded semi-annually, is 9% ?


Interest Compounded Monthly

16 total deposits

i (2) = 0.09

X denotes the level amount deposited per period

7000 = X ·(1 + (1.045)1 + ...+ (1.045)15

)= X · s16 0.045

⇒ X = 308.11

(38)


Example 2.3

Suppose that in Example 2.1, Smith’s child is born in April 1998 and thefirst payment is received in May (and deposited at the end of May.) Thepayments continue and the deposits are made at the end of the monthuntil, and including the month of, the child’s 16th birthday. The paymentsstop after the 16th birthday, but the balance continues to accumulate withinterest until the end of the month of the child’s 21st birthday. What isthe balance X in the account at that time ?

X = 1.007560 · 30 · s192 0.0075 = 20, 028.68 (39)


Some Arithmetic

After accumulating cash via a series of payments at rate i until time n, weallow this amount to grow until time n + k :

Value(n + k) = sn i · (1 + i)k =(1 + i)n − 1

i· (1 + i)k

=(1 + i)n+k − (1 + i)k

i

=(1 + i)n+k − 1

i− (1 + i)k − 1

i= sn+k i − sk i

sn+k i = sk i + sn i · (1 + i)k

(40)


Example 2.4

Suppose that in Example 2.1, the nominal annual interest rate earned onthe account changes to 7.5%, still compounded monthly, as of January2004. What is the accumulated value of the account on December 312009 ?

Value = 30 ·(s68 0.0075 · 1.0062572 + s72 0.00625

)= 6865.22 (41)


Example 2.5

Suppose 10 monthly payments of 50 each are followed by 14 monthlypayments of 75 each. If interest is at an effective monthly rate of 1%, whatis the accumulated value of the series at the time of the final payment ?

Value = 50 · s24 0.01 + 25 · s14 0.01 (42)


Present Value of an Annuity

Make a lump sum payment X now to receive periodic payments C ,starting one period from today. If the market bears a constant interest ofi , the the the Present Value of this Annuity Immediate is calculated as

X =C

1 + i+

C

(1 + i)2+ ...+

C

(1 + i)n

= C ·n∑

k=1

νk = C · ν ·n−1∑k=0

νk

= C · ν · 1− νn

1− ν

= C · 1− νn

i= C · 1− (1 + i)−n

i≡ C · an i

(43)


Example 2.7

Brown has bought a new car and requires a loan of 12000 to pay for it.The car dealer offers Brown two alternatives on the loan:

A : Monthly payments for 3 years, starting one month after purchase,with an annual interest rate of 12% compounded monthly

B : Monthly payments for 4 years, also starting one month afterpurchase, with an annual interest rate of 15% compounded monthly.

Find Brown’s monthly payment and the total amount paid over the courseof the repayment period under each of the two options.


Example 2.7

12000 = PA · a36 0.01 ⇒ PA = 398.57

12000 = PB · a48 0.0125 ⇒ PB = 333.97

TotalValue(A) = 36 · PA = 14348.52

TotalValue(B) = 48 · PB = 16030.56

(44)


Example 2.8

Suppose that in Example 2.7, Brown can repay the loan, still with 36payments under option A or 48 payments under option B, with the firstpayment made 9 months after the car is purchased in either case.Assuming interest accrues from the time of the car purchase, find thepayments required under options A and B. This is known as a deferredannuity


Example 2.8

12000 = P ′A ·(1.01−9 + 1.01−10 + ...+ 1.01−44

)= P ′A · ν8A · a36 0.01⇒ P ′A = 431.60

12000 = P ′B ·(1.0125−9 + 1.0125−10 + ...+ 1.0125−56

)= P ′B · ν8B · a48 0.0125⇒ P ′B = 368.86

(45)


Some Facts About Annuities

There is a duality between an i and sn i :

sn i = (1 + i)nan i

an i = νnsn i(46)

As n→∞, we have

a∞ i = limn→∞

an i

= limn→∞

1− νn

i=

1

i

(47)


Example 2.9

A perpetuity immediate pays X per year. Brian receives the first npayments, Colleen receives the next n payments, and Jeff receives theremaining payments. Brian’s share of the present value of the originalperpetuity is 40%, and Jeff’s share is K . Calculate K .


Example 2.9

PV (Brian) = X · an i = X · 1− νn

i

= 0.4 · X · a∞ i = 0.4X

i⇒ 1− νn = 0.4

PV (Colleen) = νn · X · an i = 0.4 · 0.6X

i= 0.24

X

i

X

i= PV = PV (Brian) + PV (Jeff ) + PV (Colleen)

= 0.4X

i+ K + 0.24

X

i

⇒ K = 0.36X

i

(48)


Example 2.9

PV (Brian) = X · an i = X · 1− νn

i

= 0.4 · X · a∞ i = 0.4X

i⇒ 1− νn = 0.4

PV (Colleen) = νn · X · an i = 0.4 · 0.6X

i= 0.24

X

iX

i= PV = PV (Brian) + PV (Jeff ) + PV (Colleen)

= 0.4X

i+ K + 0.24

X

i

⇒ K = 0.36X

i

(48)


Recall...

sn i = 1 + (1 + i) + (1 + i)2 + ...+ (1 + i)n−1 =(1 + i)n − 1

i

an i = ν + ν2 = ...+ νn =1− νn

i

(49)

Of course, sn i is the accumulation of all the payments at the final time,whereas an i is the present value of all the paymenst, one period before thefirst payment. Now, define

sn i := (1 + i) + (1 + i)2 + ...+ (1 + i)n = (1 + i)sn i

= annuity due

= accummulated value one-period after final payment

an i := (1 + i)an i

= present value at time of first payment

(50)


Level Payment Annuities: Some Generalizations.

What happens when compounding interest period and annuity paymentperiod don’t coincide? For example: Consider 4 deposits made per year,over 16 years, of 1000 each. What is the balance after the last payment ifinterest is quoted at 9% nominal annual interest rate, compoundedmonthly?

In this case, the equivalent rate j for the 3−month periods satisfies

1 + j =(1 + 0.09

12

)3. It follows that

Value = 1000s64 j = 141, 076 (51)


Level Payment Annuities: Some Generalizations.

What happens when compounding interest period and annuity paymentperiod don’t coincide? For example: Consider 4 deposits made per year,over 16 years, of 1000 each. What is the balance after the last payment ifinterest is quoted at 9% nominal annual interest rate, compoundedmonthly?In this case, the equivalent rate j for the 3−month periods satisfies

1 + j =(1 + 0.09

12

)3. It follows that

Value = 1000s64 j = 141, 076 (51)


mthly payable annuities: Some Generalizations.

Imagine that for each of n periods in the term of an annuity, 1 is paid outover m payments of 1

m each. Then the present value is computed using

j = i (m)

m , where

a(m)n i = ν · 1

msm j + ν2 · 1

msm j + ...+ νn · 1

msm j

=(ν + ν2 + ...+ νn

)· 1

m·

(1 + i (m)

m

)m− 1

i (m)

m

= an i ·

(1 + i (m)

m

)m− 1

i (m)= an i ·

i

i (m)

s(m)n i = sn i ·

i

i (m)

(52)



As m→∞, we have i (m) → ln (1 + i), and so it follows that

a(m)n i → an i ·

i

ln (1 + i)

s(m)n i → sn i ·

i

ln (1 + i)

(53)

Let’s analyze this using calculus!



Take t2 − t1 = 1m , and assume that 1 is paid out at a continuous rate.

Then, from t to t + dt, the amount paid out is dt. At time n, the value ofdt accumulated from t to n is dt · (1 + i)n−t .

Add up all of these bits andwe obtain

sn i =

∫ n

0(1 + i)n−tdt =

∫ n

0e ln (1+i)(n−t)dt

= e ln (1+i)n ·

[e− ln (1+i)t

− ln (1 + i)

]n0

=(1 + i)n − 1

ln (1 + i)= sn i ·

i

ln (1 + i)= lim

m→∞s(m)n i

(54)



Take t2 − t1 = 1m , and assume that 1 is paid out at a continuous rate.

Then, from t to t + dt, the amount paid out is dt. At time n, the value ofdt accumulated from t to n is dt · (1 + i)n−t . Add up all of these bits andwe obtain

sn i =

∫ n

0(1 + i)n−tdt =

∫ n

0e ln (1+i)(n−t)dt

= e ln (1+i)n ·

[e− ln (1+i)t

− ln (1 + i)

]n0

=(1 + i)n − 1

ln (1 + i)= sn i ·

i

ln (1 + i)= lim

m→∞s(m)n i

(54)


An Example: Daily vs Continuous deposits

Consider depositing 12 per day in 2004 and 2005 and 15 per day in 2006.The earned interest in 2004-05 is i1 = 9% effective per year, and i2 = 12%effective per year in 2006. Compute the total accumulated amount at theend of 2006 if computed via (a) daily deposits and (b) continuousdeposits. Assume no leap years!



For daily deposits, we compute the equivalent daily interest rates j1 for2004-05 and j2 for 2006 by

j1 = (1.09)1

365 − 1 = 0.00023631

j2 = (1.12)1

365 − 1 = 0.00031054(55)

So,

AccValue = 12 · s730 j1 · (1 + i2) + 15 · s365 j2= 12 · s730 0.00023631 · (1.12) + 15 · s365 0.00031054= 16502.59

(56)



For continuous deposits, we compute that 12 · 365 = 4380 is invested peryearfor the period 2004-05 and 15 · 365 = 5475 for 2006.So,

AccValue = 4380 · s2 0.09 · (1.12) + 5475 · s1 0.12

= 4380 · 1.092 − 1

ln (1.09)· (1.12) + 5475 · 1.12− 1

ln (1.12)

= 16504.75

(57)


Continuous Annuities in General

Define a(t1, t2) as the accumulated value at t2 of an amount 1 invested att1.

Hence, ∫ t2

t1

a(t, t2)dt (58)

is the accumulated value at t2 of a continuous annuity paying 1 perunit time over the interval (t1, t2),and ∫ t2

t1

1

a(t, t2)dt (59)

is the present value at t1 of a continuous annuity paying 1 per unit timeover the interval (t1, t2),



Define a(t1, t2) as the accumulated value at t2 of an amount 1 invested att1. Hence, ∫ t2

t1

a(t, t2)dt (58)

is the accumulated value at t2 of a continuous annuity paying 1 perunit time over the interval (t1, t2),and ∫ t2

t1

1

a(t, t2)dt (59)

is the present value at t1 of a continuous annuity paying 1 per unit timeover the interval (t1, t2),



If δs is the force of interest at time s, then

a(t1, t2) = e∫ t2t1δsds

an δs =

∫ n

0e−

∫ t0 δsdsdt

sn δs =

∫ n

0e∫ nt δsdsdt

(60)


Unknown Number of Payments in an Annuity

Assuming n level payments of J each at interest rate i , the accumulatedvalue is

M = J[(1 + i)n−1 + ...+ (1 + i) + 1

]= J · (1 + i)n − 1

i= J · sn i

⇒ (1 + i)n = 1 + i · M

J

⇒ n =ln(1 + i · MJ

)ln 1 + i

(61)


Example 2.13

Smith wishes to accumulate 1000 by means of semi-annual depositsearning interest at nominal rate i (2) = 0.08, with interest creditiedsemi-annually. Regular deposits of 50 are made. Find the number ofperiods and regular deposits needed, along with any additional fractionaldeposits if such a fractional deposit is made at the time of the last regulardeposit or the very next period (i.e. last regular deposit can be 0.)


Example 2.13

Here, the parameters are M = 1000, J = 50, i = 0.04. It follows that

n =ln(1 + i · MJ

)ln 1 + i

= 14.9866 (62)

Rounding down to n = 14, we obtain 50s14 0.04 = 914.60.. However, 50 isnot enough to carry over to get a final amount of 1000, so let the 914.60accrue for 1 period to 914.60 · (1.04) = 951.18 , and then add thefractional payment of 48.82. Hence 15 periods are needed


Example 2.13


n =ln(1 + i · MJ

)ln 1 + i

= 14.9866 (62)

Rounding down to n = 14, we obtain 50s14 0.04 = 914.60..

However, 50 isnot enough to carry over to get a final amount of 1000, so let the 914.60accrue for 1 period to 914.60 · (1.04) = 951.18 , and then add thefractional payment of 48.82. Hence 15 periods are needed


Example 2.13


n =ln(1 + i · MJ

)ln 1 + i

= 14.9866 (62)

Rounding down to n = 14, we obtain 50s14 0.04 = 914.60.. However, 50 isnot enough to carry over to get a final amount of 1000, so let the 914.60accrue for 1 period to 914.60 · (1.04) = 951.18 , and then add thefractional payment of 48.82. Hence 15 periods are needed


Example 2.13

Smith pays 100 per month to a fund earning i (12) = 0.09, with interestcredited on the last day of each month. At the time of each deposit, 10 isdeducted from the deposit for expenses and administration fees. The firstdeposit is made on the last day of Jan 2000. In which month does theaccumulated value become greater than the total gross contribution tothat point?


Example 2.13

We wish to solve for the time when the Value of the account exceeds thetotal amount invested, i.e. the value n that satisfies

90 ·[1 + 1.0075 + ...+ 1.0075n−1

]≥ 100n

sn i ≥100

90n

1.0075n ≥ 1 + 0.00833n

(63)

Essentially, we are solving a fixed point equation for f (x) = 0, where

f (x) = ekx − 1− 0.00833x

k = ln (1.0075)(64)

In this case, plotting the above function of x gives an intercept ofapproximately 28.6. Hence, n = 29.


Dividend Discount Model for Stock Shares

We now have a philosophy for pricing assets that bring a future stream ofpayments. Simply put, the value today of the asset is simply the NetPresent Value, opr NPV, of that stream of payments, when adjusted forinflation or (compound) growth, if applicable.



If a table, or number line of future payments looks like

Table: Example 1.5: Payments In

Week 0 1 2 ...n

Payment in 0 K K · (1 + r) K · (1 + r)n−1

then the NPV, valued with rate i , is

NPV =K

1 + i+

K · (1 + r)

(1 + i)2+ ...+

K · (1 + r)n−1

(1 + i)n

=K

1 + i·n−1∑j=0

(1 + r

1 + i

)j

=K

1 + i·

1−(1+r1+i

)n1− 1+r

1+i

(65)



As n→∞, we approach a perpetuity, and

NPV → K

1 + i· 1

1− 1+r1+i

=K

i − r(66)

In words, we have a perpetuity that is adjusted for the growth rate r . Weuse this formula to price a stock or any other asset as a perpetuity withpossible compound growth at rate r .


Example 2.13

Stock X pays dividend 50 per year with growth of 5% after the first year.John purchases X at the theoretical price corresponding to an effectiveyield of 10%. After receiving the the 10th dividend, John sells the stock forprice P. If annual yield for John was 8%, what is the fair price for P ?

The equation of value is

50

0.10− 0.05= 1000 =

50

1.08·

9∑j=0

(1.05

1.08

)j

+P

1.0810

⇒ P = 1000(1.08)10 − 50(1.08)9 ·9∑

j=0

(1.05

1.08

)j

= 1000(1.08)10 − 50(1.08)9

(1−

(1.051.08

)101−

(1.051.08

)1)

= 1275.54.

(67)


Increasing Annuities

Consider now a payment schedule where the payment, at time k , wasKm = m .

Now, the net present value is

X =n∑

m=1

Kmνm =

n∑m=1

mνm =n∑

m=1

m

(1 + i)m(68)

Then

(1 + i) · X =n∑

m=1

mνm−1 =n∑

m=1

m

(1 + i)m−1

i · X = 1 + ν + ν2 + ...+ νn−1 − nνn

=1− νn

1− ν− nνn

X =an i − nνn

i=: (Ia)n i

(69)



Consider now a payment schedule where the payment, at time k , wasKm = m .Now, the net present value is

X =n∑

m=1

Kmνm =

n∑m=1

mνm =n∑

m=1

m

(1 + i)m(68)

Then

(1 + i) · X =n∑

m=1

mνm−1 =n∑

m=1

m

(1 + i)m−1

i · X = 1 + ν + ν2 + ...+ νn−1 − nνn

=1− νn

1− ν− nνn

X =an i − nνn

i=: (Ia)n i

(69)



An increasing perpetuity is defined as

limn→∞

(Ia)n i = limn→∞

1−νn1−ν − nνn

i

=1 + i

i2

(70)

Notice that the level payment perpetuity of Kl ≡ 1 has present value ofonly 1

i . Since i ≈ 0, we have 1i2>> 1

i .

HW: apply same analysis for decreasing annuities, with Km = n −m + 1,where m ∈ 1, 2, .., n


Example 2.22

Consider perpetuities X and Y with payment

Table: Example 2.22: Payments Schedule X

Week 1 2 3 ..n..

Payment in 1 2 3 ..n..

Table: Example 2.22: Payments Schedule Y

Week 1 2 3 4 ..2k − 1 2k ..

Payment in q q 2q 2q ..kq kq..

at the end of each year. It is known that NPV (X ) = NPV (Y ) whenvalued at annual interest rate i = 0.10. Solve for q.


Example 2.22

If we take νeq := ν2 =(

11.1

)2= 1

1.21 , and so ieq = 0.21. It follows that

NPV (X ) =1 + i

i2=

1.1

0.01= 110

NPV (Y ) = qν + qν2 + 2qν3 + 2qν4 + 3qν5 + 3qν6 + ...

= q ·(ν + 2ν3 + 3ν5 + ..

)+ q ·

(ν2 + 2ν4 + 3ν6 + ..

)=(

q +q

ν

)·(ν2 + 2ν4 + 3ν6 + ..

)=(

q +q

ν

)·(νeq + 2ν2eq + 3ν3eq + ..

)= 2.1 · q · 1 + ieq

i2eq= 2.1 · q · 1.21

0.212= 57.62q

q = 1.91

(71)


Continuous Payment Schemes

What if the payments Km were just discrete samples at time tm of acontinuous payment scheme? For example, the payment at time t = t1was Ki = h(ti )dt.

Then all payments must be summed to obtain the NPV and AccValue:

NPV =

∫ n

0

1

(1 + i)th(t)dt

AccVal =

∫ n

0(1 + i)n−th(t)dt

(72)

A special case is when h(t) = t. Then

(I a)n i

=

∫ n

0

1

(1 + i)ttdt

(I s)n i

=

∫ n

0(1 + i)n−ttdt

(73)



Integration by parts allows us to obtain

(I a)n i

=

∫ n

0

1

(1 + i)ttdt =

an i − nνn

δ(I s)n i

=

∫ n

0(1 + i)n−ttdt =

sn i − n

δ

δ := ln (1 + i)

(74)

where δ is the constant force of interest.

If the force of interest varies, then

(I a)n δ∗

=

∫ n

0e−

∫ t0 δsdstdt(

I s)n δ∗

=

∫ n

0e∫ nt δsdstdt

(75)



Integration by parts allows us to obtain

(I a)n i

=

∫ n

0

1

(1 + i)ttdt =

an i − nνn

δ(I s)n i

=

∫ n

0(1 + i)n−ttdt =

sn i − n

δ

δ := ln (1 + i)

(74)

where δ is the constant force of interest.If the force of interest varies, then

(I a)n δ∗

=

∫ n

0e−

∫ t0 δsdstdt(

I s)n δ∗

=

∫ n

0e∫ nt δsdstdt

(75)


Unique Interest Rates for Annuity Valuation: Anapplication of the Intermediate Value Theorem

Returning to the discrete setting, consider the following important theorem

Theorem

Assume that an annuity X pays amount Kj > 0 at time tj for all times0 < t1 < t2.. < tn. Suppose that you are given L > 0. Then there exists aunique i > −1 such that NPV (X )(i) :=

∑nl=1

Kl

(1+i)tl= L under rate i .

Without this theorem, we would not be able to uniquely price returns onassets.


Unique Interest Rates for Annuity Valuation: Anapplication of the Intermediate Value Theorem

Proof.

It follows that since (Kl , tl)nl=1 ⊂ R2+, we have NPV (X )′(i) < 0. This

also follows from financial reasoning as a higher rate i would demand alower up front payment L. Since

limi→∞

NPV (X )(i) = 0

limi→−1

NPV (X )(i) =∞(76)

it follows from the IVT that there exists a unique i (due to strictdecreasing nature of NPV (X )(i)) such that NPV (X )(i) = L


Yield and Reinvestment Rates

Assume a loan amount L for a term of n years, and a total value M of allpayments after the n years. Then it must be that for level payments K theyield rate i satisfies

L =M

(1 + i)n=

K

1 + i+

K

(1 + i)2+ ..+

K

(1 + i)n= K · an i (77)

On a loan, the Internal Rate of Return, or IRR, is the rate of interest forwhich the loan amount upront is equal in value to the NPV of all loanpayments. Also known as the loan rate. Sometimes, the entity that loansL can reinvest these loan repayments at a higher rate.


Ex. 2.25

Smith owns a 10000 savings bond that pays i(12)Bond = 0.06. Upon receipt of

an interest payment, he immediately deposits it into an account earning

interest, payable monthly, at a rate of i(12)X = 0.12. Find the accumulated

value of this account just after the 12th, 24th, and 36th deposit. In each

case, find the average annual yield i(12)avg based on his initial investment of

10000. Assume that the savings bond may be cashed in any time for10000.


Ex. 2.25

The interest paid each month is at a rate of i (12)

12 = 0.0612 = 0.5%. On an

investment of 10000, this means payments of 50 each month. The valueof his reinvestment of these payments at rate 0.01 is

Table: Example 2.25: Accumulated Value of Reinvestment

Month 12 24 36

Acc Val 50 · s12 0.01 50 · s24 0.01 50 · s36 0.01= 634.13 1348.67 2153.84


Ex. 2.25

After 12 months, the equivalent rate j needed to accumulate10000 + 50 · s12 0.01 = 10634.13 satisfies

10000(

1 +i(12)avg

12

)12= 10634.13

⇒ i(12)avg = 0.0616

(78)

Do this for the other cases.


Ex. 2.25

In general, we have

10000(

1 +i(12)avg (n)

12

)n= 10000[1 + 0.005sn 0.01]

= 10000[1 + 0.005

(1.01)n − 1

0.01

]⇒ i

(12)avg (n) = 12

[(1

2+

1

2(1.01)n

) 1n − 1

] (79)


Sinking Fund Method of Valuation

It often happens that interest payments returned to an investor at rate ican only be reinvested at ”market” rate j . If j 6= i , then the yield rate tovalue the investment is also 6= i and we consider another way to determinethe value. Instead of using the present value method, we now consider thevalue at the end of the term n of the investment.

When an investment is made, it can be considered a loan (principal) to theperson receiving the money up front. By the end of the term, the entireprincipal has been repaid and the investor has enjoyed (consumed) periodicinterest payments for making the initial investment.

Consider, for example, the fair purchase price P of an annuity with levelpayments K over n periods.



Receive periodic payments K in return for initial investment P.

The interest payment is P · i , received every unit of time over intervalof length n.

These interest payments are consumed, and the excess K − P · i everyperiod is reinvested at a rate j into a sinking fund.

The fair value P is now determined via the recursive equation thatsays investor recovers P at end of term, and this is equivalent to theamount accumulated by reinvesting the excess K − P · i per period:

P = (K − P · i) sn j

P =Ksn j

1 + i · sn j=

Ksn j

1 + ij · ((1 + j)n − 1)

limj→i

P =Ksn i

1 + (1 + i)n − 1= Kan i

(80)







P = (K − P · i) sn j

P =Ksn j

1 + i · sn j

=Ksn j

1 + ij · ((1 + j)n − 1)

limj→i

P =Ksn i

1 + (1 + i)n − 1= Kan i

(80)







P = (K − P · i) sn j

P =Ksn j

1 + i · sn j=

Ksn j

1 + ij · ((1 + j)n − 1)

limj→i

P =Ksn i

1 + (1 + i)n − 1= Kan i

(80)


Amortization

When repaying a loan L with periodic payments Km at time m, we havethe equation of value

L =n∑

m=1

Km

(1 + i)m(81)

In this method, we track the outstanding balance on a loan, denoted byOB(i) or sometimes OBi .


Amortization

OB0 = L

OB1 = L · (1 + i)− K1

OB2 = OB1 · (1 + i)− K2

OBt+1 = OBt · (1 + i)− Kt+1 = OBt − PRt+1

PRt+1 = Kt+1 − i · OBt

It+1 = i · OBt

OBn = 0 = OBn−1 · (1 + i)− Kn

(82)

Also note that the total interest and cash paid out are, respectively,∑nt=1 It and

∑nt=1 Kt −

∑nt=1 It .


Retrospective and Prospective Forms

At time t, if we look backwards to balance out how the original loanamount has grown and subtract form that our total payments, then

OBt = L · (1 + i)t −t∑

m=1

Km · (1 + i)t−m (83)

If we take the NPV viewpoint, then the outstanding balance today issimply the present value of all future payments until the the loan is paidoff:

OBt =n∑

m=t+1

Km

(1 + i)m−t(84)


Example 3.2

A loan of 3000 at an effective quarterly rate of j = 0.02 is amortized bymeans of 12 quarterly payments, beginning one quarter after the loan ismade. Each payment consists of a principal repayment of 250 plus interestdue on the previous quarters outstanding balance. Construct theamortization schedule.


Example 3.2

In this case, we have level princiapl repayments, and so PRt = 250 for all0 ≤ t ≤ 12. Since It+1 = 0.02 · OBt , it follows that

OBt+1 = OBt − 250

⇒ OBt = 3000− 250t

It = 0.02 · (3000− 250 · (t − 1)) = 65− 5t

Kt = It + PRt = 250 + 65− 5t = 315− 5t

(85)


Amortization of a Loan with Level Repayments

In this case, which is common, the payments are held constant, say atK, but the proportion of interest paid versus principal paid varies withtime.

Examples such as mortgages, car payments, etc..

Can use retrospective method to value outstanding balance at time t.


Amortization of a Loan with Level Repayments

Here, the outstanding balance is of course OBt = K · an−t i , and the totalcash paid out by time t is

∑tm=1 Kt = K · t. It follows that, with

δ = ln (1 + i)

It = i · OBt−1 = i · K · an−t+1 i = K ·(1− (1 + i)t−n−1

)PRt = Kt − It =

K

(1 + i)n+1−t

OBt =K

i·(

1− e−δ·(n−t)) (86)


Sinking Fund Method of Valuation Revisited

Even though the outstanding balance is paid off only at the end of theterm of the loan, we can still interpolate and say that the outstandingbalance decreases as

OBt = L− L

sn j· st j (87)

where the balance decreases by the amount paid into the account earningat a rate j.


Sinking Fund Method of Valuation Revisited

Note that we can also calculate the net interest payment each month via

Kt = L · i +L

sn j

PRt = OBt−1 − OBt = L ·(

st j − st−1 jsn j

)It = Kt − PRt = L ·

(i − j ·

st−1 jsn j

) (88)


Sinking Fund Example from previous SOA Exam FM/1

Ronaldo borrows L for 10 years at an annual effective interest rate of100 · i%. At the end of each year, he pays the interest on the loan anddeposits the level amount necessary to repay the principal to a sinkingfund earning an annual effective interest rate of 100 · j%. The totalpayments made by Ronaldo over the 10-year period is X. Calculate X .


Sinking Fund Example from previous SOA Exam FM/1

Answer:

Annual interest payment = L · iAnnual sinking fund deposit is L

s10|j

Total annual payment = L · i + Ls10|j

Total of all payments = 10 ·(

L · i + Ls10|j

)


Makeham’s Formula

Consider the following scenario: An investor makes a loan at interest ratei, with only interest payments for the term of the loan, and then a lumpsum L at the end of term. The investor then sells the loan to a speculator,who values the expected cash flows at a rate j. Then the value thespeculator puts on the investment is

A =L

(1 + j)n+ L · i · an j =

L

(1 + j)n+

i

j·(

L− L

(1 + j)n

)(89)


Makeham’s Formula

This idea generalizes to the case where the principal L is paid over a seriesof lump sum payments L =

∑nm=1 Lm and periodic repayments

K =∑n

m=1 Km.

Each of these lump sum payments can be viewed as corresponding to anindividual loan:

As =Ls

(1 + j)ts+ Ls · i · ats j =

Ls

(1 + j)ts+

i

j·(

Ls −Ls

(1 + j)ts

)(90)


Makeham’s Formula

We can find the total value now by summing up the individual loan values

A =n∑

s=1

As =n∑

s=1

Ls

(1 + j)ts+

i

j·(

Ls −Ls

(1 + j)ts

)

=n∑

s=1

Ls

(1 + j)ts+

i

j·

(n∑

s=1

Ls −n∑

s=1

Ls

(1 + j)ts

)

= K +i

j· (L− K )

(91)


Mortgage Repayment from previous SOA Exam FM/1

A loan L is being repaid with n annual level payments of K each. With themth payment, m < n, the borrower pays an extra amount A, and thenagrees to repay the remaining balance over l years with a revised annualpayment. The effective rate of interest is 100 · i%. . Calculate the amountof the revised annual payment.


Mortgage Repayment from previous SOA Exam FM/1

With n −m yealy payments left right before the extra repayment, theremaining balance is K · a(n−m)|i . Right after the extra payment A, theremaining balance is K · a(n−m)|i − A. Hence, the new payment over thenew 10−year schedule is

Knew =K · a(n−m)|i − A

al |i(92)


Bonds

A Bond is a financial contract where an up front payment is exchanged forcoupon payments (periodic interest paid to the debt-holder) plus a faceamount the maturity date (end of term). The Bond is valued using a yieldrate used to value future cash flows. The notation used with bondvaluation is

j = the Yield rate per period

F = the Face amount

r = the Coupon rate

C = the redemption amount, usually equal to the face value

N = the number of coupon payments until maturity

P = the price of the bond, where


Bonds

The bond can now be valued as

P =F

(1 + j)n+ F · r ·

(1

1 + j+ ..+

1

(1 + j)n

)= Fνn + F · r · an j

= F + F · (r − j) · an j

= Fνn +r

j· (F − Fνn)

(93)

Note that Bond prices are usually listed, and the investor determines yieldrate from listed price. To do so, must invert our Bond pricing formula.This requires some calculation.


Example 4.1

Price the following bond with face 100:

Issue Date: March 1,2004

Maturity Date: Feb 28, 2006

r = 1.625%

j = 1.675%

Term is 2 years (4 six month periods)

P = 100 · 0.5 · 0.01625 ·(

1

1.008375+ ...+

1

1.0083754

)+

100

1.0083754= 99.02

(94)


Bonds

At time t, we define Pt to be the price of the bond, just after the couponpayment. This is equal to the NPV of the remainin payments:

Pt =F

(1 + j)n−t+ F · r · an−t j (95)

Now, since the initial price P satisfies the relation P − F = F · (r − j)an j ,we say that a bond is bought at premium if r > j , at par if r = j , and at adiscount if r < j .


Bond Pricing Between Coupon Dates

Let

u :=the number of days since last coupon payment

number of days in coupon period(96)

Then Pt = Pm · (1 + j)u − u · F · r , where m is the number of couponpayments left and Pm · (1 + j)u is known as the price plus accrued.


Amortization of Bond

Sometimes need to determine the Book Value of a bond for tax or investorpurposes. The Book Value at time t is the outstanding balance at time tright after the coupon payment. This is the present value of the remainingcash payments of that the bond holder is entitled to receive:

Pt = OBt = F + F · (r − j) · an−t j (97)


Amortization of Bond

We can use the idea that we are loaning money to bond-issuer, and thatwe price this as a mortgage. So,

Pt = OBt = F + F · (r − j) · an−t jKt = F · r for t ∈ 1, 2, 3, .., n − 1Kn = F · r + F

It = Kt − PRt

PRt = Kt − (OBt−1 − OBt)

= F · (r − j) · νn−t+1

(98)


Bond Pricing Lemma

Lemma

Consider a bond with face F , coupon rate r , and term T = 2n, n ∈ N,purchased by A for a price PA. Just after the nth coupon, A sells the bondto B who desires a yield j and correspondingly pays PB . If PB > PA and

j < r or

j > r and PB >F+PA

2

then i > j , where i is the non-zero yield rate calculated on A′s investment.


Bond Pricing Lemma: Proof 1

Proof.

We prove each case separately, both of which reveal something about thesituation described above.Case 1 First, if j < r

PA = F · r · an i +PB

(1 + i)n

>Fr

i·(

1− 1

(1 + i)n

)+

PA

(1 + i)n

PB = F + F · (r − j) · an j

(99)

We can see that this implies

Fr

i< PA < PB = F + F · (r − j) · an j < F + F · r − j

j=

Fr

j(100)

and so i > j


Bond Pricing Lemma: Proof 2

Proof.

Case 2 If j > r then by Makeham’s formula F > PB > PA and

PA = F · ran i +PB

(1 + i)n

= F · ran i +F

(1 + i)n+

[PB − F

(1 + i)n

]PB = F · ran j +

F

(1 + j)n

(101)

For the sake of contradiction, assume i ≤ j . Then it follows by ourassumptions that

F − PB < PB − PA ≤F − PB

(1 + i)n(102)

which is impossible.


Exotic Bonds

Callable Bonds: Bond can be recalled by issuer over a window ofredemption dates. Price investor pays is the minimum of these possibleprices. So, if the bond was purchased at a premium, then price at earliestpossible date. If purchased at a discount, price at latest possibleredemption date


Exotic Bonds

Serial Bonds: Face amount F is redeemed over a collection of datest1, .., tn with staggered payments F1 + ..+ Fn = F , coupon ratesr1, .., rn, tailored to yield rates j1, .., jn. One can view this as the sum of mbonds with the corresponding payments, and can use Makehams formulato price this

P =n∑

m=1

Pm

Pt =Ft

(1 + j)nt+

rtjt·(

Ft −Ft

(1 + jt)nt

) (103)


Measuring The Rate of Return Of An Investment

Yield rate can be derived from cash-flows from an investment

When presented with more than one investment opportunity, wouldlike to be able to quantitatively measure their value to the investor

There are many ways to do this some work in situations that othersdont


Internal Rate of Return

In an investment, at any time may have cash flows out and in saypayments to buy a license for a taxicab, with monthly maintenance feesversus fare payments in.

Notation is such that Ak is a payment received and Bk is a payment madeout, each at the same time tk . The net payment is thus Ck .

f (j) :=n∑

k=1

Ck

(1 + j)tk(104)

The Internal Rate of Return j is one that balances out all net payments sothat their net present value is 0.


Example 5.1

Smith buys 1000 shares of stock at 5.00 per share and pays a commissionof 2%. Six months later he receives a cash-dividend of 0.20 per share,which he immediately reinvests commission free in shares at a price of 4.00per share. Six months after that he buys another 500 shares at a price4.50 per share, along with a commission of 2%. Six months after that hereceives another cash dividend of 0.25 per share and sells his existingshares at 5.00 per share, again paying a 2% commission. Find Smithsinternal rate of return for the entire transaction in the form i (2)


Example 5.1

Let 0 represent the time of the original share purchase, t = 1 at 6 months,t = 2 at 12 months and t = 3 at 18 months after the original purchase.Then

(A0,B0) = (0, 5000 + [0.02× 5000]) = (0, 5100)

(A1,B1) = (200, 200)

(A2,B2) = (0, [1.02× 4.50× 500]) = (0, 2295)

(A3,B3) = ([0.25 + 0.98× 5]× 1550, 0)

= (7982.50, 0)

∴ (C0,C1,C2,C3) = (−5100, 0,−2295, 7982.5)

⇒ 0 = −5100 + 0 · ν − 2295ν2 + 7982.50ν3

(105)

Solving this polynomial, we obtain

ν =1

1 + j⇒ i (2) = 2j = 0.0649 (106)


Uniqueness? Guaranteed Solution?

If C0 < 0 and Cj > 0 for all j ∈ 1, 2, ..n, then by the IntermediateValue theorem and simple calculus, we have a unique solution for theinternal return rate j . (i.e. Mortgages or other loans paid off byperiodic payments)

BUT, we can have situations where the above isnt satisfied, and wecompute more than solution for j , or worse, no real solution.

What can we use to measure the value of an investment in this case?


Net Present Value

Assume that among all possible investment alternatives, all have samemeasure of risk and fixed interest rate i that investor proposes(Utility theory!)

In this incomplete case, use i , also known as cost of capital, to findpresent value of cash flows of each investment on the table

The one with the largest present value, given investors rate i, is mostpreferable. In symbols, for investments a, b, c , with NPVfa(i), fb(i), fc(i), choose the biggest

HW: Can NPV curves ever intersect for different values of i ?

Of course, if i = j =yield rate, then NPV = 0


Profitability Index

Again, if the investor proposes his own interest rate i , then we can useanother measure, the Profitability Index:

I =PV [ Cash Flows In]

PV [ Cash Flows Out ](107)

Note that if i = j = IRR, then I = 1 by design,as yield rate is what we useto balance out flows with in flows.


Profitability Index: Example

Compare

(a) lending 1000 and being repaid 250 per year for 5 years withi = 0.05 and

(b) lending 1000 and being repaid 140 per year for 10 years withi = 0.05

We compute

Ia =250a5 0.05

1000= 1.0824

Ib =140a10 0.05

1000= 1.0810

(108)


MIRR

Goal: Find j such that Accumulated Value at end of project of paymentsout at rate j matches Accumulated Value of payments received atinvestor’s cost of capital i .

Notation:

Ak is the payment in at time k

Bk is the payment out at time k

Then j solves the equation

N∑n=0

An · (1 + i)n =N∑

m=0

Bm · (1 + j)m (109)


MIRR: Example

Compare

(a) lending 1000 and being repaid 250 per year for 5 years withi = 0.05 and

(b) lending 1000 and being repaid 140 per year for 10 years withi = 0.05

We compute

1000(1 + ja)5 = 250s5 0.05 = 1381.41

⇒ ja = 0.0668

1000(1 + jb)10 = 140s10 0.05 = 1760.90

⇒ jb = 0.0582

(110)


Project Return Rate

There are times during a project where investor is borrower and timeswhen she is lender. Cost of financing at rate i , and project return rate is jwhen she is net lender. Solving j in terms of i would give minimum returnrate needed for project to break even if set NPV to 0 at end of project.

Goal: Set up an equation of value such that Net Project Value = 0 whent = N. If the value at t = n is > 0, then earn at rate j . If balance < 0,then ”earn” (pay) at rate i (cost of capital). As an example, consider

S0 = 1→ S1− = (1 + j)

S1 = 1 + j − 2.3→ S2− = (1 + j − 2.3)(1 + i)

S2 = (1 + j − 2.3)(1 + i) + 1.33 = 0

⇒ j = 1.3− 1.33

1 + i

(111)


Dollar Weighted Rate of Return

Goal: Calculate measure of return over 1 year.

Use Simple Interest forfraction of years. Define interest gained as the difference between the totalamount paid out during year to term and the total amount paid in duringyear to term.

idollar weigted =Total Interest Gained

Average Amount on deposit during year(112)


Dollar Weighted Rate of Return

Goal: Calculate measure of return over 1 year. Use Simple Interest forfraction of years. Define interest gained as the difference between the totalamount paid out during year to term and the total amount paid in duringyear to term.

idollar weigted =Total Interest Gained

Average Amount on deposit during year(112)


Example 5.3

A Pension fund receives contributions and pays benefits from time to time.The fund began the year 2005 with a balance of 1, 000, 000. There werecontributions to the fund of 200, 000 at the end of February and again atthe end of August. There was a benefit of 500, 000 paid out of the fund atthe end of October. The balance remaining at the start of the year 2006was 1, 100, 000. Find the dollar weighted return on the fund, assumingeach month is 1

12 of a year


Example 5.3

1, 100, 000 = 1, 000, 000(1 + i) + 200, 000 ·(

1 +10

12i

)+ 200, 000 ·

(1 +

4

12i

)− 500, 000 ·

(1 +

2

12i

)T .I .G . = 1, 100, 000 + 500, 000− 1, 000, 000

− 200, 000− 200, 000 = 200, 000

A.A.o.d .d .y = 1, 000, 000 + 200, 000 · 10

12

+ 200, 000 · 4

12− 500, 000 · 2

12= 1, 150, 000

⇒ i =200, 000

1, 150, 000= 0.1739

(113)


Even More Methods of Valuation

Time Weighted Rate of Return: Compound over fraction of years, butthe time length doesnt matter

Portfolio Method: Contributions to a fund, once opened, aresegregated from initial investment and accrue interest at a differentrate for a fixed term

Interest Preference Rates for Borrowing and Lending: If accumulatedamount falls below zero, pay interest at a different rate than interestcredited on a positive balance. This is useful to compare twoalternatives when one may have a negative balance for some time


Continuous Withdrawal and Investment

Begin with the equation for accumulated value

F (t2) = F (t1)·(1+i)t2−t1+n∑

k=1

ct∗k (1+i)t2−t∗k +

∫ t2

t1

c(t)(1+i)t2−tdt (114)

As an example, consider t1 = 0, t2 = 1, t∗k = kn , ct∗k = 1, c(t) = N − n.

Then

F (1) = F (0) · (1 + i) +n∑

k=1

(1 + i)1−kn + (N − n) ·

∫ 1

0(1 + i)1−tdt

= F (0) · (1 + i) +i

(1 + i)1n − 1

+(N − n) · iln (1 + i)

(115)

Now, given F (0),F (1), n,N, solve for i .


Term Structure of Interest Rates

The relationship of the yield of a bond (or other fixed incomeproduct) to its time to maturity is the Term Structure

The graph of this relationship is the Yield Curve

Term Structure is dynamic ; reflects new information gained asmarkets evolve

A yield curve that corresponds to Normal Term Structure is one thatsees yield increase when the time to maturity increases. When theyield decreases as term increases, we say the yield curve is Inverted,and if it is roughly constant, we say it is a Flat curve.


Law of One Price

Bond Prices are linear. If we have a coupon paying bond, it musthave the same value as the sum of the prices of each coupon as anindividual zero coupon bond, as well as the lump sum paid atmaturity.

If this does not hold, then we have observed a case of arbirtrage. Thismeans one could construct a portfolio that consists of purchasingindividual pieces of a coupon bond while selling the equivalent wholebond, making a net profit with no risk involved.


Example 6.1

Suppose that the current term structure has the following yields onzero-coupon bonds, where all yields are nominal annual rates of interestcompounded semi-annually

Table: Example 6.1: Term Structure

Term (Years) 0.5 1 1.5 2

Zero Coupon Bond Rate 8% 9% 10% 11%

Find the price per 100 face amount and yield to maturity of each of thefollowing 2-year bonds (with semi-annual coupons)

a.) Zero coupon bond

b.) 5% annual coupon rate

c .) 10% annual coupon rate


Example 6.1

a.) PV = 100

(1+ 0.112 )

4 = 80.72, which implies j (2) = 0.11

b.)

PV =2.5

1.04+

2.5

1.0452+

2.5

1.053+

102.5

1.0554= 89.59

=2.5

1 + j(2)

2

+2.5(

1 + j(2)

2

)2 +2.5(

1 + j(2)

2

)3 +102.5(

1 + j(2)

2

)4⇒ j (2) = 0.109354

(116)


Example 6.1

c .) Similarly,

PV =5

1.04+

5

1.0452+

5

1.053+

105

1.0554

= 98.46

=5

1 + j(2)

2

+5(

1 + j(2)

2

)2 +5(

1 + j(2)

2

)3 +105(

1 + j(2)

2

)4 (117)

and so j (2) = 0.108775


Spot Rates

The Spot Rate st is the effective annual interest rate for a zerocoupon bond maturing t years from now.

There is a relationship between the spot rate and the yield tomaturity of a bond.

If term structure of spot rates is increasing, then yield for bonds withsame maturity decreases as coupons increase


Spot Rates and Yield Rates

The general equation of value here to determine the yield rate yr on abond with face 1 is

PV = r ·(

1

1 + s1+

1

(1 + s2)2+ ..+

1

(1 + sn)n

)+

1

(1 + sn)n

= r ·(

1

1 + yr+

1

(1 + yr )2+ ..+

1

(1 + yr )n

)+

1

(1 + yr )n

(118)



Solving for r , we obtain

r = −1

(1+yr )n− 1

(1+sn)n∑nk=1

1(1+yr )k

−∑n

k=11

(1+sk )k

(119)



Q: Is there anything we can say about dyrdr given a normal term structure?

A: Implicit differentiation leads to

dyrdr

=1

r·

1(1+sn)n

− 1(1+yr )n

r∑n

k=1k

(1+yr )k+1 + n(1+yr )n+1

< 0

(120)

if we have a normal term structure as we expect sn > yr and so1

(1+sn)n< 1

(1+yr )n.


Forward Rates

In keeping with Law of One Price (No Arbitrage), can set up therelationship between spot rates and forward rates

Forward Rate is the interest rate to be charged starting at time t for afixed period

We will talk about 1−year Forward Rates, i.e. interest rate chargedfrom to t + 1


Exact Form

If we set up a deal such that 1 is to be invested at time 0, and the 1−yearspot rate s1 = 0.08 , and the 2−year spot rate is s2 = 0.09, then at t = 1,if the rate to borrow and lend is the same, the rate from year 1 to year 2,i1,2 , should be defined via

1.092 = (1 + i1,2) · 1.08

i1,2 = 0.1001(121)


General Relationship between Spot Rates and ForwardRates

1 + s1 = 1 + i0,1

(1 + s2)2 = (1 + i0,1)(1 + i1,2)

... =...

(1 + sn)n = Πnk=1(1 + ik−1,k)

(122)

Inverting this relationship, we obtain

ik−1,k =(1 + sk)k

(1 + sk−1)k−1− 1 (123)

Q: For a normal term structure where sk > sk−1, what can we say aboutthe relationship between sk and ik−1,k ?


General Relationship between Spot Rates and ForwardRates

From the previous slide, we can compute the following table:


Spot Rate s1 = 0.05 s2 = 0.1 s3 = 0.15

Forward Rate i0,1 = 0.05 i1,2 = 0.1524 i2,3 = 0.2596


Force of Interest vs YTM

Define αt as the continuously compounded yield rate. Then

PV (t = 0) = Fe−tαt

= Fe−∫ t0 δsds

⇒ tαt =

∫ t

0δsds

⇒ αt + tdαt

dt= δt

(124)



Summarizing, we have the dual relationship

αt =1

t

∫ t

0δsds

δt = αt + tdαt

dt

(125)



Recall that in the discrete case,

1 + sn = [(i + i0,1) · (1 + i1,2) · ... · (1 + in−1,n)]1n (126)

Hence, 1 + sn is the geometric mean of the forward rates.

In the continuous case, the YTM αt is the arithmetic mean of the force ofinterest over the interval [0, t]. It follows that δt can be seen as a forwardrate in the continuous case.

Q: Again, if the term structure is increasing, ie. if dαtdt > 0, then how does

αt relate to δt ?


Example 6.5

Suppose that the yield to maturity for a zero-coupon bond maturing attime t is αt = 0.09− 0.08 · 0.94t , a continuously compounded rate.Forward Rate is the interest rate to be charged starting at time t for afixed period

(a) Find the related forward rate

(b) A borrower plans to borrow 1000 in one year and repay the loanwith a single payment at the end of the second year. Determine theamount that will have to be paid back based on the stated termstructure


Example 6.5

a.) We find that

δt = αt + tdαt

dt= 0.09− 0.08 · 0.94t − 0.08 · 0.94t · ln (0.94) · t (127)

b.) We find that

Value = 1000e∫ 21 δtdt = 1024.11

= 1000e2α2−α1 (128)


Example 6.5

a.) We find that

δt = αt + tdαt

dt= 0.09− 0.08 · 0.94t − 0.08 · 0.94t · ln (0.94) · t (127)

b.) We find that

Value = 1000e∫ 21 δtdt = 1024.11 = 1000e2α2−α1 (128)


At Par Yield

Measure of Bond Yield: Find the Coupon rate rt such that the BondYield is also rt .

In this case, the Bond price is at par.

For an n−year bond,

1 = rn ·

(n∑

k=1

(1 + sk)−k

)+

1

(1 + sn)n

⇒ rn =1− 1

(1+sn)n∑nk=1

1(1+sk )k

(129)


Interest Rate Swaps

Used to convert Floating Rate Liability to Fixed Rate Liability

Used between two borrowers that have access to different interestmarkets

Example

(FixedRate(A),FloatingRate(A)) = (0.08,Prime + 0.01)

(FixedRate(B),FloatingRate(B)) = (0.09,Prime + 0.015)(130)


Interest Rate Swaps

Swap: A borrows at FixedRate(A), and loans to B at rate 0.085. Bborrows at FloatingRate(B), and loans to A at rate prime + 0.0125.

Net effect:

A pays (0.08− 0.085 + prime + 0.0125) = prime + 0.0075

B pays (prime + 0.015− (prime + 0.0125) + 0.085) = 0.0875

Notice both of these are smaller than what they could have gotten on theirown. Usually, there is a question of default risk, so an intermediary can bebrought in, for a small fee, depending on the spread between availablerates for A and B


Interest Rate Swaps

In general, a swap is a series of coupons exchanged upon delivery.

Imagine a bond with face value F , and term structure (impliedforward rates) for period k − 1 to k : ik−1,knk=1.

Then the implied swap rate Rn for term n is the rate that equatesthe present value all the coupon exchanges, forward rate to constantRn, to 0:

0 =n∑

k=1

(Rn − ik−1,k)F

(1 + sk)k(131)


Interest Rate Swaps

We can interpret Rn in two ways. The first is to look at Rn as an averageforward rate:

Rn =n∑

j=1

pj × ij−1,j

pj =(1 + sj)

−j∑nm=1(1 + sm)−m

(132)

Note that 0 < pj < 1 and∑m

j=1 pj = 1 for positive spot rates.


Interest Rate Swaps

The second way to interpret Rn is to look at Rn as an at par yield rate :

Rn =n∑

j=1

( (1 + sj)−j∑n

m=1(1 + sm)−m

)×

((1 + sj)

j

(1 + sj−1)j−1− 1

)

=

∑nj=1

(1

(1+sj−1)j−1 − 1(1+sj )j

)∑n

m=1(1 + sm)−m=

1− (1 + sn)−n∑nm=1(1 + sm)−m

= rn

(133)

as the numerator above is a telescoping series.


Deferred and Non-Level Notional Interest Rate Swaps

Keeping with the analysis above, we have a Deferred Swap beginning ink periods with value

R(k)n =

n∑j=k

p(k)j × ij−1,j

p(k)j =

(1 + sj)−j∑n

m=k(1 + sm)−m.

(134)


Deferred and Non-Level Notional Interest Rate Swaps

Similarly, if the notional used to calculate the coupon varies with time, andis defined as Fj at time j , then

0 =n∑

k=1

(Rn − ik−1,k)Fk

(1 + sk)k(135)

then

Rn =n∑

j=k

qj × ij−1,j

qj =Fj × (1 + sj)

−j∑nm=k Fm × (1 + sm)−m

.

(136)


Example: Calculating the Swap Curve

Recall the table from Example 6.3:


Spot Rate s1 = 0.05 s2 = 0.1 s3 = 0.15

Forward Rate i0,1 = 0.05 i1,2 = 0.1524 i2,3 = 0.2596


Example: Calculating the Swap Curve

We can now add to this table:


Spot s1 = 0.05 s2 = 0.1 s3 = 0.15

Fwd i0,1 = 0.05 i1,2 = 0.1524 i2,3 = 0.2596

Prob p1 =1

1.051

1.05+ 1

1.102+ 1

1.153

p2 =1

1.1021

1.05+ 1

1.102+ 1

1.153

p3 =1

1.1531

1.05+ 1

1.102+ 1

1.153

∴ r3 =1

1.05 × 0.051

1.05 + 11.102

+ 11.153

+1

1.102× 0.1524

11.05 + 1

1.102+ 1

1.153

+1

1.153× 0.2596

11.05 + 1

1.102+ 1

1.153

= 0.1413.(137)

HW: Compute r1, r2. Can you set up a spreadsheet to do this?


Susceptibility: Single Factor Model

How does the PV of the cashflow change as the yield rate (termstructure) changes? (1st order changes)

Once quantified, this can be used as a measure of risk associated withinvesting in this cashflow.

Consider the case where we have continuous compounding with a force ofinterest δ. For a stream of payments, we have the present value, or upfront price, determined by P(δ). The only risk factor that could changethe value then is δ. If we were to equate this value with a zero-couponbond of unknown Duration D, then

P(δ) = P(0)e−δD

D = − 1

P

dP

dδ

(138)

This of course implies that dPdδ < 0.


Susceptibility and Sensitivity to Market Changes: SingleFactor Model

If the environment is one of annual compounding, then δ = ln (1 + i) andwe define

DM := limh→0

1

h· P(i)− P(i + h)

P(i)

= − d

diln (P(i))

D := (1 + i) · DM

(139)

For P(i) = (1 + i)−n, we obtain

DM =n

1 + i

D = n(140)


Intuition

Increasing yield decreases present value. Thus, we take the negativeof derivative w.r.t yield.

The Modified Duration DM measures the ratio of rate of change of Pwrt i to the actual value P.

The Macaulay Duration D := (1 + i) · DM says that the longer thetime maturity n, the more susceptible it is to yield change. For ann−year zero coupon bond, D = n. Hence, the name duration

Can apply the linearization formula from calculus:

P(i + h) ≈ P(i) + P ′(i) · h = P(i)− DM · P(i) · h

= P(i)− D

1 + i· P(i) · h

(141)


Duration of General Cashflows

P =n∑

m=1

Km

(1 + i)m

DM =− d

diln (P(i)) = − 1

P

dP

di=

∑nm=1

m·Km(1+i)m+1∑n

m=1Km

(1+i)m

D = (1 + i) · DM =n∑

m=1

m · pm = E[m]

pm =

Km(1+i)m∑nl=1

Kl

(1+i)l

(142)


Duration of General Cashflows

Note also that

dD

di=

d

di

n∑m=1

m ·Km

(1+i)m∑nl=1

Kl

(1+i)l

= − 1

1 + i

n∑m=1

m2 · pm −

(n∑

m=1

m · pm

)2

= −Var [m]

1 + i< 0

(143)


Duration of a Perpetuity

Even though it has no finite term, we can still find the duration of aperpetuity. In this case

D = −(1 + i)1

P

dP

di

= −(1 + i) · 11i

· d

di

(1

i

)= 1 +

1

i

(144)

An equivalent way to calculate this is to see that

D =

∑∞k=1

k(1+i)k∑∞

k=11

(1+i)k

=(Ia)∞ i

a∞ i

=1+ii2

1i

= 1 +1

i

(145)


Example 7.3 Duration of a Coupon Bond

In the case of a coupon bond, we have K1 = K2 = .. = Kn−1 = F · r , andKn = F + F · r , and so

D =

∑nm=1

m·F ·r(1+i)m + n·F

(1+i)n∑nm=1

F ·r(1+i)m + F

(1+i)n

=

r ·(

(1+i)an i−n

(1+i)n

i

)+ n

(1+i)n

r · an i + 1(1+i)n

(146)


Bonds and Macaulay Duration: Example from previousSOA Exam FM

A bond with Face F will pay a coupon of F · r at the end of each of thenext three years. It wil also pay the face value of F at the end of thethree-year period. The bond’s duration (Macaulay duration) when valuedusing an annual effective interest rate of 20% is X. Calculate X .


Bonds and Macaulay Duration: Example from previousSOA Exam FM

Recall that the Macaulay Duration over n time periods is defined as anexpection of sorts:

D =

∑ni=1 tiPV (i)∑nj=1 PV (j)

(147)

where the PV (i) are the present values of the revenue received at time ti .

For our present case, we have

D =1 · F ·r1.2 + 2 · F ·r

(1.2)2+ 3 · F ·(1+r)

(1.2)3

F ·r1.2 + F ·r

(1.2)2+ F ·(1+r)

(1.2)3

=1 · r

1.2 + 2 · r(1.2)2

+ 3 · (1+r)(1.2)3

r1.2 + r

(1.2)2+ (1+r)

(1.2)3

(148)


Portfolio Duration

Imagine that P(i) is the price of a portfolio of income streams:P(i) =

∑nk=1 Pk(i). Then

D = −(1 + i) · P ′(i)

P(i)= −(1 + i) ·

∑nk=1 P ′k(i)

P(i)

= −(1 + i) ·n∑

k=1

P ′k(i)

P(i)= −(1 + i) ·

n∑k=1

Pk(i)

P(i)·

P ′k(i)

Pk(i)

=n∑

k=1

Dk · qk

(Dk , qk) =

(−(1 + i) ·

P ′k(i)

Pk(i),

Pk(i)

P(i)

)(149)

and so the portfolio duration is the weighted average of the individualdurations.


Portfolio Duration

Smith holds a portfolio of a 20− year 100 bond bought at par and withduration of 8, a stock purchased for 50 with duration 7 and an 30− yearannuity purchased for 50 with duration 5. What is the portfolio duration?

D = 8 · 100

200+ 7 · 50

200+ 5

50

200= 7 (150)


Cashflow Immunization and Convexity

An institution such as a bank or insurance company will have both assetsand liabilities which may come due at different times. These may bestructured so that the net present value of the institution is zero. Hence,for a given i ,

P(i) ≈ 0 (151)

If the rate i is perturbed slightly, the institution may find itself with anegative NPV. To counter this risk, the institution may structure its assetsand liabilities such that P is a local minimum at i .


Cashflow Immunization and Convexity

To explain this further, define the Convexity

Γ :=P ′′(i)

P(i)(152)

and so

P(i + h)− P(i)

P(i)≈ −DM · h + Γ · h2 (153)

If P ′(i) = 0 and Γ(i) > 0, then DM = 0 and we have a local minimum


Cashflow Immunization Example

Suppose a bank has promised its investors a one-year 5% return ondeposits of 100. The bank has 50 per investor on hand, is able to purchaseperpetuities that pay one per year and one year zero-coupon bonds thatyield 10%. How should the bank structure it’s assets so they areimmunized from interest rate risk?


Cashflow Immunization Example

At rate i , we have the NPV of x units perpetuity and y units of bondminus the NPV of the liability of the one-year withdrawal as

P(i) = 50 +x

i+

1.1 · y1 + i

− 105

1 + i

P ′(i) = − x

i2− 1.1 · y − 105

(1 + i)2

P ′′(i) = 2 · x

i3+ 2 · 1.1 · y − 105

(1 + i)3

(154)

For whatever rate i we fix, if we set

x = 50i2

y =105− 50 · (1 + i)2

1.1

(155)

then P(i) = P ′(i) = 0 and P ′′(i) = 100i ·(1+i) > 0 so we have a local

minimum.Albert Cohen (MSU) MATH 360: Theory of Investment and Credit MSU Spring 2014 178 / 223

Some Comments

Book, or Amortized, value is calculated using fixed rate

Market value determined via market term structure

The spot rate used in this term structure varies with time

This leads to difference between Book and Market value of cashflow

These values converge as time approaches maturity; investor can holdon to his investment to realize original yield rate

Before maturity, however, the cashflow is valued at market rate

Previous analysis is for flat term structure. To generalize to normal orinverted term structure, require new definitions and multivariablecalculus


Basic Ideas and Beyond

As we have seen, interest rate fluctuation can lead to possibleshortfalls that banks may wish to immunize their portfolio against.Currency fluctuation also leads to operational risk, and shouldfactor into planning. For example, should a band tour Europe orAmerica this summer ? Click here for an insightful article by NeilShah in the Wall Street Journal TM, with comment from themanager of a very prominent rock band

Interest rates are only one risk factor. Another very real factor isknown as longevity risk, which is due to the possibility that apensioner may live longer than expected. Hedging against such apossibility is extremely important, and a topic we hope to cover inSTT 455− 456. In the meantime, please consult the paper by Tsai,Tzeng, and Wang on Hedging Longevity Risk When Interest RatesAre Uncertain


http://online.wsj.com/article/SB10001424052970204630904577056123331660042.html

http://www.soa.org/library/journals/north-american-actuarial-journal/2011/no-2/naaj-2011-vol15-no2-tsai.pdf



Probability Space

Let us define an event as a point ω in the set of all possible outcomes Ω.This includes the events ”The stock doubled in price over two tradingperiods” or ”the average stock price over ten years was 10 dollars”.

In our initial case, we will consider the simple binary spaceΩ = H,T for a one-period asset evolution. So, given an initialvalue S0, we have the final value S1(ω), with

S1(H) = uS0,S1(T ) = dS0 (156)

with d < 1 < u. Hence, a stock increases or decreases in price,according to the flip of a coin.

Let P be the probability measure associated with these events:

P[H] = p = 1− P[T ] (157)


Arbitrage

Assume that S0(1 + r) > uS0

Where is the risk involved with investing in the asset S ?

Assume that S0(1 + r) < dS0

Why would anyone hold a bank account (zero-coupon bond)?

Lemma Arbitrage free ⇒ d < 1 + r < u


Derivative Pricing

Let S1(ω) be the price of an underlying asset at time 1. Define thefollowing instruments:

Zero-Coupon Bond : V B0 = 1

1+r ,VB1 (ω) = 1

Forward Contract : V F0 = 0,V F

1 = S1(ω)− F

Call Option : V C1 (ω) = max(S1(ω)− K , 0)

Put Option : V P1 (ω) = max(K − S1(ω), 0)

In both the Call and Put option, K is known as the Strike.Once again, a Forward Contract is a deal that is locked in at time 0 forinitial price 0, but requires at time 1 the buyer to purchase the asset forprice F .

What is the value V0 of the above put and call options?


Derivative Pricing



1+r ,VB1 (ω) = 1


1 = S1(ω)− F



In both the Call and Put option, K is known as the Strike.

Once again, a Forward Contract is a deal that is locked in at time 0 forinitial price 0, but requires at time 1 the buyer to purchase the asset forprice F .



Derivative Pricing



1+r ,VB1 (ω) = 1


1 = S1(ω)− F



In both the Call and Put option, K is known as the Strike.Once again, a Forward Contract is a deal that is locked in at time 0 forinitial price 0, but requires at time 1 the buyer to purchase the asset forprice F .



Put-Call Parity

Can we replicate a forward contract using zero coupon bonds and put andcall options?

Yes: The final value of a replicating strategy X has value

V C1 − V P

1 + (K − F ) = S1 − F = X1(ω) (158)

This is achieved (replicated) by

Purchasing one call option

Selling one put option

Purchasing K − F zero coupon bonds

all at time 0.Since this strategy must have zero initial value, we obtain

V C0 − V P

0 =F − K

1 + r(159)

Question: How would this change in a multi-period model?


Put-Call Parity

Can we replicate a forward contract using zero coupon bonds and put andcall options?Yes: The final value of a replicating strategy X has value

V C1 − V P

1 + (K − F ) = S1 − F = X1(ω) (158)






V C0 − V P

0 =F − K

1 + r(159)



Put-Call Parity


V C1 − V P

1 + (K − F ) = S1 − F = X1(ω) (158)





all at time 0.

Since this strategy must have zero initial value, we obtain

V C0 − V P

0 =F − K

1 + r(159)



Put-Call Parity


V C1 − V P

1 + (K − F ) = S1 − F = X1(ω) (158)






V C0 − V P

0 =F − K

1 + r(159)



Put-Call Parity


V C1 − V P

1 + (K − F ) = S1 − F = X1(ω) (158)






V C0 − V P

0 =F − K

1 + r(159)

Question: How would this change in a multi-period model?Albert Cohen (MSU) MATH 360: Theory of Investment and Credit MSU Spring 2014 184 / 223

General Derivative Pricing -One period model

If we begin with some initial capital X0, then we end with X1(ω). To pricea derivative, we need to match

X1(ω) = V1(ω) ∀ ω ∈ Ω (160)

to have X0 = V0, the price of the derivative we seek.

A strategy by the pair (X0,∆0) wherein

X0 is the initial capital

∆0 is the initial number of shares (units of underlying asset.)

What does the sign of ∆0 indicate?


Replicating Strategy

Initial holding in bond (bank account) is X0 −∆0S0

Value of portfolio at maturity is

X1(ω) = (X0 −∆0S0)(1 + r) + ∆0S1(ω) (161)

Pathwise, we compute

V1(H) = (X0 −∆0S0)(1 + r) + ∆0uS0

V1(T ) = (X0 −∆0S0)(1 + r) + ∆0dS0

Algebra yields

∆0 =V1(H)− V1(T )

(u − d)S0(162)


Risk Neutral Probability

Let us assume the existence of a pair (p, q) of positive numbers, and usethese to multiply our pricing equation(s):

pV1(H) = p(X0 −∆0S0)(1 + r) + p∆0uS0

qV1(T ) = q(X0 −∆0S0)(1 + r) + q∆0dS0

Addition yields

X0(1 + r) + ∆0S0(pu + qd − (1 + r)) = pV1(H) + qV1(T ) (163)


If we constrain

0 = pu + qd − (1 + r)

1 = p + q

0 ≤ p

0 ≤ q

then we have a risk neutral probability P where

V0 = X0 =1

1 + rE[V1] =

pV1(H) + qV1(T )

1 + r(164)

with

p =1 + r − d

u − d

q =u − (1 + r)

u − d


If we constrain

0 = pu + qd − (1 + r)

1 = p + q

0 ≤ p

0 ≤ q

then we have a risk neutral probability P where

V0 = X0 =1

1 + rE[V1] =

pV1(H) + qV1(T )

1 + r(164)

with

p =1 + r − d

u − d

q =u − (1 + r)

u − dAlbert Cohen (MSU) MATH 360: Theory of Investment and Credit MSU Spring 2014 188 / 223

Example: Pricing a forward contract

Consider the case of a stock with

S0 = 100

u = 1.2

d = 0.8

r = 0.05

Then the forward price is computed via

0 =1

1 + rE[S1 − F ]⇒ F = E[S1] (165)


This leads to the explicit price

F = puS0 + qdS0

= (0.625)(1.2)(100) + (0.375)(0.8)(100) = 105

Homework Question: What is the price of a call option in the caseabove,with strike K = 95?


General one period risk neutral measure

We define a finite set of outcomes Ω ≡ ω1, ω2, ..., ωn and anysubcollection of outcomes A ⊂ Ω an event.

Furthermore, we define a probability measure P, not necessarily thephysical measure P to be risk neutral if

P[ω] > 0 ∀ ω ∈ Ω

X0 = 11+r E[X1]

for all strategies X .



We define a finite set of outcomes Ω ≡ ω1, ω2, ..., ωn and anysubcollection of outcomes A ⊂ Ω an event.Furthermore, we define a probability measure P, not necessarily thephysical measure P to be risk neutral if

P[ω] > 0 ∀ ω ∈ Ω

X0 = 11+r E[X1]

for all strategies X .



The measure is indifferent to investing in a zero-coupon bond, or arisky asset X

The same initial capital X0 in both cases produces the same”‘average”’ return after one period.

Not the physical measure attached by observation, experts, etc..

In fact, physical measure has no impact on pricing


Example: Risk Neutral measure for trinomial case

Assume that Ω = ω1, ω2, ω3 with

S1(ω1) = uS0

S1(ω2) = S0

S1(ω3) = dS0

Given a payoff V1(ω) to replicate, are we assured that a replicatingstrategy exists?Homework: Try our first example with

S0 = 100

r = 0.05, u = 1.2, d = 0.8

V1(ω) = χ (S1(ω) > 90)




S1(ω1) = uS0

S1(ω2) = S0

S1(ω3) = dS0

Given a payoff V1(ω) to replicate, are we assured that a replicatingstrategy exists?

Homework: Try our first example with

S0 = 100

r = 0.05, u = 1.2, d = 0.8

V1(ω) = χ (S1(ω) > 90)




S1(ω1) = uS0

S1(ω2) = S0

S1(ω3) = dS0

Given a payoff V1(ω) to replicate, are we assured that a replicatingstrategy exists?Homework: Try our first example with

S0 = 100

r = 0.05, u = 1.2, d = 0.8

V1(ω) = χ (S1(ω) > 90)


Existence of Risk Neutral measure

Let P be a probability measure on a finite space Ω. The following areequivalent:

P is a risk neutral measure

For all traded securities S i , S i0 = 1

1+r E[S i1

]Proof: Homework (Hint: One direction is much easier than others. Also,strategies are linear in the underlying asset.)






1+r E[S i1

]

Proof: Homework (Hint: One direction is much easier than others. Also,strategies are linear in the underlying asset.)






1+r E[S i1

]Proof: Homework (Hint: One direction is much easier than others. Also,strategies are linear in the underlying asset.)


Complete Markets

A market is complete if it is arbitrage free and every non-traded asset canbe replicated.

Fundamental Theorem of Asset Pricing 1: A market is arbitrage freeiff there exists a risk neutral measure

Fundamental Theorem of Asset Pricing 2: A market is complete iffthere exists exactly one risk neutral measure


2-period pricing

Consider the case

r = 0.05,S0 = 100

S1(H) = 1.2S0

S1(T ) = 0.8S0

S2(HH) = 1.2S1(H)

S2(HT ) = 0.8S1(H)

S2(TH) = 1.2S1(T )

S2(TT ) = 0.8S1(T )

Now price a digital option that has payoff V2 := χ(S2 ≥ 100)


We can do this for 2-period problems

Case by Case, or..

by developing a general theory for multi-period asset pricing

In the latter method, we need a general framework to carry out ourcomputations


Multi-Period Pricing - Introduction

Again, we define the finite set of outcomes Ω and any subcollection ofoutcomes A ⊂ Ω an event.

How does this relate to the case of 2 consecutive coin flips

Ω ≡ HH,HT ,TH,TTSet of events includes statements like at least one head

= HH,HT ,TH





Ω ≡ HH,HT ,TH,TT

Set of events includes statements like at least one head

= HH,HT ,TH





Ω ≡ HH,HT ,TH,TTSet of events includes statements like at least one head

= HH,HT ,TH


σ−algebras

Given a non-empty set Ω of outcomes, a σ−algebra F is a collection ofsubsets of Ω that satisfies

∅ ∈ FA ∈ F ⇒ Ac ∈ FA1,A2,A3, .... ∈ F ⇒ ∪∞n=1An ∈ F

Some Examples

F0 = ∅,ΩF1 = ∅,Ω, HH,HT, TT ,THF2 = ∅,Ω, HH, HT, TT, TH, ....

F2 is completed by taking all unions of ∅,Ω, HH, HT, TT, TH.


σ−algebras


∅ ∈ F

A ∈ F ⇒ Ac ∈ FA1,A2,A3, .... ∈ F ⇒ ∪∞n=1An ∈ F

Some Examples




σ−algebras


∅ ∈ FA ∈ F ⇒ Ac ∈ F

A1,A2,A3, .... ∈ F ⇒ ∪∞n=1An ∈ F

Some Examples




σ−algebras


∅ ∈ FA ∈ F ⇒ Ac ∈ FA1,A2,A3, .... ∈ F ⇒ ∪∞n=1An ∈ F

Some Examples




σ−algebras


∅ ∈ FA ∈ F ⇒ Ac ∈ FA1,A2,A3, .... ∈ F ⇒ ∪∞n=1An ∈ F

Some Examples

F0 = ∅,Ω

F1 = ∅,Ω, HH,HT, TT ,THF2 = ∅,Ω, HH, HT, TT, TH, ....



σ−algebras


∅ ∈ FA ∈ F ⇒ Ac ∈ FA1,A2,A3, .... ∈ F ⇒ ∪∞n=1An ∈ F

Some Examples

F0 = ∅,ΩF1 = ∅,Ω, HH,HT, TT ,TH

F2 = ∅,Ω, HH, HT, TT, TH, ....



σ−algebras


∅ ∈ FA ∈ F ⇒ Ac ∈ FA1,A2,A3, .... ∈ F ⇒ ∪∞n=1An ∈ F

Some Examples




Notice that F0 ⊂ F1 ⊂ F2.

Correspondingly, given an Ω, we define a

Filtration as ..

a sequence of σ−algebras F0,F1,F2, ...,Fn, ... such that

F0 ⊂ F1 ⊂ F2 ⊂ ... ⊂ Fn ⊂ ...and F = σ(Ω) as the σ−algebra of all subsets of Ω.

Given a pair (Ω,F), we define a Random Variable X (ω) as a mappingX : Ω→ R




Filtration as ..


F0 ⊂ F1 ⊂ F2 ⊂ ... ⊂ Fn ⊂ ...

and F = σ(Ω) as the σ−algebra of all subsets of Ω.





Filtration as ..


F0 ⊂ F1 ⊂ F2 ⊂ ... ⊂ Fn ⊂ ...and F = σ(Ω) as the σ−algebra of all subsets of Ω.



Given a pair (Ω,F) and random variable X ,

σ(X ) = the collection of all sets ω ∈ Ω | X (ω) ∈ A ⊂ R

Given a G ⊂ F , we say that X is G−measurable if

A ∈ σ(X )⇒ A ∈ G

A Probability Space for us will be the triple (Ω,F ,P), where


Given a pair (Ω,F) and random variable X ,

σ(X ) = the collection of all sets ω ∈ Ω | X (ω) ∈ A ⊂ RGiven a G ⊂ F , we say that X is G−measurable if

A ∈ σ(X )⇒ A ∈ G

A Probability Space for us will be the triple (Ω,F ,P), where


P : F → [0, 1]

P[∅] = 0

For any countable disjoint sets A1,A2, ... ∈ FP [∪∞n=1An] =

∑∞n=1 P[An]


And so

P[A] :=∑

ω∈A P[ω]

E[X ] :=∑

ω X (ω)P[ω] =∑n

k=1 xkP[X (ω) = xk]

with Variance := E[(X − E[X ])2

]


Conditional Expectation

Let us return to the two flip model.

If we are give a probability measure P, and know the value S1, can weestimate the value S2 given (conditional on) this information?

An Example:

S0 = 100,S1(H) = 120,S1(T ) = 80

p = 0.4, q = 0.6 for each flip

ω = (ω1, ω2) - each flip is independent, and a path ω is the total pathgenerated by both flips

Given this set-up, compute

E [S2 | S1] (ω1) (166)


Conditional Expectation

Let us return to the two flip model.

If we are give a probability measure P, and know the value S1, can weestimate the value S2 given (conditional on) this information?

An Example:

S0 = 100, S1(H) = 120,S1(T ) = 80

p = 0.4, q = 0.6 for each flip

ω = (ω1, ω2) - each flip is independent, and a path ω is the total pathgenerated by both flips

Given this set-up, compute

E [S2 | S1] (ω1) (166)


This means if we take X = S1, for any A ∈ σ(S1)∑ω∈A

S2(ω)P[ω] =∑ω∈A

E [S2 | S1] (ω)P[ω] (167)

Can extend this to any sub σ−algebra of σ(S2)


Assume X ∈ F and G ⊂ F

Tower Property: If G1 ⊂ G2, then

E [X | G1] = E[E [X | G2] | G1

](168)

If X is G−measurable, then E [X | G ] = X


Taking out what is known: If X is dependent only on the informationcontained in G , then

E [XY | G ] = X E [Y | G ] (171)

Example: If G = F1, and X = S1(ω1), Y = S2, then

E [S1(ω1)S2 | F1] = S1(ω1)E [S2 | F1] (172)

Example: What is E[S2 | S0] ?

Definition: When conditioning on the σ−algebra Fn, we take thenotation

E [X | Fn] = En [X ] (173)

This is true, of course, for any measure we use.


Martingales

Observe a random process Mn that depends only on the first n coin flips.The history of the random variable is encapsulated in the filtration Fn itgenerates.

If Mn = En [Mn+1] then Mn is called a Martingale

If Mn ≤ En [Mn+1] then Mn is called a Submartingale

If Mn ≥ En [Mn+1] then Mn is called a Supermartingale

By the definition and the Tower property above, we have for all k ≥ 0

Mn = En [Mn+k ] (174)

if Mn is a Martingale


Martingale Pricing Theorem Under our model with risk neutralprobabilities p = 1+r−d

u−d , q = u−1−ru−d , the process

Mn :=Sn

(1 + r)n(175)

is a martingaleProofWe use the Tower property again:

En

[Sn+1

(1 + r)n+1

]= En

[Sn

(1 + r)n1

1 + r

Sn+1

Sn

]=

Sn

(1 + r)nEn

[1

1 + r

Sn+1

Sn

]=

Sn

(1 + r)npu + qd

1 + r

=Sn

(1 + r)n

QED




Mn :=Sn

(1 + r)n(175)

is a martingale

ProofWe use the Tower property again:

En

[Sn+1

(1 + r)n+1

]= En

[Sn

(1 + r)n1

1 + r

Sn+1

Sn

]=

Sn

(1 + r)nEn

[1

1 + r

Sn+1

Sn

]=

Sn

(1 + r)npu + qd

1 + r

=Sn

(1 + r)n

QED




Mn :=Sn

(1 + r)n(175)

is a martingaleProof

We use the Tower property again:

En

[Sn+1

(1 + r)n+1

]= En

[Sn

(1 + r)n1

1 + r

Sn+1

Sn

]=

Sn

(1 + r)nEn

[1

1 + r

Sn+1

Sn

]=

Sn

(1 + r)npu + qd

1 + r

=Sn

(1 + r)n

QED




Mn :=Sn

(1 + r)n(175)

is a martingaleProofWe use the Tower property again:

En

[Sn+1

(1 + r)n+1

]= En

[Sn

(1 + r)n1

1 + r

Sn+1

Sn

]=

Sn

(1 + r)nEn

[1

1 + r

Sn+1

Sn

]=

Sn

(1 + r)npu + qd

1 + r

=Sn

(1 + r)n

QEDAlbert Cohen (MSU) MATH 360: Theory of Investment and Credit MSU Spring 2014 210 / 223

Let us return to the multiperiod model. For each time n, we willdynamically redistribute our wealth Xn by deciding to hold ∆n shares of Sn

and invest the rest in the bank account at rate r . At time n + 1, thismeans we have the value

Xn+1 = (Xn −∆nSn) (1 + r) + ∆nSn+1 (176)

If we discount this recursive process to form

Mn :=Xn

(1 + r)n(177)

then Mn is a martingale


Risk Neutral Pricing Formula

Assume now that we have the regular assumptions on our coin flip space,and that at time N we are asked to deliver a path dependent derivativevalue VN . Then for times 0 ≤ n ≤ N, the value of this derivative iscomputed via

Vn = En

[Vn+1

1 + r

](178)

and so

V0 = E0

[VN

(1 + r)N

](179)


Markov Processes

If we use the above approach for a more exotic option, say a lookbackoption that pays the maximum over the term of a stock, then we find thisapproach lacking. There is not enough information in the tree or thedistinct values for S3 as stated. We need more. Consider our generalmulti-period binomial model under P

Definition We say that a process X is adapted if it depends only on theflips ω1, ..., ωn

Definition We say that an adapted process X is Markov if for every0 ≤ n ≤ N − 1 and every function f (x) there exists another function g(x)such that

En [f (Xn+1)] = g(Xn) (180)


This notion of Markovity is essential to our state-dependent pricingalgorithm. Indeed, since our stock process evolves from time n to timen + 1, using only the information in Sn, we can in fact say that for everyf (s) there exists a g(s) such that

g(s) = En [f (Sn+1) | Sn = s]

g(s) =pf (2s) + qf (0.5s)

1 + r

(181)

So, for any f (s) := VN(s), we can work our recursive algorithm backwardsto find the gn(s) := Vn(s) for all 0 ≤ n ≤ N − 1


Returning to our example of a lookback option, we see that the problemwas that Mn := max0≤i≤nSi is not Markov by itself, but the pair (Mn,Sn)is. Why?

Let’s generate the tree!

Homework Can you think of any other processes that are not Markov?That are Martingales, but not Markov?


The Interview Process

Consider the following scenario: After graduating, you go on the jobmarket, and have 4 possible job interviews with 4 different companies. Sosure of your prospects that you know that each company will make anoffer, with an identically, independently distributed probability attached tothe 4 possible salary offers -

P [Salary Offer=x1] = p1




p1 + p2 + p3 + p4 = 1

(182)


The Interview Process

How should you interview?

Specifically, when should you accept an offer and cancel theremaining interviews?

How does your strategy change if you can interview as many times asyou like, but the distribution of offers remains the same as above?


Let’s review the basic contracts we can write:

Forward Contract Initial Value is 0, because both buyer and sellermay have to pay a balance at maturity

(European) Put/Call Option Initial Value is > 0, because both onlyseller must pay balance at maturity.

(European) ”Exotic” Option Initial Value is > 0, because both onlyseller must pay balance at maturity.

During the term of the contract, can the value of the contract ever fallbelow the intrinsic value of the payoff? Symbolically, does it ever occurthat

vn(s) < g(s) (183)

where g(s) is of the form of g(S) := max S − K , 0, in the case of a Calloption, for example.


For Freedom! (we must charge extra...)

What happens if we write a contract that allows the purchaser to exercisethe contract whenever she feels it to be in her advantage? By allowing thisextra freedom, we must

Charge more than we would for a European contract that is exercisedonly at the term N

Hedge our replicating strategy X differently, to allow for thepossibility of early exercise

Extend the notion of the replicating strategy as a Martingale, whenproperly discounted


Some examples:

”American Bond:” g(s) = 1

”American Digital Option:” g(s) = 16≤s≤10 and

p =1

2= q

r =1

4

S0 = 4, u = 2, d =1

2

(184)


Using the state-variable version of our Risk-Neutral pricing algorithm, wewere able to compute the fair price of the European option as V0.

How would we compute this V0 for the corresponding Americanoption?

What would our tree look like?

What would our strategy be?

Certainly, for any strategy X we enact, we must have

Xn ≥ max K − Sn, 0 (185)

to reflect the early payoff the option holder can take.


Stopping Times

The question of when exactly an investor will choose to ”prune the tree”and take her payoff must be asked. In precise language, we define aStopping Time τ as an adapted random variable on our discreteprobability space (Ω,F ,P) with filtration FkNk=0

τ : Ω→ 0, 1, 2, ...Nω ∈ Ω | τ(ω) = k ∈ Fk ∀k = 0, 1, 2, ..,N

(186)


Stopping Times:Example

Consider the case of

p =1

2= q

r =1

4

S0 = 4, u = 2, d =1

2,N = 2

V2 := max K − S2, 0

(187)

Then for τ := min m | vm(Sm) = K − Sm, 0, we have

ω | τ(ω) = 0 = φ ∈ F0

ω | τ(ω) = 1 = TH,TT ∈ F1

ω | τ(ω) = 2 = HH,HT ∈ F2

(188)

and so τ is a stopping time. Can you come up with a random time that isnot a stopping time?


MATH 360 (Mathematics of Investment)

Documents

Transcript of MATH 360 (Mathematics of Investment)