chapter 2 ms&e 242

Investment Science

Chapter 2

Dr. James A. Tzitzouris<[email protected]>

2.1

(a) ($1)(1.033)227 = $1, 587.70

(b) ($1)(1.066)227 = $1, 999, 300.00

2.2

We are given that (1+r)n = 2, so that taking the log of both sides, we have n ln (1 + r) = ln 2 ≈ 0.69.Using the first suggested approximation, we have that nr ≈ n ln (1 + r) ≈ 0.69. Since i = 100r, wemust have that ni ≈ 69. Thus n ≈ 69/i.

If we use the more accurate approximation, we have that nr(1 − 0.5r) ≈ 0.69. Now, if r ≈ 0.08,then (1− 0.5r) ≈ 0.96 and so we must have 0.96n · (100r) = 0.96ni ≈ 69 and we have n ≈ 72/i.

2.3

Note that the rates calculated below are also commonly refered to as the “Annual PercentageRates” (APRs), for example, on your monthly credit card statement.

(a) (1 + 0.03/12)12 − 1 = 3.04%

(b) (1 + 0.18/12)12 − 1 = 19.56%

(c) (1 + 0.18/4)4 − 1 = 19.25%

1

2.4

Iteration λ f(λ) f ′(λ)0 1 1 31 2/3 1/9 7/32 13/21 377/441 47/213 78/329 ... ...

2.5

First, denote the present value of the annual payment in the nth year (for n = 0, . . . , 19) by PVn.Since the interest rate is 10%, we must have

PVn(1 + 0.10)n = $500, 000,

so thatPVn = $500, 000/(1.1)n.

Summing each yearly payment from n = 0 (since payment starts immediately) to n = 19, we arriveat the net present value of the lottery, denoted by PV and given by

PV =19∑

n=0

PVn =19∑

n=0

$500, 000/(1.1)n.

Recognizing the series on the right as a geometric series, we arrive at

PV = ($500, 000)(11)(1− (1/1.1)20) ≈ $4, 682, 460.

2.6

First we consider the six month analysis. For simplicity, assume that “Plan A” is to remain inthe first apartment and that “Plan B” is to switch to the second apartment. Under Plan A, themonthly cash flows are given as follows:

(−1000,−1000,−1000,−1000,−1000,−1000),

and the present value of these cash flows are given by

PVA = −10005∑

n=0

1(1 + 0.12/12)n

.

2

Note the fact that we divide the annual interest rate by 12 to arrive at the monthly interest rate.Then solving the geometric series on the right, we have

PVA = −$5, 853.43.

Under Plan B, the monthly cash flows are given as follows:

(−1900,−900,−900,−900,−900,−900),

and the present value of these cash flows are given by

PVB = −1000− 9005∑

n=0

1(1 + 0.12/12)n

.

Note that this implies the following relationship between PVA and PVB:

PVB = 0.9PVA − 1000 = −$6, 268.09,

so we conclude that the couple should not switch.

Now we consider the one year analysis. The cash flows for Plan A are given by:

(−1000,−1000,−1000,−1000,−1000,−1000,−1000,−1000,−1000,−1000,−1000,−1000),

and the cash flows for Plan B are given by

(−1900,−900,−900,−900,−900,−900,−900,−900,−900,−900,−900,−900),

so the only thing that changes when calculating PVA is that we sum from n = 0 to n = 11 sincethe window of time is now twelve months. Keeping this fact in mind, we arrive at

PVA = −100011∑

n=0

1/(1 + 0.12/12)n = −$11, 367.63,

and as before,PVB = 0.9PVA − 1000 = −$11, 230.87,

so we conclude that the couple should switch.

The break-even point is about eleven months, though you were not expected to determine that.

2.7

3

In Example 2.4, we determined that the biggest NPV (using a 10% discount rate) was attainedby waiting two years to cut the trees. The cash flow stream for this option was (−1, 0, 3). Weknow that waiting three years is not better from an NPV perspective, so that NPV (−1, 0, 3) ≥NPV (−1, 0, 0, x) for some cash flow x. This means

NPV (−1, 0, 3) = −1 + 3/1.12 ≥ −1 + x/1.13 = NPV (−1, 0, 0, x),

so that x ≤ $3.3.

2.8

First, we construct the cash flow stream corresponding to the change from Plan A to Plan B.In the first year we spend $30, 000 to buy the copier (Plan B) instead of spending $6, 000 tolease. Thus, in first year the incremental charge is −$24, 000. For each of the next four years,we spend only −$2, 000 instead of −$8, 000 for an incremental benefit of $6, 000. Finally, thereis a resale value of $10, 000 when we buy the copier. Thus, our cash flow stream is given by(−24000, 6000, 6000, 6000, 6000, 10000). The IRR is the rate r that satisfies the following equation:

24 = 6/(1 + r) + 6/(1 + r)2 + 6/(1 + r)3 + 6/(1 + r)4 + 10/(1 + r)5.

We can solve the above equation using Newton’s method to obtain r = 11.84%. Since this is higherthan the prevailing rate (10%), we should buy instead of lease (choose Plan B).

2.9

You have a choice: fix the roof now or wait for five years. The roof must be replaced every 20years regardless. Clearly, you will save money by waiting. The question is, how much? If you don’twait, you will pay $20, 000 every 20 years for the rest of eternity. This cash flow stream is given by(20000, 20000, 20000, ...) where the period is 20 years and the rate is 1.0520 − 1. The present valueof that stream must be equal to

20000∞∑

k=0

(1/1.0520

)k.

Instead, if we wait five years, we have a cash flow stream that looks exactly the same, but does notbegin for five years. The present value of that stream is(

20000/1.055) ∞∑

k=0

(1/1.0520

)k.

Taking the difference of the two cash flow streams, we have

20000(1− 1/1.055

) ∞∑k=0

(1/1.0520

)k = 20000(1− 1/1.055

) 11− 1/1.0520

≈ $6, 948.17.

4

2.10

(a)

Year Option 1 Option 2 Depletion Allowance Taxable Income Tax After-Tax Income1 352,000 400,000 400,000 800,000 360,000 840,0002 308,000 350,000 350,000 650,000 292,500 707,5003 220,000 250,000 250,000 250,000 112,500 387,5004 100,000 150,000 150,000 50,000 22,500 177,5005 25,000 50,000 50,000 0 0 50,000

(b) NPV and IRR using after-tax income:

NPV (0, 840000, 707500, 387500, 177500, 50000)@20% ≈ $1, 521, 000

IRR(−1000000, 840000, 707500, 387500, 177500, 50000) ≈ 52.79%

2.11

Year Project #1 Project #20 -100 -1501 30 422 30 423 30 424 30 425 30 42

IRR 15.24% 12.38%NPV@5% $29.88 $31.84

According to the IRRs, Project # 1 is better. However, the opposite is true when consideringNPVs. To explain the difference note the following. Project #2 costs $50 more than Project #1.Thus, one could borrow $150, invest $100 in Project #1, earn $29.88 plus the original $50 to have$79.88, which is better than Project #2.

2.12

We must have that

5

A1 = B1

n∑k=1

ck1,

and

A2 = B2

n∑k=1

ck2.

Note that by the IRR Theorem, there are unique positive values c1 and c2 that satisfy the twoequations above. Now observe that

A1/B1 =n∑

k=1

ck1 <

n∑k=1

ck1 = A2/B2.

Since c1 and c2 are both positive, we must have that c1 < c2. Therefore, the IRRs, denoted by r1

and r2, respectively, must satisfy

r1 = 1/c1 − 1 > 1/c2 − 1 = r2.

2.13

(a) We need to prove that there exists a value c > 0 such that Px(c) = Py(c). That is,

n∑k=0

xkck =

n∑k=0

ykck.

Rearranging terms, we have

P (c) =n∑

k=0

xkck −

n∑k=0

ykck = 0.

Since x0 < y0, we must have that P (0) = x0 − y0 < 0. Furthermore, since

n∑k=0

xk >n∑

k=0

yk,

we must have that

P (1) =n∑

k=0

(xk − yk) > 0.

6

Since the continuous function P satisfies the following: P (1) > 0 and P (0) < 0, we must have thatthere exists a c ∈ (0, 1) such that P (c) = 0. This means that r = 1/c− 1 > 0.

(b) We solve

−100 + 30n∑

k=1

ck = −150 + 42n∑

k=1

ck,

which yields c = 0.946, so that r = 5.7%.

2.14

Let the amount of the purchase be denoted by A. The ACRS method gives us the following NPV:

NPV1 = A(0.25 + 0.38/(1 + r) + 0.37/(1 + r)2

),

while the alternate ACRS method give us:

NPV2 = (A/3)(1 + 1/(1 + r) + 1/(1 + r)2

).

Taking the difference, and doing some algebra (left to the reader), we arrive at

NPV2 −NPV1 = A (1/12r + 12/100) r/(1 + r)2,

which is always positive for r > 0. Thus, the alternate ACRS method (“straight-line”) is alwaysbetter.

2.15

First, without inflation (all dollar amounts in thousands):

Year Pre-Tax Income Costs Depreciation Taxable Income Tax After-Tax Income0 0 10000 0 0 0 -100001 3300 310 2000 990 337 26532 3300 310 2000 990 337 26533 3300 310 2000 990 337 26534 3300 310 2000 990 337 26535 3300 310 2000 990 337 2653

NPV@12%=-435

Now, with 4% inflation (all dollar amounts in thousands):

7

Year Pre-Tax Income Costs Depreciation Taxable Income Tax After-Tax Income0 0 10000 0 0 0 -100001 3300 310 2000 990 337 26532 3432 322 2000 1110 377 27323 3569 335 2000 1234 420 28144 3712 349 2000 1363 464 29005 3861 363 2000 1498 509 2989

NPV@12%=89

8

Investment Science

Chapter 3


3.1

UseA =

rP

1− 1(1+r)n

with r = 7/12 = 0.58%, P = $25, 000, and n = 7× 12 = 84, to obtain A = $377.32.

3.2

Observe that since the net present value of X is P , the cash flow stream arrived at by cycling X isequivalent to one obtained by receiving payment of P every n + 1 periods (since k = 0, . . . , n). Letd = 1/(1 + r). Then

P∞ = P∞∑

k=0

(dn+1)k.

Solving explicitly for the geometric series, we have that

P∞ =P

1− dn+1.

Denoting the annual worth by A, we must have

A =rP

1− dn,

so that solving for P as a function of P∞ and substituting the result into the equation for A, wearrive at

A = r

(1− dn+1

1− dn

)P∞.

1

That is, A is directly proportional to P∞.

3.3

(a) To find the life expectancy, we multiply each age of death by its probability. Thus the lifeexpectancy is

L = 90× 0.07 + 91× 0.08 + · · ·+ 101× 0.04 = 95.13 years.

(b) To find the present value of an annuity that ends at age 95.13, we calculate the values for ages95 and 96. From the standard formula

P =A

r

(1− 1

(1 + r)n

),

with n = 5 and n = 6, we find that P95 = $39, 927 and P96 = $46, 228. Then, taking P =0.87×P95 +0.13×P96 (the average of the two values, weighted by fraction of the interval), we haveP = $40, 746.

(c) To find the expected present value fo the annuity, we calculate the probabilities qi of survivalto various ages i. For example, q90 = 1.0, q91 = q90 − 0.07 = 0.93, etc. For each year greater than90, we evaluate $10, 000× qi/1.08i−90. Hence the expected net present value is

NPV = $10, 000101∑i=91

qi

1.08i−90= $38, 387.

Note that the expected present value of the annuity is less than the present value evaluated at theexpected lifetime. This will always be the case.

3.4

First, find the monthly payment M at the APR of 8.083% using the annuity formula, as

M =0.08083

12 (1 + 0.0808312 )360

(1 + 0.0808312 )360 − 1

$203, 150 = $1, 502.41.

Next, find the intial balance for this monthly payment at the interest rate of 7.875%, given by

B =12

0.07875

(1− 1

(1 + 0.0808312 )360

)$1, 502 = $207, 209.13.

2

The total fees are the difference between the initial balance and the amount of the loan: $207, 209.13−$203, 150 = $4, 059.13.

3.5

After five years, the payment the company needs to make if excercising the call provision is

PC5 = (1 + 0.05)× Face Value = 105.

Exercising the call provision is advantageous, so

105 < P5 =100

(1 + λ)15+

10λ

(1− 1

(1 + λ)15

).

Therefore, the YTM then is lower than 9.366%.

3.6

(a) Monthly payment:

m =0.112 (1 + 0.1

12 )360

(1 + 0.112 )360 − 1

× $100, 000 = $877.57

Total interest is 360m− $100, 000 = $215, 925.20.

(b) Bi-weekly payment m/2 = $438.79. Let n be the number of periods. Then

$100, 000 = $438.79×(1 + 0.1

26 )n − 10.126 (1 + 0.1

26 )n

Hence n = 545 or 20.95 years. Total interest is 545 × $438.79 − $100, 000 = $139, 140.55. Thesavings in total interest over the monthly program is $76, 784.65 or 35.6%.

3.7

First, amortize the present value of $22,487 over four years which gives AA = $9042 per year.Second, amortize the present value of $37,582 over four years which gives AB = $9914 per year.Since AA < AB, car A should be purchased.

3.8

3

(a) A = $100, 000× 0.08(1.08)30

1.0830−1= $8.882.74

(b) P5 = $8882.74× (1.08)25−10.08(1.08)25

= $94, 821.26

(c) A′ = $94, 821.26× 0.09(1.09)25

1.0925−1= $9, 653.40

(d) 94, 821.26 = 8882.74 × (1.09)n−10.09(1.09)n yields n ≈ 38 years, which means that the total life of the

mortgage is 43 years.

3.9

Straightforward use of the formula for a bond price (assuming coupons every six months) gives 91.17.

3.10

Use the formula to obtain 6.84 years.

3.11

Using NPV = A/r, we have

D = −(1 + r)NPV

dPV

dr,

= −r(1 + r)A

(−A2

r

)=

1 + r

r.

Hence,

DM =D

1 + r=

1r.

3.12

(a)

PA = 885.84,

PB = 771.68,

PC = 657.52,

PD = 869.57

4

(b)

DA = 2.72,

DB = 2.84,

DC = 3.00,

DD = 1.00

(c) C is most sensitive to a change in yield.

(d)

VA + VB + VC + VD = NPV,

DAVA + DBVB + DCVC + DDVD = 2NPV,

where NPV is the present value of the obligation.

(e) Use bond D.

VC + VD = NPV,

DCVC + DDVD = 2NPV,

whereNPV =

2, 0001.152

= $1, 512.29.

Solving VC = $756.15 and VD = $756.15.

(f) None

3.13

dP

dλ= −

n∑k=0

e−λtktkCk = −DP

3.14

5

This follows directly from the Macaulay duration formula by setting λ = my and noting that thesecond term in the formula goes to zero as n goes to infinity.

3.15

We begin with

C =1

P (1 + (λ/m))2n(n + 1)P

m2,

and define T = n/m to arrive at

C =1

P (1 + (λ/m))2T (T + (1/m)),

so that as m →∞, we find that C → T 2.

3.16(a) We have that

P (λ) =∑

t

ctdt(λ)− dt̄(λ),

P ′′(λ) =∑

t

ctt(t + 1)dt(λ)d2(λ)− t̄(t̄ + 1)dt̄(λ)d2(λ).

Therefore,

P ′′(0)(1 + r)2 =∑

t

ctt2dt +

∑t

cttdt − t̄2dt̄ − t̄dt̄,

=∑

t

ctt2dt − t̄2dt̄ (by the second condition).

Since ∑t

ctdt − dt̄ =∑

t

βctdt − βdt̄ = 0 (for all β),

and ∑t

tctdt − t̄dt̄ =∑

t

(αt)ctdt − (αt̄)dt̄ = 0 (for all α),

6

it follows that

P ′′(0)(1 + r)2 =∑

t

ct(t2 + αt + β)dt − (t̄2 + αt̄ + β)dt̄.

(b) Let α = −2t̄ and β = t̄2 + 1. Then t2 + αt + β = (t − t̄)2 + 1, which has a minimum at t̄ andhas a value of 1 there.

P ′′(0)(1 + r)2 =∑

t

ct(t2 + αt + β)dt − (t̄2 + αt̄ + β)dt̄,

≥∑

t

ctdt − dt̄ = 0.

Therefore, P ′′(0) ≥ 0.

7

Investment Science

Chapter 4

Solutions to Suggested Problems


4.1

(One forward rate)

f1,2 =(1 + s2)2

(1 + s1)− 1 =

1.0692

1.063− 1 = 7.5%

4.2

(Spot Update)Use

f1,k =[(1 + sk)k

1 + s1

]1/(k−1)

− 1

.Hence, for example,

f1,k =[(1.061)6

1.05

]1/5

− 1 = 6.32%

.All values are

f1,2 f1,3 f1,4 f1,5 f1,6

5.60 5.90 6.07 6.25 6.32

1

4.3

(Construction of a zero)Use a combination of the two bonds: let x be the number of 9% bonds, and y teh number of 7%bonds. Select x and y to satisfy

9x + 7y = 0,

x + y = 1.

The first equation makes the net coupon zero. The second makes the face value equal to 100. Theseequations give x = −3.5, and y = 4.5, respectively. The price is P = −3.5× 101.00 + 4.5× 93.20 =65.90.

4.5

(Instantaneous rates)(a) es(t2)t2 = es(t1)t1eft1,t2 (t2−t1) =⇒ ft1,t2 = s(t2)t2−s(t1)t1

t2−t1

(b) r(t) = limt→t1s(t)t−s(t1)t1

t−t1= d[s(t)t]

dt = s(t) + s′(t)

(c) We have

d(lnx(t)) = r(t)dt,

= s(t)dt + s′(t)dt,

= d[s(t)t].

Hence,

lnx(t) = lnx(0) + s(t)t,

and finally that

x(t) = x(0)es(t)t.

This is in agreement with the invariance property of expectation dynamics. Investing continuouslygive the same result as investing in a bond that matures at time t.

4.6

(Discount conversion)

2

The discount factors are found by successive multiplication. For example,

d0,2 = d0,1d1,2 = 0.950× 0.940 = 0.893.

The complete set is 0.950, 0.893, 0.770, 0.707, 0.646.

4.7

(Bond taxes)Let t be the tax rate, xi be the number of bond i purchased, ci be the coupon of bond i, pi be theprice of bond i. To create a zero coupon bond, we require, first, that the after tax coupons match.Hence

x1(1− t)c1 + x2(1− t)c2 = 0,

which reduces to

x1c1 + x2c2 = 0.

Next, we require that the after tax final cash flows match. Hence

p0 = x1p1 + x2p2.

Using this last relation in the equationfor final cash flow, we find

x1 + x2 = 1.

Combining these equations, we find that

p0 =c2p1− c1p2

c2 − c1.

After plugging in the given values, we find that

p0 = 37.64.

4.8

(Real zeros)We assume that with coupon bonds there is a capital gains tax at maturity. We replicate the zero-coupon bond’s after-tax cash flows using bonds 1 and 2. Let xi be the amount of bond i required

3

(for i = 1, 2). We require:

100c1(1− t)x1 + 100c2(1− t)x2 = −(

100− p0

n

)t,

(100− (100− p1)t + 100c1(1− t))x1 + (100− (100− p2)t + 100c2(1− t))x2 = 100−(

100− p0

n

)t,

p1x1 + p2x2 = p0.

Setting

p1 = 92.21,

p2 = 75.84,

t = 30%,

c1 = 0.10,

c2 = 0.07,

n = 10,

we find that x2 = 5.25428 and p0 = 32.767.

4.9

(Flat forwards)For i < j,

(1 + r)i(1 + fi,j)j−i = (1 + r)j .

Hence,

(1 + fi,j)j−i = (1 + r)j−i,

which implies that fi,j = r.

4.10

(Orange County blues)We make the following three assumptions. First, the portfolio is restructured annually to maintaina duration of 10 years. Second, the value of money borrowed is maintained at $12.5 billion everyyear. Finally, Orange County makes interst on deposit at the rate which prevailed at the beginningof the given year.

4

Year 1 P = 20(1.085) +−10(20)(−0.005)

1.085− 12.5(0.07) = 21.75,

r =21.75− 2020− 12.5

= 23.33%,

Year 2 P = (21.75)(1.08) +−10(21.75)(−0.005)

1.08− 12.5(0.065) = 23.68,

r =23.68− 21.7521.75− 12.5

= 20.86%,

Year 3 P = (23.68)(1.075) +−10(23.68)(−0.005)

1.075− 12.5(0.06) = 25.81,

r =25.81− 23.6823.68− 12.5

= 19.02%,

Year 4 P = (25.81)(1.07) +−10(25.81)(−0.005)

1.07− 12.5(0.055) = 28.14,

r =28.14− 25.8125.81− 12.5

= 17.51%,

Year 5 P = (28.14)(1.065) +−10(28.14)(−0.02)

1.065− 12.5(0.05) = 24.06,

r =24.06− 28.1428.14− 12.5

= −26.09%,

Year 6 P = (24.06)(1.06) +−10(24.06)(−0.02)

1.06− 12.5(0.045) = 20.80,

r =20.80− 24.0624.06− 12.5

= −28.20%,

Money left after 6 years = 20.79−12.5−7.5 = 0.79. If invested in a bank = 7.5(1.06)(1.055)(1.045)(1.4)(1.06)−7.5 = 2.64.

4.11

(Running PV examples)

5

(a)d0,1 d0,2 d0,3 d0,4 d0,5 d0,6

0.9524 0.9018 0.8492 0.7981 0.7472 0.7010=⇒ NPV = 9.497

(b)

Year 0 1 2 34 5 6Cash Flow −40 10 10 10 10 10 10Discount 0.9524 0.9469 0.9416 0.9399 0.9362 9381PV(n) 9.497 51.970 44.324 36.453 28.144 19.381 10.000

4.12

(Pure duration)

P (λ) =n∑

k=0

xk(1 + sk/m)−k =n∑

k=0

xk

((1 + s0

k/m)eλ/m)−k

,

dP (λ)dλ

=n∑

k=0

xk

(−k

m

)((1 + s0

k/m)eλ/m)−k−1

(1 + s0k/m)eλ/m,

=n∑

k=0

xk

(−k

m

)(1 + sk/m)−k,

− 1P (λ)

dP (λ)dλ

=∑n

k=0 xk

(km

)(1 + sk/m)−k∑n

k=0 xk(1 + sk/m)−k≡ D.

This D exactly corresponds to the original definition of duration as a cash flow weighted averageof the times of cash payments. No modification factor is needed even though we are working indiscrete time.

4.14

(Mortgage division)(a)

P (k) = B − rM(k − 1) = B − r

((1 + r)k−1M(0)−

((1 + r)k−1 − 1

r

)B

),

= (1 + r)k−1(B − rM),

6

We are looking for

V =n∑

k=1

P (k)(1 + r)k

,

=n∑

k=1

(1 + r)k−1(B − rM)(1 + r)k

,

=n(B − rM)

1 + r.

(b) Plug the expression for B into the result found in part (a) to obtain

V =n

1 + r

[r(1 + r)nM

(1 + r)n − 1− rM

],

=n

1 + r

[rM

(1 + r)n − 1

].

(c)

W =n∑

k=1

I(k)(1 + r)k

,

=n∑

k=1

B − P (k)(1 + r)k

,

= B

(n∑

k=1

1(1 + r)k

)− V,

=(

r(1 + r)nM

(1 + r)n − 1

)((1 + r)n − 1r(1 + r)n

)− V,

= M − V.

(d) It should be clear that V → 0. (Use L’Hopital’s rule if it is not obvious.)

(e) We know P (k) = (1 + r)k−1(B − rM). Clearly B − rM > 0. (This follows from part (a).) SoP (k) is increasing in k and I(k) = B−P (k) must be decreasing in k. Remember that duration is a

7

weighted sum of the times k, with the weights being proportional to the cash flows at those times.Hence the duration of the stream determined by P (k), which increases in k, should be the larger,because more relative weight is given to higher values of k.

4.15

(Short-rate sensitivity)In general,

Pk−1(λ) = ck−1 +Pk(λ)

1 + rk−1 + λ.

Differentiation at λ = 0 leads to

Sk−1 = − Pk

(1 + rk−1)2+

Sk

1 + rk−1.

Hence,

ak =1

(1 + rk−1)2,

bk =1

1 + rk−1,

and this process, together with

Pk−1 = ck−1 +Pk

1 + rk−1,

is initiated with Pn = cn and Sn = 0; and the two processes are worked backward to k = 0. S0 isthe final result.

8

Investment Science

Chapter 6



6.1

The money invested is X0. The money received at the end of a year is X0 −X1 + X0. Hence,

R =2X0 −X1

X0.

6.3

For solution method, see solution to Problem 6.4 in this solution set.

(a) α = 19/23

(b) The minimum standard deviation is approximately 13.7%.

(c) The expected return of this portfolio is approximately 11.4%.

6.4

Let α and β equal the percent of investment in stock 1 and stock 2, respectively. The problem is:

minα,β α2σ21 + 2αβσ12 + β2σ2

2

s.t. α + β = 1.

1

Setting up the Lagrangian, L, we have:

L = α2σ21 + 2αβσ12 + β2σ2

2 − λ(α + β − 1).

The first order necessary conditions are:

0 =∂L

∂α= 2ασ2

1 + 2βσ12 − λ,

0 =∂L

∂β= 2βσ2

2 + 2ασ12 − λ,

1 = α + β,

which imply

α =σ2

2 − σ12

σ21 + σ2

2 − 2σ12.

The mean rate of return is simply αm1 + βm2.

6.5

(a) The expected rate of return equals

(0.5)(3)(106) + (0.5)u106 + 0.5u

.

(b) By inspection, it can be seen that buying 3 million units of insurance eliminates all uncertaintyregarding the return. So, 3 million units of insurance results in a variance of 0 and a correspondingexpected rate of return equal to

32.5

− 1 = 20%.

6.6

(a) The three assets are on a single horizontal line. The efficient set is a single point on the sameline, but to the left of the left-most of the three original points.

2

(b) Let wi be the percentage of the total investment invested in asset i. The since the assets areuncorrelated, we have

var(total investment) =n∑

i=1

w2i σ

2i ,

where the weights sum to 1. Setting up the Lagrangian, we have

L =n∑

i=1

w2i σ

2i − λ

(n∑

i=1

wi − 1

),

so that the first-order necessary conditions imply

wiσ2i =

λ

2,∀i = 1, . . . , n,

or wj = λ2σ2

j. Since the weights sum to 1, we have

λ =2∑n

i=11σ2

i

,

which implies

wj =σ̄2

σ2j

,∀j = 1, . . . , n,

where

σ̄2 =1∑n

i=11σ2

i

.

The minimum variance is

σ2min =

n∑i=1

w2i σ

2i = σ̄2.

6.7

(a) First solve for the vi’s from

2v1 + v2 = 1,v1 + 2v2 + v3 = 1,

+ v2 + v3 = 1.

3

This yields v1 = 0.5, v2 = 0, and v3 = 0.5. This solution happens to be normalized, so alsow1 = 0.5, w2 = 0, and w3 = 0.5.

(b) In this case, we solve2v1 + v2 = 0.4,v1 + 2v2 + v3 = 0.8,

+ v2 + v3 = 0.8.

This yields v1 = 0.1, v2 = 0.2, and v3 = 0.3. This solution must be normalized, to arrive atw1 = 1/3, w2 = 1/6, and w3 = 1/2.

(c) We find the vi’s by the formula vi = vbi −rfva

i where the vbi ’s and the va

i ’s are the solutions fromparts (b) and (a) above, respectively. Thus, v1 = 0, v2 = 0.2, and v3 = 0.2. This solution must benormalized, to arrive at w1 = 0, w2 = 1/2, and w3 = 1/2.

6.8

(a)

var(r − rM ) = var(r)− 2cov(r, rM ) + var(rM ),

=n∑

i,j=1

αiαjσij − 2n∑

i=1

αiσiM + σ2M .

So, to minimize var(r − rM ) subject ton∑

i=1

αi = 1,

set up the Lagrangian

L =n∑

i,j=1

αiαjσij − 2n∑

i=1

αiσiM + σ2M + λ

(n∑

i=1

αi − 1

).

The first order necessary conditions implyn∑

j=1

αjσij − 2σiM + λ = 0,∀i = 1, . . . , n,

n∑i=1

αi = 1.

4

(b) Similar to (a) with the added constraint that

n∑i=1

αiri = m.

So, the first order necessary conditions imply

n∑j=1

αjσij − 2σiM + λ + µri = 0,∀i = 1, . . . , n,

n∑i=1

αiri = m,

n∑i=1

αi = 1.

5

Investment Science

Chapter 7



7.1

(a) r̄ = 0.07 + 0.23−0.070.32 σ = 0.07 + σ

2

(b) i. σ = 0.64, ii. Solve 0.07w + 0.23(1−w) = 0.39, yielding w = −1. Hence, borrow $1000 at therisk-free rate; invest $2000 in the market.

(c) $1182

7.2

(a)

σ2M =

14(σ2

A + 2σAB + σ2B),

σ2AM =

12(σ2

A + σAB),

σ2AB =

12(σAB + σ2

B),

βA =σ2

A + σAB

σ2M

,

βB =σ2

B + σAB

σ2M

.

1

(b)

r̄A = 0.1 +54(0.18− 0.1) = 20%,

r̄B = 0.1 +34(0.18− 0.1) = 16%.

7.3

(a) Using the two-fund theorem and noting that the market portfolio cannot contain assets innegative amounts, we have

12w +

12v =

(0.7 0 0.3

)T with a return of 0.1,

2w − v =(

0.4 0.6 0)T with a return of 0.16,

so the expected rate of return of the market portfolio r̄M is bounded as follows: 0.1 ≤ r̄M ≤ 0.16.

(b) Since r̄M ≥ r̄min var portfolio, we have 0.12 < r̄M ≤ 0.16.

7.4

From (6.9) we have

n∑i=1

σMiλwi = λσ2M = r̄M − rf .

Hence,

λ =r̄M − rf

σ2M

.

Furthermore,

n∑i=1

σkiλwi = λσkM = r̄k − rf .

Substituting the expression for λ into the above equation, we arrive at our objective.

2

7.5

We have

βi =σiM

σ2M

=xiσ

2i∑n

j=1 xjσ2j

for all i = 1, . . . , n, since the assets are uncorrelated.

7.6

The market consists of $150 in shares of A and $300 in shares of B. Hence, the market return is

rM = (150/450)rA + (300/450)rB =13rA +

23rB.

(a) 0.13

(b) 0.09

(c) σAM = 13σ2

A + 23ρABσAσB = 0.0105, βA = 1.2963

(d) Since Simpleland satisfies the CAPM exactly, stocks A and B plot on the security market line.Specifically, r̄A − rf = βA(r̄M − rf ). Hence, rf = 0.0625.

7.7

(a) Let p be a portfolio such that p = (1−α)w0+αw1. Then, σ2p = (1−α)2σ2

0 +2(1−α)ασ01+α2σ21.

So, since 0 = dσ2p

dα

∣∣∣α=0

, we have 0 = −2σ20 + 2σ01 which implies that A = 1.

(b) 0 = σ1z = (1− α)σ01 + ασ21 implies (using (a)), α =

(σ20

σ20−σ2

1

)< 0.

(c) The zero-beta portfolio is on the minimum variance set but below the minimum variance point.

(d) ρ = σiMσiσM

=⇒ r̄i = r̄z + ρσi

σM(r̄M − r̄z) = 0.09 + (0.5)(0.5/0.15)(0.15− 0.09) That is, r̄i = 10%.

3

Investment ScienceChapter 10



10.1

We are given that S = $412, M = 3 (quarters), and r = 9% (compounded quarterly). Thestorage cost is $2 per ounce per year, so ck = $0.50 for each quarter k = 0, 1, and 2 (at thebeginning of the quarter). We use the formula:

along with the following intermediate results:

to arrive at

10.2

There is a typo in the text. The exponent “M” should be “-M” in the formula.

Suppose at time zero you:

1. borrow S(0)2. buy 1 unit of the asset for S(0)3. take a short position of (1-q)M units at a forward price of F per unit4. at the beginning of each period, sell q units of the asset to pay the cost of carry

Note that the total cash outlay for these actions is zero.

F= Sd 0,3

∑k=0

M−1 ck

d k , M

d 0,3 = 1

10.094

3 = 0.9354

d 1,3 = 1

10.094

2 = 0.9565

d 2,3 = 1

10.094

1 = 0.9780

F = $ 440.45$ 0.5345$ 0.5227$ 0.5112 = $ 442.02

10.2 (continued)

At the time of delivery, you have (1-q)M units of the asset remaining. Then you do thefollowing:

1. make delivery and receive F(1-q)M as payment2. repay your loan with S(0)/d(0,M)

Your total profit is F(1-q)M – S(0)/d(0,M). If this amount is anything but zero (negativeor positive), then you have made a profit (or received a loss) without making anyinvestment (i.e., without taking any risk). Thus, consistent with our assumption thatarbitrage opportunities do not exist, the profit must be zero. Here, we have glossed overthe case where the profit is negative. Technically speaking, a negative profit is possiblein this example, however if you were to reverse each position in above analysis, youwould have a new investment strategy that has a positive arbitrage return. Thus, theprofit must be exactly zero and we arrive at the answer.

10.3

In general, the spot and forward prices are related by:

If the yield curve is flat, and with forward contracts being settled at the end of the month,this formula can be written for any maturity date M as:

So, taking any two maturities, we can solve for the spot price and the interest rate. Inparticular, by taking the forward prices for April and July we have the equations:

F=S 0

d 0, M ⋅1−qM

S=Fd 0, M −∑k=0

M −1

ck d 0, k

S= F

1 r12

M −∑k=0

M −1 ck

1 r12

k

S= 406.50

1 r12

− 2012

S= 409.3

1 r12

4−2012 ∑k=0

M −1 1

1 r12

k

Solving the equations simultaneously we obtain:

Other contract dates give the same results.

10.4

There is a typo in the text. The exponent “r-q” should be “r+q” in the formula.

In order to solve this problem, we assume that the interval of time from 0 to T is dividedinto n discreet time intervals of length T/n. Next we proceed as in Exercise 2.

Suppose at time zero you:

1. borrow S(0)2. buy 1 unit of the asset for S(0)3. take a short position of e-qT units at a forward price of F per unit4. at the beginning of each period, sell qT/n units of the asset to pay the cost of carry

Notice that the cost of carry at the beginning of the kth period is qT/n, not q. This isbecause here (unlike in Exercise 2), q is defined as the cost of carry per unit time. This isa subtle but significant difference. If xk denotes the number of units of the asset held attime tk=Tk/n, then taking limits as n becomes increasingly large, we must have

Note that the total cash outlay for these actions is zero.

At the time of delivery, you have e-qT units of the asset remaining. Then you do thefollowing:

1. make delivery and receive Fe-qT as payment2. repay your loan with S(0)erT

Your total profit is Fe-qT – S(0)erT. If this amount is anything but zero (negative orpositive), then you have made a profit (or received a loss) without making any investment(i.e., without taking any risk). Thus, consistent with our assumption that arbitrageopportunities do not exist, the profit must be zero. Here, we have glossed over the casewhere the profit is negative. Technically speaking, a negative profit is possible in thisexample, however if you were to reverse each position in above analysis, you would havea new investment strategy that has a positive arbitrage return. Thus, the profit must beexactly zero and we arrive at the answer.

S=403.15, r=0.05

F=S 0erqT

x=xk−xk−1 , t=Tn

,

xk=xk−1−qx k−1Tn

⇒ x t

=−qx ⇒ dxdt

=−qx ⇒ x T =e−qT

10.7

We solve first for Ft, the current forward price of the bond:

Now we solve for the value of the forward contract:

10.12

We obtain the mean-variance hedge formula by maximizing:

which is the same as maximizing

Taking the derivative with respect to h we obtain the first order condition that:

Solving for h we find the mean-variance hedge formula:

10.13

The minimum variance hedge is

Hence the farmer should short 131,250 pounds of orange juice. To check how effectivethis hedge is, we note that

F t =S

d 0,2∑

k=1

2 d 0, k ck

d 0,2= $ 831.47

f t=F t−F 0d 0,2=−$ 100.34

E [ x ]hF T−F 0−r var [ x ]2 hcov [ x , F T ]h2 var [F T ]

E [ xh F T−F 0]−r var [ xhF T ]

F T−F 0−2 r cov [ x , F T ]−2 hr var [F T ]=0

h=F T−F 0

2 r var [F T ]−

cov [ x , F T ]var [F T ]

h=−w=−G

O

SG

SO×150,000=−131,250

new=1−2 old = 0.714 old

10.14

We have the future cashflow y = ST W + (FT – F0) h.

(a) The equal and opposite hedge h is given by an opposite equivalent dollar value of thehedging instrument. Therefore h = -kW where k is the price ratio between the asset andthe hedging instrument. By taking this hedge position, the cashflow received at the futuredate can be written as

where its variance is given by

Hence

(b) In example 10.12, we have k = K/M = 0.262, thus the equal and opposite hedge forthe receivable W = 1,000,000 Danish Krone is h = -262,000 Deutsche Marks and thestandard deviation of the future cashflow y is given by

so that

As expected, the standard deviation with the equal and opposite hedge is greater than withthe minimum variance hedge for which we determined

y=W STk F 0−F T

y2 = W 2 var [ST−kF TkF 0 ] = W 2S

2−2 k STk 2F2

y = W S×1−2 kST

S2 k 2 F

2

S2

y = W K×1−2 kKM

K2 k 2 M

2

K2 = x

2×1−2 KM

M

K K

M 2 M

2

K2

y = 0.7211 x

y = 0.6 x




11.1

We are given that

so that

and the binomial lattice is given by

11.2

Each movement in k corresponds to a month, and each movement in K corresponds to ayear. Let kK denote the first month of year K. Then

So

S 0=$ 100,=0.12,=0.20, t=0.25,

u=e t=1.105,

d=1u=0.905,

p=121

2 t=0.65,

100110.50

122.10134.90

149.10

90.50100

110.50100

122.10

81.90

67.10

90.50

74.1081.90

Probability = 17.9%

Probability = 38.4%

Probability = 31.1%

Probability = 11.1%

Probability = 1.5%

W K =∑i=0

11

w k K−1i

E [W K ]=E[∑i=0

11

w k K−1i ]=12 , Var[∑i=0

11

w k K−1i 2]=12

11.3

(a) Proof: For n = 2,

which implies

That is,

so

(b) Since r1=50% and r2=-20%, the arithmetic mean is

and the geometric mean is

(c) The arithmetic mean rate of return essentially assigns a return based on simpleinterest, while the geometric mean rate of return is a measure of compound interest.Usually, the geometric mean rate of return is the most appropriate for measurement ofinvestment performance.

11.4

So

1−22≥0,

122122

2≥412.

14 12

2≥12,

A≥G.

12 r1r2=0.15

[1r11r2]12−1=0.0954

w−w−w 2

22 = − 122 [−22 ww2−2 w ww222 w−22 w4−4]

= − 122 [w− w2]2w

2

2

u = − 1

22 ∫−∞

∞

ew e−w−w 2/22

dw

= ew

2

2 1

22 ∫−∞

∞

e−[w− w2]2/22

dw

= ew

2

2

11.5

Suppose that u = ew, where w is distributed as

then

and following the method in exercise 11.4, we have

so that

and

11.6

We have

so

11.7

We have G(t) = F(s, t) = S1/2(t), then

and therefore according to Ito's Lemma, we must have that

N w , σ 2

E [u2 ] = E [e2 w ]

= 1

22 ∫−∞

∞

e2 w e−w−w 2 /22

dw

2 w−w−w 2

22 = − 122 [w− w22]22 w22

E [u2 ] = e222 w

var [u ] = E [u2 ]−u2 = e22 w e2

−1

= −122 = 0.2−1

2×0.16 = 0.12

E [ ln S 1] = 0.12, Stdev [ ln S 1] = 0.40E [S 1] = 1.22, Stdev [S 1] = 0.51

∂ F∂ S

= 12S

, ∂2 F∂ S 2 = − 1

4S 3

dG t = 12S

a S− 18S 3

b2 S 2dt 12S

b S dz

= 12

a−18

b2G dt12

bG dz




12.1

The initial cost of the spread is nonnegative since C(K1) ≥ C(K2) for K1 < K2 (see 12.4).

12.2

Use the same portfolio as in the text:

1. buy one call2. sell one put3. lend an amount d K

This will reproduce the payment of the stock, except that it will be short by an amountwith present value D. Hence: C – P + dK = S – D.

12.3

Q = max {0, S−K }−max {0, K−S }K ,

= {S−K −0K , if S≥K0−K−S K , if S≤K} = S.

12.4

1. Assume K2 > K1, and suppose to the contrary that C(K2) > C(K1). Buy option 1 andsell option 2. Use option 1 to cover the obligations of option 2, since we have thatmax{0, S- K1} ≥ max{0, S- K2} for all S. Keep profit of C(K2) – C(K1).

2. Assume K2 > K1, and suppose to the contrary that K2 – K1 < C(K1) – C(K2). Buyoption 2 and short option 1 to obtain K2 – K1 + ε profit (where ε>0). Use option 2 andprofits K2 – K1 to cover option 1 since we may conclude that

3. Assume K3 > K2 > K1 and suppose to the contrary that

Buy (K3 – K2)/(K3 – K1) of option 1 and (K2 – K1)/(K3 – K1) of option 3 and short oneunit of option 2. The profit is some ε>0. Notice that

max {0, S−K 2}K 2 – K 1 = max {K 2 – K 1, S−K 1} ≥ max {0, S−K 1}.

C K 2 K 3−K 2

K 3−K 1C K 1 K 2−K 1

K 3−K 1C K 3 .

and that

Therefore it is possible to cover option 2 and make a profit of ε>0. This is anarbitrage opportunity, so the original inequality cannot hold.

12.6

The payoff of (A) purchase one call is always greater than or equal to the payoff of (B)purchase one share of stock and sell K bonds. Hence, the price of (A) must be greaterthan or equal to the price of (B). Thus, C ≥ S – KB(T). However, the value of a call isalways greater than or equal to zero. Thus, C ≥ max{0, S – KB(T)}.

12.7

From Exercise 6, C ≥ max{0, S – KB(T)}. Since the limit of B(T) is zero as T approachesinfinity, the limit of C(S, T) is the greater than or equal to that of max{0, S – KB(T)} as Tapproaches infinity, which is S. Clearly, the price of a call option must be less than orequal to that of the underlying. Thus, in the limit, C = S.

12.9

This is like the example in the text. Draw a lattice with three stages. The payoffs of thefour final nodes are 27, 27, 0, and 0. Roll back one stage. The three nodes there haveimplied values 27, 9, and 0. Roll back one more stage. The implied values there are 15and 3, which can be written as 12+3 and 0 + 3. Hence, the implied value of the initialnode is 4 + 3 = 7. The result can also be found by direct risk-neutral valuation (withoutrolling back the lattice) using q = 1/3 and R = 1. The risk-neutral probability of the twonodes with 27 are 1/27 and 6/27. Hence, the total value is 7.

K 3−K 2

K 3−K 1C K 1 K 2−K 1

K 3−K 1C K 3 ≥ K 3−K 2

K 3−K 1S−K 1 K 2−K 1

K 3−K 1S−K 3

= S−K 2

K 3−K 2

K 3−K 1C K 1 K 2−K 1

K 3−K 1C K 3 0




13.1

The equation is best implemented on a computer. The answer for the call stated in theexercise is C = $2.57.

13.2

(a) For the expression P(S) = a1S + a2S-γ we have

Substituting in the Black-Scholes equation and canceling terms we find that

Hence, γ = 2r/σ^2 satisfies the equation. Since a1 and a2 are arbitrary, this represents twoindependent solutions to the second-order differential equation; and hence is the generalsolution.

(b) P(∞) = 0 implies a1 = 0. P(G) = K – G implies a2G-γ = K – G leading to a2 = (K – G)/G-γ. Hence, P(S) = (K – G)(S/G)-γ

(c) It makes sense to maximize P(S) since the maximization is independent of S. Wemaximize (K-G)G-γ. Differentiation with respect to G gives the condition

Thus, G = γK/(γ+1) and

13.3

Using the spreadsheet implementation of the previous exercise, we adjust σ by trial anderror to obtain the given call premium. The result is σ = 0.251.

P ' S = a1−a2 S−−1

P ' ' S = 1a2 S−−2

122 1−r −r=0

−G− K−G G−1 = 0.

P S =K 1−

11 SK

−

.

13.4

For

we find

Hence

For delta, we have

For the call of the example, we have

The value at that value of S is (by the above approximation) C = 2.972. The value of Δ atthe base point is 0.5258. The difference in S is 62 – 57.55 = 4.44. Hence, the final valueis 2.972 + 4.44 * 0.5258 = 5.30.

13.5

We calculate the price of the call at $63 to be $6.557; hence, Δ ≈ 6.557 – 5.798 = 0.759. Setting the expiration time T as follows, T = 5/12 + 0.1, we obtain a call price of $6.490.Hence, θ ≈ (6.490 – 5.798)/0.1 = 6.02.

13.6

This is just the Black-Scholes equation.

S=K e−rT

d 1 =ln e−rT r2/2T

T

= T

2

d 2 = − T

2

C ≈ S 12

T 22−S erT e−rT1

2−

T 22

=S T

2≈ 0.4 S t

=N d 1 ≈12

t 22

≈ 120.2 t

K e−rT = 60 e−0.5/12 = 57.55

13.7

For theta, use Exercise 6 to write

= ∂2 C∂ S 2 =

∂ ∂ S

=∂ N d 1

∂ S= N ' d 1

∂ d 1

∂ S

=N ' d 1

S T

= rC−rS −122 S 2

= rSN d 1−rK e−rT N d 2−rSN d 1−12 S N ' d 1/T

= −SN ' d 12T

−rK e−rT N d 2

chapter 2 ms&e 242

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