Investment Science AMA532, S2, 2012/13, Solution Outline to Assignment One

1. Let pj , j = 1, · · · , k be the probability distribution. Then by Cauchy-Scharz inequality,we have

|σ12| = |E((x1 − x̄1)(x2 − x̄2))| ≤k∑

j=1

|xj1 − x̄1|

√pj |xj

2 − x̄2|√

pj

≤

√

√

√

√

√

k∑

j=1

(xj1 − x̄1)2pj

√

√

√

√

√

k∑

j=1

(xj2 − x̄2)2pj = σ1σ2.

2. a. The arithmetic mean of stock T is (0.29 + 0.18 − 0.12 − 0.15 + 0.21)/5 = 0.082.

The arithmetic mean of stock B is (0.18 − 0.03 − 0.15 + 0.12 + 0.09)/5 = 0.042.

By the measure of the mean, stock T is preferred.

b. The standard deviation of stock T is

σ =(

[(0.29 − 0.082)2 + (0.18 − 0.082)2 + (−0.12 − 0.082)2 + (−0.15 − 0.082)2+

(0.21 − 0.082)2]/5).5

= 0.1810

The standard deviation of stock B is

σ =(

[(0.18 − 0.042)2 + (−0.03 − 0.042)2 + (−0.15 − 0.042)2 + (0.12 − 0.042)2+

(0.09 − 0.042)2]/5).5

= 0.1179

By the measure of standard deviation, stock B is preferred.

The coefficients of variation (or shape ratios) are 0.0820−0.020.1810

= 0.3425 and 0.042−0.020.1179

=0.1866. By this measure, stock B is more preferable.

[Note: the solutions with sample mean, that is, when the denominator n = 5 above ischanged to n − 1 = 4, are also acceptable.]

3. For 6 months rent: apt A has cash flow (6000, 6000, 6000, 6000, 6000, 6000) and apt Bhas cash flow (11000, 5000, 5000, 5000, 5000, 5000, 5000).

PV of apt A =5∑

k=0

6000/(1 + 0.0025)k = 35776.31

PV of apt B =

6000 +5∑

k=0

5000/(1 + 0.0025)k = 35813.59

6

PV of apt A < PV of apt B, they should not switch.

For 12 months rent: apt A has cash flow (6000, 6000, 6000, 6000, 6000, 6000, 6000, 6000,6000, 6000, 6000, 6000) and apt B has

CFS(11000, 5000, 5000, 5000, 5000, 5000, 5000, 5000, 5000, 5000, 5000, 5000, 5000, 5000).

PV of apt A =11∑

k=0

6000/(1 + 0.0025)k = 71020.63

PV of apt B =

11000 +11∑

k=1

5000/(1 + 0.0025)k = 65183.86

PV of apt A > PV of apt B, they should switch.

4. For Project 1, the IRR equation is

0 = −100 +30

(1 + r)+

30

(1 + r)2+

30

(1 + r)3+

30

(1 + r)4+

30

(1 + r)5

By trial and error method, r ≈ 0.15.

[or by Newton method.]

For Project 2, the IRR equation is

0 = −150 +42

(1 + r)+

42

(1 + r)2+

42

(1 + r)3+

42

(1 + r)4+

42

(1 + r)5

By trial and error method, r ≈ 0.12.

[or by Newton method.]

By IRR evaluation, the recomendation is project 1.

For Project 1,

PV = −100+30

(1 + 0.05)+

30

(1 + 0.05)2+

30

(1 + 0.05)3+

30

(1 + 0.05)4+

30

(1 + 0.05)5= 29.88.

For Project 2,

PV = −150+42

(1 + 0.05)+

42

(1 + 0.05)2+

42

(1 + 0.05)3+

42

(1 + 0.05)4+

42

(1 + 0.05)5= 31.84.

By PV, the recommedation is project 2.

7

5. It is known that there is no advantage in buying either the stock or the forward contractif we can borrow to buy a stock today (so both strategies do not require any initialcash) and if the profit from this strategy is the same as the profit of a long forwardcontract. The profit of a long forward contract with a price for delivery of $53 is equalto: $ ST − $53, where ST is the (unknown) value of one share of CLP at expirationof the forward contract in one year. If we borrow $50 today to buy one share of CLPstock (that costs $50), we have to repay in one year: $50(1 + r). Our total profit inone year from borrowing to buy one share of CLP is therefore: $ST − $50(1 + r). Nowwe can equate the two profit equations and solve for the interest rate r:

$ST − $53 = $ST − $50(1 + r)

$53 = $50(1 + r)

$53

$50− 1 = r

r = 0.06

Therefore, the 1-year effective interest rate that is consistent with no advantage toeither buying the stock or forward contract is 6 percent.

6. Please note that we have given the continuously compounded rate of interest as 6%.Therefore, the effective annual interest rate is exp(0.06) − 1 = 0.062. In this exercise,we need to find the future value of the put premium. For the $0.95-strike put, it is:$0.0178 × 1.062 = $0.02.

(a) The diagram for the profit of the company is as follows.

Company’s profit = copper price− cost

= copper price − 0.90

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................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

......................

unhedged profit ($)

00.9 copper price

.........................................................................................................................................................................................................................................................................................................................................................................................................................

8

(b)

Purchased put profit = max{0, strike price − copper price}= max{0, 0.95 − copper price}

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................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

......................

Profit ($)

0

0.95Strike price

copper price

Purchased put position

..

...............................

...............................................

...............................

................................

...............................

...............................

................................

...............................

...............

(c) Net income

= Company’s profit + Purchased put profit − FV(Put premium)

= copper price − 0.90 + max{0, 0.95 − copper price} − 0.0178 × e0.06

= copper price + max{0, 0.95 − copper price} − 0.9189

= +max{copper price − 0.95, 0} + 0.95 − 0.9189

= +max{copper price − 0.95, 0} + 0.0311

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................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

......................

Net income ($)

00.0311

0.95Strike price

copper price

...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

9

7. Our initial cash required to put on the collar, i.e. the net option premium, is as follows:−$51.873 + $51.777 = −$0.096. Therefore, we receive only 10 cents if we enter intothis collar. The position is very close to a zero-cost collar.

The profit diagram of holding a long index, long 950 put, and short 1107 call looks asfollows:

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................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

......................

Profit

0 950 1107 Index price

−69.9021

87.0979

.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

8. The profit is equal to

2max{S − 950, 0} − 3max{S − 1050, 0} − 2 × 120.405 × 1.02 + 3 × 71.802 × 1.02

= 2max{S − 950, 0} − 3max{S − 1050, 0} − 25.9121

=

−25.9121, S ≤ 9502S − 1925.91, 950 ≤ S ≤ 1050−S + 1224.088, 1050 ≤ S,

where S is the strike price.

9. The owner of the stock is entitled to receive dividends. As we will get the stock onlyin one year, the value of the prepaid forward contract is todays stock price, less thepresent value of the four dividend payments:

F P0,T = $50 −

4∑

i=1

$1e0.06× 3

12i

= $50 − $0.985 − $0.970 − $0.956 − $0.942

= $50 − $3.853 = $46.147

b) The forward price is equivalent to the future value of the prepaid forward. With aninterest rate of 6 percent and an expiration of the forward in one year, we thus have:

F0,T = F P0,T × e0.06×1 = $46.147 × e0.06×1 = $46.147 × 1.0618 = $49.00

10

10. a) We plug the continuously compounded interest rate and the time to expiration inyears into the valuation formula and notice that the time to expiration is 9 months, or0.75 years. We have:

F0,T = S0 × er×T = $1,100 × e0.05×0.75 = $1,100 × 1.0382 = $1,142.02

b) We engage in a reverse cash and carry strategy. In particular, we do the following:

Description Today In 9 monthsLong forward, resulting 0 ST − F0,T

from customer purchaseSell short the index +S0 −ST

Lend +S0 −S0 S0 × erT

TOTAL 0 S0 × erT − F0,T

Specifically, with the numbers of the exercise, we have:

Description Today In 9 monthsLong forward, resulting 0 ST − $1,142.02from customer purchaseSell short the index $1,100 −ST

Lend $ 1,100 −$1,100 $1,100 × e0.05×0.75

= $1,142.02TOTAL 0 0

Therefore, the market maker is perfectly hedged. She does not have any risk in thefuture, because she has successfully created a synthetic short position in the forwardcontract.

c) Now, we will engage in cash and carry arbitrage:

Description Today In 9 monthsShort forward, resulting 0 F0,T − ST

from customer purchaseBuy the index −S0 ST

Borrow +S0 +S0 −S0 × erT

TOTAL 0 F0,T − S0 × erT

Specifically, with the numbers of the exercise, we have:

11

Description Today In 9 monthsShort forward, resulting 0 $1, 142.02 − ST

from customer purchaseBuy the index −$1,100 ST

Borrow $1,100 $1,100 −$1,100 × e0.05×0.75

= −$1,142.02TOTAL 0 0

Again, the market maker is perfectly hedged. He does not have any index price risk inthe future, because he has successfully created a synthetic long position in the forwardcontract that perfectly offsets his obligation from the sold forward contract.

11. Suppose that the amount deposited in the bank is X0 and the value of the stock afterone year is X1. Then R = X1

X0

. The total return of the investment is thus equal to1.3X0+X1

2X0

= 0.65 + 0.5R.

12. ρ = 0.3, σA = 0.15, σB = 0.05.

(a) The minimum variance problem is

minα2σ2A + 2α(1 − α)σAB + (1 − α)2σ2

B := f(α).

Sodf(α)

dα= 2ασ2

A + (2 − 4α)σAB − 2(1 − α)σ2B = 0.

Thus

α =−2σAB + 2σ2

B

2σ2A − 4σAB + 2σ2

B

=−σAB + σ2

B

σ2A − 2σAB + σ2

B

= 0.0122,

1 − α = 0.9878.

(b) Minimum SD =√

α2σ2A + 2α(1 − α)σAB + (1 − α)2σ2

B = 0.05.

(c) Expected return is = αr̄A + (1 − α)r̄B = 0.0203.

13. (Rain insurance) Let u be the # of units of insurance bought. So the possible outcomesof return are (.4, u) and (.6, 4 × 106).

(a) The expected rate of return is .4u+.6×4×106−(106+.5u)

106+.5u.

(b) Let X be his return. His mean return is E(X) = (.4u + .6 × 4 × 106).

The variance of his return is V ar(u) = E(X2)−E(X)2 = .4u2 + .6(4× 106)2 − (.4u +.6 × 4 × 106)2.

12

dV ar(u)

du= .8u − 2(.4u + .6 × 4 × 106) × .4 = .48u − 1.92 × 106 = 0.

Then u = 4 × 106. So V ar(u) = 0.

When u = 4 × 106, the expected rate of return = 4×106

106+2×106 − 1 = 43− 1 = 1

3. That is,

33%.

or

Rate of return Prob.$3M−$.5u$1M+$.5u

.6

$.5u−$1M$1M+$.5u

.4

Variance is minimized (=0) when the rate of return is unchanged, i.e.,

$3M − $.5u = $.5u − $1M.

Thus u = $4M.

14. (Wild cats)

(a)

(b) The problem is

min Var =∑

i,j

wiwjσij

s.t.∑

j

wj = 1.

13

Since σij = 0, i 6= j, Var =∑

j w2jσ

2j . Let

L(w, λ) =∑

j

w2jσ

2j − λ

∑

j

wj − 1

.

Then∂L(w, λ)

∂wj

= 2σ2i wi − λ = 0

wj = λ2σ2

j

. So∑

jλ

2σ2

j

= 1. We have

λ =(

∑

j1

2σ2

j

)−1

. So wj = 1σ2

j

(

∑

j1σ2

j

)−1

, and Var =∑

j1σ4

j

(

∑

j1σ2

j

)−2

σ2j =

(

∑

j1σ2

j

)−1

.

The portfolio for the minimum variance point is

(

1σ2

1

(

∑

j1σ2

j

)−1

, · · · , 1σ2

n

(

∑

j1σ2

j

)−1)

,

while the minimum variance point is

(

(

∑

j1σ2

j

)

−1

, r̄

)

.

15. (Markowitz fun)

(a) The problem is

min1

2w>Σw, s.t.

∑

j

wj = 1.

Let L(w, µ) = 12w>Σw − µ(

∑

j wj − 1). Thus

w1 + w2 − µ = 0

w1 + 2w2 + w3 − µ = 0

w2 + 2w3 − µ = 0

w1 + w2 + w3 = 1.

Eliminating µ, we have

−w2 − w3 = 0

w1 − 2w3 = 0

w1 + w2 + w3 = 1.

Thus−w2 − 3w3 = −1, −2w3 = −1

We have w3 = 0.5 and w2 = −0.5. So w1 = 1. The minimum variance solution isw1 = 1, w2 = −0.5 and w3 = 0.5. Here µ = 0.5.

14

(b) Solve the following system of equations

w1 + w2 − .4λ − µ = 0

w1 + 2w2 + w3 − .8λ − µ = 0

w2 + 2w3 − .8λ − µ = 0

0.4w1 + 0.8w2 + 0.8w3 = 0.7

w1 + w2 + w3 = 1.

From the second and third equations, we have w1 + w2 − w3 = 0. Thus together withw1 + w2 + w3 = 1, we have 2w3 = 1, so w3 = 1

2.

Multiplying w1 + w2 − w3 = 0 by 4 and adding it to 4w1 + 8w2 + 8w3 = 7, we have4w2 + 12w3 = 7. So w2 = 1

4(7 − 12 × 1

2) = 1

4. Finally we have w1 = 1 − 1

4− 1

2= 1

4.

The optimal portfolio solution is w1 = 14, w2 = 1

4, w3 = 1

2.

(c) Let λ = 1 and µ = 0. The system of linear equations is

v1 + v2 = 0.4

v1 + 2v2 + v3 = 0.8

v2 + 2v3 = 0.8

From the first two equations, we have −v2 − v3 = −0.4. This together with the thirdequation gives v3 = 0.4 and v2 = 0. From the first equation again, we have v1 = 0.4.The normalized solution is w1 = 0.5, w2 = 0 and w3 = 0.5 with the expected rate ofreturn being equal to 0.6. The other optimal portfolio w1 = 1, w2 = −0.5 and w3 = 0.5has the expected rate of return 0.4.

By the two-fund theorem, we have

0.4(1 − α) + 0.6α = 1, thus α =10 − 4

6 − 4= 3.

The optimal portfolio for the expected rate of return being equal to 0.7 is

(1 − 3)(1,−0.5, 0.5) + 3(0.5, 0, 0.5) = (−0.5, 1, 0.5).

(d) Let rf = 0.1. The solution is found by the system of equations:

v1 + v2 = 0.4 − 0.1

v1 + 2v2 + v3 = 0.8 − 0.1

v2 + 2v3 = 0.8 − 0.1.

or by the formula:

v1 = 0.4 − 0.1 × 2 = 0.2

v2 = 0 − 0.1 × 1 = 0.1

v3 = 0.4 − 0.1 × 1 = 0.3.

The normalized solution is w1 = 0.20.2+0.1+0.3

= 13, w2 = 1

6and w3 = 1

2.

15

16. We know that the efficient portfolio required in One-Fund Theorem satisfies

n∑

i=1

σkivi = r̄k − rf , k = 1, 2, ..., n. (1)

Indeed, from the equations for the efficient set, we have

n∑

j=1

σijwj − λr̄i − µ = 0, i = 1, · · · , n.

Thus, by the definition of v1 with (µ = 1, λ = 0) and v2 with (µ = 0, λ = 1),

n∑

j=1

σijv1j − 1 = 0, i = 1, · · · , n. (2)

n∑

j=1

σijv2j − r̄i = 0, i = 1, · · · , n. (3)

It is clear that (1) can be obtained from (3) − rf (2).

17. (i) Let λ = 0, µ = 1.

Equations are

1v1 + 2v2 = 12v1 + 2v2 + v3 = 1

v2 + 2v3 = 1

Solutions are

v11 = −.2, v1

2 = .6, v13 = .2.

Normalized solutions are w1 = −13, w2 = 1, w3 = 1

3.

Let λ = 1, µ = 0.

Equations are

1v1 + 2v2 = .12v1 + 2v2 + v3 = .2

v2 + 2v3 = .3

Solutions are

v21 = −.02, v2

2 = .06, v23 = .12.

Normalized solutions are w1 = −.125, w2 = .375, w3 = .75.

16

(ii) Let rf = 0.1. The solution is found by solving

v1 = v21 − rfv

11 = 0

v2 = v22 − rfv

12 = 0

v3 = v23 − rfv

13 = .1

Normalized solutions are w1 = 0, w2 = 0, w3 = 1.

18. (Without shorting) The Lagrangian is

L(w, λ, µ, ξ) =1

2[w2

1 + 2w22 + 2w2

3 + 2w1w2]

−λ(3w1 + w2 + w3 − 2) − µ(w1 + w2 + w3 − 1) − ξ1w1 − ξ2w2 − ξ3w3.

The KKT condition is

w1 + w2 − 3λ − µ − ξ1 = 02w2 + w1 − λ − µ − ξ2 = 02w3 − λ − µ − ξ3 = 03w1 + w2 + w3 − 2 = 0w1 + w2 + w3 − 1 = 0ξ1w1 = 0, ξ1 ≥ 0, w1 ≥ 0,ξ2w2 = 0, ξ2 ≥ 0, w2 ≥ 0,ξ3w3 = 0, ξ3 ≥ 0, w3 ≥ 0.

Case 1. Let ξ1 > 0, w1 = 0. As r̄2 = r̄3 = 1, this case cannot find the solution.

Case 2. Let ξ2 > 0, w2 = 0. Then

3w1 + w3 − 2 = 0, w1 + w3 − 1 = 0,

=⇒ w1 =1

2, w3 =

1

2,

=⇒ ξ1 = 0, ξ3 = 0,

3λ + µ =1

2, λ + µ + ξ2 =

1

2, λ + µ = 1

=⇒ λ = −1

2, µ = 2, ξ2 = −1 < 0, (impossible)

Case 3. Let ξ3 > 0, w3 = 0. Then w1 = 12, w2 = 1

2. So ξ1 = 0, ξ2 = 0. Thus

3λ + µ = 1, λ + µ =3

2, λ + µ + ξ3 = 0

=⇒ λ = −1

4, µ =

5

2, ξ3 = −2 < 0, (impossible)

17

Case 4. Let ξ1 = ξ2 = ξ3 = 0, w1 > 0, w2 > 0, w3 > 0. Then λ = −w2

2, µ = w1 + 5

2w2.

So

−w1 − 2w2 + 2w3 = 0, 3w1 + w2 + w3 = 2, w1 + w2 + w3 = 1

=⇒ −7w1 − 4w2 = −4, 2w1 = 1,

=⇒ w1 =1

2, w2 = (−7

2+ 4)/4 =

1

8, w3 =

3

8, λ = − 1

16, µ =

13

16.

The optimal portfolio for r̄ = 2 is (12, 1

8, 3

8).

19. N/A

18