111

Investment Science

Chapter 3

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

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

Investment Science

Chapter 4

Solutions to Suggested Problems

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

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

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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

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(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

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文本框

4.11 (Running PV examples)

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Solution of HW3

Problem 6.1 X0 = outlay (in this case it is equal to the deposit.) X1 = amount received, equal to the returned deposit plus the profit from shorting. Thus, X1= (X0 – X1) + X0 The total return, R = X1/X0 = ((X0 – X1) + X0)/X0 Thus, R = 2X0 – X1 X0

Problem 6.2 Let a and b be the outcomes of two die rolls. Then Z=ab. By independence, we know:

[ ]2

21 2 3 4 5 6[ ] [ ] 3.5 12.256

E ab E a E b + + + + +⎛ ⎞= = = =⎜ ⎟⎝ ⎠

[ ] ( ) [ ]( ) ( ) ( ) [ ] [ ]( )

( ) ( ) [ ]( ) [ ]( ) ( )

2 22 2 2

22 22 2 4

var

1 4 9 16 25 36 3.5 79.976

Z E ab E ab E a b E a E b

E a E b E a E b

⎡ ⎤ ⎡ ⎤= − = − =⎣ ⎦ ⎣ ⎦

+ + + + +⎛ ⎞⎡ ⎤ ⎡ ⎤= − = −⎜ ⎟⎣ ⎦ ⎣ ⎦ ⎝ ⎠≈

Problem 6.3 Using the answer of the Problem 6.4: a) 19 / 23α =

b) ( ) ( ) ( ) ( ) ( )2 2 2 2 22 2 21 21 19 / 23 0.15 1 19 / 23 0.3 13.4%α σ α σ+ − = + − =

c) ( )1 21 11.4r r rα α= + − ≈

Problem 6.4 Let α equal the percent of investment in stock 1, then the percent in stock 2 is (1-α). Now, problem becomes Minimize σ2

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删划线

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−=

=−++−=

=+−−+=

+−+−+=

−+−+=

++=

= ∑=

2212

21

1222

2212

2212

21

2

22

221212

21

2

222

22

22

21212

221

22212

21

2

22

221221

21

21

2

1,

2

2

022)242(/

022422/

222

)1()1(**2

2

σσσσσ

α

σσασσσασ

ασσασσασασ

ασασσασασασ

ασαασσα

σωσωωσω

σωωσ

dd

dd

jiijji

: take derivatives respect to α

Thus percentage of asset 2 = 1-α = ⎟⎟⎠

⎞

⎝

⎛+−

−2212

21

1221

2 σσσσσ

⎜⎜

The mean rate of return of this portfolio is )1(** 21 αα −+= rrr

⎟⎟⎠

⎞⎜⎜⎝

⎛

+−+−+

=

+−−+−=

+−−++−−=

2212

21

1221212

221

2212

2112

21212

221

2212

2112

212

2212

2112

221

2)(

)2/()]()([

)]2/()[()]2/()[(

σσσσσσ

σσσσσσσ

σσσσσσσσσσ

rrrr

rr

rrr

Problem 6.5 Money invested for concert 1 year from now= 1 million Revenue expected unless it rains= 3 million Chances of rain=50% Rain Insurance = $ 0.5 per unit Money obtained from insurance if it rains= $ 1 per unit Let no. of units of insurance bought= u a.)Now total money invested, X0 = money invested + no. of units of insurance bought = (1*10^6+ 0.5*u)

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删划线

Revenue obtained X0 = 3*10^6+ 1*u Now since there are 50% chances of rain we have total revenue:- X1= (3*10^ 6+u) *0.5 = (1.5*10^6+0.5u) E (Total Return) = X1/X0=(0.5*3,000,000+0.5*u)/ (1,000,000+0.5u) Now rate of return E(rate of return)= X1/X0-1= = ((1.5*10^6+0.5u) - (1*10^6+ 0.5*u))/ (1*10^6+ 0.5*u) = (500,000)/ (1,000,000+0.5*u) b.) To get the minimum variance we will have to buy all the 3 million units which will give us a variance of 0. We have to Minimize Var(r) Hence u= 3*10^ 6 units. Var(r) = 0 Therefore r = (10^6(1.5-1))/ (1*10^6+0.5*3*10^6) = (0.5)/(1+1.5) = 0.2 = 20%

Problem 6.6 (A)

r

σ

The efficient set (at minimum-variance point)

assets

(B)

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Since assets are uncorrelated,

2

12

12

12

1

2

2

2

11

22

1

22

21

2

21

12

Then,2

2

02

1

σ

σ

µ

µσ

σµω

σµω

σωµ

µσωω

ωµσω

σω

==

=

==

=

=

=−=∂∂

⎟⎠

⎞⎜⎝

⎛−−=

=

∑

∑

∑∑

∑∑

∑

=

=

==

==

=

n

i i

n

i i

n

i i

n

ii

ii

ii

iii

n

ii

n

iii

n

iii

L

L

Var

Therefore,

∑∑∑===

===

=

n

i i

n

i i

n

iii

ii

Var1

24

12

4

1

22

2

2

1σ

σσσσω

σσω

2σ=Var

Problem 6.7 Covariance matrix

2 1 0 1 2 1 0 1 2

V=

Expected rates of return E(r1)= 0.4 E(r2)= 0.8 E(r3)= 0.8

a) Minimum variance portfolio (consider that w1=w3 by symmetry) Markowitz formulation:

1

1

1

1 3

0

1

n

iij jj

n

iiin

ii

r

w r

w r

ww w

λ µσ=

=

=

− − =

=

=

=

∑

∑

∑

(1.1)

For solving the system of equations (1.1) we omit the last normalizing constraint. Then, we normalize the obtained solution { }1 2 3, ,v v v to the solution { }1 2 3, ,w w w that satisfies the constraint. System with 6 variables and 6 equations:

w1 w2 w3 λ µ E(r) = 2 1 0 -0.4 -1 0 01 2 1 -0.8 -1 0 00 1 2 -0.8 -1 0 0

0.4 0.8 0.8 0 0 -1 01 1 1 0 0 0 11 0 -1 0 0 0 0

By using calculator:

w1 w2 w3 λ µ E(r) 0.5 0 0.5 0 1 0.6

b) λ=1, µ=0 System with 3 variables and 3 equations:

v1 v2 v3 = 2 1 0 0.4 1 2 1 0.8 0 1 2 0.8

By using calculator:

v1 v2 v3 w1 w2 w3 E(r) 0.1 0.2 0.3 0.167 0.333 0.5 0.733

c) Given that there is a risky-free part, there is a single fund of risky assets such that any efficient portfolio can be constructed as a combination of the fund and the risk-free instrument. By using equation 6.10, we get a system of 3 equations with 3 variables.

2.0 where

,..,1 ,1

=

=−=∑ =

f

fkin

i ki

r

nkrrvσ

v1 v2 v3 =

2 1 0 0.2 1 2 1 0.6 0 1 2 0.6

v1 v2 v3 w1 w2 w3

0 0.2 0.2 0 0.5 0.5

Problem 6.8 a)

1( )

. . 1

n

i i Mi

n

ii

Min Var r r

S T

α

α

=

−

=

∑

∑

2

1 1 1

1

1

1

1

2 (1

2 2 0

1 0

,

2

1

n n n n

i j ij i iM M ii j i i

n

j ij iMji

n

ii

n

j ij iMj

n

ii

L

L for each i

L

So

for each i

)α α σ α σ σ µ α

α σ σ µα

αµ

µα σ σ

α

= = =

=

=

=

=

= − + + −

∂= − − =

∂

∂= − =

∂

= +

=

∑∑ ∑ ∑

∑

∑

∑

∑

b)

1

~

( )

. . 1

n

i i Mi

n

iin

i ii

Min Var r r

S T

r r

α

α

α

=

−

=

=

∑

∑

∑

~2

1 1 1

1

1

~

1

1

~

2 (1 ) (

2 2 0

1 0

0

,

2

1

n n n n n

i j ij i iM M i i ii j i i i

n

j ij iM iji

n

ii

n

i ii

n

j ij iMj

n

iin

i ii

L r

L r for each i

L

L r r

So

r for each i

r r

)rα α σ α σ σ µ α λ α

α σ σ µ λα

αµ

αλ

µα σ σ λ

α

α

= = =

=

=

=

=

= − + + − + −

∂= − − − =

∂

∂= − =

∂∂

= − =∂

= + +

=

=

∑∑ ∑ ∑ ∑

∑

∑

∑

∑

∑

∑

Problem 6.9 (Beeting Wheel) Consider a general betting wheel with n segments. The payoff for a $1 bet on a segment i is Ai. Suppose you bet an amount Bi = 1/Ai on segment i for each i. Show that the amount you win is independent of the outcome of the wheel. What is the risk-free rate of return for the wheel? Apply this to the wheel in Example 6.7 Sol:

Data: n segments (from segment i =1 to segment n) Amount received from segment i = Ai for each $1 invested Fraction of segment i invested = Bi = 1/Ai (Wi) a) Expected return for segment i, Ri = Ai / $1 Amount to win if the outcome equal to segment i:

Expected return for segment i times the fraction invested in segment i

1$1*1

** =⇒⇒Ai

AiBiRiWiRi

For any outcome on the wheel the amount we will win will be $1. Risk-free rate of return for the wheel: For any outcome of the wheel, the amount that we will win will be = $1., X0 = $1 (amount received)

The amount invested will be X1= ∑ Ai1

∑==

AiXXR

11$

10

1111

−⎟⎠⎞

⎜⎝⎛=−=

−

∑ AiRr

Problem 6.10

Derive fk

n

iiki rrw −=∑

=1

λσ .

Solution

2/1

1,

1

)(tan

⎟⎟⎠

⎞⎜⎜⎝

⎛

−=

∑

∑

=

=

n

jijiij

n

ifii

ww

rrw

σ

θ

0

)()(

tan

1,

2/1

1,

1

1

2/1

1,

=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛ −−⎟⎟⎠

⎞⎜⎜⎝

⎛−

=∂∂

∑

∑

∑∑∑

=

=

=

==

n

jijiij

n

jijiij

n

jjkjn

ifii

n

jijiijfk

k ww

ww

wrrwwwrr

w σ

σ

σσ

θ , nk ,1=

Since , we have: 02

1,

>=∑=

σσn

jijiij ww

0)()(111,

=⎟⎠

⎞⎜⎝

⎛ −−⎟⎟⎠

⎞⎜⎜⎝

⎛− ∑∑∑

===

n

jjkj

n

ifii

n

jijiijfk wrrwwwrr σσ

)()(

1

1,

1fk

n

jjkjn

jijiij

n

ifii

rrwww

rrw−=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠

⎞⎜⎝

⎛ −

∑∑

∑=

=

= σσ

lets denote

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠

⎞⎜⎝

⎛ −=

∑

∑

=

=

n

jijiij

n

ifii

ww

rrw

1,

1

)(

σλ

fk

n

iiki rrw −=∑

=1

λσ

Problem 7.1 7.1 Assume that the expected rate of return on the market portfolio is 23% and the rate of return on T-bills(the risk-free rate) is 7%.The standard deviation of the market is 32%.Assume that the market portfolio is efficient a) What is the equation of the capital market line? formulation expected rate of return =[ [ risk-free rate + (expected rate of return from market-risk free

rate)]/ standard deviation of market ] * standard deviation

expected rate of return from market = 0.23 risk-free rate = 0.07 standard deviation of market = 0.32 expected rate of return = 0.07+ (0.23-0.07)/0.32 * (standard deviation) = 0.07+0.5 (standard deviation)

b) i) if expected return of 39% is desired, what is the standard deviation of this position? expected return = 0.39

�standard deviation position = 0.64

ii) If you have $1,000 to invest, how should you allocate it to achieve the above position?

� Expected return is 0.39

� we can get 1000(1+0.39) = 1390

1390 = x (1+0.07) +(1000-x)(1+0.23) x= -1000 risk free asset

� -1000

maket portfolio �

2000 c) If you invest $300 in the risk-free asset and $700 in the market portfolio, how much money should you expect to have at the end of the year? 300(1+0.07)+700*(1+0.23)=1182

Problem 7.2 A small world

Consider a world in which there are only two risky assets, A and B, and a risk free asset F. Te two risky assets are in equal supply in the market; that is, M= ½ (A+B). The following information is known: rF = .10 �

A2 = .04 �

AB = .01 �B

2 = .02 �M = .18

(a) Find a general expression for �

M2, � A, � B

Since the market has only two risky assets A and B, then the expected rate of return and the variance of the market depend solely on the expected rate of return and the variance of the assets A and B. �

M2 = � �

A2 +2� (1- � ) �

AB + (1- � )2�B

2

Since A and B are in equal amounts in the market � = 1- � ( � =0.5)

� A = [cov (rA,rM)]/ �M

2 cov(rA,rM) = cov (rA, � rA + (1- � )rB) = � �

A2 + (1- � ) �

AB

� A = [ � �A

2 + (1- � ) �AB]/[ � �

A2 +2� (1- � ) �

AB + (1- � )2�B

2]

= [ �A

2 + �AB]/ � ( �

A2 +2�

AB + �B

2)= =2 [ �

A2 + �

AB]/( �A

2 +2�AB + �

B2

� B = [cov (rB,rM)]/ �M

2 cov(rB,rM) = cov (rB, � rA + (1- � )rB) = � �

B2 + (1- � ) �

AB

� B = [ � �B

2 + (1- � ) �AB]/[ � �

A2 +2� (1- � ) �

AB + (1- � )2�B

2]

= [ �B

2 + �AB]/ � ( �

A2 + 2�

AB + �B

2)= = 2[ �

B2 + �

AB]/( �A

2 + 2�AB + �

B2

(b)According to CAPM, what are the numerical values of

�A and

�B �

i – rf = � i (�M-rf)

�A = � A (

�M- rf) + rf

=[ [ �A

2 + �AB]/ � ( �

A2 2�

AB + �B

2)] (�M- rf) + rf

= [(.04+.01)/(.5(.04+2(.01)+.02))][.18-.10] + .10 = .2

�B= � B(

�M- rf) + rf

=[ [ �B

2 + �AB]/ � ( �

A2 2�

AB + �B

2)] (�M- rf) + rf

= [(.02+.01)/(.5(.04+2(.01)+.02))][.18-.10] + .10 = .16

Problem 7.3 7.3 (Bounds on returns) Consider a universe of just three securities. They have expected rates of return of 10%, 20%, and 10%, respectively. Two portfolios are known to lie on the minimum-variance set. They are defined by the portfolio weights w = [.60, .20, .20], v = [.80, -.20, .40]. It’s also known that the market portfolio is efficient. (a) Given this information, what are the minimum and maximum possible values for the expected rate of return on the market portfolio? (b) Now suppose you are told that w represents the minimum-variance portfolio. Does this changer your answers to part (a)? (a) Market portfolio can’ t contain negative amount of security. ��

��

�� ! #"$�&% ��' ��

( ��)�* �� ,+�� -#.0/ -#.01 -#.01�%�-#. � � 2 - �#3�4( ��)�* �� ,+!� -#. � %�-#. � %�-#. �� -#. 4�� 2 - �#2 -#.05( ��)�* �� ,+�6 -#. � -#. 4 -#. 4 %�-#. � � 2 - �#3 �

78� ��9� �� -#.05 3��#3 � Expected rate of return on the market portfolio:

�� :�� * � 9

��

�:;=<��) � ��> �� :�?� � � * � 9 �:�� " � �� ! �"$�!%��' �� '

( ��)�* �� ,+�� !-#@ -#.0/ -#.01 -#.0-�1�%�-#. -�� ( ��)�* �� ,+!� ��-#@ -#. � %�-#. � %�-#.0- 4 �-#.0-�1 �( ��)�* �� ,+�6 �!-#@ -#. � -#. 4 -#.0- 4 %�-#. -�� 7�� -#.0-�1� �-#.0- 4��

From Mr = 0.08+0.04� and 0.53 � 3 2, we know 0.13 Mr 3 0.16.

(b) If given portfolio1 is the minimum-variance portfolio, rate of return of portfolio1 is the minimum rate of return of market portfolio. Rate of return of portfolio1=0.06+0.04+0.02=0.12.

Therefore, the expected rate of return of market portfolio became 0.123 Mr3 0.16.

Problem 7.4

(Quick CAPM derivation) Derive the CAPM formula for fk rr −_

by using Equation (6.9)

in Chapter 6. [Hint: Note that �=

=n

niMkiik rrw ).],cov(σ

Apply (6.9) both to asset k and to the market itself. Solution Equation 6-9 from the textbook

fk

n

niiki rrw −=

�=

_

λσ k = 1,2,….n

We apply the above equation both to asset k and to market M. So we get

[ ] fiMiMii rrww −=+_

2 σσλ and

[ ] fMMMiiM rrww −=+_

2σσλ

From the hint:

[ ] iMMiMiMii rrww σσσ ==+ ),cov(2 and

[ ] 22 ),cov( MMMMMiiM rrww σσσ ==+

Substituting in above equations we get:

fiiM rr −=_

λσ

fMM rr −=_

2λσ

Eliminating λ from the above two equations and solving, we get:

2

_

M

fM rr

σλ

−= . Therefore,

fiiMM

fM rrrr

−=− _

2

_

σσ . From the textbook, i

M

iM βσσ =

2 . Substituting, we obtain

fiifM rrrr −=−__

)( β Which is the required capital asset pricing (CAPM) formula.

Problem 7.5 2

2 2 2

1

2

2 2 2

1

/

cov 0

/

j jM M

n

M i ii

jM j j

n

j j j i ii

x

x i contributes according to its weight

Therefore x x

β σ σ

σ σ

σ σ

β σ σ

=

=

=

= =

=

=

�

�

Problem 7.6 Problem 7.6 – Simpleland In Simpleland, there are only two risky stocks, A and B, whose details are listed below:

- Number of shares outstanding: o A: 100 o B: 150

- Price per share o A: $1.50 o B: $2.00

- Expected rate of return o A: 15% o B: 12%

- Standard deviation of return o A: 15% o B: 9%

Furthermore, the correlation coefficient between the returns of stocks A and B is � ab=1/3. There is also a risk free asset, and Simpleland satisfies the CAPM exactly.

a. Expected rate of return of the market portfolio

Market Capitalization

- Stock A: 100*1.50=150 - Stock B: 150*2.00=300 - Total: 450

We can deduce the respective weights in the market portfolio: - Stock A: 1/3 (=150/450) - Stock B: 2/3

So the expected rate of return of the market portfolio is: E(rm)=0.15*1/3+0.12*2/3 E(rm)=13%

b. Standard deviation

Var(rm)= � a2* �

a 2+ � b

2* �b2+2* � a* � b* � ab* �

a* �b

Var(rm)=(1/3)2*0.152+(2/3) 2*0.092+2*1/3*2/3*1/3*0.15*0.09 Var(rm)=0.0081 � (rm)=0.09

c. Beta of stock A

� a=Cov(ra,rm)/ �m

2 And we know that: rm=1/3*ra+2/3*rb Hence: Cov(ra,rm)= Cov(ra, 1/3*ra+2/3*rb) Cov(ra,rm)= 1/3*Cov(ra,ra)+2/3*Cov(ra,rb) Cov(ra,rm)= 1/3* �

a2+2/3* � ab* �

a* �b

Cov(ra,rm)= 1/3*0.152+2/3*1/3*0.15*0.09 Cov(ra,rm)=0.0105

So the Beta of stock A is: � a=0.0105/(0.09)2 � a =1.296

d. Risk Free Asset in Simpleland Relation (7.2) gives us: E(ra)-rf = � a*(E(rm)- rf) So: rf =(E(ra)- � a*E(rm)) / (1- � a) rf =(0.15-1.296*0.13) / (1-1.296) rf =6.25%

Problem 7.7

(Zero-beta assets) Let w0 be the portfolio (weights) of r isky assets corresponding the minimum-variance point in the feasible region. Let w1 be any other portfolio on the efficient frontier. Define r0 and r1 to be the corresponding returns. a) There is a formula of the form � 01= A � 0². Find A. [Hint: Consider the portfolios (1- � )w0 + � w1, and consider small variations of the variance of such portfolios near

� =0.

012

0

0

22

2101

20

0

22

2101

20

0

22

0

22

21

201

20

222

10

10

22)(

)0(2))0(42()01(2)(

2)42()1(2)(

)(

)1(2)1(

. wand on w

based dconstructe becan portfolio A third frontier.efficient theon the portfolios twobe wand Let w

σσα

σ

σσσα

σ

ασσασαα

σ

ασ

σασαασασ

α

α

α

α

+−=∂

∂

+−+−−=∂

∂

+−+−−=∂

∂

∂∂

+−+−=

=

=

=

=

1

2001

012

0

==

=

A

Aσσ

σσ

b) Corresponding to the portfolio w1 there is a portfolio wz on the minimum-variance set that has zero beta with respect to w1; that is, � 1,z=0. This portfolio can be expressed as wz= (1- � )w0 + � w1. Find the proper value of � .

20

212

1

20

21

20

20

21

21

21

20

20

21

20

21

21

20

212

02

1

21

20

20

21

21

20

20

21

21

202

221

221

221

22022

12

421

2412

0

21

2

21

2

22

2

21

2

212

0

21

2

21

2

22

2

21

2

22

0

21

2

2

21

2

1

211

22

0

0

21

21

2220

21

201

20

22

01z0

)1(1

)(*

1*

)()()(

1

2)1(20

0 Because

2)42()1(200 :Set

: thenMVP, theis wSince1

1-1

1

: thatconcluded becan it ,portfolios three theofn combinatiolinear By

)1(2)1(

)1(2)1(

wand wofn combinatio weighted then thezero, is If portfolio.point varianceminimum thebe Let w

σαασσσ

αα

σσβ

σσσσσσσ

σσσσ

σσσ

σσσσσσβ

σσσσσ

σσσσσσσσσ

σσσσσ

σσσ

σσσσ

σσ

σσσ

σσσσ

σσ

σσσβ

βσσβ

σ

βσσβσββ

σ

ααββ

α

σβσββσβσ

σασαασασ

σ

−=−�=−

−�=

−+−

−=

+−

−=�

−=

+=+�++

=

��

++��

��+

=

��

++��

��+

−=�+

=

+−−=

=

+−+−−=�=∂

∂

−−=�−=

+−+−=

+−+−=

z

zzz

z

zz

z

zz

z

z

zz

z

z

z

z

z

z

zz

zz

z

21

20

20

σσσα

−=

c) Show the relation of the three portfolios on a diagram that includes the feasible region.

d) If there is no risk-free asset, it can be shown that other assets can be priced according to the formula

)( zMiMzi rrrr −−−−====−−−− ββββ where the subscript M denotes the market portfolio and zr is the expected rate of return on the portfolio that has zero beta with the market portfolio. Suppose that the expected returns on the market and the zero-beta portfolio are 15% and 9% respectively. Suppose that a stock i has a correlation coefficient with the market of .5. Assume also that the standard deviation of the returns of the market and stock i are 15% and 5% respectively. Find the expected return of stock i.

)09.015.0(15.0

00375.009.0

)(

00375.005.0*15.0*5.05.0

%5

%15

%9

%15

2−

��

+=

−=−

==�===

=

=

i

zMiMzi

iMiM

i

M

z

M

r

rrrr

r

r

βσρ

σσ

%101.0 ==ir

r

σ

w0

w1

wz

0σ

1σzσ

0r

1r

zr

Problem 7.8 Electron Wizards, Inc. has a new idea for producing TV sets, and it is planning to

enter the development stage. Once the product is developed (which will be at the end of 1 year), the company expects to sell its new process for a price p, with expected

value Mp 24$= . However, this sale price will depend on the market of TV sets at the time. By examining the stick histories of various TV companies, it is determined that the

final sales price p is correlated wth the market return us ( )( )[ ] 220$ Μ=−− σMrrppE MM . To develop the process, EWI must invest in a research and development project. The

cost c of the project will be known shortly after th project is begun. The current estimate is that the cost will be either c=$20M or c=$16M, and each of these is equally likely. (This uncertainty is uncorrelated with the final price and is also uncorrelated with the market). Assume that the risk free rate is rf=9% and the expected return of the market

%33=Mr . (a) What is the expected rate of return of this project? (b) What is the beta of this project?

Hint: ( )( ) ( )( )[ ]MMMM rrppE

cErr

c

ppE −−

��

=�� −− 1

c) Is this an acceptable project based on a CAPM criterion? In particular what is the excess rate of return (+ or -) above the predicted by the CAPM?

Solution:

a) ( ) 11 −��

��=�−=

−=

c

pErE

c

p

c

cpr . Due to the fact that p,c are independent we

have ( ) ( ) ( ) 35.0124*05625.011 = −=−

��

= rEpEc

ErE

b)

( )( )[ ] ( ) ( ) ( ) ( ) ( )

( )( )[ ] 2125.11

Μ=−−��

��=

�� −−=��

�� −−−−=−−

σMM

MMMMMM

rrppEc

E

rrc

ppErr

c

cp

c

cpErrrrE

Then: 125.1125.1

2

2

2===

M

M

M

rM

σσ

σσβ

c) Based on the CAPM the expected return is:

( ) ( ) ( ) 36.009.009.033.0125.1 =�+−=+−=�−=− rrrrrrrrr ffMfMf ββThe expected excess rate of return is:

27.0+=− frr

Thus we conclude that, based on the CAPM model, the project is not acceptable since it gives smaller return rate than CAPM. Nevertheless, the difference is only 0.01 therefore the final decision should not be based only on the CAPM criterion.

Problem 7.9 Formulation: (Gavin’s problem) Prove to Gavin Jones that the results he obtained in egs. 7.5 and 7.7 were not accidents. Specifically, for a fund with return � rf + (1- � ) rm show that both CAPM models give the price of $ 100 worth of fund assets as $100. We have to prove that the formulas, “ certainty equivalent form of the CAPM” & the “ CAPM as a pricing formula” both will give the same results for pricing an asset P by appropriately discounting the final return Q. we have to do this for the case mentioned for Gavin Jones in egs. 7.5 & 7.7 of the book, i.e. for an asset combination of two securities with weights � and (1- � ). The return for an asset mixture (Q) with above weights is given by (please note that Q or rM with a bar on top is represented as Q or rM i.e., boldface) Q = P ( � rf + (1- � ) rM + 1)………………………………………………………(1) Also for the covariance we can write, cov (Q, rM ) = cov ( P ( � rf + (1- � ) rM + 1), rM ) …………………………….…(2) = P(1- � ) � 2

M from certainty equivalent form of the CAPM, we have, P = [1/ (1+ rf )] * [ Q- { cov (Q, rM )* ( rM - rf ) / � 2

M } Substituting (1) and (2) from above, P = [1/ (1+ rf )]* [ P ( � rf + (1- � ) rM + 1) – { P(1- � ) � 2

M )* ( rM - rf ) / � 2 M } ]

= [1/ (1+ rf )]* P [ ( � rf + (1- � ) rM + 1) – { (1- � ) * ( rM - rf )} ]

expanding and canceling out common terms, P = [1/ (1+ rf )]* (1+ rf ) = P. hence the proof.

Problem 8.1 (A simple portfolio) Someone who believes that the collection of all stocks satisfies a single-factor model with the market portfolio serving as the factor gives you information on three stocks which makes up a portfolio. In addition, you know that the market portfolio has an expected rate of return of 12% and a standard deviation of 18%. The risk-free rate is 5%. (a) What is the portfolio’s expected rate of return? (b) Assuming the factor model is accurate, what is the standard deviation of this

rate of return?

Market Rate of return

Standar deviation

12% 18% Risk free rate 5%

Stock Beta Standar deviation of

random error term

Weight in

porfolio

A 1.1 7.00% 20% B 0.8 2.30% 50% C 1 1.00% 30%

a) The equation for a single-factor model for stock returns is:

So, solving we have: R1 = 1.1*(0.12- 0.05)+0.05 = 0.127 = 12.7% R2 =0.8*(0.12- 0.05)+0.05 = 0.160= 10.6% R3 =1.0*(0.12- 0.05)+0.05 = 0.120 = 12.0% Since Rportfolio =

Rportfolio = 0.127*.20+0.160*0.50+0.120*0.30 = 0.1144 = 11.44% b) Standard Deviations: We know that:

Solving, we have: B2 = (0.20*1.1+0.50*0.8+0.30*1)^2 = 0.846

)( fmifi rrrr −×=− β

i

n

iiportfolio wrr ×=

�=1

2222efb σσσ +×= i

ii bwb ×= �

=

3

1

23

1

22ei

iie w σσ ×= �

=

� 2M = 0.18^2 = 0.324 = 3.24% � 2e = 0.2^2*0.07^2+0.5^2*0.023^2+0.3^2*0.01^2 = 0.03%

Portfolio's standard deviation = (84.6%*3.24%+0.03%)^0.5 = 16.66%

Problem 8.2 Two stocks are believed to satisfy the two-factor model r1 = a1 + 2f1 + f2 r2 = a1 + 3f1 + 4f2 In addition there is a risk-free asset with a rate of return on 10%. It is known that r-bar1 = 15% and r-bar2 = 20%. What are the values of

�0,

�1, and

�2 for this model?

Since there is a risk-free asset with rate-of-return of 10%:

�0 = .10

Using the relationship: if ri = ai + � bij fj then r-bari =

�0 + � bij

� j

Yields: .15 = .10 + 2

� 1 +

� 2

.20 = .10 + 3

� 1 + 4

� 2

� 1 = .02

� 2 = .01

Problem 8.3 Suppose there are n random variables x1, x2, … xn and let V be the corresponding

covariance matrix. An eigen vector of V is a vector v = (v1, v2, … vn) such that Vv = λv

for some λ (called an eigenvalue of V). The random variable v1x1 + v2x2 + … + vnxn is a principle component. The first principle component is the one corresponding to the largest eigenvalue of V, the second to the second largest, and so forth. A good candidate for the factor in a one-factor model of n asset returns, is the first principle component extracted from the n returns themselves: that is, by using the principle eigenvector of the covariance matrix of the return. Find the first principle component for the data of example 8.2. Does this factor (when normalized) resemble the return on the market portfolio? [Note: For this part, you need an eigenvector calculator as available in most matrix operation packages.]

From the example 8.2 we have the following data

From the above first we calculate the averages, variance and finally the covariance for the above data In the covariance we have to divide the value that is obtained from the excel solver by 9*10 to adjust the bias. Using the above Covariance values the Covariance matrix is constructed From the excel we get it as Covariance matrix From the above matrix we have to calculate the eigen values

Year Stock 1 Stock 2 Stock 3 Stock 4 Market Riskless

1 11.91 29.59 23.27 27.24 23 6.2 2 18.37 15.25 19.47 17.05 17.54 6.7 3 3.64 3.53 -6.58 10.2 2.7 6.4 4 24.37 17.67 15.08 20.26 19.34 5.8 5 30.42 12.74 16.24 19.84 19.81 5.9 6 -1.45 -2.56 -15.05 1.51 -4.39 5.2 7 20.11 25.46 17.8 12.24 18.9 4.9 8 9.28 6.92 18.82 16.12 12.78 5.5 9 17.63 9.73 3.05 22.93 13.34 6.1

10 15.71 25.09 16.94 3.49 15.31 5.84 Average 15.00 14.34 10.90 15.09 13.83 5.854 Var 90.26 107.23 162.20 68.25 72.12 Cov 65.09 73.62 100.78 48.99 72.12

β 0.9 1.02 1.4 0.68 1

α 1.95 0.34 -6.11 3.82 0 e-var 31.54 32.09 21.37 34.99

1 2 3 4 1 90.26 50.89 79.02 40.18 2 50.89 107.23 105.38 30.99 3 79.02 105.38 162.20 56.54 4 40.18 30.99 56.54 68.25

To find the eigen values we solve using Excel the following equation:

( )det 0V Eλ− = .

This equation has several roots. We are interested in the largest eigen value. We get the largest eigen value = 311.16 The corresponding eigen vector is found to be V = [0.217 0.263 0.360 0.153] Since this is the normalized form we have the eigen vector as V = 0.217r1 + 0.263r2 + 0.360r3 + 0.153r4

Problem 8.4 Let ri, for i = 1,2,…,n, be independent samples of a return r of meanµ and variance 2σ . Define the estimates

.)ˆ(1

1

1ˆ

2

1

2

1

��

=

=

−−

=

=

n

ii

n

ii

rn

s

rn

µ

µ

Show that 22)( σ=sE . Solution.

n)expectatio theoflinearity (by the)ˆ()(1

1

ˆˆ21

1

)ˆˆ2(1

1

n)expectatio theof properties(by )ˆ(1

1

)ˆ(1

1)(

2

1

2

1

222

1

22

2

1

2

1

2

�� −

−=

�� +−

−=

�� +−

−=

��

−−

=

��

−−

=

=

=

=

=

=

µ

µµ

µµ

µ

µ

nErEn

nnrEn

rrEn

rEn

rn

EsE

n

ii

n

ii

n

iii

n

ii

n

ii

Since 22 )()()( YEYVYE += , then

.

)(1

1

)(1

1)(

2

22

22

1

22

222

σ

µσµσ

µσµσ

=

��

�� +−+−

=

��

�� +−+−

= �=

nnn

n

nn

nsE

n

i

q.e.d.

Problem 8.5

a) Show that )ˆ(rσ is independent of n

n

n

rr

n

n

22 σσ =

= and

22

ˆˆ

n

n

n

rnr

σσ =

=

)ˆvar(*)ˆvar()ˆvar( 2nn rnrnr == ……………………….(1)

since n

r nn

2

)var(σ

= and nn

22ˆ

σσ =

2

22

*)var(

nnnrn

σσ ==

From (1), 22

22 *)var( σσ ==

nnrn

It shows that )ˆ(rσ is independent of n b) Show that how )ˆ( 2σσ depends on “n”

)2...(....................).........ˆvar(*)ˆ*var()ˆvar( 2222nn nn σσσ ==

Assume that the return are normally distributed,

)1(*

2

1

2)var(

2

442

−=

−=

nnnn

n

σσσ

Therefore, 1

2)ˆvar(

42

−=

n

σσ

Then, )1(

2)ˆ(

22

−=

n

σσσ

More data does not help to estimate the mean more precisely but it improves the estimation of the volatility.

Problem 8.6 A record of annual percentage rates of return of the stock S is shown in the following table. Record of Rates of Return: Month Percent rateof return Month Percent rateof return 1 1.0 13 4.2 2 0.5 14 4.5 3 4.2 15 -2.5 4 -2.7 16 2.1 5 -2.0 17 -1.7 6 3.5 18 3.7 7 -3.1 19 3.2 8 4.1 20 -2.4 9 1.7 21 2.7 10 0.1 22 2.9 11 -2.4 23 -1.9 12 3.2 24 1.1 a) Estimate the arithmetic mean rate of return, expressed in percent per year. b) Estimate the ariithmetic standard deviation of these returns, again as percent per year. c) Estimate the accuracy of the estimates found in parts (a) and (b). d) How do you think the answers to © would change is you had 2 years of weekly data instead of monthly data? (See exercise 5.)

n = number of periods = 24

Period i (month) ri

1 1.0 4.93038E-32 2 0.5 0.25 3 4.2 10.24 4 -2.7 13.69 5 -2.0 9 6 3.5 6.25 7 -3.1 16.81

( )2rri −

8 4.1 9.61 9 1.7 0.49

10 0.1 0.81 11 -2.4 11.56 12 3.2 4.84 13 4.2 10.24 14 4.5 12.25 15 -2.5 12.25 16 2.1 1.21 17 -1.7 7.29 18 3.7 7.29 19 3.2 4.84 20 -2.4 11.56 21 2.7 2.89 22 2.9 3.61 23 -1.9 8.41 24 1.1 0.01

a)

1.0 165.4

12.0%

b)

7.191304 sy = 9.289545%

c) The accuracy of the estimators is determined by taking the following standard deviations: Accuracy of the mean estimator

6.5687% where n = 2 since we are dealing with yearly standard deviations Accuracy of the variance estimator

2.1206= 0.021206%

25.4472=0.254472% where n = 24 and σ is the monthly σ. Since we are quaring the standard deviation with units as percent, then we need to multiply by 100 to get the answer as a percent and not percent squared

�=

==n

iirn

r1

1ˆ � =

=

−=24

1

2)ˆ(n

ii rr

=−−

= � =

=

24

1

22 )ˆ(1

1 n

ii rr

ns

== rryˆ12ˆ

==nr

σσ ˆ

=−

=1

2)(

22

nsstdev monthly

σ

== monthlyyearly sstdevsstdev )(*12)( 22

d) Exercise 8-5 proves that the accuracy of the mean estimator is independent of the number of periods, n. As a result, the accuracy of the mean estimator would not change by using 2 years of weekly data instead of 2 years of monthly data. The accuracy of the standard deviation estimator however, is dependent on the number of periods. The accuracy would increase with more periods to estimate the standard deviation.

Problem 8.7 Gavin Jones figured out a clever way to get 24 samples of monthly returns in just over one year instead of only samples; he takes overlapping samples; that is, the first sample covers Jan. 1 to Feb. 1 and the second sample covers Jan. 15 to Feb. 15, and so forth. He

figures that error in his estimate of r , the mean monthly return, will be reduces by this method. Analyze Gavin’s idea. How does the variance of his estimate compare with that of the usual method of using 12 nonoverlapping monthly returns? Solution �

===

n

ii rr

nErHeadE

1

)1

()(

Monthly rate of return=r Half monthly rate of return =r/2

22

22

24/12/1)(

12/1)(

24/12/)(

12/1)(

yi

yi

yi

yi

V

rV

rrE

rrrE

σσρ

σσ

ρ

==

==

==

==

From the hint: 1++= iiir ρρ

rrEEEE

rrE

rErErrErrCov

iiiiiii

iiii

iiiiii

*)])([(][][][

*)])([(

)()()()(

212

121

211

11!,

−+++=

−++=−=

+++++

+++

+++

ρρρρρρρρρρρ

From the hint:

iρ are uncorrelated- � independent. 222

12 )2/(2/1)()()( rEVE ii +=+= σρρρ

From the formula above

2/

2/4/4/*3

2/*2/)2/(2/2/*2/2/*2/

][][][][][][][

2

2222

222

221

2121

σσ

σρρρρρρρ

=

−++=−++++=

−+++= +++++

rrr

rrrrrrrr

rEEEEEEE iiiiiii

From the page 215 and the hint:

2

222

2

24

12

24

1

12

1

)]2/(24*224[24

1

)]2423cov(2...)32cov(2)2,1cov(2)24(....)2()1([24

1

)(24

1)

24

1()(

σ

σσ

=

+=

+++++++++=

== ��==

rrrrrrrVrVrV

rVrVrHeadVi

ii

i

T

he result is equal to page 215 (8.11)

Problem 8.8 We use the general model with p=Pr+e where P is an mxn matrix, r is an n-dimensional vector, and P and e are m-dimensional vectors. The vector p is a set of observation values and e is a vector of errors having mean 0. The error vector has covariance matrix Q. Then the estimate r’ of r is

r’=inv(tr(P)inv(Q)P)tr(P)inv(Q)p

where inv is the inverse of a matrix and tr is the transpose of a matrix. a) If there is a single asset an just one measurement of the form p=r+e, we must show that r’=p. Solution: Suppose that p=r+e, then P and Q are scalars with tr(P)=P=1. Thus

r’=inv(Pinv(Q)P)Pinv(Q)p=Qinv(Q)p=p.

b) Suppose we have two uncorrelated measurements with values p1 and p2 having variances � 1 and � 2 respectively. We need to show that

r’=[1/(1/ � 1^2 +1/ � 2^2)](p1/ � 1^2 + p2/ � 2^2).

Solution: Here we have P=tr(1 1) p=tr(p1 p2) r=(r) and Q is the matrix with entries Q11= � 1^2, Q12=0, Q21=0, Q22= � 2^2 with the inverse of Q, inv(Q), given by the entries

inv(Q)11=1/ � 1^2 inv(Q)12=0 inv(Q)21=0

inv(Q)22=1/ � 2^2 Putting these into the above formula we get

r’=[1/(1/ � 1^2 +1/ � 2^2)](p1/ � 1^2 + p2/ � 2^2).

c) We consider example 8.5. There are measurements of the form

r1=p1+e1 r2=p2+e2 r3=p3+e3 r4=p4+e4

r1=rf+�1*fM

r2=rf+�2*fM

r3=rf+�3*fM

r4=rf+�4*fM

where the ei’s are uncorrelated and where cov(ei,fM)=0.25 � i^2. Assuming the

�i’s are

known exactly, find the best estimates of the ri’s. Solution: First we note that the inverse of a 2x2 matrix A is given with entries A11=a, A12=b, A21=c, A22=d.

Inv(A)=[1/(ad-bc)]M

Where M is the matrix with entries M11=d, M12=-b, M21=-c, and M22=a. For part (c) we use the above formula stated at the beginning with

P=tr(1 1).

We also use the formula

r_ei=r_f + �i(r_M-r_f)

where r_ei and r_M are the expected values of the return for security i (using the equilibrium model) and the market respectively, r_f is the risk free rate, and

� _ei= �i � _M/( � 10)

for the variance of the error in the measurement of r_ei. We have as well that

cov(e_i, f_M)=0.25 � _i^2

where the errors e_i (for the measurements r_h) are uncorrelated. For stock 1 we have r_h=15.00 and r_e=13.05 as computed in example 8.5. So

p=tr(15.00 13.05)

and Q is the matrix with entries Q11=3.00^2, Q12=Q21=0.25(3.00)^2, and Q22=2.42^2. Inserting these into the above formula we get

r’=13.72%.

Note that the first entry for Q is � /( � 10)

For stock 2 we have r_h=14.34% and r_e=13.99% using �2=1.02 and we also compute

p=tr(14.34 13.99)

and Q is the matrix with entries Q11=3.275^2, Q12=Q21=0.25(3.275)^2, and Q22=2.74^2 where

2.74=1.02 � _M/( � 10).

Inserting these into the above formula we get

r’=14.12%.


p=tr(10.90 17.03)


3.76=1.40 � _M/( � 10).


r’=14.25%.


p=tr(15.09 11.27)


1.826=0.68 � _M/( � 10).


r’=12.19%.

Problem 1 An investor has a utility function U(x) = x^1/4 for salary. He has a new job offer which pays $80,000 with a bonus. The bonus will be $0, $10,000, $20,000, $30,000, $40,000, $50,000, or $60,000, each with equal probability. What is the certainty equivalent of this job offer? Certainty Equivalent U(x) = x^1/4 Salary = $80,000 with bonus Bonus = $0, $10,000, $20,000, $30,000, $40,000, $50,000, $60,000 p of each = 1/7 Find the certainty equivalent Salary = $80,000, $90,000, $100,000, $110,000, $120,000, $130,000, $140,000 E(U(x)) = 1/7 ($16.82 + $17.32 + $17.78 + $18.21 + $18.61 + $18.99 + $19.34) E(U(x)) = 1/7($127.077)

E(U(x)) = $18.154 Need to find A value C such that U(C) = $18.15 Cx^1/4 = $18.15 C = $18.15^4 C = $108,610.04 The value C above is the Certainty Equivalent.

Problem 2 Suppose an investor has exponential utility function U(x) = -e-ax and an initial wealth level of W. The investor is faced with an opportunity to invest an amount w

� W and obtain a random payoff x. Show that his evaluation of this incremental

investment is independent of W. To evaluate the incremental investment, we will compare the investment versus not making the investment.

[ ])( xwWUE +− is the expected value of the utility function if the investment is

made, and [ ])(WUE is the expected value of the utility function of wealth if the investment is not made.

[ ] [ ]( )[ ] [ ]

( )[ ] [ ][ ] ( )[ ] [ ]

( )[ ] 1

)()(

<−>⋅−

−>−

−>−>+−

−−

−−−−

−−−−

−+−−

wxa

aWwxaaW

aWwxaaW

aWxwWa

eE

eEeEeE

eEeeE

eEeE

WUExwWUE

The initial wealth, W, is no longer part of the equation; only the investment w and the payoff x are. This shows that the evaluation is independent of W.

Problem 3 3) Suppose U(x) is utility function with Arrow-Pratt risk aversion coefficient a(x)

Let V(x) = C + bU(x). What is the risk aversion coefficient of V? a(x) = -U”(x)/U’(x) V’(x) = bU’(x) V”(x) = bU”(x) a(V) = bU”(x)/bU’(x) a(V) = U”(x)/U’(x) Therefore the risk aversion coefficient = -a(x)

Problem 4 The Arrow-Pratt relative risk aversion coefficient is

� (x) = x * U”(x) / U’(x) Show that the risk aversion coefficients are constant for U(x)1 = ln x and U(x)2 = � * x� U’(x)1 = 1 / x U”(x)1 = -1 / x2 So, � (x)1 = x * (-1 / x2) / (1 / x) = -(1 / x) / (1 / x) = -1 = constant U’(x)2 = � 2 * x( � -1) U”(x)2 = ( � 3 – � 2) * x( � -2) = � 2 ( � -1)* x( � -2) So, � (x)2 = � -1= constant

Problem 5 A young woman uses the first procedure described in Section 9.4 to deduce her utility function U(x) over the range A<=x<=B. She uses the normalization U(A)=A, U(B)=B. To check her result, she repeats the whole procedure over the range A’<=x<=B’, where A<A’<B’<B. The result is a utility function V(x), with V(A’)=A’, V(B’)=B’. If the results are consistent, U and V should be equivalent; that is, V(x)=aU(x)+b for some a>0 and b. Find A and B. Given: U(A)=A, U(B)=B from A to B V(A’)=A’, U(B’)=B’ from A’ to B’ V(x)=aU(x)+b a>0 Find: a and b Solution:

( )

)'()'(

)'(')'('

)'()'()'(

)'(')'(

)'()'(

)'('

)'(

)'()'(

)'(

)'(1

)'(

)'(''

)'(

)'(

)'(

)'(')'(

)'(

''

)'(

'

)'(')'(

)'(')'(

BUAU

BUAAUBb

BUAUAU

BUAAU

BUAU

AUBb

AU

BUAUb

AU

BUb

AU

BUAB

bAU

BbU

AU

BUAbBU

AU

bAB

AU

bAa

bBaUBBV

bAaUAAV

−−=

−−

−=

−=−=−

+−=+−=

−=

+==+==

( )

)'()'(

''

)'()'(''

)'(')'('

)'('

)'(')'(

)'(')'(

BUAU

BAa

BUAUaBA

BaUBAaUA

BaUBb

bBaUBBV

bAaUAAV

−−=

−=−−+=

−=+==+==

Problem 6 The HARA ( for hyperbolic absolute risk aversion) class of utility functions is defined by

γ

γγγ ��

�� +−

−= bax

xU1

1)( , b>0.

Show how the parametersγ , a and b can b chosen to obtain the following special cases (or an equivalent from). (a) Linear or risk neutral: U(x) =x Let λ =1 so we have ax=x then a=1, b=0.

(b) Quadratic: 2

2

1)( cxxxU −=

Let λ =2 so we will have 0.5ax^2+0.5b^2+abx/(1-b) So a=b=1 (c) Exponential: axexU −−=)( λ =- ∞

)1

lim( bax

x+

−∞→ γ= x

x

x1

)1(lim +∞→

xbax =��

�� +−

−� γ

γγγ

1

1

then b=1 a=1 (d) Power: γcxxU =)(

γ

γγγ �

�� +−

−= bax

xU1

1)(

γcxxU =)( then a=1 b=1 γγ = (e) Logarithmic: xxU ln)( =

γ

γγγ �

�� +−

−= bax

xU1

1)( =

( )( )( ) ( ) γγ

γλ

γ γγγ

γγγ

γγγ

xbaxbax−=−+−=−+

−− − 1

))1(()1(

11

1*

1 1

( ) ( ) 1111

11)( −−−

−

−=−=∂

∂ γγγγ

γγγγ

xxx

xU

xxthenUba ln)(0,1,0 ====γ The Arrow-Pratt risk aversion coefficient is,

a

bxbx

aa

bax

a

bax

a

xU

xUxa

bax

axU

bax

axU

+−

=+

−

=�� +

−

�� +

−=

′′′

−=

�� +

−−=′′

�� +

−=′

−

−

−

−

γγγ

γ

γ

γ

γ

γ

γ

γ

1

11

11

1

)()(

)(

1)(

1)(

1

2

2

2

2

1

Problem 7 (The venture capitalist) A venture capitalist with a utility function U(x)=Sqrt(x) carried out the procedure of Example 9.3. Find an analytical expression for C as a function of e, and for e as a function of C. Do the values in Table 9.1 of the example agree with these expressions? Venture capitalist U(x)=Sqrt(x) Lottery outcomes, either $1M or $9M p(receiving $1M) varies For p(either two outcomes)=0.5, E[U(x)]=$5M. Certainty equivalent, C=$4M U(C)=e p 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 e 9 8.2 7.4 6.6 5.8 5 4.2 3.4 2.6 1.8 1 C 9 7.84 6.76 5.76 4.84 4 3.24 2.56 1.96 1.44 1 Analytical expression for C as a function of e is found: For this problem, we can solve U(C) for C. From this, we will have an equation with an unknown, variable probability p. Since we know the probabilities in the table, we can calculate the expected values in terms of probability and vice versa. Substituting probabilities as a function of expected values into the certainty equivalent equation yields the desired certainty equivalent as a function of expected values. To find the second part of the problem, we simply solve the equation from the first part for expected values to get expected value as a function of certainty equivalent. Finally, we compare these two equations to the table by substituting in values from the table for C and e to determine that the equations remain true statements. If they were not true, the table and equations would not agree. U(C)=Sqrt(C)=e=E[U(x)] For an undetermined probability value for the $1M lottery outcome, U(C)=p*U(1)+(1-p)*U(9) From the table� e=p*1+(1-p)*9 e=p+9-9*p=9-8*p Since we are looking for an expression for C as a function of e, we can solve the last equation for p and substitute this into the equation for C to find the answer. e=9-8*p e-9=-8*p

p=(9-e)/8 an equation for C � U(C)=p*U(1)+(1-p)*U(9)=Sqrt(C) C=[p*U(1)+(1-p)*U(9)]^2 U(1)=Sqrt(1)=1 U(9)=Sqrt(9)=3 C=[p*1+(1-p)*3]^2=[p+3*(1-p)]^2=(p+3-3*p)^2=(3-2*p)^2=4*p^2-12*p+9 Substituting for p � C=4*[(9-e)/8]^2-12* [(9-e)/8]+9 =4* [(9-e)^2]/64-12*(9-e)/8+9 =(4/64)* [e 2-18*e+81]-(108+12*e)/8+9 =(1/16)*(e 2-18*e+81)-(216/16)+(24*e)/16+(144/16) =(1/16)* e 2-(18/16)*e+(81/16)-(216/16)+(24*e)/16+(144/16) =(1/16)* e 2+(6/16)*e+(9/16)=(1/16)*(e 2+6*e+9)=(1/16)*(e+3)^2 So, C(e)=[(e+3)^2]/16 Analytical expression for e as a function of C is found: U(C)=Sqrt(C)=e=E[U(x)] U(C)=p*U(1)+(1-p)*U(9)=Sqrt(C) U(1)=Sqrt(1)=1 U(9)=Sqrt(9)=3 p=(9-e)/8 U(C)=Sqrt(C)=(9-e)/8+[1-(9-e)/8]*3=(9-e)/8+3-[3*(9-e)]/8 =(9-e)/8+3-(27/8)+3*e/8=(9-e+24-27+3*e)/8 =(2*e+6)/8=(e+3)/4 Sqrt(C)=(e+3)/4 4*Sqrt(C)-3=e So, e(C)=4*Sqrt(C)-3 Do the values of Table 9.1 in Example 9.3 agree with these expressions? Substituting values from table into these equations� C(6.6)=[(6.6+3)^2]/16=5.76, the same value as the table e(5.76)=4*Sqrt(5.76)-3=6.6, the same value as the table From these observations, the values in the table agree with these equations.

Problem 8 There is a useful approximation to certainty equivalent that is easy to derive. A second-

order expansion near _

x=E(x) gives

)()(''2

1)()]([

))((''2

1))((')()(

__

2_____

xVarxUxUxUE

XxxUXxxUxUxU

+≈

−+−+≈

On the other hand, if we let c donate the certainty equivalent and assume it is close to _

x , we can use the first-order expansion

))((')()(___

XCXUXUCU −+=

Using these approximations, show that

)('

)()(_

__

XU

XUCUXC

−+=

As the definition

)('

)()("

2

1

)('

)()]()("2

1)([

)()("2

1)()(

)]([)(

_

__

_

___

_

__

XU

XVarXUXC

XU

XUXVarXUXUXC

XVarXUXUCU

XUECU

+=�

−++=�

+≈

=

Problem 9 An investor with unit wealth maximizes the expected value of the utility function U(x) = ax – bx2/2 and obtains a mean-variance efficient portfolio. A friend of his with wealth W and the same utility function does the same calculation, but gets a different portfolio return. However, changing b to b’ does yield the same result. What is the value of b’? In general; E[U(x)] = E[ax – 1/2bx2] = aE[x] –1/2bE[x2] = aE[x] – 1/2b(var[x] + E[x] 2) In this situation, if the random payoff of the portfolio of the investor with unit wealth is R, it would maximize: E[U(R)] = aE[R] – 1/2b(var[R] + E[R] 2) Similarly, if the investor with wealth W purchases the same portfolio, the payoff will be WR and R should maximize: E[U(x2)] = aE[RW] – 1/2b2(var[RW] + E[RW] 2) = aWE[R] – 1/2b2(W

2var[R] + W2E[R] 2) = W[aE[R] – 1/2b2W(var[R] + E[R] 2)] If b2 = b’= b/W is substituted in the final equation for the second investor, the same R will solve the expected value of the utility function as the R using unit wealth.

Problem 10 Suppose an investor has utility function U. There are n risky assets with rate of return r i=1, 2, …, n, and one risk-free asset with rate of return rf. The investor has initial wealth W0. Suppose that the optimal portfolio for this investor has random) payoff x* . Show that

E[U�(x* )(r i-rf)]=0

for i = 1, 2, …, n. 1. From (9.4) p. 243 we know that E[U

�(x* )di ]=

�Pi. If there is a risk-free asset with rate

of return R, then di = R and Pi = 1. Thus, �= E[U

�(x* )] R =>

�= E[U

�(x* )] (1+ rf )

2. If there is a asset i with total return Ri, then di = Ri and Pi = 1. Thus, �= E[U

�(x* ) Ri] =>

�= E[U

�(x* )(1+ r i )]

=> �= E[U

�(x* )(1+ r i )] = E[U

�(x* )] (1+ rf )

E[U�(x* )(1+ r i)] - E[U

�(x* )(1+ rf)] = 0

E[U�(x* )(1+ r i )-U

�(x* )(1+ rf)]=0

E[U�(x* )( r i - rf)]=0

Problem 11

15006.0

2502.0

1250

2.0

2.16.0

2.18.0

300062.138.1

1

2.1

3.0

2.1

4.0

300062.13

8.1

2.1

36.0

2.1

48.0

300062.13

36.0

2.1

4.0

300062.13

9.0

4

3

2

1

4321

42

421

321

4321

42421

321

42421321

421321

+=−=

−==

=++++=

++=+++=

=

=+++

=+

+++

=+++

=+

+++

++++

=++

++++

W

W

W

thus

W

W

W

WW

W

θθθθ

θθθθθθ

θθθθθθ

λ

θθθθ

λθθθθθ

λθθθ

λθθθθθθθθ

λθθθθθθ

The price of money back guaranteed investment P = $1,500

Problem 12

Formulation: The following is a general result from matrix theory: Let A be mxn matrix. Suppose that the equation pAx = can achieve no 0≥p except 0=p . Then there is a vector 0>y with 0=yAT . Use this result to show that if there is no arbitrage, there are positive state prices; that is, prove the positive state price theorem in Section 9.9. [Hint: If there are S states and N securities, let A be an appropriate ( )xNS 1+ matrix] Solution:

Let construct a matrix ��

�

�

��

�

�

−−−−

=

N

SNSSS

N

N

pppp

dddd

dddd

dddd

A

........

........

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

........

321

321

2232221

1131211

. Here ijd is

dividend of the security j in the state i . Let take vector ( )Nθθθθθ .......321= as vector of weights of the securities in the portfolio.

T

P

D

D

D

pppp

dddd

dddd

dddd

A

NNN

SNSSS

N

N

=��

�

�

�

−

=��

�

�

�

��

�

�

�

−−−−

=..........

........

........

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

........

3

2

1

3

2

1

321

321

2232221

1131211

θ

θθθ

θ

(*) Here iD is dividend in state i and NP is price of the portfolio. Now lets assume that 0=NP or 0<NP and there is some k s.t. 0>kD . In this case we have arbitrage because with 0 or negative price there is a possibility to get dividend in one of the states. In order to avoid the arbitrage we need to conclude that if 0≤NP then for all i 0=iD . It means that system (*) can achieve with 0=T . According to the algebra we have that 0>∃ iy s.t.

0

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

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

.......

.......

1

3

2

1

21

222212

112111

=��

�

��

�

�

��

�

��

�

�

−

−−

=

+SNSNNN

S

S

T

y

y

y

y

pddd

pddd

pddd

yA (**)

Because 01 ≠+Sy we can divide all iy by 1+Sy and define them as state price. Expression (**) will looks like:

0

1

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

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

.......

.......

3

2

1

21

222212

112111

=��

�

�

��

�

�

��

�

�

��

�

�

−

−−

= ψψψ

ψ

NSNNN

S

S

T

pddd

pddd

pddd

A where 0>iψ

By solving this we have for each state that �=

=S

jjjii dp

1

ψ which means that for

each state we constructed a positive state price iψ s.t. �=

=S

jjjii dp

1

ψ .

Problem 13 From the above exercise, we have ( ) ( )[ ] 0' * =− xUErrk , where r is the risk free rate. In

the4 quadratic case, we have U’(x)=1-cx. We denote by the MR the return of the

portfolio, and using the fact that initial capital is W we get ( ) ( )[ ] 0' =− Mk WRUErr ,

equivalently ( )( )[ ] 01 =−− RRcWRE iM , so

( )[ ] ( )[ ])(,cov RRRRRcWRRRREcWRR iMiMMiMi −+=−=− , so,

( )[ ]iMiMi RRcWRRRcWRR ,cov)( =−−− , and equivalently ( )iMi RRCovRR ,γ=− ,

where MRCW−= 1γ . If we apply this relation to the portfolio, we obtain

( ) )(, MMMM RVarRRCovRR γγ ==− , so ( ) ( )RRRR

RVar

RRCovRR MiM

M

iMi −=−=− β)(

)(

,,

and so, the problem is solved.

Problem 9.14. (At the track) At the horse race one Saturday afternoon Gavin Jones studies the racing form and concludes that the horse No Arbitrage has a 25% chance to win and is posted at 4 to 1 odds. (For every dollar Gavin bets, he receives $5 if the horse wins and nothing if it loses.) He can either bet on this horse or keep money in his pocket Gavin decides that he has a square-root utility for money.

(a) What fraction of his money should Gavin bet on No Arbitrage? (b) What is the implied winning payoff of a $1 bet against No Arbitrage?

Solution:

(a) If we denote by α fraction of his money m G. J. should bet we need

1 3max [ ] 4 (1 )

4 4E U m m mα α= + + −

So we need the 1st derivative to be equal to zero, that is, 1 3

02 84 (1 )

dE m m

d m m mα α α= − =

+ −

Solving this equation we obtain 7

0.134652

α = =

(b) Implied wining payoff of a $1 bet against No Arbitrage is 5

1.254

=

Problem 15 (General risk-neutral pricing) We can transform the log-optimal pricing formula into a risk-neutral pricing equation. From the log-optimal pricing equation we have

( )*

dP E

R=

Where R* is the return on the log-optimal portfolio. We can then define a new expectation operation �E by

�

*( ) ( )

RxE x E

R= .

This can be regarded as the expectation of an artificial probability. Note that the usual rules of expectation hold. Namely:

a) If x is certain, then �

( )E x x= . This is because *

1 1( )E

RR= .

b) For any random variables x and y, there holds � � �

( ) ( ) ( )E ax by aE x bE y+ = + .

c) For any nonnegative random variable x, there holds �

( ) 0E x ≥ . Using this new expectation operation, with the implied artificial probabilities, show that the price of any security d is

�( )E d

PR

= .

This is risk neutral pricing.

From the rules b), we know � �1

* ( ) ( )d

E d ER R

= (1)

According to the definition of the operation �E , for any variable x,

�*

( ) ( )Rx

E x ER

= , assume

d/R as a variable, we can get

* *

*( ) ( ) ( )

dRd dRE E E

R R R= = (2)

From the log-optimal pricing equation we have: ( )*

dP E

R= (3)

So,

( ) (3)*

*( ) ( ) (2)

*1

* ( ) (1)

dP E By

Rd

R dRE E ByR R

E d ByR

=

= =

=

♦

111

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