book solutions

David Ruppert

Statistics and Finance: An

Introduction

Solutions Manual

July 9, 2004

Springer

Berlin Heidelberg NewYork

Hong Kong London

Milan Paris Tokyo

2

Probability and Statistical Models

1. (a)

E(0.1X + 0.9Y ) = 1.

Var(0.1X + 0.9Y ) = (0.12)(2) + 2(0.1)(0.9)(1) + (0.92)(3) = 2.63.

(b)

Var{wX + (1 − w)Y } = 3w2 − 4w + 3.

The derivative of this expression is 6w − 4. Setting this derivativeequal to 0 gives us w = 2/3. The second derivative is positive so thesolution must be a minimum. In this problem, assets X and Y havesame expected return. This means that regardless of the choice ofw, that is, the asset allocation, the expected return of the portfoliodoesn’t change. So by minimizing the variance, we can reduce the riskwithout reducing the return. Thus the ratio w = 2/3 corresponds tothe optimal portfolio

2. (a) Use (2.54) with w1 = (1 1)T and w2 = (1 1)T.(b) Use part (a) and the facts that Cov(α1X,α2X) = α1α2σ

2X , Cov(α1X,Z) =

0, Cov(Y, α2X) = 0, and Cov(Y,Z) = 0.(c) Using (2.54) with w1 and w2 the appropriately-sized vectors of ones,

it can be shown that

Cov

(n1∑

i=1

Xi,

n2∑

i′=1

Yi′

)=

n1∑

i=1

n2∑

i′=1

Cov(Xi, Yi′).

3. The likelihood is

L(σ2) =

n∏

i=1

1√2πσ2

e−1

2σ2(Yi−µ)2 .

Therefore the log-likelihood is

2 2 Probability and Statistical Models

log L(σ2) = − 1

2σ2

n∑

i=1

(Yi − µ)2 − n log(σ2)/2 + H

where H consists of all terms that do not depend on σ. Differentiating thelog-likelihood with respect to σ2 and setting the derivative1 with respectto σ2 equal to zero we get

1

2(σ2)2

n∑

i=1

(Yi − µ)2 − n

2σ2= 0

whose solution is

σ2 =1

n

n∑

i=1

(Yi − µ)2.

4. Rearranging the first equation, we get

β0 = E(Y ) − β1E(X). (2.1)

Substituting this into the second equation and rearranging, we get

E(XY ) − E(X)E(Y ) = β1{E(X2) − E(X)2}.

Then usingσXY = E(XY ) − E(X)E(Y )

andσ2

X = E(X2) − E(X)2

we get

β1 =σXY

σ2X

,

and substituting this into (2.1) we get

β0 = E(Y ) − σXY

σ2X

E(X).

5.

E(wTX) = E

(N∑

i=1

wiXi

)=

N∑

i=1

wiE(Xi) = wT{E(X)}.

Next

Var(wTX) = E

{w

TX − E(wT

X)}2

= E

[N∑

i=1

wi{Xi − E(Xi)}]2

1 The solution to this problem is algebraically simpler if we treat σ2 rather than σ

as the parameter.

2 Probability and Statistical Models 3

=

N∑

i=1

N∑

j=1

E [wiwj{Xi − E(Xi)}{Xj − E(Xj)}] =

N∑

i=1

N∑

j=1

wiwjCov(Xi,Xj).

One can easily check that for any N × 1 vector X and N × N matrix A

XTAX =

N∑

i=1

N∑

j=1

XiXjAij ,

whence

wTCOV(X)w =

N∑

i=1

N∑

j=1

wiwjCov(Xi,Xj).

6. Since

log{L(µ, σ2)} = −n

2log(2π) − n

2log(σ2) − 1

2σ2

n∑

i=1

(Yi − µ)2

and ∑ni=1(Yi − Y )2

σ2ML

= n,

it follows that

log{L(Y , σ2ML)} = −n

2{1 + log(2π) + log(σ2

ML}.

Next, the solution to

0 =∂

∂σ2

[log{L(0, σ2)}

]=

∂

∂σ2

{−n

2log(2π) − n

2log(σ2) − 1

2σ2

n∑

i=1

Y 2i

}

= − n

2σ2+

1

2(σ2)2

n∑

i=1

Y 2i ,

solves nσ2 =∑n

i=1 Y 2i so that

σ20,ML =

1

n

n∑

i=1

Y 2i .

7. (a)

E{X − E(X)} = E(X) − E{E(X)} = E(X) − E(X) = 0.

(b) By independence of X − E(X) and Y − E(Y ) we have

E [{X − E(X)}{Y − E(Y )}] = E{X−E(X)}E{Y −E(Y )} = 0·0 = 0.


8. (a) Since

Y = E(Y ) +σXY

σ2X

{X − E(X)}.

and E{X − E(X)} = 0 by Problem 7. it follows that E(Y ) = E(Y )

so that E{Y − Y } = 0 − 0 = 0.(b)

E(Y − Y )2 = E

({Y − E(Y )}2 +

σ2XY

σ4X

{X − E(X)}2

−2σXY

σ2X

E[{(Y − E(Y )}{X − E(X)}

])

= σ2Y +

σ2XY

σ2X

− 2σ2

XY

σ2X

= σ2Y

{1 − σ2

XY

σ2Xσ2

Y

}= σ2

Y

(1 − ρ2

XY

).

9.

E(XY ) = E(X3) =

∫ a

−a

x3

2adx =

x4

8a

∣∣∣∣a

−a

= 0

and E(X) = 0 so that σXY = E(XY )−E(X)E(Y ) = 0− 0 = 0. Since Yis determined by X one suspects that X and Y are not independent. Thiscan be proved by finding set A1 and A2 such that P{X1 ∈ A1 and Y ∈A2} 6= P{X ∈ A1}P{Y ∈ A2}. This is easy. For example,

1/2 = P{|X| > a/2 and Y > (a/2)2}

6= P{|X| > a/2}P{Y > (a/2)2} = (1/2)(1/2).

10. There is an error on page 55. The MAP estimator is 4/5, not 5/6. This canbe shown by finding the value of θ that maximizes f(θ|3) = 30 θ4(1 − θ).Thus, one solves

0 =d

d θ30 θ4(1 − θ) = 30 θ3(4 − 5θ)

whose solution is θ = 4/5.11. (a) Since the kurtosis of a N(µ, σ2) random variable is 3, E(X − µ)4 =

3σ4. Therefore, for a random variable X that is 95% N(0, 1) and 5%N(0, 10), we have

E(X4) = (0.95)(3)(14) + (0.05(3)(104) = 1502.9

andE(x2) = (0.95)(12) + (0.05)(102) = 5.9500.

Therefore, the kurtosis is

1502.9

5.952= 42.45.

2 Probability and Statistical Models 5

(b) One has that E(X4) = 3p + 3(1− p)σ4 and E(X2) = p + (1− p)σ2 sothat the kurtosis is

3{p + (1 − p)σ4}p2 + 2p(1 − p)σ2 + (1 − p)2σ4

.

(c) For any fixed value of p less than 1,

limσ→∞

3{p + (1 − p)σ4}p2 + 2p(1 − p)σ2 + (1 − p)2σ4

=3

1 − p.

Therefore, by letting σ get very large and p get close to 1, the kurtosiscan be made arbitrarily large. Suitable σ and p such that the kurtosisis greater than 10,000 can be found by fixing p such that 3/(1 − p) >10, 000 that then increasing σ until the kurtosis exceeds 10,000.

(d) There is an error in the second sentence of part (d). The sentenceshould be “Show that for any p0 < 1, no matter how close to 1, thereis a p > p0 and a σ, such that the normal mixture with these valuesof p and σ has a kurtosis at least M .” This result is similar to part(c). One can always find a p > p0 such that 3/(1 − p) > M and withthis value of p, the kurtosis will converge to a value greater than Mas σ increases to ∞.

12. The conditional CDF is

P (X ≤ x|X > c) =P (c < X ≤ x)

P (x > c)

=P (X ≤ x) − P (X ≤ c)

1 − P (X ≤ x)=

Φ(x/σ) − Φ(c/σ)

1 − Φ(c/σ).

The conditional PDF is

d

dx

{Φ(x/σ) − Φ(c/σ)

1 − Φ(c/σ)

}=

(d/dx)Φ(x/σ) − Φ(c/σ)

1 − Φ(c/σ)=

φ(x/σ)

σ{1 − Φ(c/σ)} .

If one substitutes x = 0.25, c = 0.25, and σ = 0.3113 into this formula,the PDF is 4.002. If one substitutes a = 1.1, x = 0.25, and c = 0.25 intothe Pareto PDF, that is, into ac2/xa+1, the result is 4.000. Thus, the twoPDFs are equal to two decimals.


L(θ) =

n∏

i=1

(θ−1e−Xi/θ

)

so the log-likelihood is

log{L(θ)} = −n log(θ) −∑n

i=1 Xi

θ.

Thus the MLE solves


0 =d

d θ

{−n log(θ) −

∑ni=1 Xi

θ

}=

−n

θ+

∑ni=1 Xi

θ2

whose solution is X.14. (a) The CDF is

P (Y ≤ y) = P (3X − 2 ≤ y) = P

(X ≤ y − 2

3

)=

y − 2

3

for 2 < y < 3. For y ≤ 2, the CDF is 0 and for y > 5 the CDF is 1.The PDF is 1/3 for 2 < y < 5 and 0 elsewhere. The median is solutionto 1/2 = (y − 2)/3, that is, 3.5.

(b) The CDF is√

y for 0 < y < 1, 0 for y ≤ 0, and 1 for y > 1. The PDF

is (1/2)y−1/2 for 0 < y < 1 and 0 elsewhere. The 1st quartile is 1/4and the third quartile is (3/4)2.

15. (a) The CDF is (x − 1)/4 for 1 < x < 5 so the 0.1-quantile solves 0.1 =(x − 1)/4 so that x = 1.4.

(b) The 0.1-quantile of X−1 solves 0.1 = P (X−1 ≤ x) = P (X ≥ x−1) =1 − (x−1 − 1)/4 or 0.9 = (x−1 − 1)/4. Thus, the 0.1-quantile of X−1

is 1/4.6.16. There is an inconsistency in the notation: σXY should be changed to sXY .

(a)

s2d =

1

n − 1

n∑

i=1

{(Xi,1 − X1) − (Xi,2 − X2)}2

= s21 + s2

2 −2

n − 1

n∑

i=1

{(Xi,1 − X1)(Xi,2 − X2)} = s21 + s2

2 − 2s1,2.

(b) The numerators of the t-statistics are both equal to X1 − X2 − ∆0.Since s1,2 = 0 and n1 = n2,

√1/n1 + 1/n2 spool =

√2/n

√(s2

1 + s22)/2 =√

(s21 + s2

2)/n. If s1,2 = 0, then sd/√

n =√

(s21 + s2

2)/n. Thus, the de-nominators of the two t-statistics are also equal.

3

Returns

1. (a)

R2 =P2 + D2

P1− 1 =

56.2

51− 1.

(b)

R4(3) =

(P4 + D4

P3

)(P3 + D3

P2

)(P2 + D2

P1

)− 1

=

(58.25

53

)(53.25

56

)(56.2

51

)− 1.

(c)

r3 = log

(P3 + D3

P2

)= log

(53.25

56

).

2. (a) N(0.3, 1.8).(b)

Φ

(2 − 0.3√

1.8

)= 0.897.

(c) Var(r2) = 0.6.(d) Given rt−2, rt(3) is distributed N{rt−2 + (2)(0.1), (2)(0.6)} so given

rt−2 = 0.8, rt(3) is N(1.0, 1.2).3. The expected value and standard deviation of the 20-year log-return are

(0.1)(20) = 2 and (0.2)(20) = 4, respectively. According to the geometricrandom walk model, the log returns are normally distributed, so the 0.05-quantile of the 20-year log return is 2−(1.654)(4) = 4.58 and therefore the0.05-quantile of the 20-year return is e−4.58 = 0.0103. The 0.95-quantile ofthe log return is 2 + (1.645)(4) = 8.56 and the 0.95-quantile of the returnis e8.56 = 5321.

8 3 Returns

4. (a) Two examples of measurable uncertainty are random sampling from apopulations and computer simulation. Of course, there are many more.In random sampling each possible sample has a known probability ofbeing chosen. In computer simulation, random variables have knowndistributions specified by the programmer.

(b) Again, there is an almost infinite choice of examples. One exampleis the uncertainty about whether in will rain tomorrow. In this case,meteorologists estimate the probably from past data and models ofthe atmosphere.

5. We have Xk = X0 exp(Z1 + · · · + Zk) where Z1, . . . , Zk are iid N(µ, σ2).Therefore, X2

k is lognormal, in particular, X2k = X2

0 exp(W ) where Wis N(2kµ, 4kσ2). [Note: It is assumed here that X0 is a fixed constant.If instead X0 is a random variable, then the expectations and quantilesbeing found here are conditional expectations and quantiles given X0.The unconditional expectations and quantiles cannot be found since thedistribution of X0 has not been specified.](a) It follows that E(X2

k) = X20 exp{2kµ + (4kσ2)/2} by the formula for

the mean of a lognormal distribution.(b) W has a N(kµ, kσ2) density which is

fW (w) =1√

2πkσ2exp

{− 1

2kσ2(w − kµ)2

}.

First note that Xk = g(W ) where g(w) = X0 exp(w) so that W =h(Xk) where h(y) = log(y/X0) is the inverse of g. Also, h′(y) = 1/y,so by the change of variables formula (2.10) the density of Xk is

fXk(x) =

1

x√

2πkσ2exp

(− 1

2kσ2[log(x) − {kµ + log(X0)}]2

).

(c) The third quartile of Xk is the solution x to

.75 = P{Xk ≤ x} = P{log(W ) ≤ log(x/X0)} = Φ

{log(x/X0) − kµ√

kσ

}.

Therefore,log(x/X0) − kµ√

kσ= Φ−1(.75) = 0.6745.

(In MATLAB, norminv(.75) = 0.6745.) Therefore,

x = X0 exp{√

kσ(0.6745) + kµ}.

[Another method of solution is to notice that since W is N(kµ, kσ2)the third quartile of W is kµ +

√kσ 0.6745 (see Section 2.8.2, subsec-

tion “normal quantiles”). Then since Xk = X0 exp(W ) is a monoton-ically increasing function g(w) = X0 exp(w) of W , the third quartileof Xk is x = g(kµ +

√kσ0.6745) = X0 exp{

√kσ(0.6745) + kµ}.]

3 Returns 9

6. Data before 1998 should not be used to test the Super Bowl theory, sincethese data were already used to formulate the hypothesis. To test thehypothesis one would need to collect data for a number of years past1998. Once this is done, there are a number of ways to test the SuperBowl theory. One way would be the independent samples t-test, the firstsample being the years when the a former NFL wins the Super Bowl andthe second sample being the years when a former AFL team wins. Thenull hypothesis would be H0: µ1 = µ and the alternative would be H1:µ1 > µ2 since the Super Bowl theory predicts that µ1 > µ2.Defining bull and bear markets can be tricky, as Malkiel discusses. Hementions the possibility that the market could be down for most of theyear but then recovers at the end of the year. Is this a bull or bear market?One possibility is to define a bull (bear) market to occur if the net returnfor the year is positive (negative).Let p1 and p2 be the probabilities of bull markets in years when a formerNFL, respectively, former AFL team wins. To test if a former NFL teamwinning predicts a bull market, we could test the null hypotheses H0:p1 = 1/2 versus H1: p1 > 1/2. Similarly to test whether a former AFLteam winning predicts a bear market we could test H0: p2 = 1/2 versusH1: p2 < 1/2.Another set of hypotheses to test is the null hypothesis H0: p1 = p2 versusthe alternative H1: p1 > p2. These could be easily tested by a likelihoodratio test.

4

Time Series Models

1. (a) The process is stationary since φ = 0.7 so that |φ| < 1. Strictly speak-ing, the process is only stationary if it started in the stationary distri-bution. If the process has been operating for a reasonably long time(say 10 time periods) we can assume that it has converged to the sta-tionary distribution. In the remaining parts of the problem we willassume that it is in the stationary distribution.

(b)

µ =5

1 − 0.7.

(c)

γ(0) =2

1 − 0.72.

(d)

γ(h) =2(0.7|h|)

1 − 0.72.

2. (a) Var(Y1) = γ(0) = σ2ε /(1 − φ2) = 2/(1 − (.3)2) = 2.198.

(b)

Cov(Y1, Y3) = γ(2) =σ2

ε (φ)2

1 − φ2=

2(.3)2

1 − (.3)2= .1978.

(c) Var{(Y1+Y3)/2} = (1/4)Var(Y1) +(1/4)Var(Y2) + 2(1/2)(1/2)Cov(Y1, Y3)= (1/2)(2.198) + (1/2)(.1978) = 1.1978. (Note: Var(Y1) =Var(Y3) be-cause the process is stationary.)

3.

yn+1 = 102 + 0.5(99 − 102) + 0.2(102 − 102) + 0.1(101 − 102) = 100.4.

yn+2 = 102 + 0.5(100.4 − 102) + 0.2(99 − 102) + 0.1(102 − 102) = 100.6.

12 4 Time Series Models

4. ARIMA(0,1,0) can be written as ∆yt = yt − yt−1 = εt so that

yt = yt−1 + εt = yt−2 + εt−1 + εt = · · · = y + 0 + ε1 + · · · + εt

which is the definition of a random walk.5. Without loss of generality, we can assume that µ = 0 since the covariance

are independent of µ. Since Yt = εt − θ1εt−1 − θ2εt−2,

γ(0) = Var(εt) + θ21Var(εt−1) + θ2

2Var(εt−2) = (1 + θ21 + θ2

2)σ2ε .

Similarly, γ(1) = (−θ1 + θ1θ2)σ2ε , γ(2) = −θ2σ

2ε , and γ(k) = 0 for k ≥ 3.

The autocorrelation function is ρ(0) = 1, ρ(1) = (−θ1+θ1θ2)/(1+θ21+θ2

2),ρ(2) = −θ2/(1 + θ2

1 + θ22), and ρ(k) = 0 for k ≥ 3.

6. (a) Again, assume that µ = 0. Then using the hint

γ(k) = Cov(yt, yt−k) = Cov(φ1yt−1 + φ2yt−2 + εt, yt−k)

= φ1Cov(yt−1, yt−k) + φ2Cov(yt−2, yt−k) = φ1γ(k − 1) + φ2γ(k − 2).

for any k > 0. (If k = 0, then there is a non-zero covariance betweenεt and yt−k, which adds another term to the equation.)

(b) Using the result in (a) with k = 1, we get ρ(1) = φ1ρ(0) + φ2ρ(−1) =φ1 + φ2ρ(1). Using this result with k = 2 we get ρ(2) = φ1ρ(1) +φ2ρ(0) = φ1ρ(1) + φ2.

(c)

(φ1

φ2

)=

(1 0.4

0.4 1

)−1(0.40.2

)=

(0.38100.0475

).

Thenρ(3) = (0.4)(0.3810) + (0.2)(0.0476) = 0.16192.

7. The covariance between εt−i and εt+h−j is σ2ε if j = i + h and is zero

otherwise. Therefore,

Cov

∞∑

i=0

εt−iφi,

∞∑

j=0

εt+h−jφj

=

∞∑

i=1

σ2ε φi φi+h =

σ2ε φ|h|

1 − φ2.

8.

∆wt = (wt0 + Yt0 + · · · + Yt−1 + Yt) − (wt0 + Yt0 + · · · + Yt−1) = Yt.

9. (a) The series in the bottom panel is such that its first difference is non-stationary and tend to wander. In fact, its first difference is the seriesin the middle panel. We can see that the the first difference spendslong periods where it is always positive and also long periods whereit is always negative. During a period when the first difference is pos-itive the series is constantly moving upward. Similarly, when the first

4 Time Series Models 13

difference is negative the series is constantly moving downward. Thisis why the series in the bottom panel shows momentum. In contrast,the first difference of the series in the middle panel is the series in thetop panel which is stationary with mean 0 and only short term corre-lation. The series in the top panel never stays positive or negative fora long period. Thus, the series in the middle panel does not move ina constant direction for long periods.

(b) The series in the bottom panel would not be a good model for a stockprice. Under such a model it is possible to predict the direction ofprice movement which would allow one to make nearly certain shortterm profits. Such market conditions would not last very long sincetraders would rush in to take advantage of this opportunity.

5

Portfolios

1. (a)

0.03 = (0.02)w + (0.05)(1 − w) = 0.05 − 0.03w ⇒ w = 2/3.

(b) We need to find w that solves

(√

5/100)2 = w2(√

6/100)2 + (√

11/100)2(1 − w)2

+(2)(√

6/100)(√

11/100)w(1 − w)

or15.3752w2 − 20.3752w + 6 = 0.

The solutions are 0.8835 and 0.4417. We see from the equation in part(a) that the expected return is decreasing in w so that the smaller w,that is, 0.4417, gives the higher expected return.

2. 2/7 in risk-free, 3.7 in C, and 2/7 in D.3. (a)

w =(85)(300)

(85)(300) + (35)(100).

1 − w =(35)(100)

(85)(300) + (35)(100).

(b)

wj =njPj∑N

k=1 nkPk

.

4. The equationRP = w1R1 + · · · + wNRN (5.1)

is true if RP is a net or gross return, but (5.1) not in general true if RP

is a log return. However, if all the net returns are small in absolute value,

16 5 Portfolios

then the log returns are approximately equal to the net returns and (5.1)will hold approximately.Let us go through an example first. Suppose that N = 3 and the initialportfolio has $500 in asset 1, $300 in asset 2, and $200 in asset 3, so theinitial price of the portfolio is $1000. Then the weights are w1 = 0.5,w2 = 0.3, and w3 = 0.2. (Note that the number of shares being heldof each asset and the price per share are irrelevant. For example, it isimmaterial whether asset 1 is $5/share and 100 shares are held, $10/shareand and 50 shares held, or the price per share and number of shares areany other values that multiply to $500.) Suppose the gross returns are 2,1, and 0.5. Then the price of the portfolio at the end of the holding periodis

500(2) + 300(1) + 200(0.5) = 1400

and the gross return on the portfolio is 1.4 = 1400/1000. Note that

1.4 = w1(2) + w2(1) + w3(0.5) = (0.5)(2) + (0.3)(1) + (0.2)(0.5).

so (5.1) holds for gross return. Since a net return is simply the gross returnminus 1, if (5.1) holds for gross returns then in holds for net returns, andvice versa. The log returns in this example are log(2) = 0.693, log(1) = 0,and log(0.5) = − log(2) = −0.693. Thus, the right hand side of (5.1) whenR1, . . . ,RN are log returns is

(0.5 − 0.2)(0.693) = 0.138

but the log return on the portfolio is log(1.4) = 0.336 so (5.1) does nothold for log returns. In this example, equation (5.1) is not even a goodapproximation because two of the three net returns have large absolutevalues.Now let us show that (5.1) holds in general for gross returns and hencefor net returns. Let P1, . . . , PN be the prices of assets 1 through N in theportfolio. (As in the example, Pj is the price per share of the jth assettimes the number of shares in the portfolio.) Let R1, . . . , RN be the netreturns on these assets. The jth weight is equal to the ratio of the priceof the jth asset in the portfolio to the total price of the portfolio which is

wj =Pj∑Ni=1 Pi

.

At the end of the holding period, the price of the jth asset in the portfoliohas changed from Pj to Pj(1+Rj), so that the gross return on the portfoliois

∑Nj=1 Pj(1 + Rj)∑N

i=1 Pi

=

N∑

j=1

(Pj∑Ni=1 Pi

)(1 + Rj) =

N∑

i=j

wj(1 + Rj),

which proves (5.1) for gross returns.

5 Portfolios 17

5. Use the formula

(a bc d

)−1

=1

ad − bc

(d −b−c a

).

Substituting the determinant is ab − cd = σ21σ2

2(1 − ρ12) and

(a bc d

)=

(σ2

1 −ρ12σ1σ2

−ρ12σ1σ2 σ22

)

into this formula and simplifying gives (5.34).

6

Regression

1. (a)

E(Yi|Xi = 1) = 3.

σYi|Xi=1 =√

0.6.

(b)

P (Yi ≤ 3|Xi = 1) = Φ

(3 − 3√

0.6

)=

1

2.


L(β0, β1, σ2) =

n∏

i=1

1√2πσ2

exp

{− 1

2σ2(Y − β0 − β1Xi)

}

= (2π)−n/2σ−1 exp

{− 1

2σ2

n∑

i=1

(Yi − β0 − β1Xi)2

}.

Therefore, L(β0, β1, σ2) is maximized over (β0, β1) by minimizing

n∑

i=1

(Yi − β0 − β1Xi)2,

so the least-squares estimator is the maximum likelihood estimator.3. First

β1 − E(β1) =

n∑

i=1

wiεi.

Since the εi are independent of the Xi by (6.2) and since we are condition-ing on the Xi we can treat the wi as fixed weights. Therefore, by (2.55)and the independence of the εi (by (6.1)) we have that

20 6 Regression

Var(β1) =

n∑

i=1

w2i Var(εi).

Since Var(εi) = σ2ε ,

Var(β1) = σ2ε

n∑

i=1

w2i = σ2

ε

∑ni=1(Xi − X)2

{∑ni=1(Xi − X)2

}2 =σ2

ε∑ni=1(Xi − X)2

.

4. (a) The Xi are 1 + (i − 1)/29, i = 1, . . . , 30. Therefore,∑n

i=1 Xi = 165,∑ni=1 X2

i = 1124,∑n

i=1 X3i = 8562.9, and

∑ni=1 X4

i = 69, 548. Itfollows that

Corr(Xi,X2i )

=1n

∑ni=1 X3

i − ( 1n

∑ni=1 Xi)(

1n

∑ni=1 X2

i ){

1n

∑ni=1 X2

i −(

1n

∑ni=1 Xi

)2}1/2 {1n

∑ni=1 X4

i −(

1n

∑ni=1 X2

i

)2}1/2

= 0.9770.

The VIFs are both 1/(1 − 0.9770) = 43.48.(b) Since

∑ni=1(Xi − X)3 = 0 and

∑ni=1(Xi − X) = 0,

Cov((Xi − X), (Xi − X)2)

=1

n

n∑

i=1

(Xi − X)3 −(

1

n

n∑

i=1

(Xi − X)

)(1

n

n∑

i=1

(Xi − X)2

)= 0.

so the correlation between (Xi −X) and (Xi −X)2 is 0. The VIFs areboth 1.

5. (a) R2 = Corr(Y, Y ) = 0.25.(b)

0.25 = R2 = 1 − residual error SS

total SS= 1 − residual error SS

100.

Therefore residual error SS = 75.(c) Regression SS = total SS − residual error SS = 100 − 75 = 25.(d)

s2 =residual error SS

n − p − 1=

75

30 − 4= 2.885.

6. For this problem, R2 = 1 − (residual error SS)/50. Also,

σ2ε,M =

10

66 − 5 − 1= 0.1667

Therefore, the values of R2 and Cp are as in the table below. Based solelyon this information, the model to choose would be the one with the small-est value of Cp which is the model with 4 predictors. However, the modelwith all five predictors has nearly the same value of Cp and might be used.The final decision should depend upon subject matter knowledge.

6 Regression 21

Number Residual R2 Cp

of predictors error SS3 12.0 0.7600 14.04 10.2 0.7960 5.25 10.0 0.8000 6.0

7. No, we cannot accept that both β1 and β2 are zero. It is usually the casethat X and X2 are highly correlated. Therefore, either one can serve as aproxy for the other, which means that it is not necessary for both to be inthe model. The p-value for β1 tests whether X can be dropped given that

X2 is in the model, and similarly the p-value for β2 tests whether X2 canbe dropped given that X is in the model. We can conclude that either Xor X2 can be dropped, that is, that either β1 or β2 is zero. However, wecannot conclude that both X and X2 can be dropped, that is, that bothβ1 and β2 are zero. To conclude that both β1 and β2 are zero we could fitthe model Yi = β0 + β1Xi + εi and test if β1 is zero.

8. The least-squares estimator β1 solves

0 =d

dβ1

n∑

i=1

(Yi − β1Xi)2 = −

(n∑

i=1

YiXi − β1

n∑

i=1

X2i

).

Therefore

β1 =

∑ni=1 YiXi∑ni=1 X2

i

.

9. Since the fits are Yi = β0 + β1Xi, this plot has the same shape as a plotof the residuals versus Xi. The curved pattern suggest that E(Yi) is notlinear in Xi. A good remedial action would be to add a quadratic term tothe model.

10.

Source df SS MS F P

Regression 2 2.1045 1.0523 3.8853 0.05

error 12 3.2500 0.2708

total 14 5.3545

R-sq = 0.3930

11. The change in the value of the portfolio as ∆y10, ∆y20, and ∆y30 changeis approximately

F30P30DUR30∆y30−F20P20DUR20(β1∆y30+β2∆y10)+F10P10DUR10∆y10.

In order for this quantity to be zero for all values of ∆y10 and ∆y30, F10

and F30 should solve the equations

F30P30DUR30 − F20P20DUR20β1 = 0

22 6 Regression

andF10P10DUR10 − F20P20DUR20β2 = 0.

In these equations, all other quantities besides F10 and F30 are known.

7

The Capital Asset Pricing Model

1.

β =16 − 6

11 − 6= 10/5 = 2.

2. (a) Solve for w: µr = µf + w(µM − µf ) or 0.11 = 0.07 + w(0.14 − 0.07).Therefore, w = 4/7.

(b) σR = (4/7)(0.12) = 0.069.3. (a) The expected return is 0.04 + (0.10 − 0.06)(0.05)/(0.12) = 0.0567.

(b) β = 0.004/(0.122) = 0.2778.(c) The expected return is 0.04 + (0.2778)(0.10 − 0.04) = 0.1390.

4. Let X = (R1,t, . . . , RN,t)T, ω1 = (0, . . . , 0, 1, 0, . . . , 0) with the “1” in the

jth place, and ω2 = (w1,M , . . . , wN,M )T. Then ωT

1 X = Rj,t and ωT

2 X =∑Ni=1 wi,MRit. Also,

ωT

1 COV(X)ω2 =

N∑

i=1

wi,Nσi,j .

Therefore, (7.15) is a special case of (2.54).5. False. Reason: One of the most important implications of the CAPM is

the distinction between systematic (or market) risk and unsystematic (orunique) risk. The total risk is due to both of these components, but wecan expect higher returns only for higher systematic risk.

6. (a) β = 165/(152) = 0.7333.(b) The expected return is 5 + (14 − 5)β = 11.6%.(c) The percentage of the variance due to market risk is

β2σ2M

220× 100% = 55%.

7. (a) βP = (0.9 + 0.7 + 0.6)/3.(b) (βP )2(0.014) + (0.01 + 0.015 + 0.012)/9.(c) 0.01 + (0.9)(0.06 − 0.01).

8

Options Pricing

1. (a) The initial price can be found either by finding the price of a repli-cating portfolio or using risk-neutral probabilities. We will look atreplicating portfolios first. The stock price changes to 66.55 if thereare three up-steps, to 54.45 if there are two up-steps, to 44.55 if thereis one up-step, and to 35.45 if there are no up-steps. Since the exerciseprice is $55, the option is worth $11.55 if there are three up-steps and$0 otherwise. After two steps we have the following: (1) if there weretwo up-steps then the portfolio should hold 0.9545 shares and −49.44in risk-free assets. (2) if there was an up-step and a down-step or ifthere were two down-steps, then the replicating portfolio should hold0 shares and $0 in risk-free assets. After one step we have: if there wasan up-step then the portfolio should hold 0.755 shares and −35.558 inrisk-free assets but if there was a down-step then the portfolio shouldhold 0 shares and $0 in risk-free assets. The initial portfolio shouldhave 0.596 shares and $-25.51 in risk-free assets. The price of thisportfolio is (0.596)(50) − 25.51 = 4.29. Thus, the initial price of theoption is $4.29.

The risk-neutral probability of an up-step is q = {exp(0.05)−0.9}/(1.1−0.9) = 0.7564. The probability of three up-steps is q3 = 0.4327. Thusthe expected value (with respect to the risk-neutral probabilities) ofthe option is (0.4327)(11.55) = 4.9976. The discounted expected valueis 4.9976 exp{−(3)(0.05)} = 4.30. Therefore, the initial price of the op-tion is $4.30. (This differs slightly from the price of $4.29 found beforebecause of rounding errors.)

(b)

q2(66.55 − 55) exp{(2)(0.05)} = 5.98.

(c) As stated before, initially the replicating portfolio holds 0.596 sharesof stock and −25.51 dollars of the risk-free asset.

(d) If stock goes up on the first step, the replicating portfolio would thenhold 0.755 shares of stock and −35.558 dollars of the risk-free asset.

26 8 Options Pricing

Thus, 0.755 − 0.596 = 0.1590 shares of stock would be purchasedby borrowing 35.558 − 25.51exp(0.05) = 8.74 dollars. The cost ofthe 0.1590 shares is (0.1590)(55) = 8.74 which matches the amountborrowed so that the changes in the portfolio are self-financing.

(e) As given above, the risk-neutral probability of an up-step is 0.7564 oneach step. The risk-neutral probability of a down-step is 1− 0.7564 =0.2436.

2.

q =exp(0.05/20) − 0.99

1.01 − 0.99= 0.625.

The initial price of the option is

exp(0.05)

20∑

k=0

(20k

)qk(1−q)20−k

{(50)(1.01)k(0.99)20−k − 53

}+

= 0.732.

3. (a) Use the Black-Scholes formula. One finds that d1 = −0.3124, d2 =−0.4538, and C = Φ(d1)S0 − φ(d2)K exp(−rT ) = 3.16.

(b) The intrinsic value is (92 − 98)+ = 0. The adjusted intrinsic value is(92 − 97.12)+ = 0.

(c) Use put-call parity: P = C + e−rT − S0 = 8.28.(d) Use the Black-Scholes formula again. One finds that C = 2.69.

4. By using Excel’s solver or MATLAB’s interp1.m or by just plotting op-tion prices against volatility and interpolating one gets that the impliedvolatility is 0.447.

5. The price of a call option is determined as the discounted expected payoffwith respect to the risk neutral probabilities, i.e., under assumption thatthe stock price follows geometric Brownian motion with drift r, risk-freerate. If the strike price were raised by $1 with everything else unchanged,the payoff would decrease by exactly $1 if at the expiration the option wasin the money. But since the probability of being in the money is strictlyless than 1, the price of the option (i.e. expected payoff) would decreaseby amount strictly less than $1. Also, if the risk-free rate is positive, as istrue in any real situation, the discounting would decrease the change inprice.

6. One easily calculates that E(Xi) = 0.6−0.4 = 0.2 and E(X2i ) = 1 so that

Var(Xi) = 1 − 0.22 = 0.96.(a)

E(S4|S0, S1, S2) = E(S2+X3+X4|S0, S1, S2) = S2+(2)(0.2) = S2+0.4.

(Note: The practical significance of this result is that we have foundthe best predictor at time t = 2 of the random walk two steps ahead.)

(b)

Var{E(S4|S0, S1, S2)} = Var(S2) = Var(X1 + X2) = (2)(0.96) = 1.92.

8 Options Pricing 27

(c)

E{S4 − E(S4|S0, S1, S2)}2 = E(X3 + X4 − 0.4)2

= Var(X3 + X4) = (2)(0.96) = 1.92.

(Note: The practical significance of this result is that we have foundexpected squared prediction error of the best predictor at time t = 2of the random walk two steps ahead.)

(d) The risk-free rate r has not be given, so the answer must be given asa function of r. Assume that r is the simple interest rate. The optionis worth $1 at the exercise date (t = 2) if the stock price moves up onboth of the first two steps and is worth $0 otherwise. The risk-neutralprobability of an up-step is

q0 =(1 + r) − 99

101 − 99

at t = 0. Given an up-step at time 0, the probability of an up-step attime t = 1 is

q1 =(1 + r) − 100

102 − 100

The price of the call option is

q0q1

(1 + r)2.

(Note: (1 + r) would be replaced throughout by exp(r) if r was given as acontinuously compounded rate.)

7. The probability of an up-step is constant and equals

q =exp(0.03) − 0.8

1.2 − 0.8= 0.5761.

(a) At t = 2 the put is worth 0, 14, or 46 dollars if there have been twoup-steps, one up- and one down-step, or two down-steps, respectively.The value of the European put option at time t = 0 is

(14)(2)q(1 − q) + 46(1 − q)2

exp{(2)(0.03)} = 14.2226.

(b) If there is an up-step at time t = 0 then early exercise at t = 1 isworth nothing and so is clearly not optimal. If there is a down-stepat time 0, then early exercise at t = 1 is worth $30 and the value att = 1 of holding the option is worth

14q + 46(1 − q)

exp(0.03)= 26.7490.

Therefore, early exercise is optimal at t = 1 if there had just been adown-step.

28 8 Options Pricing

(c) At time t = 0 an American option is worth

5.7887q + 30(1 − q)

exp(0.03)= 15.5766.

(d) Let “UU” denote two up-steps and similarly for UD, DU, and DD. Att = 2, the option is worth 44, 10, 0, and 0 at UU, UD, DU, and DD,respectively. Thus, at time t = 0 the option is wort

44q2 + 10qp

exp{(2)(0.03)} = 16.5433.

8. Since d2 = d1 − σ√

T − t, d d1/dS = d d2/dS. Therefore, we only need toshow that

Sφ(d1) − Ke−r(T−t)φ(d1) = 0.

Next,

d22 = d2

1 − 2d1σ√

T − t + σ2(T − t) = d21 − 2 log(S/K) − 2r(T − t)

and since φ(x) = exp(−x2/2)/√

2π,

Sφ(d1) − Ke−r(T−t)φ(d1)

=exp(−d2

1/2)√2π

{S − Ke−r(T−t)elog(S/K)+r(T−t)

}= 0.

9.

Γ =∂2

∂S2C(S, T, t,K, σ, r) =

∂

∂S∆(S, T, t,K, σ, r) =

∂ Φ(d1)

∂S= φ(d1)

∂ d1

∂S

and∂ d1

∂S=

S−1

σ√

T − t.

9

Fixed Income Securities

1. (a)

y20 =1

20

∫ 20

0

(−0.022+0.0003t) dt = 0.022+(0.0003)(20)/2 = 0.0250.

(b)

y25 =1

25

∫ 25

0

(−0.022+0.0003t) dt = 0.022+(0.0003)(25)/2 = 0.0257.

The price is1000 exp(−25Y25) = 525.3188.

2. By the summation formula for a finite geometric series we have

2T∑

t=1

C

(1 + r)t=

C

1 + r

2T−1∑

t=0

1

(1 + r)t

=C

1 + r

1 − (1 + r)−2T

1 − (1 + r)−1=

C

r{1 − (1 + r)−2T }.

Substituting this result into the left-hand side of the equality and simpli-fying gives us the right-hand side.

3. (a) $50.(b)

50

0.04+

(1000 − 50

0.04

)(1 + 0.04)

−38= 1193.70.

(c) 1193.70 + 50 = 1243.70.4. (a) − log(0.848)/5 = 0.0330.

(b) 1000 exp{−(4)(0.04)} = 852.14.(c) 852.14/848 − 1 = 0.0049.

30 9 Fixed Income Securities

5. (a)

18

0.02+

(1000 − 18

0.02

)(1.02)−20 = 967.30.

(b) Below par because the current interest rate is higher than the couponrate.

6. (a) Solve for y:25

y+

(1000 − 50

y

)(1 + y)−10.

By interpolation, the yield-to-maturity is y = 0.0189.(b) 25/1100 = 0.0227.(c) The yield-to-maturity is below the current yield because the bond is

selling above par and there will be a loss of principal at maturity.7.

∫ 15

0

(0.03 + 0.001t) dt = (0.03)(15) + (0.001)(152)/2 = 0.5625.

Current value is 100 exp(−0.5625) = 56.98.8. The yield-to-maturity is

1

20

{∫ 20

0

(0.02 + 0.001t) dt +

∫ 20

10

0.0005(t − 10) dt

}

=1

20

{(0.02)(20) + (0.001)(202)/2 + (0.0005)(20 − 10)2/2

}= 0.0313.

9. (a) 1-year: price=1000(1 + 0.02/2)−2 = 980.2963-year: price=1000(1 + 0.04/2)−6 = 887.97145-year: price=1000(1 + 0.045/2)−10 = 800.5101

(b) (rates unchanged)1-year (now a 0-year): price=10003-year (now a 2-year): price=1000(1 + 0.03/2)−4 = 942.25-year (now a 4-year): price=1000(1 + 0.0425/2)−8 = 845.2

(c) (rates increase as forecasted)1-year (now a 0-year): price=10003-year (now a 2-year): price=1000(1 + 0.035/2)−4 = 933.05-year (now a 4-year): price=1000(1 + 0.0475/2)−8 = 828.8

(d) (rates up)1-year: return = 1000/980.296 − 1 = 0.02013-year: return = 933.0/887.9714 − 1 = 0.05075-year: return = 828.8/800.5101 − 1 = 0.0353

(e) (rates unchanged)1-year: return = 1000/980.296 − 1 = 0.02013-year: return = 942.2/887.9714 − 1 = 0.06115-year: return = 845.2/800.5101 − 1 = 0.0558

9 Fixed Income Securities 31

(f) No it is not correct as seen in part (e) where the 5-year bond has thehighest spot rate but has a lower return than the 3-year bond. Thereason that the returns after 1 year after not equal to the spot ratesis that the n-year bond has become an n − 1-year bond so it is nowpriced by discounting at the n−1-year spot rate. Thus, although spotrates are unchanged, the bond is not priced with the same rate asbefore and the returns will not be the spot rates.

10.

d

dδ

N∑

i=1

Ci exp{−Ti(yTi+δ)}

∣∣∣∣δ=0

= −N∑

i=1

CiTi exp{−TiyTi} = −

N∑

i=1

NPVi Ti

= −N∑

i=1

Ti ωi

(N∑

i=1

Ci exp{−TiyTi})

= −DUR

(N∑

i=1

Ci exp{−TiyTi})

.

However, (N∑

i=1

Ci exp{−TiyTi})

= bond price

sod

dδbond price = −DUR × bond price.

Therefore,

change in bond price ≈ −DUR × bond price × change in yield

where “change in yield” is δ. Rearranging give (9.29).

10

Resampling

1. In Figures 10.6 and 10.8 the expected return can be less than the targetedreturn of 12%. In such cases, the risk can be less that the optimal risk fora 12% return because the expected return is less than 12%.

2. 500 or 5% of the values of sb,boot/s were less (more) than 0.7 (1.6). There-fore, a 90% confidence interval is (0.7)(0.28) = 0.1969 to (1.6)(0.28) =0.4480.

3. (a) (0.004)w + (0.0065)(1 − w) = 0.005 so that the estimated efficientportfolio is 60% stock 1 and 40% stock 2.

(b) (0.6)2(0.012) + (0.4)2(0.023) + (2)(0.6)(0.4)(0.0051) = 0.0104.(c) Actual expected return is (0.6)(0.003) + (0.4)(0.007) = 0.046. Actual

variance of return is (0.6)2(0.01) + (0.4)2(0.02) + 2(0.6)(0.4)(0.005) =0.0092.

11

Value-at-Risk

1. Let RP , R stock1 , and R option

2 be, respectively, the returns on the portfolio,the stock, and the option on the second stock. Then

Rp = wR stock1 + (1 − w)R option

2

and therefore by (5.3) the variance of the return on the portfolio is

σ2RP

= w2(σ stock1 )2 + (1 − w)2(σ stock

1 )2 + 2w(1 − w)σ stock,option1,2 .

Here σ option2 is the standard deviation of the return on the option on

the second stock and σ stock,option1,2 is the covariance between the first stock

and the option on the second stock. Next σ option2 = L2σ

stock2 and by (8.33)

σ stock,option1,2 = L2σ

stock1,2 . Substituting these into the above expression for

σ2RP

gives (11.12).2. (Note: There is an error in Example 11.2: “s = 0.0151” should “s =

0.0141”.) As an illustration of the effect of making the mean positive,assume that the expected return is 8% per year or 0.08/253 per tradingday. Then

VaR(α) = −20, 000 × {0.08/253 + (−1.645)(0.0141)} = 458.

This value-at-risk of $458 is smaller than the value-at-risk of $471 obtainedin Example 11.2 — a change to a positive mean, with everything else heldunchanged, decreases the value-at-risk, of course.

3.

VaR(0.005) = VaR(0.05)

(0.05

0.005

)1/a

= 211

(0.05

0.005

)1/4.11

= 369.

5. (a) Use Black-Scholes. The risk-free rate should be converted to a dailyrate since σ is given for a daily return and T is in days. The price ofthe call option is $8.6576.

36 11 Value-at-Risk

(b) d1 = 0.2200 and ∆ = Φ(d1) = 0.5871.(c)

L = 0.5871115

8.6576= 7.7981.

(d)

VaR(0.05, 24 hours) = −1000{0.0001 − (1.645)(0.017)}7.7981 = 217.

(e)

VaR(0.05, 24 hours)

= −2000{0.5 + (0.5)(7.7981)}{0.0001 − (1.645)(0.017)} = 245.

(f)

µP = (0.5)(7.7981)(0.0001) + (0.5)(0.0002) = 0.00049.

and

σP =

{(0.52)(7.79812)(0.0172) + (0.52)(0.0192)

+(2)(0.5)(0.5)(7.981)(0.3)(0.017)(0.019)

}1/2

= 0.0377.

Therefore,

VaR(0.05, 24 hours) = −2000(µP − 1.645σP ) = 123.

(g)

µP = (2/3){0.5 + (0.5)(7.7981)}(0.0001) + (1/3)(0.0002) = 0.00049.

and

σP =

{(2/3)2{0.5 + (0.5)(7.7981)}2(0.0172) + (1/3)2(0.0192)

+(2)(2/3)(1/3){0.5 + (0.5)(7.7981)}(0.3)(0.017)(0.019)

}1/2

= 0.0352.

Therefore,

VaR(0.05, 24 hours) = −3000(µP − 1.645σP ) = 172.

6. For i = 1, 2, R(Pi) = −Si×{µi +Φ−1(α)σi}. Also, the standard deviationof the return on P1 + P2 is

√S2

1σ21 + 2S1S2ρσ1σ2 + S2

2σ22 ≤ S1σ1 + S2σ2

because |ρ| ≤ 1. Therefore, R(P1+P2) ≤ −(S1µ1+S2µ2)+Φ−1(α)(S1σ1+S2σ2) = R(P1) + R(P2).

11 Value-at-Risk 37

7. (a) The loss is 0 if ST < 108, 100(ST − 108) if 108 < ST ≤ 112, and 400if ST > 112.

(b) Each of the 20,000 simulated values of ST is converted into a lossusing the result of part a). Since (20, 000)(0.05) = 1000, VaR(0.05) isthe 1000th largest value of these simulated losses and ES(0.05) is theaverage of the 1000 largest simulated losses.

(c) Simulate 20,000 N(0, 1) random variables, Z1, . . . , Z20,000. The ithsimulated value of ST is 105 exp(0.1 + 5Zi). Use these as explained inpart b) to estimate VaR(0.05) and ES(0.05).

12

GARCH Models

1. The first equality is the definition of the expectation and the second equal-ity is true because the integrand is symmetric about 0. Therefore, we needonly prove the third equality.

2

∫ ∞

0

1√2π

ze−z2/2dz = −√

2

π

∫ ∞

0

(d/dz)e−z2/2dz

= −√

2

πe−z2/2

∣∣∣∣∞

0

= −√

2

π(0 − 1) =

√2

π.

2.

∫ ∞

−∞

fX(x)dx = 2

∫ ∞

0

fX(x)dx = 2

(1

4

){∫ 1

0

1dx +

∫ ∞

1

x−2dx

}

=1

2

(1 − x−1

∣∣∣∣∞

1

)= 1.

3. (a)

µ =3

1 − 0.7= 10.

(b)

Var(at) =α0

1 − α1=

1

1 − 0.5= 2.

Var(ut) =Var(at)

1 − 0.7=

2

0.3= 6.667.

(c)

ρy(h) = (0.7)|h|.

40 12 GARCH Models

(d)

ρa2(h) = (0.5)|h|.

4. (a)

E(u2|u1 = 1, u0 = 0.2) = µ + φ(1 − µ) = 0.7 + 0.5(1 − 0.7) = 0.85.

(b) Given that u0 = 0.2 and u1 = 1, we have that

a1 = (u1 − µ) − φ(u0 − µ) = (1 − 0.7) − (0.5)(0.2 − 0.7) = 0.05,

and therefore

Var(a2|u1 = 1, u0 = 0.2) = Var

{ε2

√α0 + α1a2

1

∣∣∣∣a1

}

= α0 + α1a21 = 1 + (0.3)(0.05) = 1.015.

Finally,

Var(u2|u1 = 1, u0 = 0.2) = Var(a2|u1 = 1, u0 = 0.2) = 1.015.

5. (a)

E(Yt) =3

1 − 0.6= 7.5.

(b)

ρY (h) = 0.6|h|.

(c)

ρa(h) = 1 if h = 0

= 0 otherwise.

(d)

ρa2(h) = 0.5|h|.

6. (a) E(Yt|Xt = 0.1 and at−1 = 0.6) = β0 + β1(0.1) + δ√

1 + (0.5)(0.6)2 =

0.05 + (0.3)(0.1) + (0.2)√

1 + (0.5)(0.6)2

(b) Var(Yt|Xt = 0.1 and at−1 = 0.6) = 1 + (0.5)(0.6)2

(c) Normal. Conditional on Xt and at, Yt is a constant, β0 + β1Xt + δσt,

plus another constant,√

1 + 0.5a2t−1, times εt. Since εt is normal, so

also is Yt.

(d) Not normal. As√

1 + 0.5a2t−1 varies with at−1, the distribution of

√1 + (0.5)a2

t−1εt is a normal mixture, which is not normal.

13

Nonparametric Regression and Splines

1. (a) s(t) = 1 + 0.5t for 0 ≤ 5 ≤ 1. Therefore, s(0.5) = 1.25.(b) s(t) = 6 for t ≥ 3 since s is linear for t ≥ 3 and s(4) = s(5) = 6.

Therefore, s(3) = 6.(c)

∫ 4

2

s(t)dt =

∫ 3

2

{5 + (t − 2)}dt +

∫ 4

3

6dt = 5 +(t − 2)

∣∣∣3

2

2+ 6 = 11.5.

2. (a) E(rt+1 − rt|rt = 0.04) = µ(0.04) = 0.1(0.03 − 0.04) = −0.001. There-fore, E(rt+1|rt = 0.04) = 0.04 − 0.001 = 0.039.

(b) σ2(rt+1|rt = 0.02) = σ2(0.02) = {(2.1)(0.02)}2.3. (a) s is a pdf since it is non-negative and the area under the curve is one.

A function cannot be both a pdf and a cdf. A cdf f(x) has a limit of0 as x → −∞ and a limit of 1 as x → ∞. Therefore, a cdf has aninfinite area under its curve and cannot be a pdf.

(b) Solve .1 =∫ 2

y(2 − x)dx = y2/2 − 2y + 2.

Solution is

y =2 ±

√22 − (4)(1/2)(1.9)

(2)(1/2)= 2 ±

√.2.

Choose solution between 1 and 2 which is y = 2 −√

.2.An alternative solution is to notice that s(x) is the linear functions(x) = 2− x for x between 1 and 2. The area of the triangular regionunder the curve between 2 − y and 2 is (1/2)(2 − y)2 so

.1 =(2 − y)2

2

or y = 2 ±√

.2 and solution between 1 and 2 is 2 −√

.2.4. (a)

s(1.5) = 1 + (0.5)(1.5) + 1.52 + (1.5 − 1)2 = 4.25.

42 13 Nonparametric Regression and Splines

(b)

s′′(x) = 2 + 2(x − 1)0+ + 0.6(x − 2)0+

and therefores′′(2.5) = 2 + 2 + 0.6 = 4.6.

14

Behavioral Finance

1. (a) First, E(X) = E(X) so that E{X − E(X)} = 0. Therefore,

Cov(X,X − X) = E

[E(X) +

σXW

σ2W

{W − E(W )}]

×[{X − E(X)} − σXW

σ2W

{W − E(W )}]

= E [E(X){W − E(W )}] + E

[E(X)

(−σXW

σ2W

)(W − E(W ))

]

+σXW

σ2W

(−σXW ) +σ2

XW

σ4W

σ2W = 0.

since E({W − E(W )} = 0.

(b) Because X and X − X are uncorrelated and X = X + (X − X),

Var(X) = Var(X) + Var(X − X) = Var(X) + E(X − X)2.

(c) Var(X) ≤ Var(X) since E(X − X)2 ≥ 0.2. Since for each year there is a pair of portfolios, one a winner portfolio and

the other a loser portfolio, DeBondt and Thaler should have used a pairedt-test, not an independent samples t-test.

The conclusion of DeBondt and Thaler should be correct. The reason isas follows. First, the returns on the winner and loser portfolios shouldbe positively correlated, since both portfolios should have positive betasand therefore should be positively correlated with the market return. Asdiscussed in Section 2.20.3, when there is a positive correlation betweenthe two items in each pair then the incorrect use of an independent samplest-test will make the p-value larger than it should be. Since the incorrectp-value that DeBondt and Thaler obtained was small enough to reject thenull hypothesis, the correct p-value should be even smaller and thereforeshould be even stronger evidence against the null hypothesis.

44 14 Behavioral Finance

3. (a) Even though the two decisions are made concurrently, people will tendto analyze them separately. Many people will choose (A) in Decision1 in order to be certain of a gain. People hate to lose and will oftengamble in order to avoid a certain loss. For this reason, many peoplewill chose (D) in Decision 2.

(b) [A,C]: lose $5100 with probability 1.

[A,D]: lose $7600 with probability 0.75 and gain $2400 with probability0.25.

[B,C]: lose $7500 with probability 0.75 and gain $2500 with probability0.25.

[B,D]: lose $0 with probability 6/16, lose $10,000 with probability9/16, and gain $10,000 with probability 1/16. (Note: it is assumedthat the outcome in the first decision problem is independent of theoutcome in the second decision problem. If this is not true, then theprobabilities will be different.)

(c) [B,C] has the same gain and loss probabilities as [A,D], but the gain ismore and the loss is less for [B,C] compared to [A,D]. Therefore, [B,C]is always preferred to [A,D] regardless of one’s tolerance towards risk.

4. The good performance of the Behavioral Growth Fund certainly is evi-dence against the EMH, but there are other reasons besides market effi-ciency that might explain this good performance. One would need to lookat the risk of the Behavioral Growth Fund. If its risk is higher than therisks of the indices to which the Behavioral Growth Fund is being com-pared, then the higher performance of the Behavioral Growth Fund couldbe due to a risk premium. Another issue is whether this particular fundwas selected as an example because of its strong performance. One shouldinvestigate how well have other behavioral funds done.

5. If the market is truly efficient there should be no underreaction and nooverreaction. Thus, Fama’s argument suggest that the market is “unbi-ased” in the sense that it is right on average, but this is not the same asbeing efficient. Fama’s argument is consistent with the excess volatilitythat Shiller has found. If the market is correct on average as Fama states,then the market prices are the prices that a truly efficient market wouldset plus zero-mean noise. The noise would create excess volatility, exactlywhat Shiller observed.

6. The forecast of Xt+1 is Xt+1 = φXt so the forecast error is Xt+1−Xt+1 =εt+1. Therefore the forecast errors are a white noise process and are un-correlated by the definition of white noise.

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