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 derivative equal to 0 gives us w = 2/3. The second derivative is positive so the solution must be a minimum. In this problem, assets X and Y have same expected return. This means that regardless of the choice of w, that is, the asset allocation, the expected return of the portfolio doesn’t change. So by minimizing the variance, we can reduce the risk without reducing the return. Thus the ratio w = 2/3 corresponds to the optimal portfolio 2. (a) Use (2.54) with w1 = (1 1)T and w2 = (1 1)T . 2 (b) Use part (a) and the facts that Cov(α1 X, α2 X) = α1 α2 σX , Cov(α1 X, Z) = 0, Cov(Y, α2 X) = 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 ! n1 n2 n1 X n2 X X X Cov Xi , Yi0 = Cov(Xi , Yi0 ). i=1

i=1 i0 =1

i0 =1

3. The likelihood is L(σ 2 ) =

n Y

i=1

Therefore the log-likelihood is

√

1 2πσ 2

1

2

e− 2σ2 (Yi −µ) .

2

2 Probability and Statistical Models n 1 X log L(σ ) = − 2 (Yi − µ)2 − n log(σ 2 )/2 + H 2σ i=1 2

where H consists of all terms that do not depend on σ. Differentiating the log-likelihood with respect to σ 2 and setting the derivative1 with respect to σ 2 equal to zero we get n 1 X n (Yi − µ)2 − 2 = 0 2(σ 2 )2 i=1 2σ

whose solution is

n

σ2 =

1X (Yi − µ)2 . n i=1

4. Rearranging the first equation, we get

β0 = E(Y ) − β1 E(X).

(2.1)

Substituting this into the second equation and rearranging, we get E(XY ) − E(X)E(Y ) = β1 {E(X 2 ) − E(X)2 }. Then using σXY = E(XY ) − E(X)E(Y ) and 2 = E(X 2 ) − E(X)2 σX

we get β1 =

σXY 2 , σX

and substituting this into (2.1) we get β0 = E(Y ) −

σXY 2 E(X). σX

5. E(wT X) = E

N X i=1

wi Xi

!

=

N X i=1

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

Next "N #2 X  T 2 T Var(w X) = E w X − E(w X) = E wi {Xi − E(Xi )} T

i=1

1

The solution to this problem is algebraically simpler if we treat σ 2 rather than σ as the parameter.

2 Probability and Statistical Models

=

N N X X i=1 j=1

E [wi wj {Xi − E(Xi )}{Xj − E(Xj )}] =

N N X X

3

wi wj Cov(Xi , Xj ).

i=1 j=1

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

N N X X

Xi Xj Aij ,

i=1 j=1

whence wT COV(X)w =

N X N X

wi wj Cov(Xi , Xj ).

i=1 j=1

6. Since log{L(µ, σ 2 )} = − and

n n 1 X n (Yi − µ)2 log(2π) − log(σ 2 ) − 2 2 2 2σ i=1

Pn

i=1 (Yi − 2 σ bM L

it follows that

Y )2

= n,

n 2 2 log{L(Y , σ bM σM L )} = − {1 + log(2π) + log(b L }. 2

Next, the solution to

 ∂ ∂  log{L(0, σ 2 )} = 0= 2 ∂σ ∂σ 2 =− solves nσ 2 =

7. (a)

Pn

i=1

(

n n n 1 X 2 − log(2π) − log(σ 2 ) − 2 Y 2 2 2σ i=1 i

)

n 1 X 2 n Y , + 2σ 2 2(σ 2 )2 i=1 i

Yi2 so that

n

2 σ b0,M L =

1X 2 Y . n i=1 i

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.

4

2 Probability and Statistical Models

8. (a) Since

(b)

σXY Yb = E(Y ) + 2 {X − E(X)}. σX

and E{X − E(X)} = 0 by Problem 7. it follows that E(Yb ) = E(Y ) so that E{Yb − Y } = 0 − 0 = 0.  σ2 2 E(Y − Yb )2 = E {Y − E(Y )}2 + XY 4 {X − E(X)} σX i σXY h −2 2 E {(Y − E(Y )}{X − E(X)} σX   2 2
σ2 σXY σXY 2 = σY2 1 − ρ2XY . = σY2 + XY 1 − − 2 = σ Y 2 2 2 2 σX σX σX σY

9.

3

E(XY ) = E(X ) =

Z

a

−a

a x3 x4 dx = =0 2a 8a −a

and E(X) = 0 so that σXY = E(XY ) − E(X)E(Y ) = 0 − 0 = 0. Since Y is determined by X one suspects that X and Y are not independent. This can 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 can be shown by finding the value of θ that maximizes f (θ|3) = 30 θ4 (1 − θ). Thus, one solves 0=

d 30 θ4 (1 − θ) = 30 θ3 (4 − 5θ) dθ

whose solution is θ = 4/5. 11. (a) Since the kurtosis of a N (µ, σ 2 ) random variable is 3, E(X − µ)4 = 3σ 4 . Therefore, for a random variable X that is 95% N (0, 1) and 5% N (0, 10), we have E(X 4 ) = (0.95)(3)(14 ) + (0.05(3)(104 ) = 1502.9 and E(x2 ) = (0.95)(12 ) + (0.05)(102 ) = 5.9500. Therefore, the kurtosis is 1502.9 = 42.45. 5.952

2 Probability and Statistical Models

5

(b) One has that E(X 4 ) = 3p + 3(1 − p)σ 4 and E(X 2 ) = p + (1 − p)σ 2 so that the kurtosis is p2

3{p + (1 − p)σ 4 } . + 2p(1 − p)σ 2 + (1 − p)2 σ 4

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

σ→∞ p2

3 3{p + (1 − p)σ 4 } = . 2 2 4 + 2p(1 − p)σ + (1 − p) σ 1−p

Therefore, by letting σ get very large and p get close to 1, the kurtosis can be made arbitrarily large. Suitable σ and p such that the kurtosis is 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 sentence should be “Show that for any p0 < 1, no matter how close to 1, there is a p > p0 and a σ, such that the normal mixture with these values of 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 with this value of p, the kurtosis will converge to a value greater than M as 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) Φ(x/σ) − Φ(c/σ) = . 1 − P (X ≤ x) 1 − Φ(c/σ)

The conditional PDF is   φ(x/σ) (d/dx)Φ(x/σ) − Φ(c/σ) d Φ(x/σ) − Φ(c/σ) = . = dx 1 − Φ(c/σ) 1 − Φ(c/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 into the Pareto PDF, that is, into ac2 /xa+1 , the result is 4.000. Thus, the two PDFs are equal to two decimals. 13. The likelihood is n   Y L(θ) = θ−1 e−Xi /θ i=1

so the log-likelihood is

log{L(θ)} = −n log(θ) − Thus the MLE solves

Pn

i=1

θ

Xi

.

6

2 Probability and Statistical Models

0=

d dθ

Pn Pn   Xi Xi −n −n log(θ) − i=1 = + i=12 θ θ θ

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 solution to 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 1/4 and 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: σ bXY should be changed to sXY . (a) n

s2d =

1 X {(Xi,1 − X 1 ) − (Xi,2 − X 2 )}2 n − 1 i=1 n

= s21 + s22 −

2 X {(Xi,1 − X 1 )(Xi,2 − X 2 )} = s21 + s22 − 2s1,2 . n − 1 i=1

(b) The numerators of the t-statistics are both equalp to X 1p− X 2 − ∆0 . p Since s = 0 and n = n , 1/n + 1/n s = 2/n (s21 + s22 )/2 = 1,2 1 2 1 2 pool p p √ (s21 + s22 )/n. If s1,2 = 0, then sd / n = (s21 + s22 )/n. Thus, the denominators of the two t-statistics are also equal.

3 Returns

1. (a) R2 =

56.2 P2 + D2 −1= − 1. P1 51

(b) 

(c)

   P4 + D4 P3 + D3 P2 + D2 R4 (3) = −1 P3 P2 P1     58.25 53.25 56.2 = − 1. 53 56 51

r3 = log



P3 + D3 P2



= log





= 0.897.

53.25 56



.

2. (a) N (0.3, 1.8). (b) Φ



2 − 0.3 √ 1.8

(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 geometric random walk model, the log returns are normally distributed, so the 0.05quantile of the 20-year log return is 2−(1.654)(4) = 4.58 and therefore the 0.05-quantile of the 20-year return is e−4.58 = 0.0103. The 0.95-quantile of the log return is 2 + (1.645)(4) = 8.56 and the 0.95-quantile of the return is e8.56 = 5321.

8

3 Returns

4. (a) Two examples of measurable uncertainty are random sampling from a populations and computer simulation. Of course, there are many more. In random sampling each possible sample has a known probability of being chosen. In computer simulation, random variables have known distributions specified by the programmer. (b) Again, there is an almost infinite choice of examples. One example is the uncertainty about whether in will rain tomorrow. In this case, meteorologists estimate the probably from past data and models of the atmosphere. 5. We have Xk = X0 exp(Z1 + · · · + Zk ) where Z1 , . . . , Zk are iid N (µ, σ 2 ). Therefore, Xk2 is lognormal, in particular, Xk2 = X02 exp(W ) where W is N (2kµ, 4kσ 2 ). [Note: It is assumed here that X0 is a fixed constant. If instead X0 is a random variable, then the expectations and quantiles being found here are conditional expectations and quantiles given X0 . The unconditional expectations and quantiles cannot be found since the distribution of X0 has not been specified.] (a) It follows that E(Xk2 ) = X02 exp{2kµ + (4kσ 2 )/2} by the formula for the mean of a lognormal distribution. (b) W has a N (kµ, kσ 2 ) density which is   1 1 2 (w − kµ) exp − . fW (w) = √ 2kσ 2 2πkσ 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, h0 (y) = 1/y, so by the change of variables formula (2.10) the density of Xk is   1 1 2 [log(x) − {kµ + log(X0 )}] . fXk (x) = √ exp − 2kσ 2 x 2πkσ 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 √ = Φ−1 (.75) = 0.6745. kσ (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, kσ 2 ) the third quartile of W is kµ + kσ 0.6745 (see Section 2.8.2, subsection “normal quantiles”). Then since Xk = X0 exp(W ) is a monotonically increasing function g(w) = X0 exp(w)√of W , the third quartile √ of 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, since these data were already used to formulate the hypothesis. To test the hypothesis one would need to collect data for a number of years past 1998. Once this is done, there are a number of ways to test the Super Bowl theory. One way would be the independent samples t-test, the first sample being the years when the a former NFL wins the Super Bowl and the second sample being the years when a former AFL team wins. The null 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. He mentions the possibility that the market could be down for most of the year 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 return for the year is positive (negative). Let p1 and p2 be the probabilities of bull markets in years when a former NFL, respectively, former AFL team wins. To test if a former NFL team winning 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 AFL team winning predicts a bear market we could test H0 : p2 = 1/2 versus H1 : p2 < 1/2. Another set of hypotheses to test is the null hypothesis H0 : p1 = p2 versus the alternative H1 : p1 > p2. These could be easily tested by a likelihood ratio test.

4 Time Series Models

1. (a) The process is stationary since φ = 0.7 so that |φ| < 1. Strictly speaking, the process is only stationary if it started in the stationary distribution. If the process has been operating for a reasonably long time (say 10 time periods) we can assume that it has converged to the stationary distribution. In the remaining parts of the problem we will assume that it is in the stationary distribution. (b) µ= (c)

5 . 1 − 0.7

γ(0) =

2 . 1 − 0.72

γ(h) =

2(0.7|h| ) . 1 − 0.72

(d)

2. (a) Var(Y1 ) = γ(0) = σ2 /(1 − φ2 ) = 2/(1 − (.3)2 ) = 2.198. (b) Cov(Y1 , Y3 ) = γ(2) =

3.

2(.3)2 σ2 (φ)2 = = .1978. 2 1−φ 1 − (.3)2

(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 ) because the process is stationary.) ybn+1 = 102 + 0.5(99 − 102) + 0.2(102 − 102) + 0.1(101 − 102) = 100.4.

ybn+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 = t so that yt = yt−1 + t = yt−2 + t−1 + t = · · · = y + 0 + 1 + · · · + t 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 = t − θ1 t−1 − θ2 t−2 , γ(0) = Var(t ) + θ12 Var(t−1 ) + θ22 Var(t−2 ) = (1 + θ12 + θ22 )σ2 . Similarly, γ(1) = (−θ1 + θ1 θ2 )σ2 , γ(2) = −θ2 σ2 , and γ(k) = 0 for k ≥ 3. The autocorrelation function is ρ(0) = 1, ρ(1) = (−θ1 +θ1 θ2 )/(1+θ12 +θ22 ), ρ(2) = −θ2 /(1 + θ12 + 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(φ1 yt−1 + φ2 yt−2 + t , yt−k ) = φ1 Cov(yt−1 , yt−k ) + φ2 Cov(yt−2 , yt−k ) = φ1 γ(k − 1) + φ2 γ(k − 2). for any k > 0. (If k = 0, then there is a non-zero covariance between t and yt−k , which adds another term to the equation.) (b) Using the result in (a) with k = 1, we get ρ(1) = φ1 ρ(0) + φ2 ρ(−1) = φ1 + φ2 ρ(1). Using this result with k = 2 we get ρ(2) = φ1 ρ(1) + φ2 ρ(0) = φ1 ρ(1) + φ2 . (c) 

φ1 φ2



=



1 0.4

0.4 1

−1 

0.4 0.2



=



0.3810 0.0475



.

Then ρ(3) = (0.4)(0.3810) + (0.2)(0.0476) = 0.16192. 7. The covariance between t−i and t+h−j is σ2 if j = i + h and is zero otherwise. Therefore,   ∞ ∞ ∞ X X X σ 2 φ|h| i j   Cov t−i φ , t+h−j φ = σ2 φi φi+h =  2 . 1−φ i=0 j=0 i=1

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 nonstationary and tend to wander. In fact, its first difference is the series in the middle panel. We can see that the the first difference spends long periods where it is always positive and also long periods where it is always negative. During a period when the first difference is positive the series is constantly moving upward. Similarly, when the first

4 Time Series Models

13

difference is negative the series is constantly moving downward. This is 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 the top panel which is stationary with mean 0 and only short term correlation. The series in the top panel never stays positive or negative for a long period. Thus, the series in the middle panel does not move in a constant direction for long periods. (b) The series in the bottom panel would not be a good model for a stock price. Under such a model it is possible to predict the direction of price movement which would allow one to make nearly certain short term profits. Such market conditions would not last very long since traders 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) or 15.3752w2 − 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)

4. The equation

nj Pj wj = PN . k=1 nk Pk RP = w1 R1 + · · · + wN RN

(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 initial portfolio has $500 in asset 1, $300 in asset 2, and $200 in asset 3, so the initial 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 held of each asset and the price per share are irrelevant. For example, it is immaterial whether asset 1 is $5/share and 100 shares are held, $10/share and and 50 shares held, or the price per share and number of shares are any 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 period is 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 return minus 1, if (5.1) holds for gross returns then in holds for net returns, and vice 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) when R1 , . . . , 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 not hold for log returns. In this example, equation (5.1) is not even a good approximation because two of the three net returns have large absolute values. Now let us show that (5.1) holds in general for gross returns and hence for net returns. Let P1 , . . . , PN be the prices of assets 1 through N in the portfolio. (As in the example, Pj is the price per share of the jth asset times the number of shares in the portfolio.) Let R1 , . . . , RN be the net returns on these assets. The jth weight is equal to the ratio of the price of the jth asset in the portfolio to the total price of the portfolio which is Pj wj = PN

i=1

Pi

.

At the end of the holding period, the price of the jth asset in the portfolio has changed from Pj to Pj (1+Rj ), so that the gross return on the portfolio is ! PN N N X X Pj j=1 Pj (1 + Rj ) wj (1 + Rj ), = (1 + R ) = PN PN j i=1 Pi i=1 Pi j=1 i=j which proves (5.1) for gross returns.

5 Portfolios

5. Use the formula 

a b c d

−1

1 = ad − bc



d −b −c a



.

Substituting the determinant is ab − cd = σ12 σ22 (1 − ρ12 ) and     σ12 −ρ12 σ1 σ2 a b = c d −ρ12 σ1 σ2 σ22 into this formula and simplifying gives (5.34).

17

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

2. The likelihood is L(β0 , β1 , σ 2 ) =

n Y

i=1

−n/2 −1

= (2π)

σ

√

  1 exp − 2 (Y − β0 − β1 Xi ) 2σ 2πσ 2 1

(

n 1 X 2 exp − 2 (Yi − β0 − β1 Xi ) 2σ i=1

)

.

Therefore, L(β0 , β1 , σ 2 ) is maximized over (β0 , β1 ) by minimizing n X i=1

2

(Yi − β0 − β1 Xi ) ,

so the least-squares estimator is the maximum likelihood estimator. 3. First n X wi i . βb1 − E(βb1 ) = i=1

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

20

6 Regression

Since Var(i ) =

Var(βb1 ) =

σ2 ,

n X

wi2 Var(i ).

i=1

Pn (Xi − X)2 σ2 . wi2 = σ2 Pni=1 2 = Pn 2 2 i=1 (Xi − X) i=1 i=1 (Xi − X) Pn 4. (a) P The Xi are 1 + (i −P1)/29, i = 1, . . . , 30. Therefore, i=1 Xi = 165, P n n n 2 3 4 X = 1124, X = 8562.9, and X i=1 i i=1 i i=1 i = 69, 548. It follows that Var(βb1 ) = σ2

n X

Corr(Xi , Xi2 )

Pn Pn Xi3 − ( n1 i=1 Xi )( n1 i=1 Xi2 ) =n
2 o1/2 n 1 Pn
o1/2 Pn Pn Pn 1 1 1 2 4 2 2 i=1 Xi − n i=1 Xi i=1 Xi − n i=1 Xi n n 1 n

Pn

i=1

= 0.9770.

The VIFs Pn= 43.48. Pn are both 1/(1 − 0.9770) (b) Since i=1 (Xi − X)3 = 0 and i=1 (Xi − X) = 0,

Cov((Xi − X), (Xi − X)2 ) ! ! n n n 1X 1X 1X 3 2 = 0. (Xi − X) − (Xi − X) (Xi − X) = n i=1 n i=1 n i=1

so the correlation between (Xi − X) and (Xi − X)2 is 0. The VIFs are both 1. 5. (a) R2 = Corr(Y, Yb ) = 0.25. (b) residual error SS residual error SS =1− . total SS 100 Therefore residual error SS = 75. (c) Regression SS = total SS − residual error SS = 100 − 75 = 25. (d) 0.25 = R2 = 1 −

s2 =

residual error SS 75 = = 2.885. n−p−1 30 − 4

6. For this problem, R2 = 1 − (residual error SS)/50. Also, 2 = σ b,M

10 = 0.1667 66 − 5 − 1

Therefore, the values of R2 and Cp are as in the table below. Based solely on this information, the model to choose would be the one with the smallest value of Cp which is the model with 4 predictors. However, the model with 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

Number of predictors 3 4 5

Residual error SS 12.0 10.2 10.0

R2

Cp

0.7600 0.7960 0.8000

14.0 5.2 6.0

21

7. No, we cannot accept that both β1 and β2 are zero. It is usually the case that X and X 2 are highly correlated. Therefore, either one can serve as a proxy for the other, which means that it is not necessary for both to be in the model. The p-value for β1 tests whether X can be dropped given that X 2 is in the model, and similarly the p-value for β2 tests whether X 2 can be dropped given that X is in the model. We can conclude that either X or X 2 can be dropped, that is, that either β1 or β2 is zero. However, we cannot conclude that both X and X 2 can be dropped, that is, that both β1 and β2 are zero. To conclude that both β1 and β2 are zero we could fit the model Yi = β0 + β1 Xi + i and test if β1 is zero. 8. The least-squares estimator βb1 solves ! n n n X X d X 2 2 Yi Xi − β1 Xi . (Yi − β1 Xi ) = − 0= dβ1 i=1 i=1 i=1 Therefore

Pn Yi Xi b . β1 = Pi=1 n 2 i=1 Xi

9. Since the fits are Ybi = βb0 + βb1 Xi , this plot has the same shape as a plot of the residuals versus Xi . The curved pattern suggest that E(Yi ) is not linear in Xi . A good remedial action would be to add a quadratic term to the model. 10. Source Regression error total

df 2 12 14

SS 2.1045 3.2500 5.3545

MS 1.0523 0.2708

F 3.8853

P 0.05

R-sq = 0.3930 11. The change in the value of the portfolio as ∆y10 , ∆y20 , and ∆y30 change is approximately F30 P30 DUR30 ∆y30 −F20 P20 DUR20 (βb1 ∆y30 +βb2 ∆y10 )+F10 P10 DUR10 ∆y10 .

In order for this quantity to be zero for all values of ∆y10 and ∆y30 , F10 and F30 should solve the equations F30 P30 DUR30 − F20 P20 DUR20 βb1 = 0

22

6 Regression

and F10 P10 DUR10 − F20 P20 DUR20 βb2 = 0.

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

7 The Capital Asset Pricing Model

1. β=

16 − 6 = 10/5 = 2. 11 − 6

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 , ω 1 = (0, . . . , 0, 1, 0, . . . , 0) with the “1” in the T jth place, and ω 2 = (w1,M , . . . , wN,M )T . Then ω T 1 X = Rj,t and ω 2 X = PN i=1 wi,M Rit . Also, ωT 1 COV(X)ω 2 =

N X

wi,N σi,j .

i=1

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 (or unique) risk. The total risk is due to both of these components, but we can 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 2 β 2 σM × 100% = 55%. 220

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 replicating portfolio or using risk-neutral probabilities. We will look at replicating portfolios first. The stock price changes to 66.55 if there are three up-steps, to 54.45 if there are two up-steps, to 44.55 if there is one up-step, and to 35.45 if there are no up-steps. Since the exercise price 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 were two up-steps then the portfolio should hold 0.9545 shares and −49.44 in risk-free assets. (2) if there was an up-step and a down-step or if there were two down-steps, then the replicating portfolio should hold 0 shares and $0 in risk-free assets. After one step we have: if there was an up-step then the portfolio should hold 0.755 shares and −35.558 in risk-free assets but if there was a down-step then the portfolio should hold 0 shares and $0 in risk-free assets. The initial portfolio should have 0.596 shares and $-25.51 in risk-free assets. The price of this portfolio is (0.596)(50) − 25.51 = 4.29. Thus, the initial price of the option 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 q 3 = 0.4327. Thus the expected value (with respect to the risk-neutral probabilities) of the option is (0.4327)(11.55) = 4.9976. The discounted expected value is 4.9976 exp{−(3)(0.05)} = 4.30. Therefore, the initial price of the option is $4.30. (This differs slightly from the price of $4.29 found before because of rounding errors.) (b) q 2 (66.55 − 55) exp{(2)(0.05)} = 5.98. (c) As stated before, initially the replicating portfolio holds 0.596 shares of stock and −25.51 dollars of the risk-free asset. (d) If stock goes up on the first step, the replicating portfolio would then hold 0.755 shares of stock and −35.558 dollars of the risk-free asset.

26

2.

8 Options Pricing

Thus, 0.755 − 0.596 = 0.1590 shares of stock would be purchased by borrowing 35.558 − 25.51exp(0.05) = 8.74 dollars. The cost of the 0.1590 shares is (0.1590)(55) = 8.74 which matches the amount borrowed 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 on each step. The risk-neutral probability of a down-step is 1 − 0.7564 = 0.2436. q=

exp(0.05/20) − 0.99 = 0.625. 1.01 − 0.99

The initial price of the option is exp(0.05)

 20  X 20

k=0

k

 q k (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 option prices against volatility and interpolating one gets that the implied volatility is 0.447. 5. The price of a call option is determined as the discounted expected payoff with respect to the risk neutral probabilities, i.e., under assumption that the stock price follows geometric Brownian motion with drift r, risk-free rate. 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 was in the money. But since the probability of being in the money is strictly less than 1, the price of the option (i.e. expected payoff) would decrease by amount strictly less than $1. Also, if the risk-free rate is positive, as is true in any real situation, the discounting would decrease the change in price. 6. One easily calculates that E(Xi ) = 0.6 − 0.4 = 0.2 and E(Xi2 ) = 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 found the 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 found expected squared prediction error of the best predictor at time t = 2 of the random walk two steps ahead.) (d) The risk-free rate r has not be given, so the answer must be given as a function of r. Assume that r is the simple interest rate. The option is worth $1 at the exercise date (t = 2) if the stock price moves up on both of the first two steps and is worth $0 otherwise. The risk-neutral probability 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 at time t = 1 is (1 + r) − 100 q1 = 102 − 100 The price of the call option is q0 q1 . (1 + r)2 (Note: (1 + r) would be replaced throughout by exp(r) if r was given as a continuously compounded rate.) 7. The probability of an up-step is constant and equals q=

exp(0.03) − 0.8 = 0.5761. 1.2 − 0.8

(a) At t = 2 the put is worth 0, 14, or 46 dollars if there have been two up-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 = 14.2226. exp{(2)(0.03)} (b) If there is an up-step at time t = 0 then early exercise at t = 1 is worth nothing and so is clearly not optimal. If there is a down-step at time 0, then early exercise at t = 1 is worth $30 and the value at t = 1 of holding the option is worth 14q + 46(1 − q) = 26.7490. exp(0.03) Therefore, early exercise is optimal at t = 1 if there had just been a down-step.

28

8 Options Pricing

(c) At time t = 0 an American option is worth 5.7887q + 30(1 − q) = 15.5766. exp(0.03) (d) Let “UU” denote two up-steps and similarly for UD, DU, and DD. At t = 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 44q 2 + 10qp = 16.5433. exp{(2)(0.03)} √ 8. Since d2 = d1 − σ T − t, d d1 /dS = d d2 /dS. Therefore, we only need to show that Sφ(d1 ) − Ke−r(T −t) φ(d1 ) = 0. Next, √ d22 = d21 − 2d1 σ T − t + σ 2 (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 ) = 9. Γ = and

o exp(−d21 /2) n √ S − Ke−r(T −t) elog(S/K)+r(T −t) = 0. 2π

∂ ∂ Φ(d1 ) ∂ d1 ∂2 C(S, T, t, K, σ, r) = ∆(S, T, t, K, σ, r) = = φ(d1 ) 2 ∂S ∂S ∂S ∂S ∂ d1 S −1 . = √ ∂S σ T −t

9 Fixed Income Securities

1. (a) y20 =

1 20

Z

1 25

Z

20

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

0

(b) y25 =

25

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

0

The price is 1000 exp(−25Y25 ) = 525.3188. 2. By the summation formula for a finite geometric series we have 2T X t=1

=

2T −1 C X C 1 = (1 + r)t 1 + r t=0 (1 + r)t

C C 1 − (1 + r)−2T = {1 − (1 + r)−2T }. −1 1 + r 1 − (1 + r) r

Substituting this result into the left-hand side of the equality and simplifying gives us the right-hand side. 3. (a) $50. (b)   50 50 −38 + 1000 − (1 + 0.04) = 1193.70. 0.04 0.04 (c) 4. (a) (b) (c)

1193.70 + 50 = 1243.70. − log(0.848)/5 = 0.0330. 1000 exp{−(4)(0.04)} = 852.14. 852.14/848 − 1 = 0.0049.

30

9 Fixed Income Securities

5. (a)   18 18 + 1000 − (1.02)−20 = 967.30. 0.02 0.02 (b) Below par because the current interest rate is higher than the coupon rate. 6. (a) Solve for y:   50 25 + 1000 − (1 + y)−10 . y y

7.

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. Z

15

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

0

Current value is 100 exp(−0.5625) = 56.98. 8. The yield-to-maturity is Z 20  Z 20 1 (0.02 + 0.001t) dt + 0.0005(t − 10) dt 20 0 10 = 9. (a)

(b)

(c)

(d)

(e)

1  (0.02)(20) + (0.001)(202 )/2 + (0.0005)(20 − 10)2 /2 = 0.0313. 20 1-year: price=1000(1 + 0.02/2)−2 = 980.296 3-year: price=1000(1 + 0.04/2)−6 = 887.9714 5-year: price=1000(1 + 0.045/2)−10 = 800.5101 (rates unchanged) 1-year (now a 0-year): price=1000 3-year (now a 2-year): price=1000(1 + 0.03/2)−4 = 942.2 5-year (now a 4-year): price=1000(1 + 0.0425/2)−8 = 845.2 (rates increase as forecasted) 1-year (now a 0-year): price=1000 3-year (now a 2-year): price=1000(1 + 0.035/2)−4 = 933.0 5-year (now a 4-year): price=1000(1 + 0.0475/2)−8 = 828.8 (rates up) 1-year: return = 1000/980.296 − 1 = 0.0201 3-year: return = 933.0/887.9714 − 1 = 0.0507 5-year: return = 828.8/800.5101 − 1 = 0.0353 (rates unchanged) 1-year: return = 1000/980.296 − 1 = 0.0201 3-year: return = 942.2/887.9714 − 1 = 0.0611 5-year: return = 845.2/800.5101 − 1 = 0.0558

9 Fixed Income Securities

10.

31

(f) No it is not correct as seen in part (e) where the 5-year bond has the highest spot rate but has a lower return than the 3-year bond. The reason that the returns after 1 year after not equal to the spot rates is that the n-year bond has become an n − 1-year bond so it is now priced by discounting at the n − 1-year spot rate. Thus, although spot rates are unchanged, the bond is not priced with the same rate as before and the returns will not be the spot rates. N N N X X d X NPVi Ti Ci Ti exp{−Ti yTi } = − =− Ci exp{−Ti (yTi +δ)} dδ i=1 δ=0 i=1 i=1 =−

N X

Ti ωi

i=1

However,

N X i=1

!

Ci exp{−Ti yTi } N X i=1

so

= −DUR !

Ci exp{−Ti yTi }

N X i=1

!

Ci exp{−Ti yTi } .

= bond price

d bond price = −DUR × bond price. dδ

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 targeted return of 12%. In such cases, the risk can be less that the optimal risk for a 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). Therefore, 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 efficient portfolio 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 , R1stock , and R2option be, respectively, the returns on the portfolio, the stock, and the option on the second stock. Then Rp = wR1stock + (1 − w)R2option and therefore by (5.3) the variance of the return on the portfolio is stock,option 2 . = w2 (σ1stock )2 + (1 − w)2 (σ1stock )2 + 2w(1 − w)σ1,2 σR P

Here σ2option is the standard deviation of the return on the option on stock,option the second stock and σ1,2 is the covariance between the first stock and the option on the second stock. Next σ2option = L2 σ2stock and by (8.33) stock,option stock σ1,2 = L2 σ1,2 . Substituting these into the above expression for 2 σRP 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 trading day. Then VaR(α) = −20, 000 × {0.08/253 + (−1.645)(0.0141)} = 458.

3.

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

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 daily rate since σ is given for a daily return and T is in days. The price of the call option is $8.6576.

36

11 Value-at-Risk

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

115 = 7.7981. 8.6576

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

(f)

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

and

µ bP = (0.5)(7.7981)(0.0001) + (0.5)(0.0002) = 0.00049. σ bP =

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

1/2 +(2)(0.5)(0.5)(7.981)(0.3)(0.017)(0.019) = 0.0377. Therefore, VaR(0.05, 24 hours) = −2000(b µP − 1.645b σP ) = 123. (g) µ bP = (2/3){0.5 + (0.5)(7.7981)}(0.0001) + (1/3)(0.0002) = 0.00049.

and

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

1/2 +(2)(2/3)(1/3){0.5 + (0.5)(7.7981)}(0.3)(0.017)(0.019) = 0.0352. Therefore, VaR(0.05, 24 hours) = −3000(b µP − 1.645b σP ) = 172. −1 6. For i = 1, 2, R(Pi ) = −Si × {µ pi + Φ (α)σi }. Also, the standard deviation 2 2 of the return on P1 + P2 is S1 σ1 + 2S1 S2 ρσ1 σ2 + S22 σ22 ≤ S1 σ1 + S2 σ2 because |ρ| ≤ 1. Therefore, R(P1 +P2 ) ≤ −(S1 µ1 +S2 µ2 )+Φ−1 (α)(S1 σ1 + S2 σ2 ) = 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 400 if ST > 112. (b) Each of the 20,000 simulated values of ST is converted into a loss using the result of part a). Since (20, 000)(0.05) = 1000, VaR(0.05) is the 1000th largest value of these simulated losses and ES(0.05) is the average of the 1000 largest simulated losses. (c) Simulate 20,000 N (0, 1) random variables, Z1 , . . . , Z20,000 . The ith simulated value of ST is 105 exp(0.1 + 5Zi ). Use these as explained in part 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 equality is true because the integrand is symmetric about 0. Therefore, we need only prove the third equality. r Z ∞ Z ∞ 2 2 1 −z 2 /2 √ ze (d/dz)e−z /2 dz dz = − 2 π 2π 0 0 r r r ∞ 2 −z2 /2 2 2 e (0 − 1) = . =− =− π π π 0

2.

Z

∞

−∞

3. (a)

fX (x)dx = 2

Z

0

  Z 1  Z ∞ 1 1dx + x−2 dx 4 0 1 ∞   1 1 − x−1 = 1. = 2 1

∞

fX (x)dx = 2

µ=

3 = 10. 1 − 0.7

(b)

(c)

Var(at ) =

α0 1 = = 2. 1 − α1 1 − 0.5

Var(ut ) =

Var(at ) 2 = = 6.667. 1 − 0.7 0.3

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   q Var(a2 |u1 = 1, u0 = 0.2) = Var 2 α0 + α1 a21 a1 = α0 + α1 a21 = 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 ) = (b)

3 = 7.5. 1 − 0.6

ρY (h) = 0.6|h| . (c) ρa (h) = 1 if h = 0 = 0 otherwise. (d) ρa2 (h) = 0.5|h| . p 6. (a) E(Yt |Xt = 0.1 and at−1 p = 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 onq Xt and at , Yt is a constant, β0 + β1 Xt + δσt ,

plus another constant, 1 + 0.5a2t−1 , times t . Since t is normal, so also is Yt . q (d) Not normal. As 1 + 0.5a2t−1 varies with at−1 , the distribution of q 1 + (0.5)a2t−1 t 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) 3 Z 4 Z 3 Z 4 (t − 2) 2 + 6 = 11.5. 6dt = 5 + {5 + (t − 2)}dt + s(t)dt = 2 3 2 2

2. (a) E(rt+1 − rt |rt = 0.04) = µ(0.04) = 0.1(0.03 − 0.04) = −0.001. Therefore, E(rt+1 |rt = 0.04) = 0.04 − 0.001 = 0.039. (b) σ 2 (rt+1 |rt = 0.02) = σ 2 (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 of 0 as x → −∞ and a limit of 1 as x → ∞. Therefore, a cdf has an infinite area R 2under its curve and cannot be a pdf. (b) Solve .1 = y (2 − x)dx = y 2 /2 − 2y + 2. Solution is p √ 2 ± 22 − (4)(1/2)(1.9) = 2 ± .2. y= (2)(1/2) √ Choose solution between 1 and 2 which is y = 2 − .2. An alternative solution is to notice that s(x) is the linear function s(x) = 2 − x for x between 1 and 2. The area of the triangular region under the curve between 2 − y and 2 is (1/2)(2 − y)2 so .1 = 4. (a)

or y = 2 ±

√

(2 − y)2 2

.2 and solution between 1 and 2 is 2 −

√

.2.

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

42

13 Nonparametric Regression and Splines

(b) s00 (x) = 2 + 2(x − 1)0+ + 0.6(x − 2)0+ and therefore s00 (2.5) = 2 + 2 + 0.6 = 4.6.

14 Behavioral Finance

b = E(X) so that E{X b − E(X)} b = 0. Therefore, 1. (a) First, E(X)   σXW b b Cov(X, X − X) = E E(X) + 2 {W − E(W )} σW   σXW × {X − E(X)} − 2 {W − E(W )} σW     σXW = E [E(X){W − E(W )}] + E E(X) − 2 (W − E(W )) σW +

σ2 σXW 2 (−σXW ) + XW 2 4 σW = 0. σW σW

since E({W − E(W )} = 0. b and X − X b are uncorrelated and X = X b + (X − X), b (b) Because X b + Var(X − X) b = Var(X) b + E(X − X) b 2. Var(X) = Var(X)

b ≤ Var(X) since E(X − X) b 2 ≥ 0. (c) Var(X) 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 paired t-test, not an independent samples t-test. The conclusion of DeBondt and Thaler should be correct. The reason is as follows. First, the returns on the winner and loser portfolios should be positively correlated, since both portfolios should have positive betas and therefore should be positively correlated with the market return. As discussed in Section 2.20.3, when there is a positive correlation between the two items in each pair then the incorrect use of an independent samples t-test will make the p-value larger than it should be. Since the incorrect p-value that DeBondt and Thaler obtained was small enough to reject the null hypothesis, the correct p-value should be even smaller and therefore should be even stronger evidence against the null hypothesis.

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14 Behavioral Finance

3. (a) Even though the two decisions are made concurrently, people will tend to analyze them separately. Many people will choose (A) in Decision 1 in order to be certain of a gain. People hate to lose and will often gamble in order to avoid a certain loss. For this reason, many people will 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 probability 0.25. [B,C]: lose $7500 with probability 0.75 and gain $2500 with probability 0.25. [B,D]: lose $0 with probability 6/16, lose $10,000 with probability 9/16, and gain $10,000 with probability 1/16. (Note: it is assumed that the outcome in the first decision problem is independent of the outcome in the second decision problem. If this is not true, then the probabilities will be different.) (c) [B,C] has the same gain and loss probabilities as [A,D], but the gain is more 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 evidence against the EMH, but there are other reasons besides market efficiency that might explain this good performance. One would need to look at the risk of the Behavioral Growth Fund. If its risk is higher than the risks of the indices to which the Behavioral Growth Fund is being compared, then the higher performance of the Behavioral Growth Fund could be due to a risk premium. Another issue is whether this particular fund was selected as an example because of its strong performance. One should investigate how well have other behavioral funds done. 5. If the market is truly efficient there should be no underreaction and no overreaction. Thus, Fama’s argument suggest that the market is “unbiased” in the sense that it is right on average, but this is not the same as being efficient. Fama’s argument is consistent with the excess volatility that 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 would set plus zero-mean noise. The noise would create excess volatility, exactly what Shiller observed. bt+1 = φXt so the forecast error is Xt+1 − X bt+1 = 6. The forecast of Xt+1 is X t+1 . Therefore the forecast errors are a white noise process and are uncorrelated by the definition of white noise.