University of Washington
Department of Economics

Fall 2006
Eric Zivot
Economics 424
Midterm Exam Suggested Solutions

This is a closed book and closed note exam. However, you are allowed one page of notes(double-sided). Answer all questions and write all answers in a blue book or on separate
sheets of paper. Time limit is 1 hours and 50 minutes. Total points = 120.
I. Return Calculations (25 pts, 5 points each)
1. Consider a one year investment in two Northwest stocks: Amazon and Costco.
Suppose you buy Amazon and Costco at the end of November 2005 atPA,t −1 = $48.46, PC ,t −1 = $49.59 and then sell at the end of November 2006 for

PA,t = $37.56, PC ,t = $52.91 . (Note: these are actual closing prices taken from Yahoo!)> pa.1 = 48.46> pa.2 = 37.56> pc.1 = 49.59> pc.2 = 52.91

a. What are the simple annual returns for the two stocks?
> ra = (pa.2 - pa.1)/pa.1> rc = (pc.2 - pc.1)/pc.1> ra[1] -0.2249> rc[1] 0.06695

b. What are the continuously compounded annual returns for the two stocks?
> log(1 + ra)[1] -0.2548> log(1 + rc)[1] 0.0648

c. Costco paid the following dividends between November 2005 and November 2006:$0.11 per share cash dividend in February, $0.11 per share cash dividend in May, and$0.13 per share cash dividend in July. What is the annual simple total return on Costco?
What is the annual dividend yield?
> rc.total = (pc.2 + 0.11 + 0.11 + 0.13 - pc.1)/pc.1> div.y = (0.11 + 0.11 + 0.13)/pc.1

> rc.total[1] 0.07401> div.y[1] 0.007058> # total return = cap gain + div yeild> rc + div.y[1] 0.07401

d. The inflation rate between November 2005 and November 2006 was about 3%. Using
this information, determine the simple and continuously compounded real annual returns
on Amazon and Costco.
> inflat = 0.03
> # simple real returns> ra.real = (1+ra)/(1+inflat) - 1> rc.real = (1+rc)/(1+inflat) - 1> ra.real[1] -0.2475> rc.real[1] 0.03587> # cc real returns> log(1+ra.real)[1] -0.2844> log(1+rc.real)[1] 0.03524

e. At the end of November, 2005, suppose you have $100,000 to invest in Amazon andCostco over the next year. Suppose you sell short $60,000 in Amazon and use theproceeds to buy $160,000 in Costco. Using the results from part a, compute the annual
simple return on the portfolio. Assume that both stocks do not pay a dividend.
> xa = -60000/100000> xc = 160000/100000> xa[1] -0.6> xc[1] 1.6> rp = xa*ra + xc*rc> rp[1] 0.2421

II. Probability Theory (25 points, 5 points each)
1. Suppose you currently hold $100,000 in Starbucks stock. That is, your initial wealth at
the beginning of the month is W0 = $100,000 . Let Rsbux denote the monthly simple returnon Starbucks stock, and assume that Rsbux ~ N (0.03,(0.20) 2 ) . Let W1 = W0 (1 + RSBUX ) be arandom variable representing your wealth at the end of the month.
a) Explain why the normal distribution may not be appropriate for describing thedistribution of simple returns.
The normal distribution is defined from −∞ to ∞ . Simple returns cannot be smaller than
-1. Also, multi-period simple returns are multiplicative (geometric average). That is, the2 period return is a geometric average (multiplicative) of two 1 period returns. If the 1period returns are normally distributed then the 2 period return will not be normal.

b) Compute E[W1 ] , var(W1 ) and SD (W1 )E[W1 ] = W0 + W0 E[ RSBUX ]= $100,000 + $100,000 × (0.03) = $103,000var(W1 ) = W02 var( RSBUX )= (100, 000)2 (0.20)2 = 4 × e+08
SD (W ) = W0 SD ( RSBUX )$100, 000 × (0.20) = $20,000c) What is the probability distribution of W1 ? Sketch the distribution, indicating thelocation of E[W1 ] and E[W1 ] ± 2 ⋅ SD (W1 ) .
Since R is normally distributed and W1 is a linear function of R, W1 is also normallydistributed. The mean and variance of W1 is given in part b above.
zd) Compute the 5% value-at-risk over the month. Hint: q.05
= −1.645 is the 5%
quantile from the standard normal distribution.
Rq.05= 0.03 + (0.20)( −1.645) = −0.299
R
VaR.05 = q.05
W0 = −$29,900e) Now assume that Rsbux denote the monthly continuously compounded return onStarbucks stock. Compute the 5% value-at-risk over the month.
rq.05= 0.03 + (0.20)( −1.645) = −0.299

VaR.05 = ( e q.05 − 1)W0 = −$25.844
rIII. Time Series Concepts (20 points)
1. Let {Yt } represent a stochastic process. Under what conditions is {Yt } covariancestationary? (5 points)

A covariance stationary process satisfies the following three conditions
E[Yt ] = μvar(Yt ) = σ 2cov(Yt , Yt − j ) = γ j (depends on j and not t )2. Consider the random walk modelt

Yt = Yt −1 + ε t = Y0 + ∑ ε j , Y0 = constant.
j =1

ε t ~ iid N (0, σ 2 )Is {Yt } a covariance stationary stochastic process? Why or why not? (5 points)
No. The random walk process is not stationary. The variance of the random walk processdepends on time:
ttj =1j =1var(Yt ) = ∑ var(ε j ) = ∑ σ 2 = tσ 23. The figure below shows annual observations on the dividend yield of the S&P 500
index over the period 1871 through 2000 along with the sample ACF. (10 points)

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Annual dividend price ratio on S&P 500

1870

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1890

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ACF

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Series : dpratio

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1015Lag20a) Does the dividend yield look like a realization from a covariance stationary timeseries? Why or why not.
There is no obvious trend in the dividend yield data. However, the average level looks abit lower at the end of the sample (3%) than at the beginning of the sample (6%). It
could be from a covariance stationary process, or it could be from a process with a meanvalue that depends on time.
b) Assume the dividend yield is covariance stationary. Based on the shape of the sample
autocorrelation function, would an MA(1) process or an AR(1) process best describe thedata? Briefly justify your answer.
The SACF decays toward zero and does not cut off at lag 1. Therefore, it looks more likean AR(1) process than an MA(1) process.
VI. Constant Expected Return Model (50 points, 5 points each)
1. Consider the constant expected return model
Rit = μi + ε it , ε it ~ iid N (0, σ i2 )cov( Rit , R jt ) = σ ij , corr ( Rit , R jt ) = ρij

for the monthly continuously compounded returns on the Vanguard extended marketindex (vexmx) and the Vanguard long-term bond index (vbltx) (subset of class projectdata) over the period September 2001 through September 2006. For this period there are
T=60 monthly observations. The data are shown in the graph below.

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VEXMX: CC Returns

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VBLTX: CC Returns

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Q4a) Do the monthly continuously compounded return data look like they come from acovariance stationary stochastic process? Why or why not?

There is no trend in the returns – they seem to fluctuate around a constant level. Also, thevariability of the returns looks reasonably constant for vbltx; however, the variability forvexmx is lower toward the end of the sample than it is in the beginning of the sample. Acovariance stationary process seems like a reasonable assumption for vbltx but not as
reasonable for vexmx since the variance of vexmx appears to change over time.
b) What are the formulas used to compute estimates for μi , σ i2 , and σ ij ?
T
1 T
1r , σˆ i2 =( rt − μˆ ) 2∑∑t =1 tTT − 1 t =1
1 Tσˆ ij = ∑ t =1 ( rit − μˆi )( rjt − μˆ j )Tμˆi =c) The following S-PLUS output gives the estimates of μi , σ i , and σ ij for vexmx andvbltx from the 5 years of monthly data.
> cbind(muhat.vals,sigmahat.vals)
muhat.vals sigmahat.vals
vexmx
0.005770
0.05308
vbltx
0.006143
0.02648> rhohat.vals
vexmx,vbltx-0.2038

Briefly discuss these estimates in light of what we have learned about the CER model sofar.
Both monthly mean estimates are positive and close to zero. The annualized average
returns are about 6.9% for vexmx and 7.4% for vbltx. Over this period an investment inbonds did better than an investment in stocks. The SD estimates are also fairly small,with the SD of vbltx smaller than vexmx. The annualized SD values are about 18% forvexmx and about 9% for bonds. This is expected since bonds are generally thought to besafer than stocks. Since vexmx is a well diversified portfolio, its small SD relative toindividual stocks is due to the diversification of risk. Interesting, the correlation between
vbltx and vexmx is negative. Typically, stock and bond returns are slightly negatively
correlated or uncorrelated. This is one reason to hold both stocks and bonds in yourportfolio.
d) Using the above output, compute estimated standard errors forμˆi , σˆ i , (i = vexmx, vbltx ) and ρˆ vexmx ,vbltx . Briefly comment on the precision of theestimates.
> nobs = numRows(projectReturns.ts)> se.muhat = sigmahat.vals/sqrt(nobs)> se.sigma = sigmahat.vals/sqrt(2*nobs)

> se.rho = sqrt( (1 - rhohat.vals^2)/nobs )> se.muhat
vexmx
vbltx
0.006853 0.003419> se.sigma
vexmx
vbltx
0.004846 0.002417> se.rho
vexmx,fmagx
0.1264For both vbltx and vexmx, the SE values for the mean estimates are about half the size ofthe mean estimates. This reflects a fair bit of uncertainty about the true value of themean. Similarly, the SE value for the estimated correlation is about half the size of theestimated correlation. This implies a fairly imprecise estimate of correlation.
e) The analytic formulas for SE (σˆ ) and SE ( ρˆ ) are approximations based on the Central
Limit Theorem and may not be accurate for small sample sizes. The bootstrap provides
an alternative means of computing these standard errors. Briefly describe how you would
use the bootstrap to compute a numerical standard error for σ̂ . (Note: I do not want you
to tell me how to do this in S-PLUS – I want you to describe the bootstrap algorithm)The bootstrap works as follows.
Step 1: create B bootstrap samples by sampling with replacement from the original data.
Each bootstrap sample has the same number of observations as the original sample
Step 2: From each bootstrap sample, compute an estimate of σ. This gives B bootstrap
estimates (σˆ1* ,… , σˆ B* )Step 3: Approximate the standard error for σ̂ by computing the sample standard
deviation of (σˆ1* ,… , σˆ B* )1/ 2B⎛ 1⎞(σ i*2 − σ *2 ) 2 ⎟SEboot (σˆ ) = ⎜∑i =11−B⎝⎠f) For vexmx, compute a 95% confidence interval for μ. Also, compute a 95%
confidence interval for ρ vexmx ,vbltx . Briefly comment on the precision of the estimates. In
particular, note if both positive and negative values are in the respective confidence
intervals.
> # 95% ci for mu> upper = muhat.vals["vexmx"]+2*se.muhat["vexmx"]> lower = muhat.vals["vexmx"]-2*se.muhat["vexmx"]

> cbind(lower,upper)
lower
upper
vexmx -0.007935 0.01948> # 95% ci for rho> upper = rhohat.vals+2*se.rho> lower = rhohat.vals-2*se.rho> upper = rhohat.vals+2*se.rho> lower = rhohat.vals-2*se.rho> cbind(lower,upper)
lower
uppervexmx,vbltx -0.4566 0.04893

The 95% CI for the mean of vexmx contains both small positive and negative numbers, so
it is not clear if the expected value is positive. Similarly, the 95% CI for the correlation
between vbltx and vexmx contains both positive and negative values so it is not clear if
the true correlation is negative.
g) Test the following hypotheses using a 5% significance level:
H 0 : μvexmx = 0 vs. H1 : μvexmx ≠ 0 ; H 0 : ρ vexmx ,vbltx = 0 vs. H1 : ρ vexmx ,vbltx ≠ 0 .

You can do the hypotheses tests in two ways: (1) reject the null hypothesis if the value
under the null hypothesis is not in the 95% confidence interval; (2) compute a t-statistic
and reject the null hypothesis of the absolute value of the t-statistic is greater than 2.
Using the confidence interval method, we see that we cannot reject the null hypotheses
that μvexmx = 0 and ρ vexmx ,vbltx = 0 at the 5% level because zero is in both 95% confidence
intervals.
For the t-statistic approach, we compute the following t-statistics> t.stat.mu0 = muhat.vals["vexmx"]/se.muhat["vexmx"]> t.stat.rho0 = rhohat.vals/se.rho> abs(t.stat.mu0)
vexmx
0.842
> abs(t.stat.rho0)
vexmx,vbltx
1.613

Since the absolute value of both t-stats are less than 2, we do not reject the nullhypotheses that μvexmx = 0 and ρ vexmx ,vbltx = 0 at the 5% level.
h) The figure below gives some graphical diagnostics of the return distribution forvexmx. Also, estimated values of the skewness and excess kurtosis for vexmx are

> kurtosis(projectReturns.ts[,"vexmx"])[1] 0.2538> skewness(projectReturns.ts[,"vexmx"])[1] -0.6503

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VEXMX monthly returns

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density estimate

6rvexmx

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Quantiles of Standard Normal

Based on this information, do you think the monthly cc returns on vexmx are normally
distributed? Briefly justify your answer.
The histogram is slightly left skewed (long-left tail) and the estimated skewness ismoderately negative. Also, the boxplot shows a negative outliers and the qq-plot departs
from linearity in the left tail. The excess kurtosis is only slightly greater than 3. Hence,
There is some evidence against the normal distribution for vexmx returns. A formal test
may be computed using the JB statistic> JB = nobs * (x.skew^2 + 0.25 * x.ekurt^2) / 6> JB[1] 4.39

Since the JB statistic is less than 6, we cannot reject the null hypothesis that the returns
on vexmx are normally distributed at the 5% level.
i) Below are the sample autocorrelation functions for vexmx and vbltx. Using theinformation in these graphs, would you say that the CER model assumption that returns
are uncorrelated over time is appropriate? Briefly justify your answer.

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Series : projectReturns.ts[, "vexmx"]

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Series : projectReturns.ts[, "vbltx"]

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For vexmx, none of the sample autocorrelations are outside of the standard error bands
so we cannot reject the null hypothesis that the returns on vexmx are uncorrelated overtime. For vbltx, we only see that the 2nd lag autocorrelation is outside the standard errorbands. Therefore, we can reject the null hypothesis that the returns on vbltx are
uncorrelated over time at the 5% level.
j) Below are graphs showing 24-month rolling estimates of μ and σ along with themonthly returns for vexmx and vbltx (in each graph, the higher dotted line represents therolling standard deviations and the lower solid line represents the rolling means). Basedon these graphs, would you say that the CER model assumption that μ and σ areconstant over time is appropriate? Briefly justify your answer.

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24-month rolling means and sds for VBLTX

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Rolling means
Rolling sds
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The 24-month rolling estimates of the mean and sd for vbltx are fairly constant over time,
supporting the CER model assumptions.

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24-month rolling means and sds for VEXMX

Rolling means
Rolling sds
vexmx returns
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For vexmx, the 24-month rolling mean and sd estimates are clearly changing over time.
The mean starts out negative and trend upward to a positive value at the end of thesample. The sd shows a downward trend over the sample. Therefore, the CER model
assumption of constant parameters is not reasonable for the returns on vexmx.