University of Washington
Department of Economics

Fall 2007
Eric Zivot
Economics 424
Midterm Exam 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 = 110.
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 September 2006 atPA,t −1 = $32.12, PC ,t −1 = $49.19 and then sell at the end of September 2007 for

PA,t = $93.15, PC ,t = $61.37 . (Note: these are actual closing prices taken from Yahoo! The
data for Amazon is not a mistake.)> pa.1 = 32.12> pa.2 = 93.15> pc.1 = 49.19> pc.2 = 61.37

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] 1.9> rc[1] 0.2476

b. What are the continuously compounded annual returns for the two stocks?
> log(1 + ra)[1] 1.065> log(1 + rc)[1] 0.2212

c. Costco paid the following per share cash dividends between September 2006 andSeptember 2007: $0.13 in November, $0.13 in February, $0.145 in April, and $0.145 in
July. What is the annual simple total return on Costco? What is the annual dividend
yield?

> rc.total = (pc.2 + 0.13 + 0.13 + 0.145 + 0.145 pc.1)/pc.1> div.y = (0.13 + 0.13 + 0.145 + 0.145)/pc.1> rc.total[1] 0.2588> div.y[1] 0.01118> # total return = cap gain + div yeild
rc + div.y[1] 0.2588

d. The annual inflation rate between September 2006 and September 2007 was about
3%. Using this information, determine the simple and continuously compounded realannual returns on Amazon and Costco. Note: for Costco, do not include the dividend
adjustments.
> inflat = 0.03
> # simple real returns
ra.real = (1 + ra)/(1 + inflat) - 1> rc.real = (1 + rc)/(1 + inflat) - 1> ra.real[1] 1.816> rc.real[1] 0.2113> # cc real returnslog(1 + ra.real)[1] 1.035> log(1 + rc.real)[1] 0.1917

e. At the end of September, 2006, suppose you have $100,000 to invest in Amazon andCostco over the next year. Suppose you sell short $60,000 in Costco and use the proceeds
to buy $160,000 in Amazon. Using the results from part a, compute the annual simple
return on the portfolio. Assume that both stocks do not pay a dividend.
> xc = -60000/100000> xa = 160000/100000> xa[1] 1.6> xc[1] -0.6> rp = xa * ra + xc * rc
> rp[1] 2.892

II. Probability Theory (30 points, 5 points each)
1. Suppose you currently hold $2M (million) in Starbucks stock. That is, your initial
wealth at the beginning of the month is W0 = $2 M . 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 a random variable representing your wealth at the end of themonth.
a) Compute E[W1 ] , var(W1 ) and SD (W1 )> w0 = 2> e.rsbux = 0.03
> sd.rsbux = 0.2> e.w = w0 + w0 * e.rsbux> e.w[1] 2.06> var.w = w0 * (w0 * sd.rsbux * sd.rsbux)> var.w[1] 0.16> sd.w = w0 * sd.rsbux> sd.w[1] 0.4b) 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 functionof R, W1 is also normally distributed with mean $2.06M andSD $0.4M.

c) Briefly explain why the normal distribution may not be appropriate for describing
the distribution 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). Thatis, the 2 period return is a geometric average(multiplicative) of two 1 period returns. If the 1 period
returns are normally distributed then the 2 period return
will not be normal.

2. Let rt denote the continuously compounded return on some asset in month t andassume
rt ~ iid N (0.02,(0.10)2 )Suppose you currently hold $2M (million) in the asset. That is, your initial wealth at thebeginning of the month is W0 = $2 M .
a. Compute the 1% and 5% value-at-risk (VaR) associated with a one-monthinvestment in the asset. Hint: the 1% and 5% quantiles of the standard normal
distribution are -2.326 and -1.645.
> mu = 0.02
> sd = 0.1> q.05 = mu + sd * (-1.645)> q.01 = mu + sd * (-2.326)> VaR.05 = (exp(q.05) - 1) * w0
> VaR.05[1] -0.2691> VaR.01 = (exp(q.01) - 1) * w0
> VaR.01[1] -0.383

b. Now consider a 12-month (one year) investment. Let rt (12) denote the 12-monthcontinuously compounded return. What are the mean and standard deviation of
the 12-month return?
> mu.a = 12 * mu
> sd.a = sqrt(12) * sd
> mu.a[1] 0.24> sd.a[1] 0.3464

c. Compute the 1% and 5% value-at-risk on a 12-month investment with initial
wealth of $2M.
> q.05 = mu.a + sd.a * (-1.645)> q.01 = mu.a + sd.a * (-2.326)

> VaR.05 = (exp(q.05) - 1) * w0
> VaR.05[1] -0.5619> VaR.01 = (exp(q.01) - 1) * w0
> VaR.01[1] -0.8641

III. Time Series Concepts (15 points)
1. Let {Yt } represent a stochastic process. Under what conditions is {Yt } covariancestationary? (5 points)
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 )2Is {Yt } a covariance stationary stochastic process? Why or why not? (5 points)
No. The random walk process is not stationary. The varianceof the random walk process depends on time:
tt

j =1

j =1var(Yt ) = ∑ var(ε j ) = ∑ σ 2 = tσ 2

3. 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. (5 points)

0.02

0.06

0.10

Annual dividend price ratio on S&P 500

1870

1880

1890

1900

1910

1920

1930

1940

1950

1960

1970

1980

1990

2000

0.4
-0.2

0.2

ACF

0.6

0.8

1.0

Series : dpratio

05101520Laga) 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 like an AR(1) process than anMA(1) process.

VI. Constant Expected Return Model (40 points, 5 points each)
Consider the constant expected return model
rit = μi + ε it , ε it ~ iid N (0, σ i2 )cov( rit , rjt ) = σ ij , corr ( rit , rjt ) = ρ ij

for the monthly continuously compounded returns on the Vanguard extended market
index (vexmx) and the Vanguard long-term bond index (vbltx) (subset of class projectdata) over the period September 2002 through September 2007. For this period there are
T=60 monthly observations. The data are shown in the graph below.

-0.05

0.00

0.05

vbltx

-0.05

0.00

0.05

vexmx

Q42002

Q1

Q2Q3

2003

Q4

Q1

Q2Q3Q42004

Q1Q2Q32005

Q4

Q1Q2Q3

2006

Q4

Q1

Q2Q3Q42007a) Do the monthly continuously compounded return data look like they come from theCER model? Why or why not?
The CER model postulates that cc returns are (covariancestationary) iid normal random variables with constantmeans, variances and covariances (correlations). The above
two return series look a bit like computer simulations fromthe CER model. The returns appear to fluctuate randomly
about a constant mean. However, the volatility appears tobe slightly higher before 2004 than after 2004 suggestingthat the variances of the returns are not constant throughtime.

b) What are the estimators (formulas used to compute estimates) for μi , σ i2 , and σ ij ?
T1 T
1( r − μˆ ) 2r , σˆ i2 =∑∑t =1 tt =1 t
TT −1
1 T
σˆ ij = ∑ t =1 ( rit − μˆi )( rjt − μˆ j )Tμˆi =c) The following S-PLUS output gives the estimates of μi , σ i , and ρ ij for vexmx andvbltx from the 5 years of monthly data.
muhat.vals sigmahat.vals
vexmx 0.0151
0.0349vbltx 0.0042
0.0253> rhohat.vals
vexmx,vbltx-0.1237

Briefly discuss these estimates in light of what we have learned about the CER model sofar.
The monthly mean estimates are positive with the stock fundaround 1.5% and the bond fund around 0.4% per month. The
annualized average returns are about 18% for vexmx and 5%
for vbltx. Over this period an investment in stocks did
much better than an investment in long-term bonds. The SD
estimates are also fairly small, with the SD of vbltx
smaller than vexmx. The annualized SD values are about 12%
for vexmx and about 9% for bonds. This is expected sincebonds are generally thought to be safer than stocks. Since
vexmx is a well diversified portfolio, its small SD
relative to individual stocks is due to the diversificationof risk. Interesting, the correlation between vbltx andvexmx is negative. Typically, stock and bond returns areslightly negatively correlated or uncorrelated. This is one
reason to hold both stocks and bonds in your portfolio.

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 = 60
> se.muhat = sigmahat.vals/sqrt(nobs)> se.sigma = sigmahat.vals/sqrt(2 * nobs)> se.rho = (1 - rhohat.vals^2)/sqrt(nobs)> se.muhat
vexmxvbltx
0.004516 0.003268> se.sigma
vexmx
vbltx
0.003193 0.002311

> se.rho
vexmx,vbltx
0.1271For both vexmx and vbltx the SE values for the mean are
fairly small, around 0.45% for vexmx and 0.33% for vbltx.
For vexmx the SE value for the mean estimate is about 1/3
the size of the mean estimate whereas for vbltx the SEvalues is almost as big as the mean estimate. However, themean estimate for vbltx is closer to zero than the meanestimate for vexmx. The SE value for the estimatedcorrelation is the same size of the estimated correlation.
This implies a fairly imprecise estimate of correlation.

e) For vexmx and vbltx, compute 95% confidence intervals 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.
Since T = 60, we can use our rule-of-thumb for calculating
a 95% confidence interval:> # 95% ci for muupper = muhat.vals + 2 * se.muhat> lower = muhat.vals - 2 * se.muhat> cbind(lower, upper)
loweruppervexmx 0.006046 0.02411
vbltx -0.002351 0.01072
Notice that the 95% confidence intervals for μ are fairly
wide. For vexmx, μ could be as low as 0.6% or as high as
2.4%. Annualized, this range is (7.3%, 29%). For vbltx, μcould be either positive or negative. Hence, μ is not
estimated very precisely for both assets.
> upper = rhohat.vals + 2 * se.rho> lower = rhohat.vals - 2 * se.rho> cbind(lower, upper)
lower uppervexmx,fmagx -0.378 0.1305
Here, the 95% confidence interval for ρ contains both
positive and negative values. We know that the correlation
is not too big but we are not sure about its sign.

f) Test the following hypotheses using a 5% significance level:
H 0 : μvexmx = 0 vs. H1 : μvexmx ≠ 0 ;
H 0 : μvbltx = 0 vs. H1 : μvbltx ≠ 0H 0 : ρ vexmx ,vbltx = 0 vs. H1 : ρ vexmx ,vbltx ≠ 0 .
Here, we can test hypotheses in two ways: (1) use tstatistics; (2) use 95% confidence intervals. Using tstatistics we have
> t.stat.mu0 = muhat.vals/se.muhat> t.stat.rho0 = rhohat.vals/se.rho> abs(t.stat.mu0)
vexmx vbltx
3.339 1.281> abs(t.stat.rho0)
vexmx,vbltx
0.9736Since T=60, our rule-of-thumb decision rule is: reject thenull that the true value is zero at the 5% level if theabsolute value of the t-statistic is greater than 2. We
reject the null only for vexmx.
Using 95% confidence intervals, our decision rule is:reject the null that the true value is zero at the 5% level
if zero is not in the 95% confidence interval. From theprevious question, we see that zero is not in the 95%
confidence interval only for vexmx.

g) The figures below gives some graphical diagnostics of the return distributions forvexmx and vbltx. Also, estimated values of the skewness and excess kurtosis for vexmx
and vbltx arevexmx
vbltx
skewness -0.1111852 -0.9670961
excess kurtosis -0.4891066 2.4082389

0-0.05240.0

680.0510VEXMX monthly returns-0.05

0.00.05

0.10

0.05
0.0

VEXMX returns

8640-0.052

density estimate

10

rvexmx

-0.05

0.0

0.05

0.10

-2

monthly return

-101

2

Quantiles of Standard Normal

0-0.085-0.021015

0.04

vbltx monthly returns

-0.10

-0.05

0.0

0.05

-0.08

-0.02

vbltx returns

1050

density estimate

150.04

rvbltx

-0.10

-0.05

0.0
monthly return0.05

-2-1

0

1

2

Quantiles of Standard Normal

Based on this information, do you think the monthly cc returns on vexmx and vbltx arenormally distributed? Briefly justify your answer.
For vexmx, the graphical diagnostics are consistent with anormal distribution: the histogram is bell shaped, the box

plot is symmetric with no outliers and the qq-plot against
the normal distribution is linear. Additionally, the sample
skewness and excess kurtosis values are close to zero. If
we compute the JB statistics we get> JB = (nobs * (vexmx.skew^2 + 0.25 * vexmx.ekurt^2))/6> JB[1] 0.7217
Since JB < 6, we do not reject the null hypothesis that thereturns on vexmx are normally distributed.
The story is different for vbltx. Here the graphicaldiagnostics indicate some departures from the normaldistribution. This histogram is negatively skewed (longleft tail), the boxplot shows one moderate outlier, and theqq-plot deviates from linearity in the lower tail. The
sample skewness is fairly negative and the excess kurtosis
is quite large. The JB statistic is> JB = (nobs * (vbltx.skew^2 + 0.25 * vbltx.ekurt^2))/6> JB[1] 23.85
which is greater than 6 so we reject the null hypothesis at
the 5% level that the returns are normally distributed.

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.

-0.2

0.2

ACF

0.6

1.0

Series : vexmx.ret

051015Lag

ACF

-0.2

0.2

0.6

1.0

Series : vbltx.ret

051015Lag

The SACF shows estimates of ρj together with error bands
equal to ± 2. If the estimate of ρj is outside of theTerror band then we can reject the null hypothesis that thetrue ρj is zero at the 5% level. For both vexmx and vbltx
all but one value of ρj are below ± 2. Also these values
Tare fairly small. Hence, it appears that the returns are
essentially uncorrelated over time.