University of Washington
Department of Economics

Spring 2006
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 = 100.
I. Return Calculations (25 pts, 5 points each)
1. Consider a one month investment in two Northwest stocks: Amazon and Costco.
Suppose you buy Amazon and Costco at the end of March atPA,t −1 = $36.53, PC ,t −1 = $54.16 and then sell at the end of April for

PA,t = $35.21, PC ,t = $54.43 . (Note: these are actual closing prices for 2006 taken fromYahoo!)a. What are the simple monthly returns for the two stocks?
> R.A = (p.A[2] - p.A[1])/p.A[1]> R.A[1] -0.03613> R.C = (p.C[2] - p.C[1])/p.C[1]> R.C[1] 0.004985

b. What are the continuously compounded monthly returns for the two stocks?
> r.A = log(1 + R.A)
> r.A
[1] -0.0368> r.C = log(1 + R.C)
> r.C
[1] 0.004973

c. Suppose Costco paid a $1 per share cash dividend at the end of April. What is themonthly simple total return on Costco? What is the monthly dividend yield?
> div.C = 1> R.total.C = (p.C[2] + div.C - p.C[1])/p.C[1]> R.total.C[1] 0.02345> div.yield.C = div.C/p.C[1]> div.yield.C[1] 0.01846

d. Suppose the monthly returns on Amazon and Costco from question (a) above are thesame every month for 1 year. Compute the simple annual returns as well as thecontinuously compounded annual returns for the two stocks.

> R.A.annual = (1 + R.A)^12 - 1> R.C.annual = (1 + R.C)^12 - 1> R.A.annual[1] -0.357> R.C.annual[1] 0.06149> r.A.annual = log(1 + R.A.annual)> r.C.annual = log(1 + R.C.annual)> r.A.annual[1] -0.4416> r.C.annual[1] 0.05967

e. At the end of March, 2006, suppose you have $10,000 to invest in Amazon and Costco
over the next month. Suppose you invest $6000 in Amazon and $4000 in Costco.
Compute the monthly simple return and the monthly continuously compounded return on
the portfolio. Assume that both stocks do not pay a dividend.
> x.A = 6000/10000> x.C = 4000/10000> x.A[1] 0.6> x.C[1] 0.4> R.port = x.A * R.A + x.C * R.C> R.port[1] -0.01969> r.port = log(1 + R.port)> r.port[1] -0.01988

II. Probability Theory (25 points, 5 points each)
1. Consider an investment in a portfolio consisting of Starbucks and Amazon stock. Let
Rsbux denote the monthly simple return on Starbucks and assume
that Rsbux ~ N (0.02,(0.20) 2 ) . Let Ramnz denote the monthly simple return on Amazon andassume that Ramnz ~ N (0.01,(0.10) 2 ) . Further, assume that corr ( Rsbux , Ramnz ) = 0.5 .

a) For a portfolio with 20% of wealth invested in Starbucks and the remaining 80% ofwealth invested in Amazon, compute E [ R p ], var( R p ), and SD ( R p ) where R p denotes thesimple return on the portfolio.
> x.sbux = 0.2> x.amnz = 0.8> mu.sbux = 0.02> var.sbux = 0.2^2> mu.amnz = 0.01
> var.amnz = 0.1^2> rho.sa = 0.5> cov.sa = rho.sa * sqrt(var.sbux * var.amnz)> cov.sa[1] 0.01> mu.p = x.sbux * mu.sbux + x.amnz * mu.amnz> mu.p[1] 0.012> var.p = x.sbux^2 * var.sbux + x.amnz^2 * var.amnz+ 2 * x.sbux * x.amnz * cov.sa> var.p[1] 0.0112> sd.p = sqrt(var.p)> sd.p[1] 0.1058b) What is the probability distribution of R p ? Sketch this distribution, indicating thelocation of E[ R p ] and E[ R p ] ± 2 ⋅ SD ( R p ) .

c) Compute the 5% quantile of the distribution for Rp. Using this quantile, compute themonthly 5% value-at-risk (VaR.05) of an initial $1,000 investment. (Hint: the 5% quantile
for a standard normal random variable is -1.645.)

> q.05 = mu.p + sd.p * (-1.645)> q.05[1] -0.1621> VaR.05 = q.05 * 1000
> VaR.05[1] -162.0904

2. Let rsbux,t denote the continuously compounded return on Starbucks stock in month t
and assume that rsbux ,t ~ iid N (0.02,(0.20) 2 ) . Define the annual continuously
compounded return as rt (12) = rt + rt −1 + " + rt −1 = ∑ j =0 rt − j .
11a) Compute E[rt (12)], var( rt (12)), and SD ( rt (12)) . Is the distribution of rt (12) normal?
> mu.a = 12 * mu.sbux> mu.a[1] 0.24> var.a = 12 * var.sbux> var.a[1] 0.48> sd.a = sqrt(var.a)> sd.a[1] 0.6928
The distribution of r(12) is normal.

b) Using the distribution of rt (12) , compute the annual 5% value-at-risk (VaR.05) of an
initial $1,000 investment.
> q.05.a = mu.a + sd.a * (-1.645)> q.05.a[1] -0.8997> VaR.05.a = (exp(q.05.a) - 1) * 1000
> VaR.05.a[1] -593.3

III. Time Series Concepts (25 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 2
cov(Yt , Yt − j ) = γ j (depends on j and not t )2. The figures below give simulated observations from three different covariancestationary time series (labeled series 1, series 2, and series 3, respectively). Each series
has mean zero. One series represents the Gaussian white noise process

Yt ~ iid N (0,1)

One series represents the MA(1) process
Yt = ε t + 0.75ε t −1 , ε t ~ iid N (0,1)

One series represents the AR(1) process
Yt = 0.75Yt −1 + ε t , ε t ~ iid N (0,1)-2024Series 1050

100150200

250-0.20.2

ACF

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SACF: Series 2051015Lag

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Series 3050

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SACF: Series 305101520Lag

Which series is the Gaussian white noise, which series is the MA(1) process, and which
series is the AR(1) process? Briefly justify your answers. (10 points)
Gaussian white noise: series 2 (SACF shows no autocorrelation)
MA(1) process: series 3 (SACF cuts off at lag 1)
AR(1) process: series 1 (SACF decays geometrically)

3. The figure below shows weekly (Friday) observations on the annualized US 3 monthT-bill rate over the period 1/05/62 through 3/31/95 along with the sample ACF. (10points)a) Does the interest rate series look like a realization from a covariance stationary timeseries? Why or why not.
Interest rates don’t really look like a stationary time series. The level of rates seems tochange over time, and the volatility of the rates also seems to change over time.
b) Assume the interest rate series is covariance stationary. Based on the shape of thesample autocorrelation function, would an MA(1) process or an AR(1) process best
describe the data? Briefly justify your answer.

46810121416

Weekly (Friday) Observations on Treasury Rates

1965

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Lag

Only the first lag autocorrelation for an MA(1) process is non-zero. For interest rates,
the autocorrelations are non-zero for up to 30 lags (weeks). This type of behavior can be
captured by an AR(1) process with φ very close to 1.

IV. Descriptive Statistics (25 points, 5 points each)
1. Consider one year of daily continuously compounded (cc) returns on Microsoft stockcomputed using daily closing prices over the period May 2, 2005 – April 28, 2006. The
drop in prices at the end of the sample is not a mistake. On April 28, 2006 Microsoft
stock’s price dropped 11.4% from the day before.

24.5

25.5

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MSFT: Prices

JunJulAug

Sep
2005

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JanFebMar2006

AprMayDec

JanFebMar2006

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MSFT: CC Returns

JunJulAug

Sep
2005

Oct

Nov

a. Do the daily cc returns appear to be a realization from a covariance stationary
stochastic process? Briefly justify your answer.
Except for the last observation, the cc returns look to be reasonably covariancestationary. There is no obvious trend in the data. The volatility seems to vary a little bitwith time.

0-0.10510-0.05

150.020250.05

MSFT monthly returns

-0.10

-0.05

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ret.mat

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density estimate

300.05

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monthly return0.00.05

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Quantiles of Standard Normal

23

b. The figure above shows various graphical diagnostics regarding the empirical
distribution of the daily cc returns on Microsoft. Based on these diagnostics, do you thinkthat the normal distribution is a good model for the underlying probability distribution ofthe daily cc returns on Microsoft? Briefly justify your answer by commenting on each of
the four plots.
The graphical diagnostics do not support the normal distribution for daily cc returns. The
histogram is not too informative about the shape of the distribution. However, thesmoothed histogram shows the impact of the large outlier in the long left tail and reveals
some negative skewness. The boxplot and qqplot show strong departures from normal
behavior. The boxplot shows several outliers besides the one at the end of the sample, thethe qqplot shows a strong departure from linear behavior at the upper and lower
quantiles.

c. Summary descriptive statistics, computed from S-PLUS, for the daily cc returns aregiven below. Which of these summary statistics indicate evidence for, or against, thenormal distribution as the probability model for the daily cc returns?
> summaryStats(msftDailyReturns)
Sample Quantiles:
min
1Q
median3Q
max-0.1208 -0.005378 -0.0004037 0.005444 0.04769
Sample Moments:
mean
std skewness kurtosis-0.0001255 0.01187-3.81245.29
Number of Observations:

250

Data from a normal distribution should have skewness close to zero and kurtosis close tothree. The daily Microsoft cc returns has a large negative skewness and a very large
kurtosis which is strong evidence against the normal distribution.

d. On April 28th, Microsoft’s stock dropped 11.4% from the previous day. This resulted
in over a $32 Billion loss in market value for Microsoft. Was this an unusual event if thecc returns on Microsoft follow a normal distribution? Briefly justify your answer.
If the Microsoft returns were normally distributed then there would be essentially a zerochance of seeing a -11.4% return (-12.08 cc return) in any given day. Recall, with anormal distribution 99% of the observations are between ± 3 standard deviations fromthe mean. The sample mean is -0.0001 and the sample SD is 0.01187. Therefore,

mean ± 3SD ≈ [−0.035, 0.035]

Hence we should see daily returns between -3.5% and 3.5% is the data were normally
distributed. The observed -11.4% return is more than 10 values of the SD from the mean.
e) The empirical 1% and 5% quantiles from the daily cc returns are given below.
> quantile(msftDailyReturns, probs=c(0.01, 0.05))
1%5%-0.02285873 -0.01236073

Using these quantiles, compute the daily 1% and 5% value-at-risk (VaR) based on an
investment of $100,000.
> q.msft = c(-0.02285873, -0.01236073)> VaR.msft = (exp(q.msft) - 1) * 100000> VaR.msft[1] -2259.945 -1228.465

V. Constant Expected Return Model (25 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 Boeing and Microsoft (same data aslab 5) over the period July 1992 through October 2000. For this period there are 100
monthly observations.
a) Based on the S-PLUS output below, give the “plug-in principle” estimates for
μi , σ i2 , σ i , σ ij and ρij for the two assets. Arrange these estimates nicely in a table.
muhat.vals sigma2hat.vals sigmahat.vals
rboeing
0.012436
0.0057945
0.076121
rmsft
0.027564
0.0114112
0.106823
covhat.vals rhohat.vals
rboeing,rmsft -0.000067409 -0.0082898

This is the “who’s buried in Grant’s tomb” question. The answer is given to you.
b) Using the above output, compute estimated standard errors forμˆi , σˆ i , (i = boeing , microsoft ) and ρˆ msft ,boeing . Briefly comment on the precision of theestimates.
The analytic formulas for these estimated standard errors are2m ( μˆ ) = σˆ , SEm (σˆ ) = σˆ , SEm ( ρˆ ) = (1 − ρˆ )SET
2TT

Using the above formulas we get> nobs[1] 100> se.muhat = sigmahat.vals/sqrt(nobs)
cbind(muhat.vals, se.muhat)
muhat.vals se.muhat
rboeing
0.01244 0.007612
rmsft
0.02756 0.010682> se.sigmahat = sigmahat.vals/sqrt(2 * nobs)> cbind(sigmahat.vals, se.sigmahat)sigmahat.vals se.sigmahatrboeing
0.076120.005383
rmsft
0.10682
0.007554> se.rhohat = (1 - rhohat.vals^2)/sqrt(nobs)> cbind(rhohat.vals, se.rhohat)rhohat.vals se.rhohatrboeing,rmsft-0.00829
0.09999

The SE values for the mean are one half the size of the mean, which is fairly large for
returns. Therefore, the mean is not estimated very precisely. The SE values for the SD areabout ten time smaller than the SD which is fairly small for returns. Therefore, the SD isestimated fairly precisely. The SE value, 0.1, for the correlation is larger than thecorrelation estimate. Since the correlation lies between -1 and 1 the uncertainty impliedby the SE value is moderate.
c) For Boeing, compute a 95% confidence interval for μ. Also, compute a 95%
confidence interval for ρ msft ,boeing . Briefly comment on the precision of the estimates. In

particular, note if both positive and negative values are in the respective confidence
intervals.
> ci.95.mu = c(muhat.vals[1] - 2 * se.muhat[1], muhat.vals[1]+ 2 * se.muhat[1])> names(ci.95.mu) = c("lower", "upper")> ci.95.mulowerupper-0.002789 0.02766> ci.95.rho = c(rhohat.vals - 2 * se.rhohat, rhohat.vals+ 2 * se.rhohat)> names(ci.95.rho) = c("lower", "upper")> ci.95.rho
lower upper-0.2083 0.1917

Both 95% confidence intervals contain positive and negative values. Because positive
and negative mean returns have very different implications, this result indicates the data
is not very informative about the true mean. For the correlation, the values in theconfidence interval indicate a weak linear association but the data is not clear about thedirection of this association.

The bootstrap can be used to compute an estimated standard error for the estimated 5%monthly value-at-risk, based on a $100,000 investment, computed using the formulaˆ .05 = ( e qˆ.05 − 1) ⋅ 100,000, qˆ.05 = μˆ + σˆ ⋅ ( −1.645)
VaR
The following S-PLUS output is based on 1000 bootstrap replications of the above valueat-risk formula from the monthly cc returns for Boeing:> VaRhat.bootCall:bootstrap(data = rboeing, statistic = Value.at.Risk)
Number of Replications: 1000
Summary Statistics:
Observed BiasMeanSErboeing-10665 147.6 -10517 1325d) What is the bootstrap estimate of the standard error for the value-at-risk estimate?
Using this standard error, compute an approximate 95% confidence interval for the true
5% value-at-risk.
The bootstrap SE is
n 0.05 ) = 1325
SEboot (VaR

and a 95% confidence interval is
n 0.05 ± 2 × SE (VaR
n 0.05 ) = [−$13,315, −$8,015]
VaRboote) The figure below (on the next page) shows a histogram and normal qq-plot of thebootstap value-at-risk values. Based on the shape of the bootstrap distribution, how good
do you think is the 95% bootstrap confidence interval you computed in part d)?

rboeing

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Quantiles of Standard Normal

The bootstrap histogram looks pretty normal, and the qq-plot is mostly linear which also
supports normality. Therefore, the boostrap estimates should be accurate.