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1. CAPM Model The CAPM model was fit to model the excess returns of
Exxon-Mobil (Y) as a linear function of the excess returns of the market (X)
as represented by the S&P 500 Index.
Yi = α + βXi + Ei
where the Ei are assumed to be uncorrelated, with zero mean and
constant
variance σ2 . Using a recent 500-day analysis period the following output
was generated in R:
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print(summary(lmfit0))
Call:
lm(formula = r.daily.symbol0.0[index.window] ~ r.daily.SP500.0[index.window],
x = TRUE, y = TRUE)
Residuals:
Min 1Q Median 3Q
Max
0.038885 -0.004415 0.000187 0.004445 0.026748
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.0004805 0.0003360 -1.43
0.153
r.daily.SP500.0[index.window] 0.9190652 0.0454380 20.23
<2e-16
Residual standard error: 0.007489 on 498 degrees of freedom
Multiple R-squared: 0.451, Adjusted R-squared: 0.4499
F-statistic: 409.1 on 1 and 498 DF, p-value: < 2.2e-16

(a). Explain the meaning of the residual standard error.
Solution: The residual standard error is an estimate of the standard devi ation
of the error term in the regression model. It is given by
It measures the standard deviation of the difference between the actual and
fitted value of the dependent variable.
(b). What does “498 degrees of freedom” mean?
Solution: The degrees of freedom equals (n − p) where n = 500 is the number
of sample values and p = 2 is the number of regression parameters being
estimated.
(c). What is the correlation between Y (Stock Excess Return) and X (Market
Excess Return)?
Solution: The correlation is
(we know it is positive because of the positive slope coefficient 0.919)

(d). Using this output, can you test whether the alpha of Exxon Mobil is zero
(consistent with asset pricing in an efficient market).
H0 : α = 0 at the significance level α = .05?
If so, conduct the test, explain any assumptions which are necessary, and
state the result of the test?
Solution: Yes, apply a t-test of H0: intercept equals 0. R computes this in the
coefficients table and the statistic value is −1.43 with a (two-sided) p-value of
0.153. For a nominal significance level of .05 for the test (twosided), the null
hypothesis is not rejected because the p-value is higher than the significance
level. The assumptions necessary to conduct the test are that the error terms
in the regression are i.i.d. normal variables with mean zero and constant
variance σ2 > 0. If the normal distribution doesn’t apply, then so long as the
error distribution has mean zero and constant variance, the test is
approximately approximately correct and equivalent to using a z-test for the
parameter/estimate and the CLT.)
(e). Using this output, can you test whether the β of Exxon Mobil is less than 1,
i.e., is Exxon Mobil less risky than the market:
H0 : β = 1 versus HA : β < 1.
If so, what is your test statistic; what is the approximate P -value of the test
(clearly state any assumptions you make)? Would you reject H0 in favor of
HA?

Solution: Yes, we apply a one-sided t-test using the statistic:
Under the null hypothesis T has a t-distribution with 498 degrees of free dom.
This distribution is essentially the N(0, 1) distribution since the degrees of
freedom is so high. The p-value of this statistic (one-sided) is less than 0.05
because P (Z< −1.645) = 0.05 for a Z ∼ N(0, 1) so P (T< −1.7841) ≈ P (Z<
−1.7841) which is smaller.
5. For the following batch of numbers:
5, 8, 9, 9, 11, 13, 15, 19, 19, 20, 29
(a). Make a stem-and-leaf plot of the batch.
(b). Plot the ECDF (empirical cumulative distribution function) of the batch.
(c). Draw the Boxplot of the batch.
Solution:
x=c(5,8,9,9,11,13,15,19,19,20,29)
 stem(x)
The decimal point is 1 digit(s) to the right of the |

0 | 5899
1 | 13
1 | 599
2|0
2|9
> plot(ecdf(x))

median(x)
[1] 13
quantile(x,probs=.25)
25%
9
quantile(x,probs=.75)
75%
19
Boxplot(x)

Note that the center of the box is at the median (19), the bottom is at the 25-th
percentile and top is at the 75-th percentile. The inter-quartile range is (19-
9)=10, so any value more than 1.5 × 9 = 13.5 units above or below the box will
be plotted as outliers. There are no such outliers.
6. Suppose X1,...,Xn are n values sampled at random from a fixed distri
bution:
Xi = θ + Ei
where θ is a location parameter and the Ei are i.i.d. random variables with
mean zero and median zero.
(a). Give explicit definitions of 3 different estimators of the location pa
rameter θ.
(b). For each estimator in (a), explain under what conditions it would be
expected to be better than the other two.
Solution: (a). Consider the sample mean, the sample median, and the
10%-Trimmed mean.

θT rimmedMean = average of {Xi} after excluding the highest 10% and the
lowest 10% values.
(b). We expect the sample mean to be the best when the data are a random
sample from the same normal distribution. In this case it is the MLE and will
have lower variability than any other estimate.
We expect the same median to be the best when the data are a random
sample from the bilateral exponential distribution. In this case it is the MLE and
will hve lower variability, asymptotically than any other esti mate. Also, the
median is robust against gross outliers in the data result ing from the
possibility of sampling distribution including a contamination component.
We expect the trimmed mean to be best when the chance of gross errors in
the data are such that no more than 10% of the highest and 10% of the lowest
could be such gross errors/outliers. For this estimate to be better than the
median, it must be that the information in the mean of the re maining values
(80% untrimmed) is more than the median. This would be the case if 80% of
the data values came from a normal distribution/model. arise from a normal
distribution with

q _0.5 q_0.75 q_0.9 q_0.95 q_0.99 q_0.999
N(0,1) 0.00 0.67 1.28 1.64 2.33 3.09
t (df=1) 0.00 1.00 3.08 6.31 31.82 318.31
t (df=2) 0.00 0.82 1.89 2.92 6.96 22.33
t (df=3) 0.00 0.76 1.64 2.35 4.54 10.21
t (df=4) 0.00 0.74 1.53 2.13 3.75 7.17
t (df=5) 0.00 0.73 1.48 2.02 3.36 5.89
t (df=6) 0.00 0.72 1.44 1.94 3.14 5.21
t (df=7) 0.00 0.71 1.41 1.89 3.00 4.79
t (df=8) 0.00 0.71 1.40 1.86 2.90 4.50
t (df=9) 0.00 0.70 1.38 1.83 2.82 4.30
t (df=10) 0.00 0.70 1.37 1.81 2.76 4.14
t (df=25) 0.00 0.68 1.32 1.71 2.49 3.45
t (df=50) 0.00 0.68 1.30 1.68 2.40 3.26
t (df=100) 0.00 0.68 1.29 1.66 2.36 3.17
t (df=500) 0.00 0.67 1.28 1.65 2.33 3.11
Percentiles of the Normal and t Distributions statisticshomeworkhelper.com

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