In class problems – some answers
January 17, 2013
1) You are told that E(X) = 8 and var(X) = 4. What are the expected values and variances of the
following expressions?
a) Y = 3X + 2
E(Y) = E(3X+2) = 3E(X) + 2 = 26
var(Y) = var(3X + 2) = 9var(X) = 36
b) Y = 0.6X – 4
E(0.6X – 4) = 0.8
var(0.6 - 4) = 1.44
c) Y = X/4
E(X/4) = 2
var(X/4) = 0.25
d) Y = aX + b (a and b are constants)
E(aX + b) = 8a + b
var(aX + b) = 4a2
e) Y = 3X2 + 2
From equation (B.18) in the text, we see that E(X2) = var(X) + [E(X)] 2 = 4 + 82 = 68. Therefore,
E(3X2 + 2) = 68 + 2 = 70
var(3X2 + 2) = 9var(X2)
2) Let X stand for the rate of return on a security, say, Apple, and Y the rate of return on another
security, say, General Foods. Let s2x = 16, s2y = 9, and ρ = -0.8. What is the variance of (X + Y)
in this case? Is it greater or smaller than var(X) + var(Y)? In this instance, is better to invest
equally in the two securities than in either security exclusively? Explain (This problem is the
essence of the portfolio theory of finance).
var(X + Y) = 16 + 9 + 2(-0.8)(4)(3) = 5.8, which is smaller than the sum of var(X) + var(Y) =
25.
If an investment is made equally in the two securities, we have:
var(0.5X + 0.5Y) = 0.25var(X) + 0.25var(Y) + (-0.8)(4/2)(3/2) = 3.85. This variance is
obviously smaller than either variance individually. Therefore it pays to diversify.
3)
Year
1984
1985
1986
1987
1988
1989
1990
1991
1992
1993
1994
1995
Number of new business
incorporations (Y)
634,991
664,235
702,738
685,572
685,095
676,565
647,366
628,604
666,800
706,537
741,778
766,988
Number of business failures
(X)
52,078
57,253
61,616
61,111
57,097
50,361
60,747
88,140
97,069
86,133
71,558
71,128
a) What is the average value of new business incorporations? And the variance?
Average value = 683,939.1; variance = (41,167.5)2
b) What is the average value of business failures? And the variance?
Average value = 67,857.6; variance = (15,200.1)2
c) What is the covariance between Y and X? And the correlation coefficient?
Covariance = 45,932,190.7; correlation coefficient = 0.0737 (Note that this is a sample
covariance.Microsoft Excel default is to calculate the population covariance. Why are the two
numbers different?)
d) Are the two variables independent?
Since the covariance between the two is positive, it seems that the two variables are not
independent.
e) If there is correlation between the two variables, does this mean that one variable causes the
other variable? That is, do new incorporations cause business failures, or vice versa?
No, because correlation does not imply causation.
4) If X~N(10,3), and Y~N(15,8), and if X and Y are independent, what is the probability
distribution of
a) X + Y (X+Y) ~ N(25,11)
b) X – Y (X-Y) ~ N(-5,11)
c) 3X 3X ~ N(30,27)
d) 4X + 5Y (4X + 5Y) ~ N(115,248)
5) Let X and Y represent the rates of return (in percent) on two stocks. You are told that
X~N(15,25) and Y~N(8,4), and that the correlation coefficient between the two rates of return is
-0.4. Suppose you want to hold the two stocks in your portfolio in equal proportion. What is the
probability distribution of the return on the portfolio? Is it better to hold this portfolio or to invest
in only one of the two stocks? Why?
Let W = 1/2(X) + 1/2(Y). In this example,
E(W) = 1/2(15) + 1/2(8) = 11.5
var(W) = (1/4)(25) + (1/4)(4) + 2(1/2)(1/2)(-0.4)(5)(2) = 5.25.
Therefore, W ~ N(11.5, 5.25). The variance, hence the risk, of this portfolio is smaller than that
of security X but greater than that of security Y. It is true that if you invest in security X, the
expected return is higher than the portfolio return, but so is the risk. On the other hand, if you
invest in security Y, the risk is smaller than that of the portfolio but so is the rate of return. Of
course, you do not have to invest equally in the two securities.
6) When a sample of 10 cereal boxes of a well-known brand was reweighted, it gave the
following weights (in ounces):
16.13 16.02 15.90 15.83 16.00 15.79 16.01 16.04 15.96 16.20
a) What is the sample mean? And the sample variance?
Mean = 15.9880 ounces; variance = 0.0158 (ounces squared), standard deviation = 0.1257.
b) If the true mean weight per box was 16 ounces, what is the probability of obtaining such a
(sample) mean? Which probability distribution did you use and why?
t = [(15.988 – 16) / (0.1257 / √10 )] = -0.3019.
For 9 d.f., the probability of obtaining a t value of -0.3019 or smaller is greater than 0.25 (onetailed), the p value being 0.3848. The t distribution is used here because the true variance is
unknown.
7) The same microeconomics examination was given to students at two different universities.
The results were as follows:
Sample mean of X1 = 75, Sample variance of X1 = 9.0, n1 = 50
Sample mean of X2 = 70, Sample variance of X2 = 7.2, n2 = 40
How would you test the hypothesis that the population variances of the test scores in the two
universities are the same? Which probability distribution would you use? What are the
assumptions underlying that distribution?
Use the F distribution. Assuming both samples are independent and come from the normal
populations and that the two population variances are the same, it can be shown that:
F =S12/S22 ~ F with (m – 1) and (n – 1) d.f.
In this example, F =9/7.2 = 1.25 . The probability of obtaining an F value of 1.25 or greater is
0.2371.