In class problems – some answersJanuary 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 / Number of new business incorporations (Y) / Number of business failures (X)1984 / 634,991 / 52,078
1985 / 664,235 / 57,253
1986 / 702,738 / 61,616
1987 / 685,572 / 61,111
1988 / 685,095 / 57,097
1989 / 676,565 / 50,361
1990 / 647,366 / 60,747
1991 / 628,604 / 88,140
1992 / 666,800 / 97,069
1993 / 706,537 / 86,133
1994 / 741,778 / 71,558
1995 / 766,988 / 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) 3X3X ~ 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 isthe 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 (one-tailed), 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.