M374G/M384G/CAM384T, Fall 2008
ASSIGNMENT #3: DUE FRIDAY, OCTOBER 10
I. Suppose the distribution of X given Y is uniform on (0,Y). In other words, the
conditional probability density function of X given Y is
1/y, if 0 < x < y
f(x|y) =
0, if x < 0 or x > y
Suppose also that Y has (marginal) probability density function
2y, 0 < y < 1
fY(y) =
0, otherwise.
a. Find the joint probability density function f(x,y) of X and Y.
b. Find the marginal probability density function of X. Be sure to include a sketch
of the region where f(x,y) ≠ 0 in your solution.
c. Find the conditional probability density function f(y|x) of Y given X.
d. Use your answer to (c) to describe in words the conditional distribution of Y
given X.
II. You may (and will probably need to) use the following facts from probability in this
problem:
• If the random variable Z is a function g(X,Y) of the random variables X and Y and
f(x,y) is the joint pdf of X and Y, then E(Z) =
!
g(x,y)f(x,y)dxdy
"#
#
$
"#
#
$
.
• If the random variable Y is a function g(X) of the random variable X and f(x) is the
pdf of X, then E(Y) =
!
g(x)f(x)dx
"#
#
$
.
a. Prove that if X and Y are independent random variables, then E(XY) = E(X)E(Y).
b. Explain how this shows that independent random variables have zero covariance
(i.e., are uncorrelated).
c. Suppose that the random variable X is uniform on the interval [-1, 1]. Let Y = X2.
(Thus X and Y are not independent.) Calculate Cov(X,Y) to show that it is zero. (You
will first have to figure out the pdf of X, then use this to calculate both E(X) and E(Y).)
Thus the converse of the statement in (b) is not true: We have produced random variables
X and Y that are uncorrelated, but not independent.
III. M 374G: Problems 4.2 and 4.3 (p. 79)
M 384G/CAM 384T: Problem 4.6 (pp. 79 - 80)
IV. Problem 4.7 (p. 80) Please note: The questions asked are about the two population
mean functions in general, not about the specific case illustrated in Figure 4.10. Be sure
to explain how you got your answers.