Page 124 - Probability, Random Variables and Random Processes

MULTIPLE RANDOM VARIABLES [CHAP 3

which is equal to the pdf of a normal r.v. with mean py + p(a,/ox)(x - ,ux) and variance (1 - p2)ay2. Thus,
we get

Note that when X and Y are independent, then p = 0 and E(Y I x) = py = E(Y).

3.52. The joint pdf of a bivariate r.v. (X, Y) is given by

(a) Find the means of X and Y.
(b) Find the variances of X and Y.
(c) Find the correlation coefficient of X and Y.
We note that the term in the bracket of the exponential is a quadratic function of x and y, and hence
fxy(x, y) could be a pdf of a bivariate normal r.v. If so, then it is simpler to solve equations for the various
parameters. Now, the given joint pdf of (X, Y) can be expressed as

where

Comparing the above expressions with Eqs. (3.88) and (3.89), we see that fxy(x, y) is the pdf of a bivariate
normal r.v. with p, = 0, py = 1, and the following equations:

Solving for ax2, ay2, and p, we get
ax2=ay2=2 and p=i
Hence
(a) The mean of X is zero, and the mean of Y is 1.
(b) The variance of both X and Y is 2.
(c) The correlation coefficient of X and Y is 4.

3.53. Consider a bivariate r.v. (X, Y), where X and Y denote the horizontal and vertical miss dis-
tances, respectively, from a target when a bullet is fired. Assume that X and Y are independent
and that the probability of the bullet landing on any point of the xy plane depends only on the
distance of the point from the target. Show that (X, Y) is a bivariate normal r.v.
From the assumption, we have

fxu(x7 Y) =fx(x)fu(y) = s(x2 + y2)
for some function g. Differentiating Eq. (3.109) with respect to x, we have

fi(~)fY(~) = 2xs'(x2 + y2)
Dividing Eq. (3.1 10) by Eq. (3.1 09) and rearranging, we get