Page 99 - Schaum's Outlines - Probability, Random Variables And Random Processes

MULTIPLE RANDOM VARIABLES [CHAP 3

Since the probability must be nonnegative, we conclude that
Fxdx2 , Y2) - Fxyix,, ~2) - Fxdx2 9 Y,) + Fxdx,, Y,) 2 0
ifx2 2 x, andy, 2 y,.

3.6. Consider a function
i Olx<co,O~y<oo
1 - e-(x+~)
F(x9 Y) = otherwise
Can this function be a joint cdf of a bivariate r.v. (X, Y)?
It is clear that F(x, y) satisfies properties 1 to 5 of a cdf [Eqs. (3.5) to (3.911. But substituting F(x, y) in
Eq. (3.12) and setting x, = y, = 2 and x, = y, = 1, we get

Thus, property 7 [Eq. (3.12)] is not satisfied. Hence F(x, y) cannot be a joint cdf.

Consider a bivariate r.v. (X, Y). Show that if X and Y are independent, then every event of the
form (a < X 5 b) is independent of every event of the form (c < Y I d).

By definition (3.4), if X and Y are independent, we have

Setting x, = a, x, = b, y, = c, and y2 = d in Eq. (3.95) (Prob. 3.9, we obtain

which indicates that event (a < X I b) and event (c < Y I d) are independent [Eq. (1,4611.

3.8. The joint cdf of a bivariate r.v. (X, Y) is given by
i ~20,~20,a,B>O
(l-e-"")(l-e-By)
Fx& Y) = () otherwise
(a) Find the marginal cdf's of X and Y.
(b) Show that X and Y are independent.
(c) Find P(X 5 1, Y 5 I), P(X 2 I), P(Y > I), and P(X > x, Y > y).
(a) By Eqs. (3.13) and (3.14), the marginal cdf's of X and Y are

(b) Since FX-(x, y) = FX(x)Fy(y), X and Y are independent.
(c) P(X I 1, Y I 1) = Fxr(l, 1) = (1 - e-")(l - e-8)
P(X 5 1) = Fx(l) = (1 - e-")
P(Y > 1) = 1 - P(Y I 1) = 1 - Fd1) = e-@