The moment generating function of X is given by MXt exp2et

The moment generating function of X is given by MX(t) = exp{2et ? 2} and that of Y by MY(t) = (3et+1)^10. If X and Y are independent, what are

(a)P{X+Y=2}?

(b) P{XY = 0}?

(c) E[XY ]?

Solution

X is a Poisson random variable with parameter 2, Y is a binomial random
variable with parameter (10, 3/4). Thus
(a)
P(X + Y = 2) = P(X = 0, Y = 2) + P(X = 1, Y = 1) + P(X = 2, Y + 0)
= P(X = 0)P(Y = 2) + P(X = 1)P(Y = 1) + P(X = 2)P(Y = 0)
= e-2 45(3/4)2(1/4)8+2e-210(3/4)1(1/4)9+2e-2(1/4)10

(b) P(XY = 0) = P(X = 0) + P(Y = 0) P(X = 0, Y = 0)

= e-2 + (1/4)10- e-2(1/4)10


(c) E[XY ] = E[X] · E[Y ] = 2 ·30/4=15

The moment generating function of X is given by MX(t) = exp{2et ? 2} and that of Y by MY(t) = (3et+1)^10. If X and Y are independent, what are (a)P{X+Y=2}? (b)

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