Suppose that the moment generating function of a random vari

Suppose that the moment generating function of a random variable X is MX(t)=exp(4e^(t)4) and that of a random variable Yis MY(t)=((3/5)e^(t)+(2/5))^6. If X and Y are independent, find each of the following:

P{X+Y = 3} =

P{XY = 0} =

E[XY] =

E[(X+Y)^2} =

Solution

X is poisson variate with E(X)=4, V(X)=4

Y is binomial variate with E(Y)=6*3/5=18/5, V(Y)=6*3/5*2/5=36/25

since X and Y is independent then

E(XY)=E(X)*E(Y)=4*18/5=72/5

E[(X+Y)^2} =E(X²+Y²+2XY)=E(X²)+E(Y²)+2E(XY)=20+370/25+144/5=1590/25

E(X²)=V(X)+(E(X))²=4+16=20

E(Y²)=V(Y)+(E(Y))²=36/25+324/25=370/25