Question

Suppose that X 1 has a Poisson distribution with mean 2, X 2has a Poisson distribution with mean 3 , X 3 has a Poisson distribution with mean 5 and that X 1 , X 2 and X 3 are independent. Define Y = X 1 + X 2 + X 3. Determine the moment-generating function for Y.

Answer 1

Let ,

X1 → Poisson(mean = 11 = 2)

Therefore , the moment generating fucntion of $X_1$ is ,

My(t) = 42(t-1) = 2(t-1)

Also ,

Let ,

X2 + Poisson (mean = 12 = 3)

Therefore , the moment generating fucntion of $X_2$ is ,

Mx,(t) = e12(et-1) = 3(et-1

Let ,

X3 → Poisson (mean = 13 = 5)

Therefore , the moment generating fucntion of $X_3$ is ,

My(t) = e3(et-1) = €5(et-1)

Therefore , the moment generationg function of is ,

Y = X1 + X2 + X3

My(t) = MX1+Xz+xz(t)

= Mx,(t) * Mx (t) * Mx (t)
; Since , $X_1$ , $X_2$ and $X_3$ are independent.

= 2(t-1)* <3(et-1) * €5(et-1)

= (2+3+5)(e-1

= 10(-1

Therefore ,

Y → Poisson (1 = 10)