Question

Question 3 (4101) Suppose that X, Y and Z are all independent of each other, with the following distributions: X Poisson (1) Y Gamma(,b) Z~ N(0,1) Define A as the sum: A = X+Y+Z a What is E[A]? b What is the MGF of A? (you don't need to re-derive the individual mgfs) c Use ma(t) to find E[A] (should match part a)

Answer 1

We are given here the distributions as:

$X \sim Poisson(\lambda )$

Y ~ Gammala, 8)

$Z \sim N(0,1)$

a) The expected value of A here is computed as:

E(A) = E(X + Y + Z) = E(X) + E(Y) + E(Z)

$E(A) = \lambda + \frac{\alpha}{\beta} + 0$

a E(A) = 1 + 8

This is the required expected value here.

b) The MGF for sum of independent variables, is the product of individual mgf for those variables. Therefore the MGF for A here is obtained as:

$M_A(t) = (1 - \frac{t}{\beta})^{-\alpha}*e^{\lambda(e^t - 1)}*e^{t\mu + 0.5\sigma^2 t^2}$

$M_A(t) = (1 - \frac{t}{\beta})^{-\alpha}*e^{\lambda(e^t - 1)}*e^{0.5t^2}$

$M_A(t) = (1 - \frac{t}{\beta})^{-\alpha}*e^{\lambda(e^t - 1) + 0.5t^2}$

This is the required MGF here.

c) The mgf for A here is differentiated with respect to t to get here:

$M_A'(t) = (1 - \frac{t}{\beta})^{-\alpha}* [\lambda e^t + t] e^{\lambda(e^t - 1) + 0.5t^2} + \alpha (1 - \frac{t}{\beta})^{-\alpha-1}*\frac{1}{\beta} e^{\lambda(e^t - 1) + 0.5t^2}$

The expected value now is computed as:

$E(A) = M_A'(0) = [\lambda] e^{\lambda(1 - 1) } + \frac{\alpha}{\beta} e^{\lambda(1 - 1) }$

$E(A) = M_A'(0) = \lambda + \frac{\alpha}{\beta}$

Which is same as obtained in part a) here.