Question

For the a Poisson random variable x generating function G(t)= exp^lambda(t-1) show that the expected value E(x) and variance (x) are given by lambda and lambda respectively.

Answer 1

$G(t)=e^{\lambda(e^t-1)}$

$E(x)=G'(0)$

$G'(t)=\lambda e^te^{\lambda(e^t-1)}=\lambda e^{\lambda t(e^t-1)}$

$G'(0)==\lambda e^{\lambda 0(e^t-1)}=\lambda$

$var(x)=E(x^2)-[E(x)]^2$

$E(X^2)=G''(0)$

$G''(t)=\lambda e^{\lambda e^t+t-\lambda}(\lambda e^t+1)$

$G''(0)=\lambda e^{\lambda e^0+0-\lambda}(\lambda e^0+1)=\lambda(\lambda+1)=\lambda^2+\lambda$

From the first part, $E(x)=\lambda$ . Thus;

$var(x)=E(x^2)-[E(x)]^2=\lambda^2+\lambda-\lambda^2=\lambda$