Answer to Question #287557 in Statistics and Probability for Brenda

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

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

We want to verify that $var(x)=E(x)=\lambda$

Now,

$E(x)=G'(t=1)$

First,

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

Setting $t=1$ in $G'(t)$,

$G'(t=1)=\lambda e^{\lambda\times0}=\lambda\times1=\lambda$

Therefore, $E(x)=\lambda$

To find $var(x)$, we first determine $E(x^2)-E(x)=G''(t)$

Now,

$G''(t)=\lambda^2e^{\lambda(t-1)}$

Setting $t=1$ in $G''(t=1)$ we get,

$G''(t=1)=\lambda^2e^{0}=\lambda^2\times1=\lambda^2$

We determine the value of $E(x^2)$ as follows.

$G''(t)=E(x^2)-E(x)\implies E(x^2)=G''(t)+E(x)=\lambda^2+\lambda$

So,

$E(x^2)=\lambda^2+\lambda$

Thus,

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

Therefore, $E(x)=var(x)=\lambda$ as required.