How to find $E[x^2] $ for a poisson distribution?

Question

For this question I'm not exactly sure how to find the $E[x^2]$. Can anyone please help me out?

Flaws in a certain type of drapery material appear on the average of one in 150 square feet. If we assume a poisson distribution,

Let X be the number of flaws appearing in 150 square feet. Find $E[x^2]$.

I know that

$E[x^2]$

= $\Sigma$ $x^2 f(x)$

= $\Sigma$ $x^2$ $e^{-\lambda}\frac{\lambda ^x}{x!}$

But I'm not sure how to go further. Can anyone help me out? Thanks

Answer 1

We know:

$Var[X]=E[X^2]-E[X]^2$

So

$E[X^2]=Var[X]+E[X]^2$

For Poisson distribution, $Var[X]=\lambda$ and $E[X]=\lambda$. Therefore:

$E[X^2]=\lambda+\lambda^2$

You can also use the Poisson's mgf: take the second derivative and set $t=0$ to also find that $E[X^2]=\lambda+\lambda^2$. Please let me know if you have further questions, or if you would like me to further elaborate! Good luck!

Answer 2

Hint: Write $x^2=x(x-1)+x$.

Then $E[x^2]=e^{-\lambda}\left[\lambda^2\sum_{x=2}^\infty\frac{\lambda^{x-2}}{(x-2)!}+\lambda\sum_{x=1}^\infty\frac{\lambda^{x-1}}{(x-1)!}\right]$

Can you recognize the summations?