Transformation of variables in Poisson distribution If $X$ is a random variable which follows $\text{Poisson}(u)$ distribution then what is the variance of $X^2$?
 A: Ok, we can use some tricks to find the variance of $X^2$ for $X$ with a $Poisson(u)$
Let $Y=X^2$
$Var(Y)=E(Y^2)-[E(Y)]^2=E(X^4)-[E(X^2)]^2$
For a Poisson distribution we know that $Var(X)=E(X^2)-[E(X)]^2=u$  and $E(X)=u$
$\therefore E(X^2)=u+u^2$
Next we need to find $E(X^4)$, we need some tricks here.
We first to find $E(X^3)$
We will first calculate the Expectation of $E[X(X-1)(X-2)]$
$E[X(X-1)(X-2)]=\sum_{i \ge 0} x(x-1)(x-2)\frac{ u^xe^{-u}}{x!}\\=u^3e^{-u}\sum_{x \ge 3}\frac{u^{x-3}}{(x-3)!}=u^3e^{-u}*e^u=u^3$ 
(can you see the tricks above?)
Next
$E[X(X-1)(X-2)]=E(X^3-3X^2+2X)=E(X^3)-3E(X^2)+2E(X)=u^3$
We already know the $E(X^2)$ and $E(X)$ so we now can calculate $E(X^3)$
$E(X^3)=u^3+3(u+u^2)-2u=u^3+3u^2+u$
Ok, use the same tricks you can calculate $E(X^4)$, I would like to leave it for yourself to calculate it. But I will tell you the result.(you will use $E(X^3)$ when you calculate $E(X^4)$)
$E(X^4)=u^4+6u^3+7u^2+u$
$\therefore Var(X^2)=Var(Y)=E(X^4)-E(X^2)^2\\=(u^4+6u^3+7u^2+u)-(u+u^2)^2\\=4u^3+6u^2+u$
