Normal distribution non linear transformation

Question

I have the following problem :

Given $X \sim N(\mu,\sigma^2)$ and $X' = h(X) = (\frac{x-\mu}{\sigma})^2$
Find $E[X']$ and $V[X']$.

My reasoning is as follow :

Since $X' \sim (\frac{x-\mu}{\sigma})^2$ if we take $Z \sim N(0,1)$ then $X' \sim Z^2$.

Also since $E[X'] = \int_{-\infty}^{\infty}x'f_{x'}(x')dx \sim \int_{-\infty}^{\infty}z^2f_{z}(z)dz$

Thus $E[X']$ is the same as the second, non centered, momentum of a normal standard distribution, is that correct?

Similarly $V[X'] = E(X'^2) - [E(X')]^2$ thus I can calculate the second momentum for $Z$ and it's mean to obtain $V[X']$

Answer 1

Your approach is correct but you have made some mistakes. $V(X')=E(X')^{2}-(EX')^{2}$ (not $(EX')^{2}-E(X')^{2}$).You will need the second moment and the fourth moment of $Z$ to find the variance.

Answer 2

You are on the right track.

If $Z\sim N(0,1)$ then to be found are $\mathbb EZ^2$ and $\mathbb EZ^4-(\mathbb EZ^2)^2$.

Mistake in $V[X']$ (see answer of Kavi).