For the standard proof that if $Z \sim N(0,1)$ than $Z^{2} \sim \chi^{2}_{1}$

We write:

$$M_{Z^{2}}(t)=\int_{\mathbb{R}}\exp(tz^{2})\frac{1}{\sqrt{2\pi}}\exp(\frac{-z^{2}}{2}) dz$$

That is, we use the standard normal density to compute the moment generating function (MFG).

Why do we not need to use the density of $Z^{2}$ but instead are able to use the density of just $Z$? I assume because it is a continuous one to one function ?


1 Answer 1


In principle, you don't know the density of $Z^2$, so you can't use it.

What's being done when you use the density of $Z$ is an application of the law of the unconscious statistician.


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