So I am rusty on moment generating functions. I found this problem (and solution) in old notes, and I can't remember why it works. The problem is to find the moment generating function of $Y$ when $Y = (X-3)^2$ and $X \sim N(0,1)$.
The solution:
$E(e^{t(X-3)^2}) = \int_{-\infty}^{\infty} e^{t(x-3)^2} \frac{-e^{x^2}}{\sqrt{2\pi}} dx \rightarrow Simplify$
So my intuition was to first transform X, and then find the MGF of Y in the usual way, which is undoubtedly more complicated. I know we can find the MGFs of linear transformations of variables fairly easily, but why can we do this for this situation?
Best Answer
If $g:\mathbb R\to[0,\infty)$ is a measurable function then we have for any random variable $X$ $$ \mathbb E[g(X)] = \int_\Omega g\circ X\ \mathsf d\mathbb P = \int_{\mathbb R} g\ \mathsf d(X_*\mathbb P), $$ where $X_*\mathbb P = \mathbb P\circ X^{-1}$, i.e. $X_*\mathbb P(B) = \mathbb P(X^{-1}(B)) = \mathbb P(X\in B)$ for any Borel set $B$. If $X_*\mathbb P$ is absolutely continuous with respect to Lebesgue measure $\mu$, that is $\mu(B)=0\implies X_*\mathbb P(B)=0$, written $X_*\mathbb P\ll \mu$, then by the Radon-Nikodym theorem $X$ has a density $f_X$ such that $X_*\mathbb P(B) = \int_B f_X\ \mathsf d\mu$ for all Borel sets $B$. In this case we have $$ \mathbb E[g(X)] = \int_{\mathbb R}g(x)f_X(x)\ \mathsf dx. $$
In this case $g(x) = e^{t(x-3)^2}$ is a continuous function of $f$, so $$ \mathbb E[g(X)] = \int_{-\infty}^\infty g(x)f_X(x)\ \mathsf dx = \int_{-\infty}^\infty e^{t(x-3)^2}\frac1{\sqrt{2\pi}} e^{-\frac12 x^2}\ \mathsf dx = \frac{e^{\frac{9 t}{1-2 t}}}{\sqrt{1-2 t}}, \quad t<\frac12. $$