Defining the domain of an MGF?

Question

Let $Y=X^2$ and let $X$ follow a distribution of $X\sim N(0,\sigma^2)$ for $\sigma > 0$. Find the MGF of $Y$ and specify its domain.

So what I did was I did a change of variables:

$$P(Y<y)=P(X^2<y)$$ $$=P(-y^{1/2} < X < y^{1/2})$$ $$=P(X<y^{1/2}) - P(X<-y^{1/2})$$ $$=F_X(y^{1/2})-F_X(-y^{1/2})$$

So as $X\sim N(0,\sigma^2)$ through a bit of rearranging I come to the answer that $Y\sim\gamma(\frac{1}{2\sigma^2},\frac{1}{2})$ correct me if I'm wrong!

So then since the MGF of a gamma function is: $$\frac{\alpha^r}{(\alpha-t)^r}$$

By plugging the values in of $\alpha$ and $r$ I get:

$$\frac{(\frac{1}{2\sigma^2})^{1/2}}{(\frac{1}{2\sigma^2}-t)^{1/2}}$$

And this is the place that I am stuck at. Have I already arrived at the MGF of Y? And if I have how would I be able to define its domain?

Also if it is not too much trouble, can someone explain to me intuitively what is happening here it would be greatly appreciated!

Answer 1

$$ \begin{align} \mathbb{E}[e^{tX^2}] & =\frac{1}{\sqrt{2\pi\sigma^2}} \int _{-\infty}^\infty e^{\left(t-\frac{1}{2\sigma^2}\right)x^2} \, \mathrm{d}x\\[10pt] & =\begin{cases} \dfrac{1}{\sqrt{1-2t\sigma^2}}, & 2\sigma^2 t<1 \\[6pt] \infty, & \text{otherwise} \end{cases} \end{align} $$

Answer 2

The mgf is $t\mapsto 1/\sqrt{1-2t\sigma^2}$, which is undefined when the thing under the radical is negative or zero.

So we need $1-2t\sigma^2>0$. That means $t<\dfrac{1}{2\sigma^2}$.