If $X\sim N(0,1)$, I need find the generating function of $Y=X^2$. The exercise says me that I should not solve the integral (which is I presume $\int e^{tx^2}f(x)dx$). It also says:
$$E(e^{tY}) = E(e^{tX^2})$$
and that any probability distribution function $f(x)$ has $\int_{-\infty}^{\infty}f(x)\ dx = 1$
How should I solve it? I know I must find $M_{X^2}(t) = E(e^{tX^2})$. I can't, however, solve the integral. Which methods are left? I don't know any.
$Y=X^2$
$$E(e^{tY})=E(e^{tX^2})=\int_{-\infty}^{\infty}e^{tx^2}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\\=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\dfrac{x^2(1-2t)}{2}}dx$$
I think you can do next steps.
Note that $$\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\dfrac{x^2}{2\sigma^2}}dx=\sigma$$
So you have to find $\sigma$ here, that's all.
Btw $Y$ follows Chi-Square distribution with 1 degree of freedom