$X_1$, $X_2$, and $X_3$ are random sample taken from normal distribution with $\mu=0$ and $\sigma^2=1$ (standard normal distribution). Let $Y=X_1^2+X_2^2+X_3^2$. Prove that $Y$ have distribution chi square with degree of freedom 3.
To prove that, I use moment generating function (MGF) of $Y$ like this.
We know that if $X$ standard normal distribution then $M_X(t)=e^{\frac{1}{2}t^2}$. Now I find the MGF of $Y$ to find the distribution of $Y$.
\begin{eqnarray} M_Y(t)&=&M_{X_1^2+X_2^2+X_3^2}(t)\\ &=& E\left(e^{t(X_1^2+X_2^2+X_3^2)}\right) \\ &=& E\left(e^{tX_1^2}e^{tX_2^2}e^{tX_3^2}\right)\\ &=& E\left(e^{3tX^2}\right)\\ \end{eqnarray} I cannot find $E\left(e^{3tX^2}\right)$ associated with MGF of standar normal distribution. Any idea?
Hint: for $|a| <\frac 1 2$ we have $\int e^{ax^{2}}e^{-x^{2}/2}dx=\frac 1 {\sqrt {1-2a}} \int e^{-y^{2}/2} dy $ by the substitution $y =x \sqrt {1-2a}$.