pdf of $\frac{(X_1-X_2)^2}{2}$ Let $X_1$ and $X_2$ be independent normally distributed random variables.
$$X_1 \sim N(0,1) \\ X_2 \sim N(0,1)$$
Find the pdf of $\frac{(X_1-X_2)^2}{2}$
We have that
By example we know that when $Y = X^2$ we have that
$$\begin{align}f_Y(y) &= \frac{e^{-\frac{(-\sqrt{y})^2}{2}}}{\sqrt{2\pi}}\left|-\frac{1}{2\sqrt{y}}\right|+\frac{e^{-\frac{(\sqrt{y})^2}{2}}}{\sqrt{2\pi}}\left|\frac{1}{2\sqrt{y}}\right| \\ &= \frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{y}}e^{-\frac{y}{2}} \end{align}$$
Which follows  achi-square distribution with 1 degree of freedom, therefore
$$\frac{(X_1-X_2)^2}{2} \sim \chi_1^2$$
However, how can I do the calculation for this using multivariate transformations, I cannot get the distribution?

Let $U=\frac{(X_1-X_2)}{\sqrt{2}}$ and $M_{x_i(t)}$ be moment generating function of $X_i$.
$M_{\frac{X_1-X_2}{\sqrt{2}}}(t)=E(e^{(\frac{X_1-X_2}{\sqrt{2}})t})=E(e^{\frac{X_1}{\sqrt{2}}t}) E(e^{\frac{X_1}{\sqrt{2}}t})=M_{X_1}(\frac{t}{\sqrt{2}})M_{X_2}(\frac{-t}{\sqrt{2}})=e^{\frac{1}{2}(\frac{t}{\sqrt{2}})^2}e^{\frac{1}{2}(\frac{-t}{\sqrt{2}})^2}=e^{\frac{t^2}{2}}$  So $U=\frac{(X_1-X_2)}{\sqrt{2}} \sim N(0,1)$ Therefore   $U^2=(\frac{X_1-X_2}{\sqrt{2}})^2 \sim \chi^2_1$
