Let $X_1$ and $X_2$ be independent $n(0,1)$ random variables. Find the pdf of $(X_1-X_2)^2/2$.

Question

I understand that $(X_1-X_2)/\sqrt2)$ ~ $n(0,1)$ since it is a linear combination of $X_1 $ and $X_2$ and hence $(X_1-X_2)^2/2$ ~ $\chi^2_1$. I'm having trouble on how to prove/show this transformation by hand.

Answer 1

Sketch the $x$-$y$ plane and note that the joint density of $(X,Y)$ has rotational symmetry since $f_{X,Y}(x,y) = \frac{1}{2\pi}\exp\left(-\frac{x^2+y^2}{2}\right)$ is a function only of $r = \sqrt{x^2+y^2}$, the distance of $(x,y)$ from the origin, and not of the slope of the straight line through $(0,0)$ and $(x,y)$. Rotate the axes by $-\pi/4$ which transforms $$\begin{align}(X,Y) \rightarrow (\hat{X},\hat{Y}) &= \left(X\cos(-\pi/4)+Y\sin(-\pi/4), -X\sin(-\pi/4)+Y\cos(-\pi/4)\right)\\ &= \left(\frac{X-Y}{\sqrt{2}}, \frac{X+Y}{\sqrt{2}}\right) \end{align}$$ Observe that the joint pdf is the same, and so $\displaystyle \hat{X} = \frac{X-Y}{\sqrt{2}}\sim N(0,1)\Rightarrow \frac{(X-Y)^2}{2}\sim \chi^2(1)$.