If $X_1, X_2$ are iid $N(0,1)$, why is it that $X_1X_2 \sim \frac{X_1+X_2}{\sqrt{2}}\frac{X_1-X_2}{\sqrt{2}}$?

Question

Suppose $X_1, X_2$ are iid $N(0,1)$, why is it that $X_1X_2 \sim \frac{X_1+X_2}{\sqrt{2}}\frac{X_1-X_2}{\sqrt{2}}$? Here, "$\sim$" is used to denote the fact that they have the same distribution, or that their cdf's are the same. I understand that it is a result from probaility theory that $X_1+X_2$ and $X_1-X_2$ are independent, and so scaling by $\sqrt{2}$ yields that $\frac{X_1+X_2}{\sqrt{2}}\frac{X_1-X_2}{\sqrt{2}}$ is the product of two independent $N(0,1)$ random variables. However, it appears strange to me as the right hand side is a function of $X_1$ and $X_2$. IT is almost like a circular argument to me. Can someone help me reconcile this? thanks.

Answer 1

$X_1 X_2$ has some distribution (say, it has cdf $F$).

$\frac{1}{2} (X_1+X_2)(X_1-X_2)$ has some distribution (say, it has cdf $G$).

The statement is just saying $F=G$.

Note that you could just as well say $X_1X_2 \sim \frac{1}{2} (Z_1+Z_2)(Z_1-Z_2)$ where $Z_1,Z_2 \sim N(0,1)$ are i.i.d.

Answer 2

$\frac{X_1\pm X_2}{\sqrt{2}}\sim N(0,1)$, and these two random variables are independent because they are jointly Gaussian (as linear transformations of the Gaussian vector $(X_1,X_2)$) and their covariance is zero. Indeed, $E [ \frac{X_1-X_2}{\sqrt{2}} \frac{X_1+X_2}{\sqrt{2}}]= \frac{1}{2} E [X_1^2 - X_2^2]=0$. Thus their product is the product of two independent $N(0,1)$ random variables, and in particular has the same distribution (function) as of $X_1X_2$.

Hope this helps.