Let $X_1$ and $X_2$ be ind, normal random variables with $\mu=0$ and $\sigma^2=1$. Let $Y_1=X_1-X_2$, $Y_2=X_1+X_2$. Find $F_{Y_1,Y_2}$

Question

Let $X_1$ and $X_2$ be ind, normal random variables with $\mu=0$ and $\sigma^2=1$. Let $Y_1=X_1-X_2$, $Y_2=X_1+X_2$. Find joint density of $Y_1,Y_2$.

I tried to solve this using $$f_{Y_1,Y_2}=\frac{d^2}{dy_1dy_2}F_{Y_1,Y_2}(y_1,y_2)$$ $$=\frac{d^2}{dy_1dy_2}P(Y_1\leq y_1, Y_2\leq y_2)$$ $$=\frac{d^2}{dy_1dy_2}P(X_1-X_2\leq y_1, X_1+X_2\leq y_2)$$

But from here I'm not sure how to do this because differentiating with $y_1$ and then $y_2$ doesn't seem to make sense. So I'm not sure if this method just doesn't work, I don't think I can use moment generating functions to solve either.

Answer 1

One simple way of answering this is to us characteristic functions. $Ee^{itY_1+isY_2}=Ee^{i(t+s) X_1} Ee^{i(s-t) X_2}$. This becomes $e^{-(t+s)^{2} /2} e^{-(s-t)^{2} /2}=e^{-t^{2}} e^{-s^{2}}$. From this you can conclude that $Y_1$ and $Y_2$ are independent normal variables with mean $0$ and variance $2$. Hence $F_{Y_1Y_2} (x,y)=\int_{-\infty} ^{x}\int_{-\infty} ^{y} \frac 1 {2\pi} e^{-u^{2}/4}e^{-v^{2}/4} dudv$.