Is there any short proof of this classical problem?

Question

Let $X,Y$ be two i.i.d. r.v.'s with zero mean and unit variance. If $X+Y$ and $X-Y$ are independent, then $X$ and $Y$ are both standard normal distributed.

Is there any short proof for this problem?

Answer 1

This is a weakened form of Bernstein's Theorem (weakened by unnecessarily assuming identical distributions having finite variances), so the proof is shorter. Here's a sketch, adapted from "On three characterizations of the normal distribution" by M. P. Quine:

Define $\quad U = X+Y, \quad V=(X-Y)^2$

and characteristic functions $$\phi(t) = E e^{itX}\\ \gamma(s,t) = E e^{isU + itV}. $$ Then, using independence, $${\partial\gamma \over \partial t} = E( iV e^{itV})\ E( e^{isU}), $$ so, in terms of $X$ and $Y$, it is found that $$ {\partial\gamma \over \partial t}\bigg|_{t=0} = 2iE(X^2 e^{isX}) E(e^{isY}) - 2i(E(Xe^{isX}))^2 = -2i \phi''(s)\ \phi(s)+2i(\phi'(s))^2. $$ However, we also have $${\partial\gamma \over \partial t}\bigg|_{t=0} = E(e^{isU}) {\partial \over \partial t} E(e^{itV})\bigg|_{t=0} = 2i(\phi(s))^2, $$ hence the equation $$ - \phi \ \phi'' + (\phi')^2 = \phi^2 $$ whose unique solution is $$\phi(s) = e^{-{1 \over 2}s^2}. $$ QED

NB: The cited article proves the stronger version: If $(X,Y)$ are independent and $(X+Y, X-Y$) are independent, then $X$ and $Y$ are normal with the same (finite) variance.