Let $X$ and $Y$ be independent random variable, such that $X-Y$ and $X+Y$ are independent.
Prove that $X$ and $Y$ are normal random variable.
Hint: Use characteristic functions to find a functional equation, and to find the modulus of $\phi_{X+Y}$ and the corresponding argument.
We have $$\phi_{(X+Y,X-Y)}(x,y)=\phi_{X+Y}(x)\phi_{X-Y}(y)=\phi_X(x)\phi_Y(x)\phi_X(y) \overline{\phi_Y}(y) \ \ \ \ (1)$$ $$\phi_{(X+Y,X-Y)}(x,y)=\phi_{(X,Y)}(x+y,x-y)=\phi_{X}(x+y)\phi_Y(x-y) \ \ \ \ (2)$$
we obtain,
$$\phi_{X}(x+y)\phi_Y(x-y)=\phi_X(x)\phi_Y(x)\phi_X(y) \overline{\phi_Y}(y) \ \ \ \ (3)$$
replacing $y$ with $-y$ in $(3)$:
$$\phi_{X}(x-y)\phi_Y(x+y)=\phi_X(x)\phi_Y(x)\phi_Y(y) \overline{\phi_X}(y) \ \ \ \ (4)$$
multiplying $(3)$ with $(4):$
$$\phi_{X+Y}(x+y)\phi_{X+Y}(x-y)=(\phi_{X+Y}(x))^2|\phi_{X+Y}(y)|^2 \ \ \ \ (5)$$
So,
$$|\phi_{X+Y}(x+y)||\phi_{X+Y}(x-y)|=|\phi_{X+Y}(x)|^2|\phi_{X+Y}(y)|^2 \ \ \ \ (6)$$
So we obtain a functional equation which need to be solved:
$$f(x+y)f(x-y)=(f(x))^2(f(y))^2$$
$f$ is continuous and positive, This functional equation was solved (see @Ravsky reply below) (beginning with integers, rational, then real numbers via density), we should obtain: $$\exists \sigma>0;|\phi_{X+Y}(x)|=|\phi_{X-Y}(x)|=e^{-\sigma^2x^2/2}.$$
If so, it remains to prove that $\exists \mu \in \mathbb{R}; \phi_{X+Y}(x)=e^{i \mu x -x^2\sigma^2/2}.$
Any suggestions?
Best Answer
Assuming that $f$ is a continuous function from $\Bbb R$ to $\Bbb R$, we can solve the functional equation $f(x+y)+f(x−y)=2(f(x)+f(y))$ as follows. Plugging into it $x=y=0$, we obtain $f(0)=0$.
Now fix any real $y$ and for each $n\in\Bbb Z$ put $b_n=f(ny)-f((n-1)y)$. Then $b_1=f(y)$ and $b_{n+1}=b_n+2f(y)$ for each $n$. Thus $b_n=(2n-1)f(y)$ for each $n$, that is $f(ny)-f((n-1)y)= (2n-1)f(y)$. The latter equality implies $f(ny)=n^2f(y)$ for each $n\in\Bbb Z$.
Thus for each integer $p$ and natural $q$ we have $q^2f(1/q)=f(1)$ and $f(p/q)=p^2f(1/q)=p^2f(1)/q^2$. By the continuity of $f$, $f(x)=x^2f(1)$ for each real $x$. On the other hand, the latter function satisfies the equation.