Normal distribution magic

Question

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?

Answer 1

Note that by Alex Ravsky answer we already have $|\varphi_{X+Y}(t)| = \exp(t^2c)$ where $c=f(1) \le 0$

Letting now $g=\varphi_{X+Y}$ we have the existence of continuous $h:\mathbb R \to \mathbb R$ with $h(0)=0$ such that $g(x) = e^{ih(x)}f(x) = e^{ih(x)}e^{x^2c}$. It would be enough to show $h(x)=xh(1)$.

Due to your equation $(5)$ we have:

$$ g(x+y)g(x-y) = g(x)^2f(y)^2 $$ $$ e^{ih(x+y)}e^{ih(x-y)} e^{(x+y)^2c}e^{(x-y)^2c} = e^{2x^2c}e^{2y^2c}e^{i2h(x)}$$

Hence $$ e^{i(h(x+y) + h(x-y))} = e^{i2h(x)}$$

Letting $x=t,y=t$ we get $e^{ih(2t)} = e^{i2h(t)} $

Inductivelly $x=(n-1)t,y=t$ we get $e^{ih(nt)} = e^{inh(t)}$

Using $h(0)=0$ and letting $x=0$ we get $e^{ih(y)}e^{ih(-y)}=1$, hence for any $k \in \mathbb Z$ $$ e^{ih(kt)} = e^{ikh(t)}$$

Then, taking $r = \frac{p}{q}$ where $p \in \mathbb Z, q \in \mathbb N_+$ we get for rationals:

$$ e^{ih(r)} = e^{iph(\frac{1}{q})} = e^{ir qh(\frac{1}{q})} = e^{irh(1)} $$ and by continuity, for any $x \in \mathbb R$ we get $$ e^{ih(x)} = e^{ixh(1)}$$ hence $g=\varphi_{X+Y}$ is a characteristic function of normal distribution (we use $c \le 0$). And by Cramer theorem the result follows

EDIT: Without Cramer's theorem, you can plug in your equation (3) pair $(-x,y)$ an then multyplying (3) with equation you'll get, getting really same functional equation for $\varphi_{X-Y}$, hence $X-Y$ is gaussian, too. But then by independence $2X = (X+Y) + (X-Y)$ and $2Y = (X+Y) - (X-Y)$ are gaussian, too.

Answer 2

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.