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Let $X,Y$ be two independent random variables having the same distribution, centred and with variance 1, $\phi$ is the characteristic function of $X$ and $Y$.

If $X+Y$ and $X-Y$ are independent, show that $\phi(t)=\phi(-t)$ $\forall t$ (consider $\rho(t):=\phi(t)/\phi(-t))$

Then I have to deduce X and Y are gaussian r.v.

Seems quite obvious (on the first point I think I have to show they're both symmetric...how?) but I can't come up with a solution

ps from the previous points of the exercise I know that the characteristic function of X and Y verifies $\phi(2t)=\phi(t)^3\phi(-t)$ and $\phi(t)$ is never equal to $0$

pcox90
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  • See this question: http://math.stackexchange.com/q/556030/ – saz Aug 05 '14 at 05:08
  • why (3) holds? (sorry I can't ask on the other thread) – pcox90 Aug 15 '14 at 11:03
  • This follows simply by iterating the preceding equality: $$\delta(t) = 2 \delta(t/2) = 4 \delta(t/4) = \ldots = 2^n \delta(t/2^n) = \frac{\delta(t/2^n)}{\frac{1}{2^n}}.$$ Now divide by $t$. – saz Aug 15 '14 at 11:11
  • thanks a lot! I wanted to use the substitution provided by my exercise:

    $\rho(t):=\frac{\phi(t)}{\phi(-t)}=\frac{\phi(t/2)^2}{\phi(-t/2)^2}={\rho(t/2^n)}^{2^n}$.

    but taking the limit of $\frac{\rho(t)}{t}$ I can't arrive to anything interesting...

     – pcox90 Aug 15 '14 at 15:14
  • Since you want to prove that $\varrho(t)=1$, you cannot expect that the limit $\frac{\varrho(t)}{t}$ exists as $t \to 0$. Actually, I don't understand the given hint to consider $\varrho(t) =\phi(t)/\phi(-t)$; the symmetry of $\phi$ follows much more easily by playing around with the independence of the random variables. – saz Aug 16 '14 at 06:48

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