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Let random variable X1, with variance one, has the following property: $\frac {X1+X2} {\sqrt2}$ has the same distribution as X1, where X2 is an independent copy of X1. Show that X1 ∼ N(0, 1).

So i have to show x1 is normal distribution; its very easy to show the other way of course, when x1 is normal distirbution, then obviously $\frac {X1+X2} {\sqrt2}$ ~ X1. But the other way is much harder

DOes this have to with characteristic function? since $\phi_{x1+x2}=\phi_x\phi_ {x2}$ i had written out that expression but how can i transofrm it to normal distirbution? in part, how do i realize the pdf of normal distribution from the equality here $\phi_{x1}=\phi_{x1/\sqrt(2)}\phi_ {x2\sqrt(2)}$? thanks in advance.

Mike Earnest
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TomWangzilin
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  • You have to prove that the mean is $0$ wich is easy since $\mu =\sqrt2 \mu$ – Kroki Apr 23 '18 at 17:15
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    An idea but I dont know how to finish it: Let $\phi(t)$ be the characteristic function of $X_1$. Then $\sqrt{2} X_1$ has characteristic function $\phi(\sqrt{2} t)$ while $X_1 + X_2$ has characteristic function $\phi(t)^2$. So you have $\phi(t)^2 = \phi(\sqrt{2} t)$. This certainly SUGGESTS that $\phi(t) = e^{a t^2}$ for some constant $a$...but I dont know how to PROVE it. – antkam Apr 23 '18 at 17:38

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Let $X_1,X_2,X_3,\dots$ be an i.i.d. series where each variable is distributed like $X_1$. By iterating the equation $X_1\stackrel{d}=(X_1+X_2)/\sqrt{2}$, we get that for any $n\in \mathbb N$,

$$ X_1\stackrel{d}=\frac{X_1+X_2+\dots+X_{2^n}}{\sqrt{2^n}} $$ The central limit theorem implies that the RHS of the above approaches an $N(0,1)$ distribution as $n\to\infty$, so we conclude that $X_1\sim N(0,1)$. In order to apply CLT, we only need $X_1$ to have mean zero (easy to prove) and variance one (given).

The comments below this question are now irrelevant, since I replaced my earlier answer with this new one. It turns out my earlier answer was mistaken; it assumed that all higher moments of $X_1$ existed without proof.

Mike Earnest
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  • To prove two distributions match, it's insufficient to check means of $X^n$ for natural $n$. See https://math.stackexchange.com/questions/1166637/do-moments-define-distributions – J.G. Apr 23 '18 at 21:35
  • @J.G In general, you are right. But if the moment generating functions of $X$ and $Y$ are equal an converge in a neighborhood of zero, then $X=_d Y$. See https://mathoverflow.net/a/3536/59232 In particular, the mgf of a normal variable does converge. That was what I meant when I said $\phi_Z$ was complex analytic, hence determined by its derivatives. – Mike Earnest Apr 23 '18 at 21:49
  • @MikeEarnest - i am not familiar with complex analysis, so i have 2 questions: (1) using your argument, would you need to first prove that $\phi_X$ is analytic (in addition to $\phi_Z$ being analytic)? (2) if it were proven (or assumed) that $\phi_X$ is analytic, can i conclude from $\phi(t)^2 = \phi(\sqrt{2} t)$ that $\phi(t) = e^{a t^2}$ for some constant $a$? – antkam Apr 24 '18 at 12:21
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    (1) As long as you know $\phi_Z$ is analytic, then proving $X$ has the same moments as $Z$ implies $\phi_X$ is analytic, so no. (2) I think so! The functional equation $\phi(t)^2=\phi(\sqrt{2}t)$ should have a unique analytic solution. You could prove this with a similar recursive argument like I did in (*). – Mike Earnest Apr 24 '18 at 14:24