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I am trying to show that $X$ is a standard normal (in distribution) by applying the Lindberg's version of the central limit theorem to a sequence always equal to $X$.

In order to do that, I need to show that Lindberg-Feller condition is satisfied, and, for that, I need $X$'s variance.

Is there an easier way to do this? (without using CLT) Can anyone give me a hint on how to calculate that variance?

Thanks a lot for reading!

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One does not assume that $X$, $Y$ and $(X+Y)/\sqrt2$ are i.i.d., only that $X$ and $Y$ are. –  Did Jun 13 '13 at 20:28
    
We solved this problem in probability class. I don't really recall the exact solution now, but I'm sure you can somehow show that the variance has to be finite. Try doing so by contradiction –  mm-aops Jun 13 '13 at 20:36
    
@Did you are right, it's X,Y i.i.d. and X have the same distribution as $(X+Y)/\sqrt(2)$. Thanks for pointing that out. –  Guilherme Salomé Jun 13 '13 at 22:00
    
@mm-aops X's variance has to be 1, but i could not show it yet. However, i managed to proove that Lindberg-Feller condition is satisfied if X has finite variance. –  Guilherme Salomé Jun 13 '13 at 22:11
    
you don't really need to check Lindeberg-Feller if X has finite variance, cause then it's obvious X has to have mean zero so it's just a simple case of the standard CLT for iid variables (you just take a subsequence $\frac{S_{2^n}}{2^{n/2}}$ where S_n is a sum of iid variables with distribution X. on one hand it has to converge to a normal distribution, on the other by your assumption it's distribution is always the same as X's. you can have any variance you want, just finite one. sorry, I don't remember how I proved it last time, if I have time I'll give it a try in the evening –  mm-aops Jun 14 '13 at 10:47
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2 Answers 2

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As @mm-aops mentioned, you don't need to use Lindeberg-Feller to do this.

The tricky part to prove $\mathbb E(X^2) < \infty$. The best I can think of is to follow the path in exercise 3.4.3 of Probability: Theory and Examples, which says:

3.4.3. Let $X_1, X_2, \ldots$ be i.i.d. and let $S_n = X_1 + \cdots + X_n$. Assume that $S_n / \sqrt{n}$ converges to a limit in distribution and conclude that $\mathbb E X_i^2 < \infty$.

I believe that instead of requiring $S_n/\sqrt{n}$ to converge in probability, it is enough to require a subsequence $S_{n(k)} /\sqrt{n(k)}$ to converge in probability. Thus it can be applied to this question. For details, please check the sketch of proof in the book.

The rest is easy.

Since $X \sim (X+Y)/\sqrt{2}$, we have $\mathbb E(X) = \sqrt{2}\mathbb E(X) $. Therefore $\mathbb E(X) = 0$.

Now assume $\mathbb E(X^2) < \infty$. Let $X_1,\ldots,X_n$ be a sequence of i.i.d. random variables with distribution $X$. Let $S_n = \sum_{m=1}^n X_m$. By Central Limit Theorem, we have $$ \frac {S_{2^b}} {\sqrt{2^b}} \to \chi $$ in probability when $b \to \infty$ through positive integers, where $\chi$ is the standard normal distribution. It follows from assumption that ${S_{2^b}} / {\sqrt{2^b}}$ actually has distribution $X$. Therefore $X$ must be standard normal.

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The characteristic function must verify

$$ F(\sqrt{2}\, t)=F(t)^2$$

Obviously, $F(t)=\exp(a \, t^2) $ is a solution. It should not be difficult to prove that it's the only solution.

Update 1: To see that it's the only solution, we can consider fixing $F(t_0)=c$ for some $t_0>0$ (we already know that $F(0)=1$ and $F(-t)=F^*(t)$. Then, the values of $F$ are determined for $F(t_0/2^n)$, which by continuity of the CF, determines the function in a neighbourhood of zero, which in turn determines $F(t)$ for all reals.

Update 2: The above is wrong, or at least need some serious work. As a counterexample:

$$F(t)=\exp \left( - a(t) \, t^2 \right)$$ with $a(t)=a+ \gamma \sin(2 \pi \log_2(t^2))$, $a>\gamma>0$, satisfies the equation. I guess that there must be some condition that precludes this as a valid characteristic function, but I'm not sure. Interesting: the first derivative goes to zero as $t\to0$, but the second derivative (related to the second moment of the random variable) oscillates.

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I suppose you can work on it, but it needs some serious argumentation. first of all - why does it determine all the values in a neighbourhood of zero and not just at 0 itself? (that's the crucial point, the rest is kinda true once you have that) and secondly - in general case (without the assumption that c.f. satisfies this functional equation) it's not true that if two characteristic functions are equal on a nbd of zero then they're equal everywhere. you can easily disprove that using Polya criterion and a graphic counterexample –  mm-aops Jun 15 '13 at 0:16
    
@mm-aops: Yes, your right. –  leonbloy Jun 15 '13 at 1:32
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