Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
up vote 2 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.