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!
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:
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.