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$X, X_1, X_2$ are i.i.d random variables with $\mathbb{E}[X] = 0$ and $\mathbb{E}[X^2] = 1$. Suppose $X$ has the same distribution as $\frac{X_1+X_2}{\sqrt{2}}$. I need to show that $X$ is standard Normal. A given hint is to establish a recurrence type relation.

So far, I have the following:

  1. Formulate $X_i = \frac{X_j+X_k}{\sqrt{2}},$ for some $1 \leq j,k \leq i-1$
  2. Then, the sequence $\{X_n\}$ will be i.i.d. Step 1 is necessary to preserve the identical distribution.
  3. By Central Limit Theorem, $\frac{S_n}{\sqrt{n}} =\frac{\sum_i^nX_i}{\sqrt{n}} \to \mathcal{N}(0,1)$ as $n \to \infty$.

Need to construct the sequence $\{X_n\}$ such that $\lim_{n\to \infty} \frac{S_n}{\sqrt{n}} = \frac{X_1+X_2}{\sqrt{2}}$ (or I think any $X_j$ would be good too, since they have identical distribution). Here is where I am stuck.

I have tried constructing many sequences but none of them have worked so far! Any help would be deeply appreciated! Much thanks in advance.

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  • $\begingroup$ If you take $X_i$ to be a function of $X_j$, $X_k$ then the independence fails. $\endgroup$ – NCh May 1 at 2:52
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Consider $$ \frac{S_4}{\sqrt{4}} = \dfrac{\frac{X_1+X_2}{\sqrt{2}}+\frac{X_3+X_4}{\sqrt{2}}}{\sqrt{2}} $$ This random variable has the same distribution as $X$. So $\frac{S_{2^n}}{\sqrt{2^n}}$ for each $n$ has the same distribution as $X$. Apply CLT and get desired.

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  • $\begingroup$ Excellent!! Thank you so much! $\endgroup$ – B.T May 1 at 14:41

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