# Showing $X$ is standard normal when $X$ has the same distribution as $\frac{X_1+X_2}{\sqrt{2}}$

$$X, X_1, X_2$$ 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.

• If you take $X_i$ to be a function of $X_j$, $X_k$ then the independence fails. – NCh May 1 at 2:52

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.