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I am trying to prove the following fact:

Let $X$ be i.i.d. real random variables such that $E[X_i] = 0$, $\mathrm{Var}[X_i] = \sigma^2$. Moreover it is know that $S_n = \sum_{i=1}^{n}X_i$ is distributed as $x_n X_1 + y_n$ for each $n \geq 1$ and some real $x_n, y_n$. I would like to prove that $X_1$ is distributed as normal distribution $N(0, \sigma^2)$.

My approach: By Central Limit Theorem, $\frac{S_n}{\sigma \sqrt{n}} = \frac{x_n X_1 + y_n}{\sigma \sqrt{n}}$ (1) converges weakly to $N(0,\sigma^2)$ as $n \to \infty$. Therefore $E\left[ \frac{x_n X_1 + y_n}{\sigma \sqrt{n}} \right] = y_n$ converges weakly to $E[N(0, \sigma^2)]$. Hence $y_n \to 0$. Similarly, applying this argument for variances (i.e. variances of weakly convergent sequence converge to the variance of the limit), I would conclude that $\frac{x_n}{\sigma \sqrt{n}} \to 1$ as $n \to \infty$. Then from (1) it would follow that $X_1$ is distributed as $N(0, \sigma^2)$.

I have feeling that I am missing/overlooked some issues here. Would be grateful for your comments or ideas whether this approach is correct or not.


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You might want to add as a last step that each random variable $U_n=\frac{S_n}{\sigma\sqrt{n}}$ is distributed as (but not equal to, your post goes too fast on this) the random variable $V_n=\frac{x_n}{\sigma\sqrt{n}}X_1+\frac{y_n}{\sigma\sqrt{n}}$, that $U_n$ converges in distribution to a standard normal random variable, that $V_n$ converges almost surely to $X_1$, and in particular that $V_n$ converges in distribution to $X_1$. Since the limit in distribution is unique, $U_n$ also converges in distribution to $X_1$, hence $X_1$ is indeed standard normal.

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