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I am working on the following problem:

Assume $X_1, X_2, \dots$ are i.i.d. with $E(X_1^2)=1$ and $E(X_1)=0.$ Let $S_n = X_1+\dots+X_n$.

a) Show that $P(S_{2n}\leq 2S_n + \sqrt{n})$ converges and find the limit.

b) Show that $\frac{|S_n|}{\sqrt{|X_1|+\dots + |X_n|}}$ converges in distribution.

I know that I want to use the Central Limit Theorem (CLT). The CLT tells me that under the above assumptions $\frac{S_n}{\sqrt{n}}\implies N(0,1)$, where $\implies$ denotes convergence in distribution and $N(0,1)$ is the normal distribution with mean $0$ and variance $1$.

I've tried a lot of manipulation of $P(\frac{S_n}{\sqrt{n}} \leq x)$ for some $x\in\mathbb{R}$, but I can't seem to get anything useful.

Any help solving this problem would be appreciated!

Thanks

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Hint for (a): Notice that $$ \begin{align} S_{2n}-2S_n &= (S_{2n} - S_n) - S_n\\ &=(X_{n+1} + X_{n+2} + \cdots + X_{2n}) - (X_1 + X_2 + \cdots + X_n)\\ &=(X_{n+1} - X_1) + (X_{n+2}-X_2) + \cdots + (X_{2n}-X_{n})\\ &=Y_1 + Y_2 + \cdots + Y_n \end{align} $$ where $Y_1,Y_2,\ldots$ are iid. So $P(S_{2n}\le 2S_n +\sqrt n) = P(Y_1 + Y_2+\cdots + Y_n\le \sqrt n)$.

Hint for (b): Divide top and bottom by $\sqrt n$. After this operation, the denominator converges almost surely to a constant.

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