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Let $X_1, X_2, ...$ be a sequence of independent random variables with $$\mathbb{P}[X_i=1]=\mathbb{P}[X_i=-1]=\frac{1}{2}$$ Let's now consider the sum $S_n=\sum_{k=1}^{n} X_k$. I need to show three things:

  1. That for every integer $1\leq k$ there exists a constant $c_k$ depending on $k$ but not on $n$ such that for all $1\leq n$ $$\mathbb{E}[S^{2k}_n]\leq c_{k}n^k$$
  2. To show that for every $\epsilon>0$ $$\frac{S_n}{n^{1/2+\epsilon}}\rightarrow 0 $$ almost surely.
  3. Show that $\frac{S_n}{n^{1/2}} $ does not converge to $0$ in probability.

Any help will be highly appreciated! Thanks!

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a) First use Newton's generalized Binomial formula : $S_n^{2k}=\sum \frac{(2k)!}{\alpha 1! \ldots \alpha _n !}X_1^{\alpha _1}\ldots X_n^{\alpha _n}$ where the sum is on every n-uplet $(\alpha _1, \ldots , \alpha _n)$ such that $\alpha _1+\ldots \alpha _n=2k$.

Now observe that by independance of the variables we have : $E(X_1^{\alpha _1}\ldots X_n^{\alpha _n})=1$ if all $\alpha _i$ are even, and if it's not the case $0$. So we can write : $E(S_n^{2k})=\sum \frac{(2k)!}{(2\beta _ 1)! \ldots (2\beta _ n)!}X_1^{(2\beta _ 1)!}\ldots X_n^{(2\beta _ n)!}$ where $\beta _1+\ldots +\beta _n=k$. Thus $E(S_n^{2k})\leq \sum \frac{(2k)!}{2^{\beta _1+\ldots +\beta _n}\beta _1! \ldots \beta _n !}\leq \frac{(2k)!}{2^kk!}\sum \frac{k!}{\beta _1!\ldots \beta _n !}=\frac{(2k)!}{2^kk!}n^k$. This proof can be used to prove Khintchine inequalities.

b) By using Borel-Cantelli it is sufficient to show that for evere $\delta >0$, $\sum _n \mathbb{P}(|S_n|\geq n^{1/2+\epsilon}\delta ) <+\infty .$ But Markov inequality implies that $\mathbb{P}(|S_n|\geq n^{1/2+\epsilon}\delta )\leq \frac{E(S_n^{2k})}{n^{k+2k\epsilon}\delta ^{2k}}\leq \frac{c_k}{\delta ^{2k}n^{2k\epsilon }}$ for every $k$. Choose $k$ such $2k\epsilon >1$.

c)You have to show that $\mathbb{P}(|S_n|>n^{1/2}t)\not\to 0$. It's equivalent to show that $\mathbb{P}(|S_n|^2>n^{1}t^2)\not\to 0$. Observe that $E(S_n^2)=n$ and use Paley-Zygmund inequality (http://en.wikipedia.org/wiki/Paley%E2%80%93Zygmund_inequality ).

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