Let $(X_i)_{i\geqslant 1}$ be a sequence of i.i.d. random variables in $\mathcal{L}^2$ with $\mathbb{E}(X_i)=\mu$, and finite standard deviation $\sigma>0$. Let $S_n:=\sum_{i=1}^n X_i$.

The Central Limit Theorem tells that $$ Z_n:=\frac{S_n-n\mu}{\sigma\sqrt{n}}\to\mathcal{N}(0,1). $$

That is, $$ \lim_{n\to\infty}P(Z_n\leq z)=\Phi(z). $$

Now, I've read that this implies that $$ P(S_n\geq n\mu+n^{\alpha})\to 0\text{ whenever }\alpha>\frac{1}{2}. $$

How does this follow?


You can rewrite $S_n$ in terms of $Z_n$ by rearranging the expression given in the CLT, so that, $$S_n=\sigma\sqrt{n}Z_n +\mu n $$ and replacing this in the expression at the end, you get, $$P(\sigma\sqrt{n}Z_n +\mu n \geq \mu n+ n^a)$$ which can be rearranged as $$P(Z_n \geq \frac{n^a}{\sqrt{n}})$$ and we can simplify the expression to the right of the inequality sign as $n^{a-1/2}$. So if $a>1/2$ the right hand side will go to infinity as $n \rightarrow \infty$ so the probability goes to 0. If $a \leq 1/2$ then the right hand of the inequality sign will go to zero, so it is possible that it will have positive probability. I'm not entirely sure that you wouldn't need to put some condition on the distribution of $Z_n$ for this argument to work, but in general it seems pretty straightforward.

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    $\begingroup$ No condition needed: since $Z_n$ converges in distribution to a centered normal distribution, $P(Z_n\ge n^c)$ converges to $0$ for every $c>0$ and to $\frac12$ for every $c<0$. $\endgroup$ – Did Sep 24 '15 at 12:52
  • $\begingroup$ Where did the $\sigma$ go in the last expression? Isn't it $P(Z_n\geq \frac{n^{\alpha}}{\sqrt{n}\sigma})$? $\endgroup$ – M. Meyer Sep 24 '15 at 12:57
  • $\begingroup$ I think a little more work is needed here. It's true that $n^{a-1/2}$ goes to $\infty$ but the result doesn't follow automatically. To make the issue explicit, let $p_n(z) = P(Z_n \le z)$. The central limit theorem as quoted asserts $p_n \to \Phi$ pointwise. In order to conclude that $\lim_{n \to \infty} p_n(n^{a-1/2}) = \lim_{n \to \infty} \Phi(n^{a-1/2}) = 0$ you need something like uniform convergence of $p_n$ to $\Phi$. $\endgroup$ – Nate Eldredge Sep 24 '15 at 15:31
  • $\begingroup$ @M.Meyer yeah sorry. It's true that sigma is still there. I dropped it since my argument dosen't change in presence of a constant. I agree however with Nate that it is not a completely rigorous argument, however I think if you know something more about the distribution of $Z$ it will be easy to apply the argument above successfully. $\endgroup$ – Jacob Westman Sep 24 '15 at 16:39

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