Show $n^{-p}*S_{n} \to 0$ in probability

Question

Suppose we have a sequence of i.i.d random variables $\{X_{n}\}_{n\in \Bbb N}$ with mean $0$ and variance $\sigma^{2} < \infty$. We want to show that $n^{-p}S_{n} \to 0$ in probability when $p>\frac{1}{2}$.

Here's my attempt

Showing $n^{-p}S_{n} \to 0$ in probability is equivalent to showing $$\Bbb P\left(\left|\frac{S_{n}}{n^{p}}\right|\leq \epsilon\right) \to 1 $$

Rearrange the expression in the bracket we finally get $$\Bbb P\left(\left|\frac{S_{n}}{\sigma \sqrt{n}}\right| \leq \frac{\epsilon}{\sigma}n^{p-\frac{1}{2}}\right).$$

By central limit theorem, we know $\frac{S_{n}}{\sigma \sqrt{n}} \to N(0,1)$ in distribution. But my question is that can we directly write the expression above as $$2\Phi \left(\frac{\epsilon}{\sigma}n^{p-\frac{1}{2}}\right)-1$$ and argue that as $n \to \infty$, we get $$\Bbb P\left(\left|\frac{S_{n}}{\sigma \sqrt{n}}\right| \leq \frac{\epsilon}{\sigma}n^{p-\frac{1}{2}}\right) \to 1$$

I don't know if central limit theorem is still valid when the expression in the bracket changes, Could someone help me with it? if not, how to prove it

Answer 1

The central limit theorem gives that for all fixed point $t$, $$\tag{*} \mathbb P\left(\left\lvert \frac{S_n}{\sigma\sqrt n}\right\rvert\leqslant t\right)\to \mathbb P(\lvert N\vert\leqslant t) $$ where $N$ has a standard normal distribution.

Actually, one can show that the convergence is also uniform on $\mathbb R$ because the limiting distribution function is continuous.

We can also conclude without using the uniform convergence: fix $R\gt 0$; for $n$ large enough, $\epsilon n^{p-1/2}\geqslant R$ hence $$ \mathbb P\left(\left\lvert \frac{S_n}{\sigma\sqrt n}\right\rvert\leqslant \epsilon n^{p-1/2} \right)\geqslant\mathbb P\left(\left\lvert \frac{S_n}{\sigma\sqrt n}\right\rvert\leqslant R \right). $$ Taking the $\liminf_{n\to+\infty}$ and using (*) with $t=R$ gives $$ \liminf_{n\to+\infty}\mathbb P\left(\left\lvert \frac{S_n}{\sigma\sqrt n}\right\rvert\leqslant \epsilon n^{p-1/2} \right)\geqslant\mathbb P\left(\left\lvert N\right\rvert\leqslant R \right). $$ Now, since $R$ the previous inequality is valid for all $R$, we derive that $$ \liminf_{n\to+\infty}\mathbb P\left(\left\lvert \frac{S_n}{\sigma\sqrt n}\right\rvert\leqslant \epsilon n^{p-1/2} \right)\geqslant1.$$

Answer 2

We can write $\frac{1}{n^p}S_n=\frac{1}{n^{p-1/2}}\frac{{S_n}}{\sqrt{n}}$

$\frac{{S_n}}{\sqrt{n}}$ converges in distribution to $N(0,\sigma^2),$ and $\frac{1}{n^{p-1/2}}$ converges in distribution to $0$ then $\frac{1}{n^p}S_n$ converges in distribution to $0$ which means convergence in probability to $0$.