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Assume $X_i$s are i.i.d. random variables with mean $\mu$ and variance $\sigma^2$. Prove: $$\lim_{n\to\infty}n^2\mathbb{P}\left(\left|\frac{\sum_{i=1}^{n} X_i}{n}-\mu\right|>n^{-1/4}\right)=0. \,\,\,\,\,\,\ (1)$$

My effort: I used Chebyshev inequality but did not work. Then, I thought about using the central limit theorem (CLT). By CLT: $$\lim_{n\to\infty}\mathbb{P}\left(\sqrt{n}\left|\frac{\sum_{i=1}^{n} X_i}{n}-\mu\right|>\gamma\right)=1-\mathrm{erf}\left(\frac{\gamma}{\sigma}\right).$$ The problem is that $\gamma$ cannot be a function of $n$. Otherwise, I could let $\gamma=n^{1/4}$ and prove what I need. Is there a trick I can use here? How can I prove (1)? If (1) only holds for certain conditions, please let me know.

Lastly, if we know $X_i=A_i^2 B_i^2$ where $A_i$ and $B_i$ are Gaussian random variables with non-zero means, can we prove (1)?

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    $\begingroup$ Somehow I doubt it's true in general, but Chebyshev works as soon as $X_i$ has finite $8+\epsilon$ moment. $\endgroup$ – Shalop Aug 31 '18 at 17:44
  • $\begingroup$ @Shalop Regarding your first comment, Chebyshev does not work since it gives me a constant over $n$. When multiplied by $n^2$, the product goes to infinity. Can you expand on the Chebyshev more? Maybe I am missing something. $\endgroup$ – Susan_Math123 Aug 31 '18 at 19:33
  • $\begingroup$ @Shalop Regarding your second comment: Can you please expand on what you said a little more? I have added that $X$ is the product of two dependent Gaussian random variables with non-zero means. Does it help? $\endgroup$ – Susan_Math123 Aug 31 '18 at 19:34
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    $\begingroup$ In my first comment, I'm saying to use the general form of Chebyshev for $p^{th}$ moments, i.e., $P(|Y|>a) \leq E|Y|^p / a^p$. One can argue as follows: from interpolation it is true that if $X_1$ is centered and has a finite $p^{th}$ moment for some $p\ge 2$, then $E\big|\sum_1^n X_i\big|^p \le C_p n ^{p/2}$ where $C_p$ is independent of $n$. Use this for $p>8$ and you get (1). A product of two squared gaussians has finite moments of all orders (by Cauchy-Schwarz, say) so you're good. $\endgroup$ – Shalop Aug 31 '18 at 21:10
  • $\begingroup$ @Shalop Many thanks. Can you please let me know what you mean by interpolation? How do you say if $X_1$ is centered and has finite $p$-th moment for some $p\geq 2$, then ...? Can you tell me where to read to understand this? $\endgroup$ – Susan_Math123 Aug 31 '18 at 21:58
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Replacing $X_i$ by $X_i-\mu$, we can assume that $\mu=0$. Let $$ p_n:=n^2\Pr\left\{\left\lvert \sum_{i=1}^nX_i\right\rvert>n^{3/4}\right\}. $$ Then $$ n^2\Pr\{\left\lvert X_1\right\rvert>2n^{3/4}\} \leqslant p_n+n^2\Pr\left\{\left\lvert \sum_{i=1}^{n-1}X_i\right\rvert>n^{3/4}\right\}\leqslant p_n+\frac{n^2}{(n-1)^2}p_{n-1}. $$ hence if (1) holds, then necessarily, $$ \tag{(C)}\lim_{t\to +\infty}t^{8/3}\Pr\{\left\lvert X_1\right\rvert>t\}=0. $$ To do the opposite direction, we use the following inequality (Theorem B.3 p. 172 in these notes by Emmanuel Rio): for each independent sequence of random variables $(Y_i)_{i=1}^N$, each $V\geqslant \sum_{j=1}^N\mathbb E\left[Y_j^2\right]$ and each $\lambda$, $x\gt 0$, $$ \Pr\left\{\max_{1\leqslant n\leqslant N}\left\lvert\sum_{i=1}^nY_i \right\rvert >\lambda\right\}\leqslant 2\exp\left(-\frac V{x^2}h\left(\frac{\lambda x}V\right)\right)+2 \sum_{i=1}^N\Pr\left\{ \left\lvert Y_i \right\rvert >x \right\}, $$ where $h(u)=(1+u)\log(1+u)-u$.

Then apply this with $N=2^n$, $\lambda =2^{3n/4}$, $x=\sigma^2 2^{3n/8}$ and $V=\sigma^22^n$ to get that $$\max_{2^{n-1}\leqslant k\leqslant 2^n}p_k\leqslant 2\exp\left(-\sigma^{-2}2^{n/4}h(2^{n/8}) \right)+2^{n+1}\Pr\{\left\lvert Y_1 \right\rvert>\sigma^2 2^{3n/8} \}$$ hence $p_n\to 0$.

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