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)?

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $p_n\to 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.