5
$\begingroup$

Let $(X_i)_i$ be a sequence of centered i.i.d. random variables with finite variance. Is it true that

$$\frac{\sum_{i=1}^{\lfloor n^{0.6} \rfloor}X_i}{\sqrt{n}}\stackrel{\mbox{a.s.}}{\longrightarrow} 0\quad ?$$

Should be a sort of corollary of the central limit theorem but I don't know how to prove the a.s. convergence. Any help?

$\endgroup$
  • $\begingroup$ You can use characteristic functions in more or less the same way you can to prove the CLT and get this result. (Assuming some adjustment to round $n^{0.6}$ to an integer of course.) $\endgroup$ – Ian Feb 26 at 16:53
  • $\begingroup$ Is the last index of the sum supposed to be $\lfloor n^{0.6}\rfloor$? $\endgroup$ – Mars Plastic Feb 26 at 16:54
  • $\begingroup$ Obviously yes… @Ian using characteristic functions don't you prove a convergence in distribution? $\endgroup$ – bojica Feb 26 at 17:11
  • $\begingroup$ Convergence in distribution to a constant is convergence in probability. To improve that to a.s. convergence you need to show that the convergence in probability is sufficiently fast (in the sense of Borel-Cantelli). $\endgroup$ – Ian Feb 26 at 17:17
4
$\begingroup$

By looking at the cases where $n$ is in an interval of the form $\left[N^{1/0.6},(N+1)^{1/0.6}\right)$, we notice that the wanted convergence holds if we manage to prove the following: for each positive $\eta$, the following convergence holds almost surely: $$ \lim_{n\to +\infty}\frac 1{n^{1/2+\eta}}\left\lvert \sum_{i=1}^nX_i \right\rvert=0. $$ One could can use the bounded law of the iterated logarithms, which says that under the conditions of the opening post, the random variable $$ M:=\sup_{n\geqslant 3}\frac 1{\sqrt{n\log\log n}}\left\lvert \sum_{i=1}^nX_i \right\rvert $$ is almost surely finite. Therefore, $$ \frac 1{n^{1/2+\eta}}\left\lvert \sum_{i=1}^nX_i \right\rvert\leqslant \frac{\sqrt{\log\log n}}{n^\eta}M. $$

An other way is to control the moments of order two of $2^{-N\left(1/2+\eta\right)}\max_{1\leqslant n\leqslant 2^N}\left\lvert \sum_{i=1}^nX_i \right\rvert$ by using Doob's inequality. This will prove finiteness of $$ \sum_{N\geqslant 1}2^{-N\left(1/2+\eta\right)}\max_{1\leqslant n\leqslant 2^N}\left\lvert \sum_{i=1}^nX_i \right\rvert, $$ which is sufficient to conclude.

$\endgroup$
3
$\begingroup$

Let $\alpha\in (0,1)$ and $m=\lfloor n^{\alpha} \rfloor$. Then $$ \frac{1}{\sqrt{n}}\left|\sum_{i=1}^{\lfloor n^{\alpha} \rfloor}X_i\right|\le\frac{1}{m^{1/(2\alpha)}}\left|\sum_{i=1}^{m}X_i\right|\to 0 \quad\text{a.s.} $$ by the Marcinkiewicz–Zygmund SLLN.

$\endgroup$

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.