# Tilted sum of independent random variables

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?

• 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.) – Ian Feb 26 at 16:53
• Is the last index of the sum supposed to be $\lfloor n^{0.6}\rfloor$? – Mars Plastic Feb 26 at 16:54
• Obviously yes… @Ian using characteristic functions don't you prove a convergence in distribution? – bojica Feb 26 at 17:11
• 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). – Ian Feb 26 at 17:17

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.
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.