Let $\{X_n\}_{n\geq1}$ be a sequence of centered independent random variables such as $E(X_n^2)=2n$

and $$Y_n=\frac1{n^\alpha}\sum_{i=1}^{i=n}X_i\quad\quad\alpha\geq1$$

I am trying to prove that for $\alpha > \frac32$, $Y_n \rightarrow 0$ almost surely.

I started by calculating $Var(Y_n)$ which I found to be equal to $\frac{n(n+1)}{n^{2\alpha-2}}$ and since $E(Y_n)=0$ we get

$$\lim_{n\to\infty} E(|Y_n-0|^2)=0$$

which means that for $\alpha > 1$, $\{Y_n\}_{n\geq1}$ converges to $0$ in quadratic mean.

I am stuck going form there to the almost sure convergence.

Any help would be greatly appreciated!

  • 1
    $\begingroup$ If $\{X_n\}$ is i.i.d then $EX_n^{2}$ does not depend on $n$. I think you should change 'i.i.d.' to 'independent'. $\endgroup$ – Kavi Rama Murthy Sep 13 '19 at 23:36

Let $S_n=\sum_{k=1}^n X_k$ and note that $V(S_n)=\sum_{k=1}^n V(X_k)= \sum_{k=1}^n 2k = n(n+1)$.

For any $\epsilon >0$, by Markov's bound, $$P\left(\frac{|S_n|}{n^\alpha}\geq \epsilon \right) = P(|S_n|\geq n^\alpha\epsilon) \leq \frac{n(n+1)}{n^{2\alpha}\epsilon^2}\sim \frac{1}{n^{2\alpha-2}\epsilon^2}$$

When $\alpha >\frac 32$, we have $2\alpha-2>1$ and $\displaystyle \sum_n \frac{1}{n^{2\alpha-2}\epsilon^2}$ converges.

A standard criterion of almost sure convergence implies that $$\frac{|S_n|}{n^\alpha}\xrightarrow[]{a.s} 0$$


Note that the $X_k$ can’t be iid if they don’t have the same second moment. In this proof we do not even use the independence of the $X_i$.

Consider $Z_n=\sum_{i=1}^n{\frac{X_i^2}{i^{2\alpha}}}$.

For $\alpha > 1$, $Z_n$ is an increasing sequence of rv and bounded in $L^1$ so it converges as to some non-negative rv $T$ which has a finite $L^1$ norm (so is as finite), therefore, almost surely, the sequence $(i^{-\alpha}X_i)$ is $\ell^2$.

Now, it is elementary (not completely obvious but that’s a purely analytic fact) to check that if $a_n \in \ell^2$ then $\sum_{k=1}^n{a_k} = o(\sqrt{n})$.

Thus, almost surely, for every $\alpha > 1$, $S_n=\sum_{k=1}^n{k^{-\alpha}|X_k|}$ is negligible before $n^{1/2}$. Now, note that $|Y_n| \leq S_n$.

  • $\begingroup$ @Kavi Rama Murphy: Yes, but this works for any $\alpha > 1$: so for any $\alpha > 3/2$, $Y_n^{(\alpha)}=\frac{1}{\sqrt{n}}Y_n^{(\alpha-1/2)} \rightarrow 0$ since $\alpha-1/2 > 1$. $\endgroup$ – Mindlack Sep 14 '19 at 10:13

Let us recall:

(1): Kronecker's Lemma: If $(x_n)_{n \in \mathbb N}$ is a sequence such that $\sum_{n} x_n $ converges, $(b_n)_{n \in \mathbb N}$ is increasing positive sequence such that $\lim_{n \to \infty} b_n = +\infty$, then $\lim_{n \to \infty} \frac{1}{b_n} \sum_{j=1}^n x_jb_j = 0$

(2): Kolmogorov's two-series theorem: If $(X_n)_{n \in \mathbb N}$ are independent, $\mathbb E[X_n], Var(X_n)$ exists and are finite for every $n \in \mathbb N$, series $\sum \mathbb E[X_n] , \sum Var(X_n)$ are convergent, then $\sum X_n$ converges almost surely.

Using those above, we'll prove:

Lemma: Let $(X_n)_{n \in \mathbb N}$ be independent rvs such that $Var(X_n)$ is finite for every $n \in \mathbb N$. Moreover, let $(b_n)_{n \in \mathbb N}$ be increasing sequence of positive numbers with $\lim b_n = \infty$. Let $S_n = \sum_{j=1}^n X_n$. If $\sum \frac{Var(X_n)}{b_n^2}$ converges then $\frac{S_n - \mathbb E[S_n]}{b_n}$ converges to $0$ almost surely.

Proof: Let $Y_j = \frac{X_j - \mathbb E[X_j]}{b_j}$, then for every $j \in \mathbb N$ we have: $\mathbb E[Y_j] = 0, Var(Y_j) = \frac{Var(X_j)}{b_j^2}$, so both $\sum \mathbb E[Y_j], \sum Var(Y_j)$ converges, so using (2): we get $\sum Y_j$ converges almost surely. So we have a set of $\mathbb P$ measure $1$ where for every $\omega$ in that set, we can use (1) with sequence $x_n = Y_n(\omega)$ to obtain: $\lim_{n \to \infty} \frac{1}{b_n} \sum_{j=1}^n Y_j(\omega)b_j = 0$

But $Y_j(\omega)b_j = X_j(\omega) - \mathbb E[X_j]$, so $\sum_{j=1}^n Y_j(\omega)b_j = S_n(\omega) - \mathbb E[S_n]$. So on the set of measure $1$ we have that convergence, so $\frac{S_n - \mathbb E[S_n]}{b_n}$ converges almost surely to $0$.

Answer: You can just use this with $S_n = \sum_{j=1}^n X_j$, then $\mathbb E[S_n] = 0$. Moreover $Var(X_n) = 2n$, so $\sum \frac{Var(X_n)}{n^{2a}} $ converges iff $2a-1 > 1$ which means $a>1$. So even for $a>1$ it holds that $Y_n = \frac{S_n - \mathbb E[S_n]}{n^a}$ converges almost surely to $0$.


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