# Almost sure convergence of a sum of independent random variables

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!

• 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'. – 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 $$|Y_n| \leq S_n$$.

• @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 $\alpha-1/2 > 1$. – 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 $$0$$.