2
$\begingroup$

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!

$\endgroup$
  • 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
1
$\begingroup$

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 $$\frac{|S_n|}{n^\alpha}\xrightarrow[]{a.s} 0$$

$\endgroup$
1
$\begingroup$

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 $|Y_n| \leq S_n$.

$\endgroup$
  • $\begingroup$ @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
1
$\begingroup$

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 $0$.

$\endgroup$

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.