5
$\begingroup$

Let $\{X_i\}_{i\geq 1}$ be IID with finite second moment, and $$ Y_n = \frac{2}{n(n+1)}\sum_{i=1}^n \,i\cdot X_i \, , \qquad n\geq 1 \, . $$ Could you please tell me how can I show that $Y_n$ converges in probability to $\mathrm{E}[X_1]$?

I'm thinking Kolmogorov Convergence criterion. But seems like I cannot prove it using that. Any suggestion?

$\endgroup$
10
  • $\begingroup$ I edited the $\LaTeX$. Please, check it out. $\endgroup$ – Zen Jun 2 '14 at 5:05
  • $\begingroup$ If you appreciated it, why did you take the LaTeX out again? Your last edit appears to have been nothing but an explicit removal of every single improvement made to your post. Why do that? $\endgroup$ – Glen_b Jun 2 '14 at 5:44
  • $\begingroup$ Please read the self-study tag wiki info and change your question to address the points raised there. Specifically, what have you tried, and what do you need help with. $\endgroup$ – Glen_b Jun 2 '14 at 5:49
  • $\begingroup$ Sorry @Zen. It got overwritten. My bad. Could you kindly edit the LaTeX again please? $\endgroup$ – Kingstat Jun 2 '14 at 6:04
  • $\begingroup$ @Glen_b I'm new here. Rookie mistake. Sorry! $\endgroup$ – Kingstat Jun 2 '14 at 6:08
4
$\begingroup$

Actually, we can even show that $\mathbb E|Y_n-\mathbb E[X_1]|^2\to 0$. Indeed, since $\sum_{j=1}^nj=n(n+1)/2$ and $\mathbb E[X_j]=\mathbb E[X_1]$ for all $j$, $$Y_n-\mathbb E[X_1]=\frac 2{n(n+1)}\sum_{j=1}^nj(X_j-\mathbb E[X_j]),$$ hence $$\tag{1}\mathbb E|Y_n-\mathbb E[X_1]|^2=\frac 4{n^2(n+1)^2}\sum_{i,j=1}^n ij\mathbb E\left[(X_i-\mathbb E[X_i])(X_j-\mathbb E[X_j])\right].$$ If $i\neq j$, then by independence $\mathbb E\left[(X_i-\mathbb E[X_i])(X_j-\mathbb E[X_j])\right]=0$ and plugging it in (1), $$\tag{2}\mathbb E|Y_n-\mathbb E[X_1]|^2=\frac 4{n^2(n+1)^2}\sum_{j=1}^n j^2\mathbb E\left[(X_j-\mathbb E[X_j])^2\right].$$ Using now the fact that $X_j$ has the same distribution as $X_1$ and bounding $\sum_{j=1}^nj^2$ by $n^2(n+1)$, equality (2) becomes $$\mathbb E|Y_n-\mathbb E[X_1]|^2\leqslant\frac 4{n+1}\mathbb E\left[(X_0-\mathbb E[X_0])\right]^2$$ and we are done.

$\endgroup$
0
1
$\begingroup$

Much later, here's an updated answer without hints. I mostly wanted to see if I could make sense of the details. This proof of almost sure convergence (which implies convergence in probability) complements the supplied proof of convergence in mean square and the direct proof using Chebyshev's inequality.

Proof outline

(1) Show that $Y_n\overset{a.s}{\to}Y$ for some random variable Y.

(2) Show that $\mathbb E Y = \mu := \mathbb E X_1$ and $\mathrm{Var}(Y)=0$

(3) Conclude from this that since (2) implies that $Y = \mu$ a.s., we by (1) have the desired result, namely: $Y_n\overset{a.s}{\to} \mu$.

Proof of 1)

Consider the sum of absolute values, viz.

$$\begin{align} S_n&:=\sum_{i=1}^n\frac{2i}{n(n+1)}|X_i| \\ & \leq \frac{1}{n}\sum_{i=1}^n|X_i| \overset{a.s,m.s}{\to} \mathbb E|X_i|\leq\sqrt{\mathbb E X_i^2}<\infty \end{align}$$, where the convergence and inequalities in the last line follows from the strong law of large numbers and the assumption of finite second moment.

This shows that on sets with total probability one, $S_n$ is an increasing and bounded sequence and thus convergent. But $S_n$ is the sum of the absolute values of the terms of the sum $Y_n$, so we thus know that also $Y_n$ converges on these sets. In other words, $Y_n$ is almost surely convergent with limit $Y$, say.

Proof of 2)

We use the following standard result (sometimes called the extended or improved dominated convergence theorem):

A dominated convergence theorem (DCT) Let $\{Z_n\}$ be a sequence of random variables on some probability space$(\Omega,\mathcal{F},P).$ If $Z_n \overset{a.s}{\to} Z$ and there exist random variables $M_n \overset{a.s}{\to}M$ such that $|Z_n|\leq M_n, \mathbb EM_n<\infty,\forall n$ and $\lim _n \mathbb E M_n = \mathbb E M <\infty$, then $\lim _{n\to \infty} \mathbb E Z_n=\mathbb EZ.$

To obtain the expectation of $Y$, take first $Z_n=Y_n$ and $M_n=\frac{1}{n}\sum_{i=1}^n|X_i|$ in the DCT. We get $\lim _{n\to \infty} \mathbb E Y_n=\lim _{n\to \infty} \mu=\mathbb EY.$

To obtain the variance, set $Z_n=(Y_n-\mu)^2$. We have $Z_n \overset{a.s}{\to}Z:=(Y-\mu)^2$ and also $$|Y_n-\mu|^2\leq (S_n+|\mu|)^2 \leq (\frac{1}{n}\sum_{i=1}^n|X_i|+|\mu|)^2=:M_n.$$

By expanding the square and using the mean square convergence of $\frac{1}{n}\sum_{i=1}^n|X_i|$ it is clear that this $M_n$ satisfies the requirement for applying the DCT. Thus, $$\mathrm{Var}(Y)=\mathbb E (Y-\mu)^2 = \lim_n \mathbb E (Y_n-\mu)^2=\lim_n \mathrm{Var}(Y_n)=0.$$ This finishes the proof.

$\endgroup$

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.