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$\newcommand{\var}{\operatorname{var}}$

Given $X_1,X_2,\ldots$ is a sequence of random variables with $\rho=0$ and finite second moments.

I am trying to show that

$$\var(\bar{X}_n)\rightarrow 0\text{ as }n\rightarrow\infty$$

The obvious bit is that $\bar{X}_n=\frac1n\sum_{i=1}^n X_i$ and the exercise allows for the assumption that

$$\frac1n\var(X_n)\rightarrow 0\text{ as }n\rightarrow\infty$$

Since there is no mention on variance being bounded, how does this change things, if at all?

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Is $\rho$ the same as $E[X]$? –  sai Jul 28 '12 at 18:12
    
no, $\rho=0$ means the RVs are uncorrelated. Hope that helps. –  Justin Jul 28 '12 at 18:15

2 Answers 2

up vote 2 down vote accepted

If all the covariances are zero, $\mathrm{var}(\bar X_n)=\frac1{n^2}\sum\limits_{k=1}^n\mathrm{var}(X_k)$. If furthermore $\frac1n\mathrm{var}(X_n)\to0$, then $\mathrm{var}(\bar X_n)\to0$.

Edit: Assume that $\frac1n\mathrm{var}(X_n)\to0$ when $n\to\infty$. Let $\varepsilon\gt0$. There exists $n_\varepsilon$ such that $\mathrm{var}(X_n)\leqslant\varepsilon n$ for every $n\geqslant n_\varepsilon$. Let $C_\varepsilon=\sum\limits_{k=1}^{n_\varepsilon}\mathrm{var}(X_k)$. Then, for every $n$, $$ \sum\limits_{k=1}^n\mathrm{var}(X_k)\leqslant C_\varepsilon+\varepsilon\sum\limits_{k=1}^nk\leqslant C_\varepsilon+\varepsilon n^2. $$ Thus, $\mathrm{var}(\bar X_n)\leqslant\frac1{n^2}C_\varepsilon+\varepsilon$ for every $n$, which implies that $\limsup\limits_{n\to\infty}\mathrm{var}(\bar X_n)\leqslant\varepsilon$. This holds for every $\varepsilon\gt0$, and $\mathrm{var}(\bar X_n)\geqslant0$ for every $n$, hence $\lim\limits_{n\to\infty}\mathrm{var}(\bar X_n)=0$.

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+1. For clarity, the part after the "hence" follows because if $a_n \to C$ then the Cesaro mean $\frac 1 n \sum_{i = 1} ^ n a_i \to C$ as well. –  guy Jul 29 '12 at 2:19
    
@guy: Yes, although in this case the relevant Césaro-like statement is rather that, if $a_n\to C$, then $\frac1{n^2}\sum\limits_{k=1}^nka_k\to \frac12C$. –  Did Jul 29 '12 at 22:18
    
@did can you explicitly show how the conclusion is implied from the assumption for cases of unbounded variance? I'm thinking about cases such as Var(X_n)= sqrt{n} –  Justin Aug 4 '12 at 19:05
1  
See Edit. $ $ $ $ –  Did Aug 4 '12 at 19:47

$\newcommand{\var}{\operatorname{var}}$ $$ \var\left( \frac{X_1+\cdots+X_n}{n} \right) = \frac{1}{n^2} \var(X_1+\cdots+X_n) = \frac{1}{n^2}\left(\var(X_1)+\cdots+\var(X_n)\right). $$ I think it's easy to show this does not go to $0$ as $n\to\infty$ if the sequence of variances grows fast enough. If they're all no bigger than $A$, then the whole thing is $\le nA/n^2 = A/n$, and that goes to $0$. One could get by with a weaker hypothesis than such boundedness, but you can't just drop the assumption.

(That $\rho\ne0$ means you don't get a bunch of covariance terms in the last expression on the first line above.)

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note that you have to assume, as @did did, that the covariances are zero. else $\mathrm{var}$ is not "linear" (which it isn't, anyway) –  akkkk Jul 29 '12 at 9:50
    
@Auke : That's what I said in my parenthetical sentence at the end. –  Michael Hardy Jul 29 '12 at 19:38
    
ah, so that's what it means - I was unfamiliar with this notation, thanks. –  akkkk Jul 29 '12 at 20:12
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@Auke : $\rho$ is conventionally used for the correlation, i.e. $\rho(X,Y)=\operatorname{corr}(X,Y)=\operatorname{cov}(X,Y)/\sqrt{\operatorname{‌​var}(X)\operatorname{var}(Y)}$. –  Michael Hardy Jul 29 '12 at 20:48
    
ah yes, I might have seen that once, thanks a lot. –  akkkk Jul 29 '12 at 20:54

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