Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For $X_i$ independent, is $\operatorname{Var}\left(\sum \limits_{i = 0}^\infty X_i \right) = \sum\limits_{i=0}^\infty \operatorname{Var}(X_i)$?

Thanks!

share|improve this question
    
I may lose simple ways, but it seems that you have to use e.g. Dominated Convergence Theorem so there may be a counterexample –  Ilya Nov 15 '11 at 17:14
    
In the finite case one can weaken the hypothesis of independence to that of uncorrelatedness: that all covariances are $0$. I'll be surprised if that is not also true of the infinite case. –  Michael Hardy Nov 15 '11 at 18:31
7  
Maybe this is more pathological than what you are looking for and not sure if it should be a comment or answer. Let $X_n = (-1)^n (2n-1)$ with probability one. Then $\{X_n\}$ are independent and $\sum_{i=1}^\infty \sigma_i^2 = 0$, but $\mathrm{Var}(\sum_i X_i)$ does not exist since $S_n = \sum_{i=1}^n X_i = (-1)^n n$ and so $S_n$ does not have a limit, and cannot have a mean, so cannot have a variance. –  cardinal Nov 15 '11 at 18:39
    
What if the sequence of variances of $\small X_i $ is $\small 1,4,9,16,25,\ldots,k^2,\ldots $ and thus their sum equals (formally?) $\small \zeta(-2) $ My point is, independently of the choice of $\small \zeta() $ as an example here, that if infinitely many terms are involved then we might have unexpected zeros which are counterintuitive and not a consequence of approximation. –  Gottfried Helms Nov 16 '11 at 7:31

1 Answer 1

up vote 7 down vote accepted

Yes, as soon as the RHS is finite and the series $\sum\limits_{n}\mathrm E(X_n)$ converges.

To see this, assume without loss of generality that the random variables $X_n$ are centered with variances $\sigma_n^2$ and that the series $\sum\limits_n\sigma_n^2$ converges, with $\sigma^2$ as its sum. Let $S_n=\sum\limits_{k\leqslant n}X_k$ and note that, for every $n\leqslant m$, $$ \mathrm E((S_m-S_n)^2)=\sum\limits_{k=n+1}^m\sigma_k^2, $$ which converges to zero when $n\to\infty$, hence $(S_n)_n$ is a Cauchy sequence in $L^2$. Let $S$ denote its limit in $L^2$. Then $\mathrm E(S_n^2)\to\mathrm E(S^2)$ when $n\to\infty$ and, for every $n$, $$ \mathrm E(S_n^2)=\sum\limits_{k\leqslant n}\sigma_k^2, $$ hence $\mathrm E(S^2)=\sigma^2$. Since $S_n\to S$ in $L^2$, a subsequence converges almost surely to $S$. Kolmogorov's inequality proves that the whole sequence converges almost surely to $S$, hence $S_n\to S$ in the almost sure sense as well and the proof is complete.

If the sum of the series $\sum\limits_n\sigma_n^2$ is infinite, $(S_n)_n$ diverges almost surely.

share|improve this answer
    
Thank you very much! –  badatmath Nov 16 '11 at 7:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.