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.

$X_{k}$ are random variables and they are not independent, I wonder if $E(\sum_{n=1}^{+\infty}X_{k}$)=$\sum_{n=1}^{+\infty}EX_{k}$. if the equation does not hold, what conditions are required in order to make it right.

share|improve this question
    
Thank you for all your help. –  Jim Jun 11 '11 at 2:18

4 Answers 4

up vote 5 down vote accepted

If the $X_i$ are nonnegative, it is always true, by either monotone convergence or Tonelli's theorem (thinking of the sum as an integral over $\mathbb{N}$ with respect to counting measure). If we have $\sum_k E|X_k| < \infty$ or equivalently $E \sum_k |X_k| < \infty$, then it is also true, by either dominated convergence or Fubini's theorem.

Otherwise it can fail: let $U \sim U(0,1)$, $Y_k = k 1_{\{U \le 1/k\}}$, so that $Y_k \to 0$ a.s. but $E Y_k = 1$ for each $k$. Let $Y_0 = 0$ and $X_k = Y_k - Y_{k-1}$ for $k \ge 1$. Then you can easily check that $$E \sum_{k=1}^\infty X_k = 0 \ne 1 = \sum_{k=1}^\infty E X_k.$$

share|improve this answer
    
Nice counterexample! But I think you must switch the 0-1 numbers. –  leonbloy Jun 10 '11 at 16:22
    
@leonbloy: No, I'm pretty sure it's correct as it is. Note that $\sum_k X_k = \lim Y_k = 0$ a.s., while $E X_1 = 1$ and $E X_k = 0$ for $k > 1$. –  Nate Eldredge Jun 10 '11 at 18:27
    
You're right! I was (in the LHS) taking limit and expectation in the wrong order. –  leonbloy Jun 10 '11 at 19:03

You can conclude this if $X_i$'s are non-negative. The proof follows immediately from monotone convergence theorem.

share|improve this answer
    
Out of interest, why do you need the $X_i$ to be non-negative? I would have thought that this was always true (as long as the sum of expectations is finite) and followed from linearity of expectation. Is this a subtlety to do with the Riemann rearrangement theorem? –  Chris Taylor Jun 10 '11 at 9:25
    
By monotone convergence theorem, you should first confirm that the partial sum of $X_{k}$ converges to a nonnegative random variable $X$ a.e. –  Jim Jun 10 '11 at 9:57
1  
The dominated convergence theorem yields the following: If $\sum E|X_k| \lt \infty$ then $\sum X_k$ is integrable and finite a.e. and $E\sum X_k = \sum EX_k$. This seems to me at least as useful. –  t.b. Jun 10 '11 at 9:58

Elaborating on Sivaram's answer.

Suppose that the $X_i$ are nonnegative random variables on a probability space $(\Omega,\mathcal{F},P)$, and let $S_n = \sum\nolimits_{i = 1}^n {X_i }$. Since $S_n$ is monotone increasing, it converges pointwise to $S \in [0,\infty]$; that is, for each $\omega \in \Omega$, $S(\omega ): = \lim _{n \to \infty } S_n (\omega ) \in [0,\infty]$. By the Monotone convergence theorem, since $0 \leq S_1 \leq S_2 \leq \cdots$ is a monotone increasing sequence of nonnegative random variables (measurable functions), the pointwise limit $S$ is also measurable and it holds $$ \mathop {\lim }\nolimits_{n \to \infty } {\rm E}(S_n ) := \mathop {\lim }\limits_{n \to \infty } \int_\Omega {S_n dP} = \int_\Omega {(\lim _{n \to \infty } S_n )dP} := {\rm E}(\lim _{n \to \infty } S_n ) \in [0,\infty]. $$ Hence $$ \sum\limits_{n = 1}^\infty {{\rm E}(X_n )} = \mathop {\lim }\limits_{n \to \infty } {\rm E}(S_n ) = {\rm E}(\lim _{n \to \infty } S_n ) = {\rm E}\bigg(\sum\limits_{n = 1}^\infty {X_n } \bigg), $$ where the first equality follows from the linearity of expectation. Note that if the left-hand side is finite, that is $\sum\nolimits_{n = 1}^\infty {{\rm E}(X_n )} < \infty$, then ${\rm E}(\sum\nolimits_{n = 1}^\infty {X_n }) < \infty$, implying that the pointwise limit $\sum\nolimits_{n = 1}^\infty {X_n } \,( = S)$ is almost surely finite (indeed, $\int_\Omega {SdP} < \infty $ implies that $S$ is almost surely finite).

share|improve this answer

If the $X_k$ are highly correlated with expectations converging fairly fast to $0$ --- $O(k^{-1.001})$ would be enough --- then the summation on the LHS could easily be almost surely infinite while the RHS is well defined. A simple case would be $X_k=X$ for all $k$, with $X$ either $-1$ or $1$ with equal probability. It isn't obvious just how to construct an optimal function of the correlations in order to maintain this condition if and only if the function satisfies a certain bound. I doubt whether there could be such a function---for a start, it would have to be independent of any changes to any finite number of $X_k$.

share|improve this answer
1  
minor point: in the simple case you mentioned, $\sum\nolimits_{k = 1}^\infty {X_k } $ is $+\infty$ or $-\infty$ with probability $1/2$ each, hence the LHS is not defined. –  Shai Covo Jun 10 '11 at 12:26
    
@Shai: Thanks! Is the edit OK now? –  John Bentin Jun 10 '11 at 17:43
    
Yes (assuming "infinite" is "infinite in absolute value"). –  Shai Covo Jun 12 '11 at 5:09

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.