# Thread: Expectation of infinite sum

1. ## Expectation of infinite sum

Hi,

My background is in engineering (not in math) so I am having trouble with this stuff... Question:

$X_i$ random vars such that $E[\Sigma _{i=1} ^\infty |X_i|] < \infty$

Show that $E[\Sigma _{i=1} ^\infty X_i] = \Sigma_{i=1} ^\infty E[X_i]$

I would like some advice of where to start; I don't need a full proof.

Thanks!

2. Originally Posted by Player1
Hi,

My background is in engineering (not in math) so I am having trouble with this stuff... Question:

$X_i$ random vars such that $E[\Sigma _{i=1} ^\infty |X_i|] < \infty$

Show that $E[\Sigma _{i=1} ^\infty X_i] = \Sigma_{i=1} ^\infty E[X_i]$

I would like some advice of where to start; I don't need a full proof.

Thanks!
Hello

Keep in mind this special property of mathematical expectation:

$E\left[X_1+X_2+\dots+X_n\right]=E\left[X_1\right]+E\left[X_2\right]+\dots+E\left[X_n\right]$

So, $E\left[\sum_{i=1}^{\infty}X_i\right]=E\left[X_1+X_2+X_3+\dots\right]=E\left[X_1\right]+E\left[X_2\right]+E\left[X_3\right]+\dots=\sum_{i=1}^{\infty} E\left[X_i\right]$

Does this make sense?

--Chris

3. Hi!

I know what you mean, but I cannot show it that way. I have been told that linearity only applies to finite sums.

I may need to use a convergence theorem (a random var. convergence theorem, maybe), but still don't know how to show this.

I should have said all this in the first place, sorry .

Thanks!

4. You're right, a theorem is needed in your case. Namely Lebesgue's dominated convergence theorem (or whatever name you give to it). Let's check it applies here:
a) for every n, $|\sum_{k=1}^n X_k|\leq \sum_{k=1}^n |X_k|\leq \sum_{k=1}^\infty |X_k|$ and the right-hand side is integrable because of your hypothesis;
b) almost surely, $\sum_{k=1}^n X_k$ converges to $\sum_{k=1}^\infty X_k$ (this is a consequence of the inequality above, the fact that absolute convergence of a series implies convergence, and the property that $E[\sum_{k=1}^\infty |X_k|]<\infty$ implies $\sum_{k=1}^\infty |X_k|<\infty$ almost-surely).
As a consequence of these two facts, you can tell, with the dominated convergence theorem, that indeed $\lim_n E[\sum_{k=1}^n X_k] = E[\sum_{k=1}^\infty X_k]$. Notice at last that $E[\sum_{k=1}^n X_k] =\sum_{k=1}^n E[X_k]$, so that you can also write the previous limit as $\sum_{k=1}^\infty E[X_k]$. And this is what you wanted.

5. Thanks!

6. Hi!

the property that $E[\sum_{k=1}^\infty |X_k|]<\infty$ implies $\sum_{k=1}^\infty |X_k|<\infty$ almost-surely.
I don't know why this is true.

I know convergence in mean implies convergence in probability and also in distribution, but not almost-sure convergence. What am I missing here?

Thanks!

7. Originally Posted by Player1
I know convergence in mean implies convergence in probability and also in distribution, but not almost-sure convergence. What am I missing here?
This is way simpler than that. If $Z$ is a random variable with values in $[0,\infty]$, then $E[Z]<\infty$ implies $Z<\infty$ almost surely. Why? Because if $Z=\infty$ with positive probability, then $E[Z]\geq \infty\cdot P(Z=\infty)=\infty$ (using the fact that $Z\geq 0$ in the inequality).

Laurent.

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