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Thread: Expectation of infinite sum

  1. #1
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    Expectation of infinite sum

    Hi,

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

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

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

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

    Thanks!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Player1 View Post
    Hi,

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

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

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

    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:

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

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

    Does this make sense?

    --Chris
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  3. #3
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    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!
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  4. #4
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    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, $\displaystyle |\sum_{k=1}^n X_k|\leq \sum_{k=1}^n |X_k|\leq \sum_{k=1}^\infty |X_k|$ and the right-hand side is integrable because of your hypothesis;
    b) almost surely, $\displaystyle \sum_{k=1}^n X_k$ converges to $\displaystyle \sum_{k=1}^\infty X_k$ (this is a consequence of the inequality above, the fact that absolute convergence of a series implies convergence, and the property that $\displaystyle E[\sum_{k=1}^\infty |X_k|]<\infty$ implies $\displaystyle \sum_{k=1}^\infty |X_k|<\infty$ almost-surely).
    As a consequence of these two facts, you can tell, with the dominated convergence theorem, that indeed $\displaystyle \lim_n E[\sum_{k=1}^n X_k] = E[\sum_{k=1}^\infty X_k]$. Notice at last that $\displaystyle E[\sum_{k=1}^n X_k] =\sum_{k=1}^n E[X_k] $, so that you can also write the previous limit as $\displaystyle \sum_{k=1}^\infty E[X_k]$. And this is what you wanted.
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    Thanks!
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  6. #6
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    Hi!

    the property that $\displaystyle E[\sum_{k=1}^\infty |X_k|]<\infty$ implies $\displaystyle \sum_{k=1}^\infty |X_k|<\infty$ 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!
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  7. #7
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    Quote Originally Posted by Player1 View Post
    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 $\displaystyle Z$ is a random variable with values in $\displaystyle [0,\infty]$, then $\displaystyle E[Z]<\infty$ implies $\displaystyle Z<\infty$ almost surely. Why? Because if $\displaystyle Z=\infty$ with positive probability, then $\displaystyle E[Z]\geq \infty\cdot P(Z=\infty)=\infty$ (using the fact that $\displaystyle Z\geq 0$ in the inequality).

    Laurent.
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