Results 1 to 7 of 7

Math Help - Expectation of infinite sum

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    10

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    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:

    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2008
    Posts
    10
    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!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2008
    Posts
    10


    Thanks!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Sep 2008
    Posts
    10
    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!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    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 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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Expectation
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: March 3rd 2010, 11:28 AM
  2. Replies: 7
    Last Post: October 12th 2009, 10:10 AM
  3. Prove: every set with an infinite subset is infinite
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: March 31st 2009, 04:22 PM
  4. Proof: infinite set minus a fintie set is infinite
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: February 24th 2009, 10:54 PM
  5. Expectation & Conditional Expectation
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: February 1st 2009, 10:42 AM

Search Tags


/mathhelpforum @mathhelpforum