Results 1 to 6 of 6

Math Help - Proof involving expectation of a random variable.

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    21

    Proof involving expectation of a random variable.

    Hi all,

    I had this problem as part of an example in class. I really didn't understand what was happening so I wondered if anybody could enlighten me as to what's going on here. Any help is greatly appreciated.



    Uploaded with ImageShack.us
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    start with the definition

    E(N) = \sum_{k=1}^{\infty}kP(N=k)= \sum_{k=1}^{\infty}\sum_i^kP(N=k)

    = \sum_{i=1}^{\infty}\sum_k^{\infty}P(N=k)

    = \sum_{i=1}^{\infty}P(N\ge i)

    tex isn't working for me lately, here
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2011
    Posts
    21
    Quote Originally Posted by matheagle View Post
    start with the definition

    E(N) = \sum_{k=1}^{\infty}kP(N=k)= \sum_{k=1}^{\infty}\sum_i^kP(N=k)

    = \sum_{i=1}^{\infty}\sum_k^{\infty}P(N=k)

    = \sum_{i=1}^{\infty}P(N\ge i)

    tex isn't working for me lately, here
    I'm trying to see what's going on in your statement, but it's tricky without the latex.

    E(N) is = the sum of [ k*P(N=k)], i think i understand, but what exactly is going on in the double summation in the first line?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    replace k with the sum from 1 to k, 1+1+...+1=k
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2011
    Posts
    21
    Awesome, thanks!

    Any ideas about the second part involving the sum [i*P(N>=i)]?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    same idea, but I tried to avoid a triple sum

    \sum_{i=1}^{\infty} i P(N\ge i)

    =\sum_{i=1}^{\infty} i \sum_{k=i}^{\infty}P(N=k)

    =\sum_{k=1}^{\infty}\sum_{i=1}^k i P(N=k)

    now use \sum_{i=1}^k i =k(k+1)/2

    =\sum_{k=1}^{\infty}{k(k+1)\over 2} P(N=k)

    =E(N(N+1))/2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Expectation of Discrete Random Variable
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: February 24th 2011, 07:01 PM
  2. Expectation of exponential random variable.
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: August 24th 2010, 03:03 AM
  3. Random Variable Expectation
    Posted in the Statistics Forum
    Replies: 1
    Last Post: April 5th 2010, 02:08 PM
  4. Discrete Random Variable and Expectation
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 27th 2008, 12:10 PM
  5. binomial random variable Expectation
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: October 8th 2008, 07:39 PM

Search Tags


/mathhelpforum @mathhelpforum