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Expected value of random sums with dependent variables

  1. Jun 16, 2010 #1
    Hi all,

    I have a question of computing the expectation of random sums.

    E(sim_{k=1}^N X_k) = E(N)E(X) if N and X_1, X_2,.....are independent and X_k's are iid. Here both N and X_k's are r.vs.

    But the condition of N and X_1, X_2,.....being independent is not true in many cases.

    How will you compute E(sim_{k=1}^N X_k) if N and X_1, X_2,.....are not independent (even weakly dependent).

    Can we use Law of iterative expectation? I am not sure what will E(sim_{k=1}^N X_k) equal to?

    Please help.....

    Thank you
    Regards

    Agrawal V
     
  2. jcsd
  3. Jun 16, 2010 #2
    Yes. Condition on the value of N, and then take the expectation with respect to N. If X_1, X_2, ... are identically distributed (but not necessarily independent), you still get E(X)E(N).

    EDIT #1: if N is not independent of the X's, then [tex]E(\sum_{k=1}^N X_k) = E(N)E(X)[/tex] is not generally true. Instead it would be something like

    [tex]E(\sum_{k=1}^N X_k}) = \sum_{n} nP\{N=n\}E(X|N=n)[/tex]

    So you would need to know the conditional expectation E(X|N=n).

    EDIT #2: And even that may not be quite general enough. It may happen that even though the X's are all identically distributed, E(X_i | N=n) does not equal E(X_j | N=n) for i and j different. So it would be

    [tex]E(\sum_{k=1}^N X_k}) = \sum_{n} P\{N=n\} \left (\sum_{k=1}^nE(X_k|N=n) \right )[/tex]
     
    Last edited: Jun 16, 2010
  4. Jun 17, 2010 #3
    Hi techmologist

    Thanks for the reply. In my case, I,m considering N is the stopping time and each X_i's act as a renewal process, i.e, each X_i is replaced by another X_j having a common distribution function F. So I was thinking more on the lines of renewal process and stopping time.

    I can across Wald's equality where N depends upon the X_i's until X_{n-1} and is independent of X_n, X_{n+1},....., because at X_n the condition (any stopping condition) is satisfied...which gives similar expression as E(sim_{k=1}^N X_k) = E(N)E(X). Do you think this will address the issue of dependence between N and the X_is..

    Also, can it take expectation with respect to N on this term as per law of iterative expectation....please suggest
    [tex]
    \sum_{n} nP\{N=n\}E(X|N=n)
    [/tex]

    Thank you
     
  5. Jun 17, 2010 #4
    Hi Agrawal,

    I had to look up Wald's equation, and I think now I see what you are getting at. By the way, the current Wikipedia article on Wald's equation is very confusing. I would give that article time to "stabilize" before I paid any attention to it. Instead of that, I used Sheldon Ross's book Introduction to Probability Models, 8th edition. On pages 462-463, he talks about Wald's equation and stopping times.

    So in the case you are talking about, the X_i 's are independent identically distributed random variables for a renewal process. To take an example from Ross's book, X could represent the time between arrivals of customers at a bank. But as you say, the stopping time N may depend on the X_i's. In the above example, the sequence could stop with the first customer to arrive after the bank has been open for an hour. Thus, if the twentieth customer arrived at 0:59:55, and the twenty-first customer arrived at 1:03:47, the stopping "time" would be N=21 and the sum of the waiting times would be 1:03:47.

    Note: Ross's definition of stopping time is that the event N=n is independent of X_{n+1}, X_{n+2},...., but generally depends on X_1, ..., X_n. It might be that he is labelling the X_i's differently than you. In his book, X_i is the waiting time between the (i-1)st and the ith event.

    I no longer think that conditioning on N is the way to do it, although it may be possible. That is what you meant by using the law of iterated expectations, right? In practice, finding E(X_i | N=n) is very difficult. Ross uses indicator variables to prove Wald's equation:

    [tex]I_i=\left\{\begin{array}{cc}1,&\mbox{ if }
    i\leq N\\0, & \mbox{ if } i>N\end{array}\right[/tex]

    Now note that I_i depends only on X_1, ..., X_{i-1}. You have observed the first i-1 events, and if you have stopped then N<i. If you have not stopped, then N is at least i.

    [tex]E\left( \sum_{i=1}^N X_i \right) = E\left(\sum_{i=1}^{\infty}X_iI_i\right) = \sum_{i=1}^{\infty}E(X_iI_i)[/tex]

    [tex]E\left( \sum_{i=1}^N X_i\right) = \sum_{i=1}^{\infty}E(X_i)E(I_i) = E(X)\sum_{i=1}^{\infty}E(I_i)[/tex]

    Now use the fact that: [tex]\sum_{i=1}^{\infty}E(I_i) = E\left( \sum_{i=1}^{\infty}I_i\right) = E(N)[/tex]
     
  6. Jun 18, 2010 #5
    Thank you techmologist......
     
  7. Jun 18, 2010 #6
    You are welcome. I got to learn something out of it, too. Wald's equation helped me solve a problem I had been wondering about for a while. Suppose Peter and Paul bet one dollar on successive flips of a coin until one of them is ahead $5. How many flips, on average, will it take for their game to end? At least I think my approach using Wald's equation will work... it involves taking a limit.
     
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