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Random walk:expectation

  1. Jul 16, 2006 #1
    I have a random function f(n) which takes the values +/- 1 with equal probability.
    Let the variable X take the sum of the values of f(n) after n steps. Then I can write,
    [tex]X(T) = \sum_{n=0}^T f(n)[/tex] where T = 0,1,2,... and X(0) = 0.
    And I can write the expectation of X as,
    [tex]<X> = < \sum_{n=0}^T f(n) > = \sum_{n=0}^T <f(n) > = 0[/tex] since <f(n)> = 0 (by definition).
    My question:
    if, instead [tex]X(n) = \sum_{k=0}^n e^{-\mu k} f(n-k)[/tex] then, for the expectaion of X, can I write,
    [tex]< X(n)> = \sum_{k=0}^n e^{-\mu k} < f(n-k) >[/tex]
    or even,
    [tex]< X(n)> = \sum_{k=0}^n < e^{-\mu k}> < f(n-k) >[/tex]
    If I can do either of the above then can I also use <f(n)> = 0 to say that <X(n)> = 0, which I'm pretty sure it would have to be to answer the rest of my question.

    TIA
     
    Last edited: Jul 16, 2006
  2. jcsd
  3. Jul 16, 2006 #2

    mathman

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    Gold Member

    You left out one important definition - what is mu? Is it a random variable or is it a constant? If it is random, how does it relate to f(n)? If it is constant or independent of f(n), then your conclusion is correct.
     
  4. Jul 16, 2006 #3
    Thanks, I should have mentined that about mu. It is a constant value.
    So I was correct after all. :smile:
    I guess that's just as well, since I'm already typing up my results based on that conclusion!
    Many thanks for the confirmation.
     
    Last edited: Jul 16, 2006
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