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Properties of Poisson Distribution

  1. Sep 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Use the Poisson distribution W=(λ^n/n!)*e^-λ to calculate <n>


    2. Relevant equations
    <n>=ƩW*n



    3. The attempt at a solution
    Since W = (λ^n/n!)*e^-λ I wind up with <n>=[(λ^n/n!)*e^-λ]*n
    But I really don't know where to go from here. Should I do a Taylor Series. I've tried crossing out the top n and ended up with


    [(λ^n/(n-1)!)*e^-λ] but this doesn't seem to help. If anyone can point me in the right direction or a general problem solving strategy that would be great. I'd like to demonstrate more work but I don't know what to do.
     
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  3. Sep 10, 2013 #2

    vanhees71

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    Evaluate the generating function
    [tex]Z(x)=\sum_{n=0}^{\infty} W(n) \exp(-n x).[/tex]
    Then you can get expectation values by taking derivatives of this function wrt. [itex]x[/itex] :-).
     
  4. Sep 10, 2013 #3

    vela

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    You need the summation.
    $$\langle n \rangle = \sum_{n=0}^\infty \frac{\lambda^n}{n!} e^{-\lambda} n = \sum_{n=1}^\infty \frac{\lambda^n}{(n-1)!} e^{-\lambda}$$ Note that the lower limit changed from n=0 to n=1, since the n=0 term is 0. You just need to recognize that ##e^{-\lambda}## is a constant, so you can factor it out of the summation. Similarly, try pulling one factor of ##\lambda## out so that the exponent of ##\lambda## inside the summation matches up with the factorial.
     
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