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Math Help - Proving a variable has a poission distribution

  1. #1
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    Proving a variable has a poission distribution

    Hi

    I am given that:

    P(Y=k)= \sum \binom{n}{k}p^k.(1-p)^(n-k).e^-x.(x^n/n!)

    where the sum is from n=k to infinity

    and I need to show that Y is poission distributued with parameter \lambdap

    I'm not really sure where to start, should I take out all the variables which are not raised to the power of n out of the sum and change \binom{n}{k} to n!/(n-k)!k!?

    Any tips to get me started would be much appreciated
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  2. #2
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    Re: Proving a variable has a poission distribution

    Hello,

    Yes you can take out the variables where n is not in the expression.

    You can also, in a first time, change the summation from n=0 to infinity (since this is the range of a Poisson distribution). It should simplify quite a lot, since in the n \choose k, the n! simplifies, the (n-k)! becomes 0!=1 and k! can be pulled out from the sum.
    Last edited by Moo; August 15th 2011 at 08:04 AM. Reason: little mistake :(
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    Re: Proving a variable has a poission distribution

    Thanks so much for your reply

    The question is:
    P(Y=k)= \sum \binom{n}{k} p^k(1-p)^(n-k) e^{-x} \frac{x^n}{n!}

    where the sum runs from n=k to infinity (sorry i'm rubbish at using latex)

    is this right?

    P(Y=k)= \sum x^n/(n-k)!k! p^k (1-p)^(n-k) e^{-x}

    P(Y=k)= p^k e^{-x} \sum x^n/(n-k)!k! . (1-p)^(n-k)

    then change the sum so that it runs from n=0 to infinity

    P(Y=k)=(p^k)/k! e^{-x} \sum
    then I'm not sure what I'll have left inside? x^n...

    any hints appreciated
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  4. #4
    Super Member girdav's Avatar
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    Re: Proving a variable has a poission distribution

    After the change of summation, we have P(Y=k) = \frac{p^k}{k!}e^{-x}\sum_{j=0}^{+\infty}\frac{x^{j+k}}{j!}(1-p)^j =  \frac{p^k}{k!}e^{-x}x^k\sum_{j=0}^{+\infty}\frac{((1-p)x)^j}{j!}.
    We recognize the series of \exp, and now you an conclude.
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  5. #5
    Moo
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    Re: Proving a variable has a poission distribution

    Quote Originally Posted by yellowcarrotz View Post
    Thanks so much for your reply

    The question is:
    P(Y=k)= \sum \binom{n}{k} p^k(1-p)^(n-k) e^{-x} \frac{x^n}{n!}

    where the sum runs from n=k to infinity (sorry i'm rubbish at using latex)

    is this right?

    P(Y=k)= \sum x^n/(n-k)!k! p^k (1-p)^(n-k) e^{-x}

    P(Y=k)= p^k e^{-x} \sum x^n/(n-k)!k! . (1-p)^(n-k)

    then change the sum so that it runs from n=0 to infinity

    P(Y=k)=(p^k)/k! e^{-x} \sum
    then I'm not sure what I'll have left inside? x^n...

    any hints appreciated
    Replace n-k by j, that's a way to change the summation index
    And if there's just an n, n=j+k
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  6. #6
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    Re: Proving a variable has a poission distribution

    yep i've done it, thank you so much for all your help
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