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Math Help - Poisson probability

  1. #1
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    Poisson probability

    X, Y are two independant poisoon processes with parameter a and b respectively. Find the conditional distribution of X given X+y=n,

    Using P(X/Y=n-x)

    I get something that is almost a binomial probability but i get an extra term that i am not sure should be there, specifically 1/((a+b)t)^n

    I have done this three times but i am not sure if i am making a mistake anywhere, is anyone able to check whether they get a binomial distribution with this extra term, or am i stuffing up somewhere
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  2. #2
    MHF Contributor matheagle's Avatar
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    can you post your work
    it would a lot easier than me doing it,
    because I think I did this problem here last year
    try a search here at this site
    I found this, which I'm not if it helps you
    http://www.mathhelpforum.com/math-he...on-167960.html

    it should just be

    P(X=k|X+Y=n)={P(X=k)P(Y=n-k)\over P(X+Y=n)}
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  3. #3
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    acording to the link you provided the distribution for the numerator is a poisson distribution for X+Y which now gives me that this probability is 1.
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  4. #4
    MHF Contributor matheagle's Avatar
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    please show your work
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  5. #5
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    Solving for this conditional probability:

    P(X=x/X+Y=n)=[n! (at)^x (bt)^(n-x)] /(at+bt)^n (x!(n-x)!
    =n choose x [(a^x)*(b^(n-x)]/(a+b)^n

    since the exponential terms cancel out
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