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    Sum of Poisson Distribution




    Hi everyone.
    Let X and Y be independent, with X~Poi(lambda) and Y~Poi(mu). Put Z = X + Y as their sum. Find the distribution X | Z = z.

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    Quote Originally Posted by woody198403 View Post
    Hi everyone.
    Let X and Y be independent, with X~Poi(lambda) and Y~Poi(mu). Put Z = X + Y as their sum. Find the distribution X | Z = z.

    We appreciate your effort on solving this problem.

    Hint:
    If and are independent, and Y = X1 + X2, then the distribution of X1 conditional on Y = y is a binomial. Specifically, . More generally, if X1, X2,..., Xn are independent Poisson random variables with parameters λ1, λ2,..., λn then


    Source: Wikipedia
    In the long run, we're all dead.

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    I am still working on it at the moment. I will post what I have come up with shortly.

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    The sum of two independent random variables which both have a Poisson distribution, also has a Poisson distribution.

    so the pmf of Z is

    Pr(Z=z) = ((lambda + mu)^z)/(z!) * e^(-(lambda + mu))


    and

    Pr(X|Z = z) = Pr(intersection of X and Z = z) / Pr(Z = z)

    but Im having trouble calculating Pr(intersection of X and Z = z). This is where Im stuck.

    I know that the conditional probability X|Z=z will return a binomial distribution.

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    You have to think slightly differently..

    When Given Z=z, X can range between 0 to z

    Calculate Pr(X=x1|Z = z) where x1 between ( 0 to z)
    In the long run, we're all dead.

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    Re: Sum of Poisson Distribution

    Hi,

    I have been trying to solve this problem too for quite some time. (This is for a computer science course, so I don't have a strong background in statistics).

    I understand everything that's been said so far and I was also trying to solve Pr(X|Z = z) = Pr(intersection of X and Z = z) / Pr(Z = z) before I found this thread.

    When vinux says "you have to think slightly differently," does that mean the above formula is incorrect?

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    Re: Sum of Poisson Distribution

    I was also looking at this thread http://www.talkstats.com/showthread....n-distribution where they solve a very similar problem.

    The second post, by BGM, makes me think that instead of using Pr(intersection of X and Z = z) / Pr(Z = z), we need to use something here like

    and then use the fact that

    is Pr(Z=z) = ((lambda + mu)^z)/(z!) * e^(-(lambda + mu)) but I know this isn't quite right.

    Any help would be greatly appreciated!! Thanks.

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    Re: Sum of Poisson Distribution

    Since the support of Poisson distribution is \{0, 1, 2, ... \}, so obviously Z = X + Y \geq X

    Therefore, as vinux hinted above , for x \in \{0, 1, ..., z\},

    \Pr\{X = x|Z = z\} = \frac {\Pr\{X = x, X + Y = z\}} {\Pr\{X + Y = z\}}
= \frac {\Pr\{X = x\}\Pr\{Y = z - X|X = x\}} {\Pr\{X + Y = z\}}

    = \frac {\Pr\{X = x\}\Pr\{Y = z - x\}} {\Pr\{X + Y = z\}}

    This is the key trick here, and the only thing left is that you have to know the pmf of X + Y

  9. The Following User Says Thank You to BGM For This Useful Post:

    xanderp123 (02-13-2012)

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    Smile Re: Sum of Poisson Distribution


    I think I understand it now. I was very close all along... Thank you so much!

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