Results 1 to 4 of 4

Math Help - Conditional Poisson Distibution

  1. #1
    Junior Member
    Joined
    May 2008
    Posts
    50

    Conditional Poisson Distibution

    Hi,
    I have the following problem.

    Let X and Y be independent, with X~Poi (\lambda) and Y~Poi (\mu) . Let Z = X + Y. Find the distribution X | Z = z.

    Now I know that the sum of two independent random variables which both have a Poisson distribution, also has a Poisson distribution.

    so the pmf of Z is

     \mathbb{P}(Z=z) = \frac{(\lambda + \mu)^z}{z!} e^{-(\lambda + \mu)}


    and

     \mathbb{P}(X|Z = z) = \frac{\mathbb{P}(X \cap Z = z)}{\mathbb{P}(Z = z)}

    but Im having trouble calculating  \mathbb{P}(X \cap Z = z). This is where Im stuck.

    I know that the conditional probability  X|Z=z will return a binomial distribution. Can anyone please help me understand how to calculate  \mathbb{P}(X \cap Z = z) and consequently obtain a binomial distribution?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    This should help, since you have the right idea.

    P(X=x|X+Y=z)={P(X=x,X+Y=z)\over P(X+Y=z)}

     ={P(X=x,Y=z-x)\over P(X+Y=z)}={P(X=x)P(Y=z-x)\over P(Z=z)}

    NOW, plug in x, z-x and z into the distribution of the three DIFFERENT Poissons and it should clean up nicely.
    Last edited by matheagle; August 23rd 2009 at 07:09 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2008
    Posts
    50
    Thank-you so much matheagle. That was so helpful, I understand conditional probability a lot better now.

    Just one thing,
    Ive managed to get the solution down to

     \begin{pmatrix} z \\ x \end{pmatrix} \lambda^x \mu^{z-x} (\lambda + \mu)^{-z}

    But I can't work out how to get from my solution down to what it should be.
    i.e.

    Bin~  ( z, \frac{\lambda}{\lambda + \mu} )

    I bet its something really obvious but I cant see it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by woody198403 View Post
    Thank-you so much matheagle. That was so helpful, I understand conditional probability a lot better now.

    Just one thing,
    Ive managed to get the solution down to

     \begin{pmatrix} z \\ x \end{pmatrix} \lambda^x \mu^{z-x} (\lambda + \mu)^{-z}

    But I can't work out how to get from my solution down to what it should be.
    i.e.

    Bin~  ( z, \frac{\lambda}{\lambda + \mu} )

    I bet its something really obvious but I cant see it.
    It's obvious if you work backwards:

    {z \choose x} \left( \frac{\lambda}{\lambda + \mu}\right)^x \left( 1 - \frac{\lambda}{\lambda + \mu}\right)^{z-x}


    = {z \choose x} \left( \frac{\lambda}{\lambda + \mu}\right)^x \left(\frac{\mu}{\lambda + \mu}\right)^{z-x}

    and from here you should see how it leads to your answer.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Normal + Poisson: conditional mgf
    Posted in the Advanced Statistics Forum
    Replies: 13
    Last Post: February 3rd 2011, 07:26 AM
  2. Conditional Poisson!
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: January 15th 2010, 11:06 AM
  3. Conditional Poisson
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: January 13th 2010, 08:17 AM
  4. Conditional probability with poisson?
    Posted in the Advanced Statistics Forum
    Replies: 14
    Last Post: December 9th 2009, 02:34 AM
  5. Conditional poisson question
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: April 4th 2009, 03:08 AM

Search Tags


/mathhelpforum @mathhelpforum