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Math Help - Joint Probability Mass Functions

  1. #1
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    Joint Probability Mass Functions

    Hi, I'm revising for an upcoming test and have a past paper with answers but I can't understand where they are coming from. If anyone could try explaining this too me it would be greatly appreciated.

    The number of eggs laid by an insect is a random variable N which has a poisson distribution with expectation \lambda so P(N=n) = \lambda^ne^{-\lambda}/n! for n=0,1,2,... Independently for each egg laid there is probability 1/2 each that it develops into a male insect or a female insect. Let X denote the number of male offspring, and Y the number of female offspring.

    i) Find the joint probability mass function of (N,X)

    ii) Show that the marginal distribution of X, the number of male offspring, is Poisson with expectation \lambda/2

    The answers I've been given are

    i) nx2^{-n}\lambda^ne^{-\lambda} /n! for non negative intergers x \leq n

    ii)For non-negative integers x,
    P_{.x}= \sum_{n=x}^{\infty}P_{nx} = [e^{-\lambda}(\lambda/2)^x/x!] \sum_{n\ge x}(\lambda/2)^{n-x}/(n-x)! = e^{-\lambda/2}(\lambda/2)^x /x! as the sum is \sum_{m\ge0}(\lambda/2)^m/m! = e^{\lambda/2}

    If anyone can explain how you get to these answers it would be a massive help,

    Cheers
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by shmounal View Post
    Hi, I'm revising for an upcoming test and have a past paper with answers but I can't understand where they are coming from. If anyone could try explaining this too me it would be greatly appreciated.

    The number of eggs laid by an insect is a random variable N which has a poisson distribution with expectation \lambda so P(N=n) = \lambda^ne^{-\lambda}/n! for n=0,1,2,... Independently for each egg laid there is probability 1/2 each that it develops into a male insect or a female insect. Let X denote the number of male offspring, and Y the number of female offspring.

    i) Find the joint probability mass function of (N,X)

    ii) Show that the marginal distribution of X, the number of male offspring, is Poisson with expectation \lambda/2

    The answers I've been given are

    i) nx2^{-n}\lambda^ne^{-\lambda} /n! for non negative intergers x \leq n

    ii)For non-negative integers x,
    P_{.x}= \sum_{n=x}^{\infty}P_{nx} = [e^{-\lambda}(\lambda/2)^x/x!] \sum_{n\ge x}(\lambda/2)^{n-x}/(n-x)! = e^{-\lambda/2}(\lambda/2)^x /x! as the sum is \sum_{m\ge0}(\lambda/2)^m/m! = e^{\lambda/2}

    If anyone can explain how you get to these answers it would be a massive help,

    Cheers
    The number of males has a Poisson distribution with mean \lambda/2, as does the number of females born.

    So:

     Pr(N,X)=p(X,\lambda/2)p(N-X,\lambda/2)

    CB
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  3. #3
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    Ah ok thanks, i think part of what was confusing me is that my answers had nx when it should have been \binom{n}{x}. With realising that and your help i think ive got it!!
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