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

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

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

    The answers I've been given are

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

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

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

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

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

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

    The answers I've been given are

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

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

    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 $\displaystyle \lambda/2$, as does the number of females born.

    So:

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

    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 $\displaystyle nx$ when it should have been$\displaystyle \binom{n}{x}$. With realising that and your help i think ive got it!!
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