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Thread: Joint and marginal distributions

  1. #1
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    Joint and marginal distributions

    Hi, i am trying to answer the following question:

    Consider the following joint distribution:

    P(X=n, Y=m)= (m+n)(e^-2x) x^(m+n)/(n+m)!
    n
    where m and n takes values in 0,1,2.......

    (the first bit should read m+n choose n)

    show that X and Y are independent but have the same distribution as each other.

    I know I should find the marginal distributions of X and Y, but Im not sure how to do so in this case, in my notes i only have examples where the joint distribution looks geometric. If i were to do it that way, to work out the marginal distribution of X, would I sum the joint distribution from m=0 to infinity and then work on simplifying it?
    Thanks for your help
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  2. #2
    Super Member girdav's Avatar
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    Re: Joint and marginal distributions

    Quote Originally Posted by yellowcarrotz View Post
    Consider the following joint distribution:
    $\displaystyle P(X=n, Y=m)=\binom{n+m}{n} \frac{e^{-2x} x^{m+n}}{(n+m)!} $
    where $\displaystyle m$ and $\displaystyle n$ takes values in $\displaystyle \mathbb{N}$.
    Show that $\displaystyle X$ and $\displaystyle Y$ are independent but have the same distribution as each other.

    Would I sum the joint distribution from $\displaystyle m=0$ to infinity and then work on simplifying it?
    Yes, by this way you will find $\displaystyle P(X=n)$, and by symmetry $\displaystyle P(Y=m)$.
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  3. #3
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    Re: Joint and marginal distributions

    ok thanks for your help i'm not quite show what to do about the (n+m) bit?
    n
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  4. #4
    Super Member girdav's Avatar
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    Re: Joint and marginal distributions

    Write $\displaystyle \binom{n+m}n\frac 1{(m+n)!}$ as $\displaystyle \frac 1{n!m!}$.
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  5. #5
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    Re: Joint and marginal distributions

    Got it thanks
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