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Math Help - joint pmf

  1. #1
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    joint pmf

    if X and Y are discrete RVs with pmf p(x,y) = a\; \dfrac{k!}{x!y!(k-x-y)!} where x and y are  non \; negative \; integers \; and\; x+y\leq k.

    I want to find the value of a.


    I know that I need to show that the pmf sums up to 1.

    \sum_x \sum_y p(x,y)=1
    since x and y are non negatve: 0<=x+y<=k
    here i am getting confused in where x and y range from.
    can anyone help me how to solve this?
    Last edited by chutiya; November 20th 2010 at 03:53 PM.
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  2. #2
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    \sum_{x=0}^{k} \sum_{y=0}^{k-x} p(x,y)=1

    so,
    a \cdot \sum_{x=0}^{k} \sum_{y=0}^{k-x} \dfrac{k!}{x!y!(k-x-y)!} = a \cdot \sum_{x=0}^{k} \dfrac{k!\cdot 2^{k-x}}{x!(k-x)!}=a\cdot 3^k =1

    Hence  a=\left(\frac{1}{3}\right)^k
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  3. #3
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    Thank you so much.

    But how do we get \sum_{y=o}^{k-x}\dfrac{1}{y!(k-x-y)!}= \dfrac{2^{k-x}}{(k-x)!}

    is this a gamma function? Could you explain please.
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  4. #4
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    Refer to the section on series of binomial coefficients
    Binomial coefficient - Wikipedia, the free encyclopedia

    a \cdot \sum_{x=0}^{k} \sum_{y=0}^{k-x} \dfrac{k!}{x!y!(k-x-y)!} = a \cdot \sum_{x=0}^{k} \sum_{y=0}^{k-x} \dfrac{k!\cdot \left(k-x\right)!}{x!y!(k-x-y)!(k-x)!} =a \cdot \sum_{x=0}^{k} \dfrac{k!}{x!(k-x)!}\sum_{y=0}^{k-x} \dfrac{\left(k-x\right)!}{y!(k-x-y)!}

    and additionally from binomial theorem
    (Y+X)^n = \sum_{k=0}^n {n \choose k} X^{n-k}Y^K

    So using this formula, we can write
    a \cdot \sum_{x=0}^{k} \dfrac{k!\cdot 2^{k-x}}{x!(k-x)!}= a \cdot  \sum_{x=0}^\infty {k \choose x} 2^{k-x}1^k=a \cdot \left(2+1\right)^k
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