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Binomial Distribution satisfies Marcoff Chain

  1. Jul 22, 2013 #1
    1. The problem statement

    Consider the Binomial Distribution in the form

    [itex]P_{N}(m)=\frac{N!}{(\frac{N+m}{2})!(\frac{N-m}{2})!}p^{\frac{N+m}{2}}q^{\frac{N-m}{2}}[/itex]

    where [itex]p+q=1[/itex], [itex]m[/itex] is the independent variable and [itex]N[/itex] is a parameter.

    Show that it satisfies the marcoff chain

    [itex]P_{N+1}\left(m\right)=pP_{N}\left(m-1\right)+qP_{N}\left(m+1\right)[/itex]

    2. The attempt at a solution

    I'm trying my solution starting from this:

    [itex]pP_{N}\left(m-1\right)+qP_{N}\left(m+1\right)[/itex]


    [itex]=p\frac{N!}{\left(\frac{N+m-1}{2}\right)!\left(\frac{N-m+1}{2}\right)!}p^{\frac{N+m-1}{2}}q^{\frac{N-m+1}{2}}+q\frac{N!}{\left(\frac{N+m+1}{2}\right)!\left(\frac{N-m-1}{2}\right)!}p^{\frac{N+m+1}{2}}q^{\frac{N-m-1}{2}}[/itex]


    [itex]=\frac{N!}{\left(\frac{N+m-1}{2}\right)!\left(\frac{N-m+1}{2}\right)!}p^{\frac{N+m+1}{2}}q^{\frac{N-m+1}{2}}+\frac{N!}{\left(\frac{N+m+1}{2}\right)!\left(\frac{N-m-1}{2}\right)!}p^{\frac{N+m+1}{2}}q^{\frac{N-m+1}{2}}[/itex]


    [itex]=p^{\frac{N+m+1}{2}}q^{\frac{N-m+1}{2}}\left(\frac{N!}{\left(\frac{N+m-1}{2}\right)!\left(\frac{N-m+1}{2}\right)!}+\frac{N!}{\left(\frac{N+m+1}{2}\right)!\left(\frac{N-m-1}{2}\right)!}\right)[/itex]


    I can't go any further. If you can help I would appreciate.
     
    Last edited: Jul 22, 2013
  2. jcsd
  3. Jul 22, 2013 #2

    TSny

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    (Edit) See what happens if you multiply by##\frac{N+1}{N+1}##.
     
  4. Jul 22, 2013 #3

    TSny

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    My previous comment was based on a particular way that I went about it which got to the result. But, on review, I see that multiplying by (N+1)/(N+1) isn't necessary.

    I would factor out the ##N!## in your expression.

    The important thing is to get the two denominators in your expression to match the denominator in ##P_{N+1}(m)##. For example, what could you multiply the numerator and denominator of ##\frac{1}{\left(\frac{N+m-1}{2}\right)!\left(\frac{N-m+1}{2}\right)!}## by to get the denominator in ##P_{N+1}(m)##?

    [By the way, welcome to PF, ppedro!]
     
  5. Jul 23, 2013 #4
    Hey TSny! Thanks for your reply. I see what you're suggesting but I'm not being able to compute it. The factorials are not helping me simplify the expression.
     
  6. Jul 23, 2013 #5

    TSny

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    ##P_{N+1}(m)## has a denominator of ##(\frac{N+m+1}{2})!(\frac{N-m+1}{2})!##.

    Compare that to your first denominator ##(\frac{N+m-1}{2})!(\frac{N-m+1}{2})!##.

    To get your denominator to match the denominator in ##P_{N+1}(m)##, you've got to somehow transform ##(\frac{N+m-1}{2})!## into ##(\frac{N+m+1}{2})!##.

    What can you multiply ##(\frac{N+m-1}{2})!## by to produce ##(\frac{N+m+1}{2})!##?
     
  7. Jul 23, 2013 #6
    [itex]\frac{(N+m+1)!}{(N+m-1)!}=\frac{(N+m+1)(N+m)(N+m-1)!}{(N+m-1)!}=(N+m+1)(N+m)
    [/itex]
     
  8. Jul 23, 2013 #7

    TSny

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    No. Note that ##\frac{N+m+1}{2} = \frac{N+m-1}{2} + 1##.
     
  9. Jul 23, 2013 #8
    Ok, I see your point. Thanks!
     
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