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Thread: Solution to a Modified Binomial Probability Distribution

  1. #1
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    Solution to a Modified Binomial Probability Distribution

    I could use help determining whether the following can be solved analytically.

    I have an equation $\displaystyle $\sum_{s=c}^{Q}\frac{s-c+1}{s+1}{Q\choose s}p^{s}(1-p)^{Q-s}=\frac{a-b}{a-d}$$.

    Notice this equation is the binomial probability distribution from c to Q (as opposed to going from 0 to Q), and it is multiplied by $\displaystyle $\frac{s-c+1}{s+1}$$.

    I understand that $\displaystyle $\sum_{s=0}^{Q}s{Q\choose s}p^{s}(1-p)^{Q-s}$$ simplifies to $\displaystyle $pn$$. Can a similar simplification be made to the equation I've presented above? Any thoughts on how I might go about it?

    Thank you for your help.

    Josh
    Last edited by aberrantProtagonist; Oct 21st 2012 at 01:40 PM.
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  2. #2
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Solution to a Modified Binomial Probability Distribution

    $\displaystyle \sum _{s=c}^Q \frac{(-c+s+1) p^s \binom{Q}{s} (1-p)^{Q-s}}{s+1} = $

    $\displaystyle =p^c \Gamma (c+2) (1-p)^{Q-c} \left(\frac{\binom{Q}{c} \, _2\tilde{F}_1\left(1,c-Q;c+2;\frac{p}{p-1}\right)}{c+1}-\frac{p \binom{Q}{c+1} \, _2\tilde{F}_1\left(2,c-Q+1;c+3;\frac{p}{p-1}\right)}{p-1}\right)$

    2F1=$\displaystyle Hypergeometric function \left._2F_1(a,b;c;z)\right/\Gamma (c)$ ref
    Last edited by MaxJasper; Oct 21st 2012 at 02:01 PM.
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