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Math Help - Expectation of functions

  1. #1
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    Expectation of functions

    Suppose X has binomial distribution with probability of success p over n trials,

    How do I show that:

    E[1/(1+X)] = [1-(1-p)^(n+1)]/p(n+1)?
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  2. #2
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    Quote Originally Posted by sweetadam View Post
    Suppose X has binomial distribution with probability of success p over n trials,

    How do I show that:

    E[1/(1+X)] = [1-(1-p)^(n+1)]/p(n+1)?
    Use the definition, like...

    E[\frac{1}{1+X}] = \sum_x \frac{1}{1+x} p_X = \sum_x \frac{1}{1+x} {n\choose x} p^x (1-p)^{n-x}

    Now you want to manipulate this so that you can use the fact that \sum p_X=1.

    Some tips:
    a) Play with the {n\choose x} to cancel with the \frac{1}{1+x}.
    b) Set j=x+1.

    Show me where you are stuck and I will help you proceed. To get started:

     \frac{1}{1+x} {n\choose x} = \frac{1}{1+x} \cdot \frac{n!}{x!(n-x)!} =\frac{n!}{(1+x)!(n-x)!} = \frac{n!}{(1+x)!(n-x)!}= \frac{1}{n+1}\frac{(n+1)!}{(1+x)!((n+1)-(x+1))!} = \frac{1}{1+n} {n+1 \choose x+1}

    So we have that,

     \sum_x \frac{1}{1+x} {n\choose x} p^x (1-p)^{n-x} =  \frac{1}{1+n}\sum_x {n+1 \choose x+1} p^x (1-p)^{n-x} = \frac{1}{p(1+n)}\sum_x {n+1 \choose x+1} p^{x+1} (1-p)^{(n+1)-(x+1)}

    Now set j=x+1, then

    \frac{1}{p(1+n)}\sum_{j=2}^n {n+1 \choose j} p^{j} (1-p)^{(n+1)-j}
    Last edited by Anonymous1; April 15th 2010 at 09:14 AM.
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    Moo
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