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Math Help - Bernoulli parameter expectation

  1. #1
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    Bernoulli parameter expectation

    I'm looking at this page: Bernoulli Distribution -- from Wolfram MathWorld

    I think I'm missing something simple, but how do you get from equation 24 to 25?

    Thanks!
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  2. #2
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    Yes, it looks a little... nonobvious.

    The easiest thing is to immediately apply the Binomial theorem: \displaystyle p\sum_{n=0}^N{N\choose n}p^n(1-p)^{N-n}=p(p+1-p)^N=p. Maybe they factored (1-p)^N first: \displaystyle\sum_{n=0}^N{N\choose n}p^n(1-p)^{N-n}=(1-p)^N\sum{N\choose n}\left(\frac{p}{1-p}\right)^n=(1-p)^N\left(1+\frac{p}{1-p}\right)^N= \displaystyle(1-p)^N\frac{1}{(1-p)^N}=1.
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  3. #3
    MHF Contributor matheagle's Avatar
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    yes, just pull out the p, the rest sums to one.
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  4. #4
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    Thanks!

    Since it's an expectation of p-hat, shouldn't the first p (or all the p's) be a p-hat term?
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  5. #5
    MHF Contributor matheagle's Avatar
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    P is a constant, usually unknow, that's why we estimate it with a random variable... p-hat

    E(\hat p)=p and we like that p-hat is unbiased for p.
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