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Math Help - Issue with derivation of ML parameter estimation for binomial distribution

  1. #1
    Junior Member
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    Issue with derivation of ML parameter estimation for binomial distribution

    Hi, I have an issue with this step in the derivation of the ML solution to the binomial parameter estimation for p. Found here


    https://www.projectrhea.org/rhea/ind...utions_OldKiwi



    f(x)=\left(\frac{n!}{x!\left(n-x \right)!} \right){p}^{x}{\left(1-p \right)}^{n-x}

    L(p)=\prod_{i=1}^{n}f({x}_{i})=\prod_{i=1}^{n} \left( \frac{n!}{{x}_{i}!\left(n-{x}_{i} \right)!} \right){p}^{{x}_{i}}{\left(1-p \right)}^{n-{x}_{i}

    L(p)=\left( \prod_{i=1}^{n}\left(\frac{n!}{{x}_{i}!\left(n-{x}_{i} \right)!} \right)\right){p}^{\sum_{i=1}^{n}{x}_{i}}{\left(1-p \right)}^{n-\sum_{i=1}^{n}{x}_{i}



    Why is the last term {\left(1-p \right)}^{n-\sum_{i=1}^{n}{x}_{i} and not {\left(1-p \right)}^{n^{2}-\sum_{i=1}^{n}{x}_{i}?


    Taking a concrete example, suppose 1-p = q, n = 5 and X = {2, 4, 5, 3, 1}

    then \prod_{i}^5 (1-p)^{n - x_{i}} = q^{5-2} q^{5-4} q^{5-5} q^{5-3} q^{5-1}=q^{3} q^{1} q^{0} q^{2} q^{4}= q^{10} = q^{5^2 - (2  + 4 + 5 + 3 + 1)} \not= q^{5 - (2  + 4 + 5 + 3 + 1)} = q^{-10}


    Right?

    Would really appreciate someone telling me where, if, I am going wrong here.

    Thanks in advance, MD
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  2. #2
    MHF Contributor
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    Re: Issue with derivation of ML parameter estimation for binomial distribution

    Hey Mathsdog.

    The reason for n - blah instead of n^2 - blah is that you have a total of n observations. If you have x of those observations corresponding to X = 1 then you have n - x corresponding to X = 0.
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