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Math Help - Expected value of the MLE of a binomial distribution

  1. #1
    Newbie noname's Avatar
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    Expected value of the MLE of a binomial distribution

    Hi

    I've this question:

    X1,...,Xn it's a sample from a binomial distribution b(n,a). Find the maximum likelihood estimator of a and the expected value of the estimator.

    I can find the MLE

    Ignoring the binomial coefficient i have

    <br />
f(x_1,...,x_n|a)=a^{\sum_{1}^{n}x_i}(1-a)^{n^2-\sum_{1}^{n}x_i}<br />

    I take logs and differentiate to obtain:

    <br />
\frac{d}{da}logf(x_1,...,x_n|a)=\frac{\sum_{1}^{n}  x_i}{a}-\frac{n^2-\sum_{1}^{n}x_i}{1-a}<br />

    Upon equating to zero and solving i obtain that:

    <br />
a=\frac{\sum_{1}^{n}x_i}{n^2}<br />

    Now i must find the expected value of a but I'm very confused on how to do this. Any hint?

    Thanks
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  2. #2
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    Quote Originally Posted by noname View Post
    Hi

    I've this question:

    X1,...,Xn it's a sample from a binomial distribution b(n,a). Find the maximum likelihood estimator of a and the expected value of the estimator.

    I can find the MLE

    Ignoring the binomial coefficient i have

    <br />
f(x_1,...,x_n|a)=a^{\sum_{1}^{n}x_i}(1-a)^{n^2-\sum_{1}^{n}x_i}<br />

    I take logs and differentiate to obtain:

    <br />
\frac{d}{da}logf(x_1,...,x_n|a)=\frac{\sum_{1}^{n}  x_i}{a}-\frac{n^2-\sum_{1}^{n}x_i}{1-a}<br />

    Upon equating to zero and solving i obtain that:

    <br />
a=\frac{\sum_{1}^{n}x_i}{n^2}<br />

    Now i must find the expected value of a but I'm very confused on how to do this. Any hint?

    Thanks
    E \left( \sum_{i = 1}^{n} x_i \right) = \sum_{i = 1}^{n} E(x_i).
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  3. #3
    Newbie noname's Avatar
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    Thank you!

    So if I'm correct then

    <br />
E(a)=E(\frac{\sum_{i=1}^{n}x_i}{n^2})=\frac{1}{n^2  }\sum_{i=1}^{n}E(x_i)=\frac{1}{n^2}\sum_{i=1}^{n}n  a=\frac{1}{n^2}n^2a=a<br />
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  4. #4
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    Quote Originally Posted by noname View Post
    Thank you!

    So if I'm correct then

    <br />
E(a)=E(\frac{\sum_{i=1}^{n}x_i}{n^2})=\frac{1}{n^2  }\sum_{i=1}^{n}E(x_i)=\frac{1}{n^2}\sum_{i=1}^{n}n  a=\frac{1}{n^2}n^2a=a<br />
    There's a difference between the two a's ....

    Personally I'd use the notation \hat{a} for the MLE of a. Then you have E(\hat{a}) = a. So the MLE of a is an unbiased estimator of a.
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