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Math Help - Consistent Estimator

  1. #1
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    Consistent Estimator

    If y has a binomial distribution with n trials and success probability p, show that Y/n is a consistent estimator of p.

    Can someone show how to show this. I appreciate it any and all help. thanks.
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  2. #2
    Moo
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    Hello,

    Yn=X1+...+Xn, where the Xi's are independent Bernoulli distributions B(p).
    Law of large numbers : Yn/n converges to E[X1] almost surely, and hence in probability -> consistent.
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  3. #3
    Senior Member Sambit's Avatar
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    Quote Originally Posted by jzellt View Post
    If y has a binomial distribution with n trials and success probability p, show that Y/n is a consistent estimator of p.

    Can someone show how to show this. I appreciate it any and all help. thanks.
    To show that Y/n is a consistent estimator of p, it is sufficient to show that \lim_{n\to\infty}E(\frac{Y}{n})=p and \lim_{n\to\infty}Var(\frac{Y}{n})=0.
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    Moo
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    Quote Originally Posted by Sambit View Post
    To show that Y/n is a consistent estimator of p, it is sufficient to show that \lim_{n\to\infty}E(\frac{Y}{n})=p and \lim_{n\to\infty}Var(\frac{Y}{n})=0.
    But here it may be much easier to prove convergence in probability, since we obviously have the law of large numbers ?
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  5. #5
    Senior Member Sambit's Avatar
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    Quote Originally Posted by Moo View Post
    But here it may be much easier to prove convergence in probability, since we obviously have the law of large numbers ?
    Basically both the procedures need same labour which is very little indeed. Since Y\sim Bin(n,p), Obviously E(Y)=np which gives E(\frac{Y}{n})=p and Var(Y)=np(1-p) , so Var(\frac{Y}{n})=\frac{np(1-p)}{n^2} = \frac{p(1-p)}{n}\to\0 as n\to\infty. No tough job
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  6. #6
    MHF Contributor matheagle's Avatar
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    there are a lot of meanings to consistency, strong, weak...
    I was curious as to what was meant here.
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