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Thread: Asymptotic distributions of MLEs

  1. #1
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    Asymptotic distributions of MLEs

    Hello all! Having a problem with general MLE concepts. See if you can help me with this.

    Let X1,...Xn be a random sample of size n where Xi~BIN(1,p). How do I find the Asyptotic distribution of the MLE of p?

    Thanks for any help!
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  2. #2
    MHF Contributor matheagle's Avatar
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    the MLE is $\displaystyle \hat P$ which is an sample mean of Bernoulli's

    So $\displaystyle {\hat P-P\over \sqrt{{pq\over n}}}\to N(0,1)$

    and by Slutsky's Theorem

    $\displaystyle {\hat P-P\over \sqrt{{\hat p\hat q\over n}}}\to N(0,1)$ as well.
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  3. #3
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    Quote Originally Posted by matheagle View Post
    the MLE is $\displaystyle \hat P$ which is an sample mean of Bernoulli's

    So $\displaystyle {\hat P-P\over \sqrt{{pq\over n}}}\to N(0,1)$

    and by Slutsky's Theorem

    $\displaystyle {\hat P-P\over \sqrt{{\hat p\hat q\over n}}}\to N(0,1)$ as well.
    That looks like the formula $\displaystyle \sqrt{n}(\hat{\theta}-\theta)\sim{N}(0,I_Y^{-1}(\theta))$ adapted to Bin(1,p) for n observations, or, $\displaystyle \sqrt{I_Y(\theta)}(\hat{\theta}-\theta)\sim{N}(0,1)$ for large n

    (with MLE=$\displaystyle \hat{p}$ and Fisher information $\displaystyle I_Y=\frac{1}{p(1-p)}$ for one observation, so $\displaystyle I_Y=\frac{n}{p(1-p)}$ for n observations)

    So, does the former ($\displaystyle \sqrt{n}(\hat{\theta}-\theta)->N(0,I_Y^{-1}(\theta))$) comes from Slutsky theorem? I just never seen this name before, and I was also given the above assymptotic formula without explaining where it comes from.
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  4. #4
    MHF Contributor matheagle's Avatar
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    The Central Limit Theorem gives you

    $\displaystyle {\hat P-P\over \sqrt{{pq\over n}}}\to N(0,1)$

    replacing the p's and q's with their sample versions also converges in distribution to
    a standard normal since

    $\displaystyle \hat P\to P$ and $\displaystyle 1-\hat P\to 1-P$ by either law of large numbers

    all you need is convergence in probability http://en.wikipedia.org/wiki/Slutsky's_theorem
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  5. #5
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    Ah! then, that assymptotic formula with Fisher information in it looks just like a version of the Central Limit Theorem. (edited: but of course it would, because it originates from CLT, I just didn't see the proof so I didn't see that it does. Sorry for off-top).
    Last edited by Volga; Mar 17th 2011 at 11:40 PM.
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