Let $X\sim\operatorname{Poisson}$. What is the estimator of $P(X=0)$ by maximum likelihood method. If it based on random sample of size $n$ Thank you
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"The" estimator? Based on what? With respect to which criterion? – Did Oct 09 '13 at 18:07
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Sorry, it base on Maximum likelihood method – user99853 Oct 09 '13 at 18:22
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Based on what? A sample of size 1? Of size n? Independent? Etc. – Did Oct 09 '13 at 18:51
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Based on random sample of size n – user99853 Oct 09 '13 at 19:06
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And an independent sample? You know, all this should have been in the question from the beginning... – Did Oct 09 '13 at 19:53
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Yes, it's independent . I say "random sample" it always mean i.i.d. – user99853 Oct 09 '13 at 19:58
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No it does not--but now we know you mean i.i.d. – Did Oct 09 '13 at 20:00
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I've voted to re-open. The question has undergone some editing since the terse initial posting. It's perfectly clear what it's about. – Michael Hardy Oct 09 '13 at 20:02
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If you observe only $X$, then the MLE for $\lambda=\mathbb EX$ is just $\widehat\lambda=X$.
MLEs have a nice equivariance property not shared by unbiased estimators, method-of-moments estimators, or other sorts of estimators: the MLE of $g(\lambda)$ is $g$ of the MLE of $\lambda$. Thus if you know that $\Pr(X=0)=e^{-\lambda}$, then you can conclude that the MLE for $\Pr(X=0)$ is $e^{-\widehat \lambda} = e^{-X}$.
You could of course do it by brute force: Let $\alpha=e^{-\lambda}$, so $\lambda=-\log\alpha$. Then $$ L(\alpha)=\Pr(X=x) = \frac{\lambda^x e^{-\lambda}}{x!} = \frac{(-\log\alpha)^x\alpha}{x!}. $$ Then $$ \ell(\alpha)=\log L(\alpha) = x\log(-\log\alpha) + \log\alpha+\text{constant}, $$ and then $$ \ell\,'(\alpha) = x\frac{-1/\alpha}{(-\log\alpha)}+\frac1\alpha = \frac1\alpha\left(\frac{x}{\log\alpha}+1\right). $$ For $\alpha>0$, this is $0$ only when $\alpha=e^{-x}$. Then there's a bit more work to do to show that this is indeed a global maximum.
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Thank you :)) if it based on random sample of size n . The answer is e^-(XBar) ? – user99853 Oct 09 '13 at 19:11
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As an aside, you may also consider the properties of the James-Stein Estimator if you ever encounter an example of this with dimensionality of three or more. – ely Oct 09 '13 at 19:19
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I based my answer on a sample of size $1$, because that's what the question appeared to say. But in case of an larger i.i.d. sample, it is $e^{-\overline X}$. ${}\qquad{}$ – Michael Hardy Oct 09 '13 at 19:26
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Thank you so much Michael , it's my final test question. I'm answered e^-(Xbar) but i'm not sure about it. Thank you again. – user99853 Oct 09 '13 at 19:53
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@EMS : I appreciate your pointing this out, but I don't think it's very relevant to the question, since the question was about an MLE. – Michael Hardy Oct 09 '13 at 20:01