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I was studying the method of moments estimation of parameters, and I encountered the following problem.

I have a geometric distribution as following:

$P(X=k) = p(1-p)^{k-1}$, and a sample size of n, all observations are independent

Hence, it can be quite easily shown that

$\hat p = \dfrac{1}{\overline{X}}$, where $\overline{X}$ is the sample mean, this is the MME estimation of p.

Now, if I want to find the approximate normal distribution for $\hat p$, I would have to find out the variance of $\hat p$

Since, $\hat p = \dfrac{n}{\sum\limits_{i=1}^n X_n}$, we see that the bottom follows a negative binomial distribution $NegativeBin(n, p)$.

Now I am stuck at this step, and I can't figure out how to calculate the expression for the PMF and the variance of $\hat p$.

Thank you very much!

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we see that the bottom follows a binomial distribution Bin(n,p)... Do we, really? Note that the $X_i$s are not 0-1 valued. –  Did Feb 10 '13 at 21:02
    
Sorry, I meant negative binomial, I will edit the post –  Enzo Feb 10 '13 at 21:15
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1 Answer

up vote 1 down vote accepted

Here is a fast Ansatz to find the approximate normal distribution for $\hat p_n$ when $n\to\infty$.

By the CLT, $\sum\limits_{i=1}^nX_i=nm+\sqrt{nv}Z_n$ where $m=\mathbb E(X_1)$, $v=\mathrm{var}(X_1)$ and $Z_n$ converges in distribution to a standard normal random variable. Thus, loosely speaking, $$ \hat p_n=\frac1m\left(1+\frac{\sqrt{v}}{m\sqrt{n}}Z_n\right)^{-1}\approx \frac1m-\frac{\sqrt{v}}{m^2\sqrt{n}}Z_n. $$ In particular, one may hope that $$ \mathbb E(\hat p_n)\approx\frac1m,\quad\mathrm{var}(\hat p_n)\approx\frac{v}{m^4n}. $$ This suggests that $$ \hat p_n=\frac1m+\frac{\sqrt{v}}{m^2\sqrt{n}}V_n, $$ where $V_n$ converges in distribution to a standard normal random variable.

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I see! Thank you very much! That's very helpful :) Also another quick question: How would I find the MME estimate of a parameter if the mean of that density is 0? Should I go on and calculate the second moment in this case? –  Enzo Feb 10 '13 at 21:39
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