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I need to find the asymptotic distribution of the MLE of a geometric distribution. I know $\overline X$ goes as $N(1/p, (1-p)/(n p^2))$. Using the delta method MLE=$1/\overline X$ goes as $N(p, (1-p)/(np^6))$. However if I use the asymptotic theory of MLE, I get MLE goes as $N(p, (1-p)p^2/n)$ where $(1-p)p^2$ is the CRLB (Cramer Rao bound).

The variances I get differ by the two methods. What am I missing?

Rodrigo de Azevedo
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revathi ananthakrishnan
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  • Try to type math in Latex, see http://math.stackexchange.com/help/notation Please check that my edits are all right. – leonbloy Sep 05 '14 at 20:16
  • The delta method actually shows that $1/\bar X_n=p+p\sqrt{1-p}Z_n/\sqrt{n}$ where $Z_n$ converges in distribution to $N(0,1)$. That is, exactly the other limit mentioned in your post. To go further, one needs to see your computations for the delta method. – Did Sep 05 '14 at 22:36

1 Answers1

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Yes, thanks, I believe the answer to my question is that in the Delta method, I need to substitute $1/p$ once I get $(g'(x))^2$ and not $p$. Once one substitutes $1/p$ in the answer for $(g'(x))^2$ where $g(x)=1/x$ then one will get $(1-p)p^2/n$ for the variance and the two techniques match.

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revathi ananthakrishnan
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