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?
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.