So I have a particular distribution where I know that the Maximum Likelihood Estimate is $hat{theta} = 0.3893814241$, and it is correct. I also know that the Information function, or the second derivative of the Likelihood function, is given by
$$I(theta) =frac{17(4theta-3)}{(-2theta^{2} + 3theta – 1)^{2}} – frac{49}{theta^{2}}$$
Now, I __want__ to find the estimated asymptotic variance. I know that by a theorem which requires regularity to hold (which holds in this case), the variance can be estimated with
$$Var(hat{theta}_{n}) approx [J^{-1}(hat{theta})]$$
where $J(theta)$ is the Fisher information function.
I figured that since $E[theta] = theta$, the Fisher information function is simply equal to the Information function itself, or in other words $J(theta) = E[I(theta)] = I(theta)$.
This leads me to find
$$ J^{-1}(theta) = left(frac{17(4theta-3)}{(-2theta^{2} + 3theta – 1)^{2}} – frac{49}{theta^{2}}right)^{-1}$$
but then $J^{-1}(hat{theta})$ gives me a really small negative value for the estimated asymptotic variance, which seems wrong because it’s negative obviously.
So I was wondering where I went wrong in finding the estimated asymptotic variance? I was convinced my approach was solid and made sense, but the result I have found doesn’t seem very logical. Thanks!
(To see LaTeX see the attached file)
