Variables $X_{1}, . . . , X_{n}$ - random sample from normal distribution, $N(0,\theta)$. There I need to find:
- the MLE of standard deviation of $X_{1}$;
- the asymptotic variance of the estimator we found in the previous step.
I made my own calculations but I'm not sure neither with the final results nor with the used methods of getting them. Here's what I did:
I found the MLE of $\theta$: $\hat{\theta} = \bar{X^2}$ (where $\bar{X^2}$ is the second raw sample moment).Then, I used the relation between $g(\hat{\theta})$ and $g(\theta)$ and assumed that the MLE of $\hat{\sigma}$ ( standard deviation of $X_{1}$) would be $\sqrt {\bar{X^2} } $. What about the asymptotic variance of $\hat{\sigma}$, I'm stuck - I found Var$(\hat{\theta})$ $= \left(\frac{2\theta^2}{n}\right)$ $= I^{-1}(\theta) $ and guessed that with the asymptotic distribution the asymptotic variance must be $ \left(\frac{2(\sqrt {\bar{X^2} })^2}{n}\right)$.
So, could someone, please, help?
Additional thoughts:
Alternatively, I tried to calculate the MLE of $\hat{\sigma}$ from scratch by using $$\sum_{i=1}^n ln[f(x_{i},\sigma^2)] = l$$ Then I took the first and second derivative to estimate the MLE of $\hat{\sigma}$ which was indeed $\sqrt {\bar{X^2} } $. However, while getting the $ I(\sigma) = -E(\frac{d^2 l}{d\sigma^2}) = -E(\frac{-2n}{\sigma^2}) = \frac{2n}{\sigma^2}$, therefore, the asymptotic variance is $ \left(\frac{(\sqrt {\bar{X^2} })^2}{2n}\right)$. Thus, either my original assumptions were incorrect in the first place, or I incorrectly calculated in the second.