Let $X_1, . . . , X_n$ be a random sample such that $\log X_i$ is distributed as $N(θ, θ),$ $θ > 0$ is unknown.
I've calculated the MLE and I got $$\hat\theta = \frac{-1 + \sqrt{1 + 4n^{-1} \sum_{i=1}^{n} Y_i^2}}{2} $$ where $Y_i = \log X_i$ for $i = 1, …, n$.
I was wondering what the asymptotic distribution for this MLE would be as $n \rightarrow \infty$ . I've double-checked this MLE and I'm almost certain it is correct but I don't recognise this distribution nor what the sample mean looks like. I've tried the central limit theorem but I'm unsure whether it'd apply here. Thoughts?
Best Answer
Under certain regularity conditions (like the ones mentioned here on page 1), maximum likelihood estimators have an asymptotic normal distribution. In particular, distributions which are members of the regular exponential family satisfy these conditions.
For $Y_i=\log X_i$, joint density of $Y_1,\ldots,Y_n$ is
\begin{align} f_{\theta}(y_1,\ldots,y_n)&=\frac{1}{(\sqrt{2\theta\pi})^n}\exp\left[-\frac{1}{2\theta}\sum_{i=1}^n (y_i-\theta)^2\right] \\&=\frac{1}{(\sqrt{2\theta\pi})^n}\exp\left[-\frac{1}{2\theta}\sum_{i=1}^n y_i^2+\sum_{i=1}^n y_i-\frac{n\theta}{2}\right]\quad,\small (y_1,\ldots,y_n)\in\mathbb R^n,\,\theta>0 \end{align}
This shows that $f_{\theta}$ is a member of a regular one-parameter exponential family. So we can say that the MLE $\hat\theta$ of $\theta$ has an asymptotic normal distribution, given by
$$\sqrt n(\hat\theta-\theta)\stackrel{L}\longrightarrow N\left(0,\frac{1}{I_{Y_1}(\theta)}\right)\,,$$
where $I_{Y_1}(\theta)=-E_{\theta}\left[\frac{\partial^2}{\partial\theta^2}\ln f_{\theta}(Y_1)\right]$ is the information contained in $Y_1$.
A routine calculation gives $I_{Y_1}(\theta)=\frac{2\theta+1}{2\theta^2}$, so that the limiting distribution is eventually
$$\sqrt n(\hat\theta-\theta)\stackrel{L}\longrightarrow N\left(0,\frac{2\theta^2}{2\theta+1}\right)$$