# Asymptotic distribution of the MLE of $\theta$ such that $\log X_i$ is distributed as $N(\theta, \theta)$

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?

• No reason really, that's just the problem Commented Apr 25, 2019 at 12:54
• The problem is that $\log X_i$ is distributed as $N(\theta, \theta)$ rather than $X_i$. I guess you could then let $Y = \log X_i$ then $Y_i \sim N(\theta, \theta)$ and then it is treated pretty much the same way in terms of finding the MLE, but I wonder whether this affects the asymptotic distribution? Commented Apr 25, 2019 at 13:51
• I see what you mean. I did not read the question correctly. Commented Apr 25, 2019 at 13:52

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

$$\sqrt n(\hat\theta-\theta)\stackrel{L}\longrightarrow N\left(0,\frac{2\theta^2}{2\theta+1}\right)$$