Asymptotic distribution of MLE (log-normal)

Question

Say we have a sample $X_{1},...,X_{n}$ from a log-normal distribution with parameters $\mu$ and $\sigma^{2}$. That is, $\ln(X)$~$N(\mu,\sigma^{2})$. Let $T_{n},Z_{n}$ denote the MLE's for $\mathbb{E}(X)$ and $\mathbb{D}^{2}(X)$ respectively (based on the sample of size $n$). The question is: What are the asymptotic distributions of these estimators? I know they are both asymptotically normal with means $m_{t}=\exp(\mu+\sigma^{2}/2)$ and $m_{z}=m_{t}^2(\exp(\sigma^{2})-1)$, but what about the variances?

Answer 1

I will provide the variance of the MLE estimator of the mean of the log-normal.
Since this is normal distribution territory, we know that the (centered and scaled) estimators $\hat \mu$ and $\hat \sigma^2$ have the following joint asymptotic distribution

$$\sqrt{n}\left[ \begin{matrix} \hat\mu-\mu \\ \hat\sigma^2 - \sigma^2\\ \end{matrix}\right] \sim_a\mathcal N(0, \Sigma)$$

where $$\Sigma = \left[\begin{matrix} \sigma^2 & 0\\ 0 & 2\sigma^4 \end{matrix}\right]$$

If we left-multiply the vector of centered estimators by the row vector $\mathbf c=[1,\;\; \frac 12]$, set $\theta \equiv \mu + \frac 12\sigma^2$ and $\hat \theta \equiv \hat \mu + \frac 12 \hat \sigma^2$, we obtain, by the Delta method,

$$\sqrt{n}(\hat\theta-\theta) \sim_a \mathcal{N}(0,\,V_{\theta})$$

where $$V_{\theta} = \mathbf c\Sigma\mathbf c' = [\begin{matrix} 1 & 1/2\end{matrix}]\left[\begin{matrix} \sigma^2 & 0\\ 0 & 2\sigma^4 \end{matrix}\right]\left[\begin{matrix} 1 \\ 1/2\end{matrix}\right] = \sigma^2 + \sigma^4/2 $$

We want the asymptotic distribution of $\hat E(X) = \hat m_t$. Since $E(X) = \exp\{\mu + \frac 12\sigma^2\} = g(\theta)$, with $g'(\theta) = g(\theta)$

by applying the delta method again, we have that

$$\sqrt{n}(\hat m_t-m_t) \sim_a \mathcal{N}(0,\,V_t)$$

where $$V_t = V_{\theta}\cdot\left[g'(\theta)\right]^2 = (\sigma^2 + \sigma^4/2)\cdot \exp\left\{2(\mu + \frac 12\sigma^2)\right\}$$

So approximately and for large samples, the variance of the estimator before centering and scaling is

$$\operatorname{Var}(\hat m_t) \approx \frac {m_t^2}{n} (\sigma^2 + \sigma^4/2)$$

Answer 2

The parameter estimates are given on wikipedia: http://en.wikipedia.org/wiki/Log-normal_distribution#Maximum_likelihood_estimation_of_parameters

\begin{align} \widehat \mu = \frac {\sum_k \ln x_k} n, \widehat \sigma^2 = \frac {\sum_k \left( \ln x_k - \widehat \mu \right)^2} {n} \end{align}

Note that $\ln x_k$ are normally distributed, so you can refer to http://en.wikipedia.org/wiki/Normal_distribution#Estimation_of_parameters to see the distribution of those parameter estimates.