I'm trying to find asymptotic variances of the parameters of log-normal.
Assume $ X \sim LN(\mu,\sigma), f(x) = \frac{1}{\sqrt{2\pi\sigma^2}x}e^{-\frac{(\ln x-\mu)^2}{2\sigma^2}} $ and we have $ n $ observations
Let $ \theta = (\mu, \sigma)' $ and the Fisher information is
$
\begin{eqnarray}
I(\theta) &=& E \left[ \left\{ \frac{\partial \ln f_\theta(x)}{\partial \theta}\right\}\left\{\frac{\partial \ln f_\theta(x)}{\partial \theta} \right\} ' \right] \\ &=&E \left[ \left( \begin{array}{c} \frac{\sum_{i=1}^n (\ln x_i-\mu)}{\sigma^2} \\ \frac{-n}{\sigma} + \frac{\sum_{i=1}^n (\ln x_i-\mu)^2}{\sigma^3} \end{array} \right)
\left( \frac{\sum_{i=1}^n (\ln x_i-\mu)}{\sigma^2}\ \ \ \frac{-n}{\sigma} + \frac{\sum_{i=1}^n (\ln x_i-\mu)^2}{\sigma^3} \right) \right]
\end{eqnarray} $
My textbook says this yields
$ \left( \begin{array}{rr} \frac{n}{\sigma^2} & 0 \\ 0 & \frac{\sigma^2}{2n} \end{array} \right)$.
I'm stuck in the expected value calculation. How should I take the expected from
$ \left( \frac{\sum_{i=1}^n (\ln x_i-\mu)}{\sigma^2} \right)^2 $, for example? I also don't see why these two components are zero.
Thanks in advance!
Best Answer
Since $Z_i:=\frac{\ln X_i-\mu}{\sigma}\sim N(0,\,1)$,$$EZ_i^2Z_j^2=\left\{ \begin{array}{rl} 1 & i\ne j\\ 3 & i=j \end{array}\right.=1+2\delta_{ij},$$and you can similarly show $EZ_iZ_j=\delta_{ij},\,EZ_iZ_j^2=0$. Your outer product can be rewritten as$$\frac{1}{\sigma}\sum_{i}\left(\begin{array}{c} Z_{i}\\ Z_{i}^{2}-1 \end{array}\right)\frac{1}{\sigma}\sum_{j}\left(\begin{array}{cc} Z_{j} & Z_{j}^{2}-1\end{array}\right)\\=\frac{1}{\sigma^{2}}\sum_{ij}\left(\begin{array}{cc} Z_{i}Z_{j} & Z_{i}Z_{j}^{2}-Z_{i}\\ Z_{i}^{2}Z_{j}-Z_{j} & Z_{i}^{2}Z_{j}^{2}-Z_{i}^{2}-Z_{j}^{2}+1 \end{array}\right),$$with mean$$\frac{1}{\sigma^{2}}\sum_{ij}\left(\begin{array}{cc} \delta_{ij} & 0\\ 0 & 2\delta_{ij} \end{array}\right)=\left(\begin{array}{cc} \frac{n}{\sigma^{2}} & 0\\ 0 & \frac{2n}{\sigma^{2}} \end{array}\right).$$Your textbook appears to have the last fraction upside down.