We let $X$ follow a log-normal distribution with parameters $(\xi, \sigma^2)\in R xR_{+}$ there are unknown so $X=log Y$ with $Y \sim N(\xi, \sigma^2)$. We consider a sample $X_1,...,X_n$ from this distribution.
Now I have to find the asymptotic wald confidence interval for the median $\lambda$ and the mean $\mu$ of these distribution: $$\lambda=\phi_1(\xi, \sigma^2)=med_{\xi,\sigma^2}(X)=e^{\xi}$$
and $$\mu=\phi_2(\xi, \sigma^2)=E_{\xi,\sigma^2}(X)=e^{\xi+\sigma^2/2}$$ I have found $$I(\theta)=\frac{n}{2\theta^2}$$
I have found in my notes that the wald statistic is given by:
$$W_n=n(\hat{\beta_n}-\beta)^Ti(\hat{\beta_n})(\hat{\beta_n}-\beta)$$
witch is asymptotically distributed as $\chi^2(m)$ where $m$ is the dimension, so the corresponding area is given by:
$$C=\{\beta |n(\hat{\beta_n}-\beta)^Ti(\hat{\beta_n})(\hat{\beta_n}-\beta) \leq \gamma_{1-\alpha}(m)\}$$ where $\gamma_{1-\alpha}(m)$ is the $1-\alpha$ quantile in $\chi^2(m)$ distribution.
Then we get an ellipse centered around $MLE \hat{\beta_n}$
Then the confidence interval is: $$C=(\hat{\beta_n}-\frac{z_{1-a/2}}{\sqrt{ni(\hat{\beta_n})}},\hat{\beta_n}+\frac{z_{1-a/2}}{\sqrt{ni(\hat{\beta_n})}})$$ where $z_{1-a/2}$ is the $1-a/2$-quantile in the normal distribution.
If we let $\sigma ^ 2 = \theta $, thus $ X \sim N( \mu, \theta)$ and then I get the fisher information: $$I_{X_1,...,X_n}(\theta) = \frac{n}{2\theta ^ 2} = \frac{n}{2\sigma ^ 4}$$
How can I use this to find the asymptotic wald confidence interval for the median $\lambda$ and the mean $\mu$. Can anyone help me?