Calculating the asymptotic normality result of a MLE from a skew-logistic distribution

Question

Suppose we have $X$ with cumulative distribution function $F_X(x) = (1-e^{-x})^\frac{1}{\theta}$ where $x \geq 0, \theta > 0$. How can one find a MLE for $\theta$ from this and the asymptotic normality result?

We get the density function as $f_X(x;\theta) = \frac{1}{\theta}(1-e^{-x})^{\frac{1}{\theta}-1}e^{-x}$.

$\textbf{EDIT : } $ I found the MLE to be $$\hat{\theta} = -\frac{1}{n}\sum^n_{i=1} \ln(1-e^{-x_i}).$$ Now how do we get the asymptotic normality result? I tried calculating the Fisher information number but I'm stuck at $$I(\theta) = -E\left[\frac{\partial^2}{\partial \theta^2} \ln f(x,\theta) \right] = \frac{-1}{\theta^2}+ \frac{2}{\theta^3}E[\ln(1-e^{-x})]$$

Answer 1

The log-likelihood is given by: $$\ell (\theta) = -n\log \theta + (\theta^{-1} - 1) \sum_i \log (1-e^{-x_i}) - \sum_i x_i $$ Optimising, we get the MLE you found: $$\hat{\theta}_{ML} = - \frac{1}{n} \sum_i \log (1 - e^{-x_i})$$ Now, set $Y_i = -\log (1 - e^{-X_i})$. Applying a change of variables, we obtain that $$F_Y(y) = P(Y_i \leq y) = P(X_i \leq -\log (1-e^{-y})) = F_X(-\log (1-e^{-y}))$$ So that the density of $Y$ is given by $$f_Y(y) = \frac{d}{dY} F_X(-\log (1-e^{-y})) = \frac{1}{\theta} e^{-y / \theta}$$ Thus, we identify $Y_i \sim Exp(\theta)$. Notice that $E(Y) = \theta$ and $\mathrm{Var}(Y) = \theta^2$. From there, the central limit theorem tells us that $$\sqrt{n} \frac{\hat{\theta }_{ML}- \theta}{\theta} \Longrightarrow N(0,1)$$ which exhibits the asymptotic normality of this MLE. Having observed this, we may obtain consistent standard errors by setting: $$\mathrm{s.e.}(\theta) = \frac{\hat{\theta}_{ML}}{\sqrt{n}}$$ Thus, an asymptotic $(1-\alpha)$-confidence interval is $$\left[ \hat{\theta}_{ML} - z_{1-\alpha /2 } \frac{\hat{\theta}_{ML}}{\sqrt{n} }, \hat{\theta}_{ML} + z_{1-\alpha /2 } \frac{\hat{\theta}_{ML}}{\sqrt{n} } \right]$$