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

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 $$I(\theta) = -E\left[\frac{\partial^2}{\partial \theta^2} \ln f(x,\theta) \right] = \frac{-1}{\theta^2}+ \frac{2}{\theta^3}E[\ln(1-e^{-x})]$$

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 $$\left[ \hat{\theta}_{ML} - z_{1-\alpha /2 } \frac{\hat{\theta}_{ML}}{\sqrt{n} }, \hat{\theta}_{ML} + z_{1-\alpha /2 } \frac{\hat{\theta}_{ML}}{\sqrt{n} } \right]$$
• Great! That seemed to be the second part of the exercise, applying the transformation as you did. Is it possible to also find this normality result without transformation? Also it says to find an approximate $(1 − α)$ confidence interval for $θ$ with as little approximations as possible. I can post a screenshot of the problem if you want Commented Jan 17, 2022 at 20:44
• @Geigercounter Please review the book I've referenced for conditions that guarantee asymptotic normality (not all MLEs are asymptotically normal, for instance). In your solution, you are one step away; simply compute the expectation of $E(1- e^{-x})$, which I have already done in my solution. Commented Jan 17, 2022 at 21:12