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


1 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 $$\left[ \hat{\theta}_{ML} - z_{1-\alpha /2 } \frac{\hat{\theta}_{ML}}{\sqrt{n} }, \hat{\theta}_{ML} + z_{1-\alpha /2 } \frac{\hat{\theta}_{ML}}{\sqrt{n} } \right]$$

  • $\begingroup$ 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 $\endgroup$ Jan 17, 2022 at 20:44
  • $\begingroup$ @Geigercounter Yes, there are master theorems that allow you to conclude the asymptotic normality of the MLE by studying the density only. You may find them on Chapter 5.5 of the book Asymptotic Statistics by van der Vaart. Yes, it's possible to find an asymptotic confidence interval; simply use the asymptotic normality and plug-in estimates for the standard error. This is a separate question, though, so it's best if you post it separately. $\endgroup$ Jan 17, 2022 at 20:49
  • $\begingroup$ Would you be able to clarify how I would do that without the transformation? As for the confidence interval, the problem asked for both of these at the same time... I'd greatly appreciate it if you could also elaborate on this point! :) $\endgroup$ Jan 17, 2022 at 20:53
  • $\begingroup$ @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. $\endgroup$ Jan 17, 2022 at 21:12
  • $\begingroup$ I checked your reference and I think this MLE does satisfy the conditions for asymptotic normality! I'm however still in the dark how to do the approximate confidence interval... $\endgroup$ Jan 17, 2022 at 21:15

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