I have some work here but am not sure if it is correct. Suppose $X_1,\ldots,X_n$ are i.i.d. with density function

$$f_\theta(x) = e^{\theta - x}$$

I was able to calculate the MLE estimator for this distribution is $\hat \theta = \min(X_i, \ldots, X_n)$

Since $P(\hat \theta_n \leq t) = 1 - (e^{\theta - t})^n$

Then $P(n(\hat \theta_{n} - \theta) \leq x) = P(\hat \theta_{n} \leq \frac{x}{n} + \theta) = 1 - \left[e^{\theta - \frac{x}{n} - \theta}\right]^n = 1 - e^{-x}$

Thus $\frac{d}{dx}(1 - e^{-x}) = e^{-x}$

Therefore, my work shows that $n(\hat \theta_n - \theta)$ converges in distribution to $\exp(1)$ and the $\operatorname{var}(\hat \theta_n) = \frac{1}{n^2}.$

Consistency can be proven by Slutsky, I believe:

$$n(\hat \theta_n - \theta) \xrightarrow{d} \exp(1)$$

$$\frac{1}{n} \xrightarrow{p} 0.$$

Thus $\hat \theta_n - \theta \xrightarrow{p} 0$

And $\hat \theta_n \xrightarrow{p} \theta$.

Is this correct? I am especially uncertain about my work for the asymptotic distribution, considering that the variance decreases at a faster than standard rate. Is this a rare case for the asymptotic variance?


1 Answer 1


Your work is correct.

You can also show that $\hat \theta_n \overset{p}{\rightarrow} \theta$ by explicitly computing for any fixed $\epsilon > 0$, $$\lim_{n \to \infty} \Pr[\hat \theta_n - \theta > \epsilon] = \lim_{n \to \infty} \Pr[n(\hat \theta_n - \theta) > n\epsilon] = \lim_{n \to \infty} e^{-n\epsilon} = 0.$$

The reason why $\hat \theta_n$ converges so quickly is because it is the sampling distribution of the minimum of a rather heavily right-skewed population distribution.

  • $\begingroup$ Ahh got it. Thanks! $\endgroup$
    – YepYep123
    Nov 9, 2020 at 4:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .