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?
Best Answer
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.