# Variance and Consistency of MLE estimator for a shifted exponential distribution.

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?

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.