Determine asymptotic distribution and efficiency of an estimator I have the following problem:
Let $X_1,..., X_n$ be a sample of independent, identically distributed random variables, with density $$f_{\theta}(x)=\begin{cases}
e^{\theta-x}, & \text{if } x\geq \theta\\
0, & \text{elsewhere}\\
\end{cases}$$
Let $\hat\theta_n$ be the maximum-likelihood estimator (MLE) of $\theta$. I am asked to find the asymptotic distribution of $\sqrt n (\hat\theta_n-\theta)$ and $\hat\theta_n$ efficiency.
So far, I computed the MLE, which is $\hat\theta_n=min\{x_i\}$, for $i=0,...,n.$. I also know that if I find that $\sqrt n (\hat\theta_n-\theta)$ converges in distribution to $N(0,I(\theta))$, we would have both the asymptotic distribution and efficiency of the estimator.
Central Limit Theorem could be a way to do it, but I'm really struggling on it.
Any hints or solutions would be appreciated.
 A: To answer the question you are asked, find the distribution function of $T=\sqrt n(\hat\theta_n-\theta)$. 
Now $\hat\theta_n$ has CDF $$P(\hat\theta_n\le t)=\begin{cases} 1-e^{-n(t-\theta)} &,\text{ if }t\ge\theta \\ 0&,\text{ if }t<\theta \end{cases}$$
So CDF of $T$ should be
\begin{align}
F_n(t)&=P\left(\sqrt n(\hat\theta_n-\theta)\le t\right)
\\&=P\left(\hat\theta_n\le \frac{t}{\sqrt n}+\theta\right)
\\&=\begin{cases}1-e^{-\sqrt nt}&,\text{ if }t\ge 0 \\ 0&,\text{ if }t<0\end{cases}
\end{align}
Since $n(\hat\theta_n-\theta)\sim \mathsf{Exp}(1)$, it is clear that $T$ is exponential with mean $1/\sqrt n$.
Taking limit as $n\to\infty$, show that $F_n(t)$ converges to another distribution function. This would give you a degenerate asymptotic distribution of the MLE $\hat\theta_n$.
To arrive at a non-degenerate limiting distribution of the MLE, an appropriate scaling of $\hat\theta_n$ is $T'=n(\hat\theta_n-\theta)$. We know $T'$ is exactly distributed as $\mathsf{Exp}(1)$, which is naturally also the asymptotic distribution. 

Maximum likelihood estimators are expected to be (asymptotically) efficient in most cases.
Formally, if $T_0$ is the uniformly minimum variance unbiased estimator (UMVUE) of $\theta$, then (large sample) efficiency of $\hat\theta_n$ is defined by $\mathrm e=\frac{\operatorname{Var}(T_0)}{\operatorname{Var}(\hat\theta_n)}$. Find $T_0$ and hence conclude $\hat\theta_n$ is indeed efficient.
