Asymptotic distribution of $\sqrt{n}(\hat\theta_n-\theta)$ to determine the efficiency of $\hat\theta_n$? I want to know the asymptotic distribution of $\sqrt{n}(\hat\theta_n-\theta)$ to determine the efficiency of $\hat\theta_n$.
I know there is a theorem with lots of assumptions that immediately concludes that it is efficient and the distribution if $N(0,I(\theta)^{-1})$ but I'm wondering if there is another approach?

I have that $\hat\theta_n = X_{(1)}$ and I have computed $F_{\hat\theta_n}(x)$ and $f_{\hat\theta_n}(x)$ already.
$$f_\theta (x) = e^{\theta-x}1_{x\geq \theta}$$
$$F_{\hat\theta_n} (x) = \left[1-e^{n(\theta-x)} \right] 1_{x\geq \theta}$$
$$f_{\hat\theta_n} (x) = \left[ne^{n(\theta-x)} \right] 1_{x\geq \theta}$$
Is there a way to continue from here that uses the delta method or otherwise?

This is my attempt: We want the distribution of $Y_n = \sqrt{n}(\hat\theta_n-\theta)$. We can rewrite this as $\frac{Y_n}{\sqrt{n}}+\theta = \hat\theta_n$
$$f_{Y_n} (x) = f_{\hat\theta_n} \left(\frac{Y_n}{\sqrt{n}}+\theta\right)$$
$$= ne^{-\sqrt{n}Y_n}1_{Y_n \geq 0}$$
I don't know how to continue from here. I don't see how to arrive from this to a normal approximation so that I can figure out efficiency?! I also thought about using the Delta Method but it seems like it is not applicable in this case?!
 A: Part of the sufficient conditions for asymptotical normality of the MLE is that all models in the family have the same support.  This fails in your example because it is $(\theta,\infty)$.
In particular this means that $\hat \theta_n = X_{(1)} > \theta$ and so $Y_n > 0$.  Thus it cannot be asymptotically normal (other than being degenerate, in case you consider that "normal").
We can compute the distribution more explicitly.  As you correctly noted we have
$$
\mathbb{P}(Y_n < y) = \mathbb{P}(X_{(1)} < \theta +  y/\sqrt{n}).
$$
(Note you can't just put $\theta +  y/\sqrt{n}$ straight into the density for $X_{(1)}$ - you need to differentiate the expression above and you'd find an extra $1/\sqrt{n}$).
This comes out as
$$
\mathbb{P}(Y_n < y) = 1 - \exp(-y \sqrt{n}),
$$
i.e. Exponential($\sqrt{n}$).  This is not asymptotically normal.  The variance is $1/n$.  Assuming efficiency is defined by the ratio to the Cramer Rao lower bound, we must compute the Fisher Information $I(\theta)$ and examine the ratio
$$
n/I(\theta)
$$
The Fisher information is
$$
I(\theta) = \mathbb{E}[l'(X;\theta)^2] \\
= \mathbb{E}\left[ \left(\frac{d}{d\theta}\left(n\theta - \sum_i X_i\right)\right)^2\right] = n^2.
$$
A: If in fact $F_{\hat\theta_n} (x) = \left[1-e^{n(\theta-x)} \right] 1_{x\geq \theta}$ then defining $Y_n = n(\hat{\theta}_n - \theta)$ shows that 
$$
F_{Y_n}(y) = P(\hat{\theta}_n \le \frac{y}{n} + \theta) = 1 - e^{-y}.
$$
So $Y_n \sim \text{Exp}(1)$ exactly, not asymptotically, $\hat{\theta}$ converges faster to $\theta$ than root $n$, and you don't need to appeal to any results (just use CDFs). However, do note that I am defining $Y_n$ differently than you are.
Also, the Delta method is typically used when, say a different estimator $\hat{\theta}_n$ is asymptotically Normal. So you assume that $\sqrt{n}(\hat{\theta} - \theta)$ converges already. Then you can use it if you want the asymptotic distribution of some transformed estimator $f(\hat{\theta}_n)$, where $f$ is a function with certain requirements. Typically $\sqrt{n}(f(\hat{\theta}_n) - f(\theta))$ will also be asymptotically Normal with mean $0$ again but with a different variance depending on $f$. Even though it's very popular, I'm not sure it can be of much use here.
