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?!

• So, just to clarify- $\hat \theta_n = \min_i X_i$ is the MLE for a parameter $\theta$ in a model with density (proportional to) $e^{\theta-x}$ ($x>0$)? – P.Windridge Oct 20 '17 at 13:19
• @P.Windridge Yes, precisely so – ChineseStatistician Oct 20 '17 at 13:21