Asymptotic distribution for MLE of shifted exponential distribution Suppose we have $X_1,...,X_n$ iid the shifted exponential distribution:
$$f(x)=\lambda e^{-\lambda(x-\theta)}, x\ge \theta$$
I have figured out both the MLE for $\lambda$ and $\theta$, which are $\hat \lambda = \frac{1}{\bar X - X_{min}}$ and $\hat \theta =X_{min}$.  
I also found the asymptotic distribution of $\hat \theta$:
$$\sqrt{n}(\hat \theta-\theta) \rightarrow 0$$ 
Now I'm stuck at deriving the asymptotic distribution of $\hat \lambda$ and showing that it is a consistent estimator. How do you do this?
Thanks!
 A: For consistency, by the weak law of large numbers $\bar X_n \stackrel{\text p}\to \frac 1\lambda + \theta$ and $X_\min \stackrel{\text p}\to \theta$ so by Slutsky 
$$
\bar X_n - X_\min \stackrel{\text p}\to \frac 1\lambda.
$$
By assumption $\lambda > 0$ so the map $x \mapsto x^{-1}$ is continuous, and the continuous mapping theorem finishes the job.

For the asymptotic distribution, by the standard CLT we know $\sqrt n (\bar X_n - \theta -\lambda^{-1}) \stackrel{\text d}\to \mathcal N(0, \lambda^{-2})$. Let $Y_n = \sqrt n (\bar X_n - \theta - \lambda^{-1})$ and consider
$$
\sqrt n (\bar X_n - X_{\min,n} - \lambda^{-1}) = \sqrt n ([\bar X_n - \theta - \lambda^{-1}] - [X_{\min,n} -  \theta])\\
= Y_n - Z_n
$$
where $Z_n := \sqrt n (X_{\min,n} - \theta)$. You already worked out the asymptotic distribution of $Z_n$ so we can use that along with Slutsky again to conclude
$$
Y_n - Z_n \stackrel{\text d}\to \mathcal N(0, \lambda^{-2}).
$$
You can now finish this off with the delta method.
A: Although you are also asking about the estimator $\hat{\lambda}$, I am going to note some things about $\hat{\theta}$.  In this particular case it is quite easy to obtain the exact distribution of this estimator.  Since you have a series of shifted exponential random variables, you can define the values $Y_i = X_i - \theta$ and you then have the associated series $Y_1,Y_3,Y_3 ... \sim \text{IID Exp}(\lambda)$.  This gives the exact distribution:
$$\hat{\theta} = X_{(1)} = \theta+ Y_{(1)} \sim \theta + \text{Exp}(n \lambda).$$
Note that this gives the pivotal quantity $n(\hat{\theta} - \theta) \sim \text{Exp}(\lambda)$.  You can prove that $\hat{\theta}$ is a consistent estimator by computing the probability of a deviation larger than a specified level.  For all $\varepsilon >0$ we have:
$$\begin{aligned}
\mathbb{P}(|\hat{\theta} - \theta| < \varepsilon)
= \mathbb{P}(\hat{\theta} - \theta< \varepsilon) 
= \exp(-n \lambda \varepsilon). \\[6pt]
\end{aligned}$$
We therefore get the limiting result:
$$\begin{aligned}
\lim_{n \rightarrow \infty} \mathbb{P}(|\hat{\theta} - \theta| < \varepsilon)
= \lim_{n \rightarrow \infty} \exp(-n \lambda \varepsilon) = 0, \\[6pt]
\end{aligned}$$
which is the required condition for weak consistency (i.e., convergence in probability of the estimator to the parameter it is estimating).
