Suppose $ X_1,X_2,\ldots,X_n$ are i.i.d variables having a two-parameter exponential distribution with common location and scale parameter $ \theta$ :

$ $ f_{\theta}(x)=\frac{1}{\theta}\exp\left[-\left(\frac{x-\theta}{\theta}\right)\right]\mathbf1_{x>\theta}\quad,\,\theta>0$ $

I am wondering if it is possible to derive a maximum likelihood estimator (MLE) of $ \theta$ .

The likelihood function given the sample $ x_1,\ldots,x_n$ is

$ $ L(\theta)=\frac{1}{\theta^n}e^{-n(\bar x-\theta)/\theta}\mathbf1_{x_{(1)}>\theta}\quad,\,\theta>0$ $

, where $ \bar x=\frac{1}{n}\sum\limits_{i=1}^n x_i$ and $ x_{(1)}=\min\limits_{1\le i\le n} x_i$ .

Since $ L(\theta)$ is not differentiable at $ \theta=x_{(1)}$ , I cannot apply the second-derivative test here.

Even if I could say that $ L(\theta)$ is increasing and/or decreasing in $ \theta$ under the constraint $ \theta<x_{(1)}$ , I am not sure what choice of $ \theta$ maximises $ L(\theta)$ . Differentiation is not valid as I understand.

If MLE is unique, then it is bound to be a function of the sufficient statistic $ (\overline X,X_{(1)})$ . However I don’t see how to derive it in this particular model. Any suggestion would be great.