Maximum likelihood estimator is sample mean or minimum ordered statistic? Let $X_1,X_2,X_3....X_n$ be a random sample from a population having probability density function
$$f(x)=\dfrac{1}{\theta} e^\frac{-(x-\theta)}{\theta} ;x\geq\theta$$
where $\theta\in(0,\infty)$. Find the MLE of $\theta$.
In this question when i followed normal method of finding MLE(taking log likelihood differentiating with respect to $\theta$)I found $\hat\theta =\bar x$. 
But like in uniform distribution case we have here parameter dependent on $x$  . Now i am not sure if MLE is $\bar x$ or $x_{(1)}$. 
 A: Note that $$f(x) = \frac{1}{\theta} \exp\left( - \left(\tfrac{x}{\theta} - 1 \right) \right) \mathbb 1 (x \ge \theta > 0) = \frac{e}{\theta} e^{-x/\theta} \mathbb 1 (x \ge \theta > 0).$$  Hence the likelihood given a sample $\boldsymbol x$ is proportional to $$\mathcal L(\theta \mid \boldsymbol x) = \theta^{-n} e^{-n \bar x/\theta} \mathbb 1(x_{(1)} \ge  \theta > 0).$$  The log-likelihood is $$\ell(\theta \mid \boldsymbol x) = \begin{cases} - n \log \theta - \frac{n \bar x}{\theta}, & 0 < \theta \le x_{(1)} \\ -\infty, & \text{otherwise.}\end{cases}$$   But since the critical point satisfying $$\frac{\partial \ell}{\partial \theta} = 0 = -\frac{n}{\theta} \left( 1 - \frac{\bar x}{\theta}\right)$$ is $\theta = \bar x$, and 
 $$\bar x = \frac{1}{n} \sum_{i=1}^n x_i \ge \frac{1}{n} \sum_{i=1}^n x_{(1)} = x_{(1)},$$ with equality if and only if $x_i = x_{(1)}$ for all $i = 1, \ldots, n$, we conclude that the log-likelihood has no critical point in $\theta \in (0, x_{(1)})$; hence it is maximized when $\theta = x_{(1)}$ as $\partial \ell/\partial \theta < 0$ for all such $\theta$ (i.e., it is strictly decreasing).
