Maximum likelihood estimator of the exponential distribution with two same parameter Let $X_{1},...,X_{n}$ be a random samples from  a population having the probability density function:
$f(x)= \frac{1}{\theta}e^{\frac{-(x-\theta)}{\theta}}$ where $x\geq \theta$. 
When I solved the likelihood equation of the above density, the maximum likelihood estimator turned out to be mean of the sample but if you see the range of values that the parameter $\theta$ can take is $(0,X_{1}]$. How can this be possible that the likelihood estimator is out of the range of $\theta$. Please clear my doubt regarding this. Thanks
 A: I suspect you have solved the maximum likelihood estimation problem incorrectly, most likely by ignoring the parameter constraint.  From the stated density, your log-likelihood function is:
$$\begin{equation} \begin{aligned}
l_\boldsymbol{x}(\theta) &= -n \ln \theta - \sum_{i=1}^n \frac{x_i-\theta}{\theta} \\[8pt]
&=-n \ln \theta - \frac{n \bar{x}_n}{\theta} + n \\[8pt]
&= n \left( 1 - \ln \theta -\frac{\bar{x}_n}{\theta} \right) \text{ } \text{ } \text{ } \text{ } \text{ for all } 0 < \theta \leqslant \min (x_1 ,..., x_n).
\end{aligned} \end{equation}$$
For all $0 < \theta < \min (x_1 ,..., x_n)$ this has first derivative:
$$\frac{dl_\boldsymbol{x}}{d\theta}(\theta) = n \left( - \frac{1}{\theta} + \frac{\bar{x}_n}{\theta^2} \right) = \frac{n}{\theta^2} (\bar{x}_n - \theta).$$
Now, over the range $0 < \theta < \min (x_1 ,..., x_n)$ we have $\theta < \bar{x}_n$ which means that the first derivative is strictly positive.  Hence, the maximising value occurs at the border case:
$$\hat{\theta} = \min (x_1 ,..., x_n).$$
I suspect that you have probably ignored the parameter constraint and solved the score equation (setting the derivative of the log-likelihood to zero).  That would yield the incorrect solution of the sample mean.
