$MLE$ of $\theta$ when $X_1 , X_2 , ..., X_n$ is a sample with pdf $f(x, \theta) = e^{\theta - x}; x \ge \theta$

Question

How can we find the $MLE$ of $\theta$ when $X_1 , X_2 , ..., X_n$ is a sample with pdf $f(x, \theta) = e^{\theta - x}; x \ge \theta$ ?

$L(\theta) = \prod_{i = 1}^{n} e^{\theta - x_{i}}$

$L(\theta) = e^{n\theta}e^{-\sum_{i = 1}^{n}x_i}$

$\ln(L(\theta)) = n\theta - \sum_{i = 1}^{n}x_i$

Then I have to solve $\frac{\partial\ln(L(\theta))}{\partial\theta} = 0$. But:

$\frac{\partial\ln(L(\theta))}{\partial\theta} = n$.

I'm stuck :/

Answer 1

Hint: Show and use that the likelihood function is of the form $$ L(\theta)=\mathrm{e}^{\theta n}\mathrm{e}^{-\sum x_i}\mathbf{1}_{\min(x_1,\ldots,x_n)\geq \theta}. $$