Find the MLE of $\theta$ for $X_1, X_2, ..., X_n$ with density function $f(x|\theta) = e^{-(x-\theta)}$, $x \ge \theta$ 
Suppose that $X_1, X_2, ..., X_n$ are i.i.d with density function
$$
f(x|\theta) = e^{-(x-\theta)}, \quad \quad x\ge \theta
$$
and $f(x|\theta) = 0$ otherwise.

Find the mle of $\theta$. (Hint: Be careful, and don't differentiate before thinking. For what values of $\theta$ is the likelihood positive?


My attempt
The likelihood function is
\begin{align}
L(\theta) &= f(x_1, x_2, ..., x_n|\theta) \\
&= \prod_{i = 1}^n e^{-(x_i - \theta)}
\end{align}
for $x_1, x_2, ... x_n \ge \theta$.
Correspondingly, the log-likelihood is
\begin{align}
\ell(\theta) &= -\sum_{i=1}^n (x_i - \theta) \\
&= n\theta - n\sum_{i=1}^nx_i
\end{align}
Differentiating with respect to parameter $\theta$ gives the result $\frac {d\ell} {d\theta} = n > 0$. I am stuck here.

Two questions:

*

*How should I proceed from my final step? After differentiating the result I no longer have terms in $\theta$ to calculate a MLE.

*Why does it matter for the likelihood to be positive (as in the hint)? Is it simply because it is defined in terms of the joint PDF of $X$, which must be positive? Even so, given the likelihood function $L(\theta) = f(x_1, x_2, ..., x_n|\theta) = \prod_{i = 1}^n e^{-(x_i - \theta)}$, isn't it always positive given any $\theta \in \mathbb{R}$? I could be misunderstanding something here, so please correct me.

Any advice would be greatly appreciated!
 A: First, try to write down the likelihood as detailed as possible, you know that holds that
$$f(x|\theta) = e^{-(x-\theta)},  \ x \geq \theta$$
equivalently this can be written as $f(x|\theta) = e^{-(x-\theta)}\mathbb{I}_{x\geq \theta}$
where $\mathbb{I}_{x\geq \theta}=1$ if $x\geq \theta$ and $0$ otherwise. Based on that we would calculate the likelihood function as
$$f(x_{1},x_{2},...,x_{n}|\theta)= f(x_{1}|\theta)\mathbb{I}_{x_{1}\geq\theta}...f(x_{n}|\theta)\mathbb{I}_{x_{n}\geq \theta}$$
Then the next step is to identify when this likelihood function is zero? You can check that by looking at all the indicator functions together.
Also, because the likelihood is a function of $\theta$, if you try to maximize it for values of $\theta$ that by definition are $f(x|\theta)=0$ then there is nothing to maximize there.
And lastly when you have identified for which values of $\theta$ your likelihood expression $f(x_{1},x_{2},...,x_{n}|\theta)$ is non zero, then do not take the derivative directly try to think first which of the restricted values that you have identified of $\theta$ maximize your likelihood, you can easily make a graph.
