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!

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.
