# Finding a maximum likelihood estimator when derivative of log-likelihood is invalid

I need to find the maximum likelihood for $\theta$ given the following:

$X_1, ..., X_n$ are sampled i.i.d from a population with the following density: $$f(x | \theta) = \begin{cases} e^{-(x-\theta)} & x \geq \theta \\ 0 & \text{otherwise} \end{cases} \tag*{where \theta > 0}$$

I begin by writing the likelihood... $$L(\theta; x_1, ..., x_n) = \prod_{i=1}^{n} e^{-(xi-\theta)} = e^{n\theta - \sum_{i=1}^{n} x_i}$$ and the log likelihood... $$\ell(\theta; x_1, ..., x_n) = \log e^{n\theta - \sum_{i=1}^{n} x_i} = n\theta - \sum_{i=1}^{n} x_i$$ and setting the derivative of the log likelihood to zero... \begin{align*} 0 &= \frac{\partial}{\partial \theta} \ell(\theta; x_1, ..., x_n) \\ 0 &= \frac{\partial}{\partial \theta} n\theta - \sum_{i=1}^{n} x_i \\ 0 &= n \ \ \ \ \text{(?)} \end{align*} That I where I get confused, given that the standard procedure for finding the MLE estimator does not seem to give a valid expression. Where am I going wrong? What is the appropriate method for finding the MLE estimator in this situation?

It's clear that $L(\theta; x_1, ..., x_n) = 0$ where $\theta > min\{x_1, ..., x_n\}$, but I'm not sure if/how this fact is useful.

• Two issues: (a) a function can have a maximum at its extremes or discontinuities, and (b) you might consider using indicator functions to deal with $f(x \mid \theta)=0$ when $x \lt \theta$ – Henry Dec 12 '16 at 11:26

As you say, your expressions for the likelihood and log-likelihood are only valid when $\theta$ is less than or equal to all the observed $x_i$; otherwise the likelihood is $0$ and the log-likelihood $-\infty$
Meanwhile, as your derivative suggests, your expressions for the likelihood and log-likelihood are strictly increasing functions of $\theta$ when they are valid, so you want $\theta$ to be as large as possible
So the maximum likelihood and maximum log-likelihood both occur when $\displaystyle \theta = \min_i x_i$