0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 $x \lt \theta$ $\endgroup$ – Henry Dec 12 '16 at 11:26
1
$\begingroup$

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 $\displaystyle \theta = \min_i x_i$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.