Let $X_1,\ldots,X_n$ be a sample from the distribution whose probability density function is:

$$f(x)=\begin{cases}\frac13e^{-\frac13(x-\theta)}&, x\ge\theta \\ 0&,x<\theta\end{cases}$$

Find the maximum likelihood estimator of $\theta$.

So the PDF is equal to: $$f(x\mid \theta) = \frac{1}{3}e^{-\frac{1}{3}(x-\theta)}1_{\{x\geq \theta\}}$$

and thereby the likelihood will be:

$$\mathcal{L} = \prod_{i=1}^n f(x\mid \theta)= \prod_{i=1}^n \frac{1}{3}e^{-\frac{1}{3}(x-\theta)}1_{\{x\geq \theta\}}$$


It's a common mistake.

Your pdf is

$$f(x| \theta) = \frac{1}{3}e^{-\frac{1}{3}(x-\theta)}1_{\{x\geq \theta\}}$$

Use this pdf in the likelihood (along with the indicator) and see for yourself that the MLE estimate of $\theta$ is $\min\limits_{i=1}^{n} X_i$.


$$L(\theta|x_{1},\ldots,x_{n}) = \left(\frac{1}{3}\right)^n e^{-\frac{1}{3n}(\bar{x}-\theta)}\prod_{i=1}^{n}1_{\{x_i\geq \theta\}}$$

Now, $\theta$ should satisfy $x_i \geq \theta \ \forall i$ otherwise likelihood will be zero. Therefore, $\theta \leq \min\limits_{i=1}^{n}x_i$. As, $\theta$ decreases further the exponential part decreases due to the negative sign in its power and therefore, the likelihood decreases.

  • 1
    $\begingroup$ P.S. It is not always possible to compute the minima/maxima using derivative tests. Sometimes plotting graph or simple reasoning does the job. $\endgroup$ – Dhruv Kohli - expiTTp1z0 Aug 16 '17 at 15:23
  • $\begingroup$ so based on this the likelihood I should be able to see that the MLE estimate is $\min\limits_{i=1}^{n} X_i$ ? $\endgroup$ – Daniel Robotics Aug 16 '17 at 15:41
  • 1
    $\begingroup$ Yes. Just plot the likelihood function as a function of $\theta$ and you will see it. $\endgroup$ – Dhruv Kohli - expiTTp1z0 Aug 16 '17 at 15:47
  • $\begingroup$ Sorry if I seem annoying, but I don't know how to plot the likelihood function as a function of θ $\endgroup$ – Daniel Robotics Aug 16 '17 at 19:50
  • 1
    $\begingroup$ Check edit please. However, I strongly recommend to plot the likelihood for better vizualization. $\endgroup$ – Dhruv Kohli - expiTTp1z0 Aug 16 '17 at 20:05

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.