Maximum likelihood estimator of $\theta$ from a sample

Question

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:

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

Answer 1

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 $\min\limits_{i=1}^{n} X_i$.

Edit:

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

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.