MLE of a distribution with two parameters

Question

If I have $$f(x,\theta_1,\theta_2)=\frac{1}{\theta_2} \exp\left(-\frac{x-\theta_1}{\theta_2}\right),\quad x\geq \theta_1 $$ with $\theta_1\in\mathbb{R}$ and $\theta_2 >0$. And I;m trying to find the MLE's of $\theta_1$ and $\theta_2$. I tried the usual method of finding the log likelihood and maximising that but I end up with $\frac n{\theta_2}=0$ which clearly doesnt work.

Answer 1

For $\theta_1 \le \min_i x_i$, $$\log f(x_1,\ldots,x_n) = -n\log \theta_2 + n \frac{\theta_1}{\theta_2} - \frac{1}{\theta_2}\sum_{i=1}^n x_i,$$ otherwise the log likelihood is $-\infty$.

For $\theta_1$, we can only consider $\theta_1 \in (-\infty, \min_i x_i]$, and it is clear from the expression of the log likelihood that $\theta_1 = \min_i x_i$ is the MLE since it maximizes the only term in the log likelihood involving $\theta_1$.

Taking the derivative with respect to $\theta_2$ yields $$-\frac{n}{\theta_2} - \frac{n \theta_1}{\theta_2^2} + \frac{1}{\theta_2^2} \sum_{i=1}^n x_i,$$ which leads to $\theta_1 + \theta_2 = \frac{1}{n} \sum_{i=1}^n x_i$, so the MLE is $\theta_2 = \bar{x} - \min_i x_i$.