finding the maximum likelihood estimator of location scale family of exponential distributions

Question

Let $X_1,\dots,X_n$ be i.i.d. observations from a location-scale family of exponential distributions with the pdf $$f_X(x\mid \mu,\sigma)=\frac{1}{\sigma}\exp\left(-\frac{x-\mu}{\sigma}\right), \quad x\geq \mu \text{ and } \sigma>0$$ How do I find the maximum likelihood estimator of this PDf?

Answer 1

$$ L(\mu,\sigma) = \frac 1 {\sigma^n}\exp\left(-\frac{\sum_{i=1}^n(x_i-\mu)}\sigma\right), $$ so $$ \ell(\mu,\sigma) = \log L(\mu,\sigma) = -n\log\sigma-\frac 1 \sigma\sum_{i=1}^n (x_i-\mu). $$ Notice that $\ell$ gets bigger as $\mu$ gets bigger, but $\mu$ cannot get bigger than the smallest $x$, so $\widehat\mu=\min\{x_1,\ldots,x_n\}$. Since this does not depend on $\sigma$, you just need to put it in place of $\mu$ in the expression that says what $\ell$ is, and then find the value of $\sigma$ that maximizes that.