MLE for 2 parameter exponential distribution Let $x_1, x_2... x_n$ be a random sample from a distribution with pdf:
$$f(x;\mu,\sigma)=\frac1{\sigma}\exp\left({-\frac{x-\mu}{\sigma}}\right)\,,-\infty<\mu<\infty;\, \sigma>0;\, x\ge\mu$$
How do I find the MLE for the parameters if both parameters are unknown?
I tried using the usual MLE with likelihood function:
$$L(\mu,\sigma|x_1...x_n)=\frac{1}{\sigma^n}\exp\left({-\frac{\sum{x_i}-n\mu}{\sigma}}\right)$$ But the derivative of this with respect of $\mu$ is a dead end. 
I do know that if $\sigma$ is known, the MLE for $\mu$ is $\frac{\sum{x_i}}{n}$ and if $\mu$ is known, the MLE for $\sigma$ is $\frac{\sum{x_i}-n\mu}{n}$. 
Do these help? What should be the approach?  
 A: Given the sample, the likelihood function is given by $$L(\mu,\sigma)=\frac{1}{\sigma^n}\exp\left[-\frac{1}{\sigma}\sum_{i=1}^n(x_i-\mu)\right]\mathbf1_{\mu\leqslant x_{(1)},\sigma>0}$$
This function is not differentiable at $\mu=x_{(1)}$, so that MLE of $\mu$ has to be found using a different argument. For fixed $\sigma$, $L(\mu,\sigma)$ is an increasing function of $\mu$  $\,\forall\,\sigma$, implying that $\hat\mu_{\text{MLE}}=X_{(1)}$.
MLE of $\sigma$ can be guessed from the first partial derivative as usual.
We have $\displaystyle\frac{\partial L(\mu,\sigma)}{\partial\sigma}=0\implies\sigma=\frac{1}{n}\sum_{i=1}^n(x_i-\mu)$.
So MLE of $\sigma$ could possibly be $\displaystyle\hat\sigma_{\text{MLE}}=\frac{1}{n}\sum_{i=1}^n(X_i-\hat\mu)=\frac{1}{n}\sum_{i=1}^n\left(X_i-X_{(1)}\right)$
The second partial derivative test fails here due to $L(\mu,\sigma)$ not being totally differentiable.
So to confirm that $(\hat\mu,\hat\sigma)$ is the MLE of $(\mu,\sigma)$, one has to verify that $L(\hat\mu,\hat\sigma)\geqslant L(\mu,\sigma)$, or somehow conclude that $\ln L(\hat\mu,\hat\sigma)\geqslant \ln L(\mu,\sigma)$ holds $\forall\,(\mu,\sigma)$.
A: Note that for each fixed $\sigma > 0$, the likelihood $L(\mu,\sigma)$ is an increasing function of $\mu$, provided that $\mu\leq x_{(1)}$ ($x_{(1)}$ being the smallest value of $x$). If $\mu> x_{(1)}$, the likelihood is $0$. Consequently, the MLE of $\mu$ is $\hat{\mu}=x_{(1)}$. Here is a plot of the log-likelihood for a specific example als @Glen_b suggested in the comments ($\sigma = 1, x = \{1.13, 1.56, 2.08\}$):

As for the MLE of $\sigma$, take  the first derivative of the log-likelihood, set it to zero and solve for $\sigma$
\begin{align}
\log L(x;\mu,\sigma) &=-n\log{(\sigma)}-\frac{1}{\sigma}\sum_{i=1}^{n}{(x_i-\mu)} \\
\frac{\partial \log L(x;\hat{\mu},\sigma)}{\partial \sigma}&= \frac{-n}{\sigma}+\frac{1}{\sigma^2}\sum_{i=1}^{n}{(x_i-\hat{\mu})}\\
\frac{-n}{\sigma}+\frac{1}{\sigma^2}\sum_{i=1}^{n}{(x_i-\hat{\mu})} &= 0\\
\hat{\sigma} &=\frac{1}{n}\sum_{i=1}^{n}(x_i-x_{(1)}) = \bar{x}-x_{(1)} 
\end{align}
where $\bar{x}$ is the sample mean.
These results can be found in the following references.


*

*Rahman M & Pearson LM (2001): Estimation in two-parameter exponential distributions. Journal of Statistical Computation and Simulation, 70(4), 371-386.

*Krishnamoorthy K (2016): Handbook of Statistical Distributions with Applications. 2nd ed. Chapman and Hall/CRC.

