MLE of $\delta$ for the distribution $f(x)=e^{\delta-x}$ for $x\geq\delta$.

Let $X_1,X_2,\dots, X_n$ be a random sample form a distribution $f(x)=e^{\delta-x}$ for $x\geq\delta$. Find MLE of $\delta$.

My solution:

$$L(\hat\gamma)=\prod_{i=1}^nf(x_i\mid \delta)=\prod_{i=1}^ne^{\delta-x} =e^{n\delta}e^{\sum_{i=1}^nx_i}$$

The log likelihood is $l(\hat\delta)=n\delta-\sum_{i=1}^n x_i$. To find the $\max$ we take the 1st derivative with respect to $\delta$: $$l'(\hat\delta)=n$$ the first derivative cannot be zero. I am not sure how to proceed.

Solution 1:

The Likelihood function is given by $$ \mathcal{L}(\delta)=\prod_{i=1}^n e^{\delta - X_i}I\{\delta\le X_i\}=e^{n\delta}e^{-X_1-X_2-\cdots-X_n}I\{\delta\le X_{(1)}\}, $$ where $$X_{(1)}=\inf\{X_1,X_2,\ldots,X_n\}.$$ Let us take a look at the parametric space that is given by $\Delta = (-\infty, X_{(1)}]$. Note that $\mathcal{L}(\delta)$ is a monotonic increasing function w.r.t. $\delta$, so the maximum over $\Delta$ is a boundary solution, i.e., $$\hat{\delta} = X_{(1)}.$$