I want to ensure my reasoning is correct.
The sample is distributed with the pdf: $$f_a(x) = e^{-(x-a)}\cdot \chi_{[a, +\infty)}(x)$$ where $a>0$. Find the maximum likelihood estimator for $a$.
The likelihood function would be: $$L(a, x_1, ..., x_n) = e^{-x_1+a}\cdot \chi_{[a, +\infty)}(x_1) \cdot \dots \cdot e^{-x_n+a}\cdot \chi_{[a, +\infty)}(x_n)$$
$$=e^{na-\sum\limits_{i=1}^n x_i} \prod\limits_{i=1}^n \chi_{[a, +\infty)}(x_i)$$
$$= \begin{cases} e^{na-\sum\limits_{i=1}^n x_i} & x_1, \dots, x_n \in [a, +\infty) \\ 0, & \text{otherwise} \end{cases}$$
$$= \begin{cases} e^{na-\sum\limits_{i=1}^n x_i} & \forall i \in \{1, \dots, n\}: x_i \geq a \\ 0, & \exists j \in \{1, \dots, n\}: x_j < a \end{cases}$$
which, if plotted, would probably look like this:
Since all $x_i$ have to be greater than $a$ and the maximum value is reached at $a$, my guess is that the MLE would be:
$$\hat{a}_n = \min{(x_1, \dots, x_n)}$$
Is this correct?
What you graphed is the density function, and that for only a single observation rather than a joint density for $n$ observations, not the likelihood function. You have $$ L(a) = \left.\begin{cases} e^{n(a-\bar x)} & \text{if for } i = 1, \ldots,n,\ a\le x_i \\ 0 & \text{otherwise} \end{cases}\right\} = \begin{cases} e^{n(a-\bar x)} & \text{if } a \le \min\{x_1,\ldots,x_n\}, \\ 0 & \text{otherwise.} \end{cases} $$ Thus $L(a)$ increases as $a$ increases, until $a$ gets as big as $\min\{x_1,\ldots,x_n\}.$
Note that $\min\{x_1,\ldots,x_n\}$ is a point on the $a$-axis.