Let $X_1, X_2, \dots , X_n$ be a random sample from the distribution with with pdf: $$ f(x;\theta) = e^{\theta-x} I_{[\theta, +\infty)}, \, \theta \in \mathbb{R} $$ For the sample we have: $$ f(x_1, \dots , x_n;\theta) = e^{n\theta} \cdot e^{-\sum_{i=1}^{n} x_i} \cdot I(\min x_i \geq\theta) $$ Thus, from the factorization theorem, $T = \min x_i$ is sufficient for $\theta$.
To find the distribution of $T$, we use the argument that the probability of all $x_i$'s being greater than $x$ is $[1-F(x)]^n$ and thus the CDF of $T$ is $1-[1-F(x)]^n$.
However, what if the random variable involved the parameter $\theta$? Could you give me a hint on how to find the distribution of $Y = T - \theta$?