How to find the sufficient statistics for the shifted exponential distribution $f_{\theta, k}(y) = \theta e^{-\theta (y - k)}, y \ge k, \theta \gt 0$? How to find the sufficient statistics for the shifted exponential distribution $f_{\theta, k}(y) = \theta e^{-\theta (y - k)}, y \ge k, \theta\gt 0$?
If
a) $k$ is known
b) $k$ is unknown
I believe we can use factorization theorem here.
So the likelihood
$$L(\theta, k; y) = \prod_{i = 1} \theta e^{-\theta (y_i - k)} = \mathbb{1}_{\min\{Y\} \ge k} \theta^n e^{-\theta \sum_{i=1}^n(y_i - k)} = \mathbb{1}_{\min\{Y\} \ge k} \theta^n e^{-\theta \sum_{i = 1}^n y_i + n \theta k}.$$
So here we have $\mathbb{1}_{\min{Y}\ge k} \theta^n e^{-\theta \sum_{i = 1}^n y_i + n \theta k}$ as $g(T(y), (\theta, k))$ and $h(y) = 1$. But then I am confused what's the difference between we know $k$ and we do not know $k$. Could someone please explain?
 A: 1) If $k$ is known, thus the only unknown parameter is the decay rate, i.e., $\theta$, hence
$$
\mathcal{L}(\theta; x_1,..x_n, k) = \theta^n \exp\{-\theta\sum(x_i - k)\}\prod_{i=1}^nI_{\{x_i\ge k\}} = g\left(\theta;\sum(x_i-k)\right)h(X),
$$
where $h(X) = \prod_{i=1}^nI_{\{x_i\ge k\}}$. 
2) If the $k$ is also unknown, then you should have a sufficient statistic whose dimension is at least $2$, namely,
$$
\mathcal{L}(\theta, k; x_1,..x_n) = \theta^n \exp\{-\theta\sum(x_i - k)\} I_{\{x_i\ge k\}}\prod_{i=2}^nI_{\{x_i\ge x_{(1)}\}} = g\left((\theta, k);(\sum x_i, \min\{x_1,...,x_n\} \right)h(X),
$$
where in this time 
$
h(X) = \prod_{i=2}^nI_{\{x_i\ge x_{(1)}\}}.
$
A: \begin{align}
& f(y_1,\ldots,y_n) \\[6pt]
= {} & \prod_{i=1}^n e^{-\theta(y_i-k)} 1_{\min\{y_1,\ldots,y_n\}\ge k} \\[6pt]
= {} & \exp\left( -\theta\sum_{i=1}^n (y_i-k)\right) 1_{\min\{y_1,\ldots,y_n\}\ge k}.
\end{align}
The factor $1_{\min}$ does not depend on $\theta.$ The other factor, the exponential function, depends on $y_1,\ldots,y_n$ only through the given sum.
Therefore the sum $\sum_{i=1}^n (y_i-k)$ is sufficient if $k$ is known.
If $k$ is unkown, then we can write
$$
\sum_{i=1}^n (y_i-k) = \sum_{i=1}^n ((y_i-\min)+(\min - k)) = \left( \sum_{i=1}^n (y_i - \min) \right) + n(\min-k).
$$
This then depends on $y_1,\ldots,y_n$ only through the pair
$$
\left( \sum_{i=1}^n (y_i-\min), \min \right).
$$
Therefore that pair is a sufficient statistic.
