How to find the sufficient statistics for the shifted exponential distribution $f_{\theta, k}(y) = \theta e^{-\theta (y - k)}, y \ge k, \theta \gt 0$?

Question

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?

Answer 1

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)}\}}. $

Answer 2

\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.