Sufficient statistic for $\theta$ from $f_Y(y) = e^{−(y − \theta)}$?

Question

Let $Y_1,\ldots,Y_n$ be an i.i.d. sample from $f_Y(y) = e^{−(y − \theta)}$, where $y \ge \theta$.

(a) Write down the likelihood for the $n$ samples from this distribution including an indicator function for the range of $y$. Simplify as much as possible.
Use this likelihood to show that $Y_{(1)} = \min(Y_1, \dots, Y_n)$ is a sufficient statistic for $\theta$. Specify the functions $g(u, \theta)$ and $h(y)$ you use in the factorization criterion.

(b) Use your sufficient statistic $Y_{(1)}$ to find a MVUE for $\theta$.

So, I figured that $L(y_1, \dots, y_n \mid \theta)=f(y_1, \dots, y_n \mid \theta) = f(y_1 \mid \theta) \dots f(y_n \mid \theta)$ given the fact they're all i.i.d.. From there, I get that $e^{-y_1 + \theta} \dots e^{-y_n + \theta} \mathbb{1}_{\{y_1, \dots, y_n \geq \theta\}}$, but I don't know where to go from there.

I suppose I could simplify to $e^{n\theta} \mathbb{1}_{\{Y_{(1)} \geq \theta\}} e^{-\sum\limits_{i = 1}^{n} y_i}$, but how do I get a sufficient statistic from that? I'm very fuzzy on the Rao-Blackwell Theorem and MVUE. Any ideas?

Answer 1

Presumably you are trying to factor the likelihood into something like

$$f_\theta(y)=h(y) \, g(u(y),\theta)$$ where $u(y)$ is the sufficient statistic (I would use a slightly different notation so may have made an error in understanding yours).

In that case you need the $e^{n\theta}\mathbb{1}_{\{Y_{(1)}\geq\theta\}}$ in the $g$ as they involve $\theta$, but not the $e^{-\sum\limits_{i=1}^{n}y_i}$. So you can choose

• $h(y)=e^{-\sum\limits_{i=1}^{n}y_i}$
• $g(u(y),\theta) = e^{n\theta}\mathbb{1}_{\{Y_{(1)}\geq\theta\}}$
• $u(y) = y_{(1)}$ since it is part of the $g$ and has all the useful information from $y$ for $g$ but does not involve $\theta$