$$f(x|\theta) = \begin{cases}
\frac{1}{\theta} \exp \left\{-\frac{(x-\mu)}{\theta} \right\} & \text{if } \theta > 0 \text{ and } x > \mu, \\
0 & \text{otherwise;}
\end{cases}$$
Given: A distribution belongs to the regular 1-parameter exponential family if among other regularity conditions its probability density or mass function has the form:
$$f(x|\beta) = g(x) \exp \{\beta t(x) - \psi(\beta)\}, \beta \in \Omega \subset (-\infty, \infty).$$
Furthermore, for this distribution $E[t(X)] = \psi'(\beta)$ and $Var[t(X)] = \psi''(\beta)$.
(a) Show that $f(x|\theta)$ belongs to the 1-parameter exponential family by expressing it in the form $f(x|\beta)$ above. Do not forget to identify $\beta$, $t(x)$ and $\psi(\beta)$.
(b) What is the complete sufficient statistic for $\theta$? Justify your answer.
(c) Find the mean and the variance of the complete sufficient statistic for $\theta$.
