Let $X_1, ..., X_n$ be iid $f(x; \theta, \lambda) = \dfrac{\lambda e^{-\lambda x}}{1-e^{-\lambda \theta}}$ for x $\in [0, \theta]$.
I want to find
- A minimally sufficient statistic for $\theta$ when $\lambda$ is known and
- A minimally sufficient statistic for $\lambda$ when $\theta$ is known.
My Work
I have
$$\dfrac{L(x; \theta, \lambda)}{L(y; \theta, \lambda)} = \dfrac{\dfrac{\lambda^n e^{-\lambda \sum x} }{(1-e^{-\theta \lambda})^n} }{ ~~ \dfrac{\lambda^n e^{-\lambda \sum y} }{(1-e^{-\theta \lambda})^n} ~~}$$ for $x \in [0, \theta]$ and $y \in [0, \theta]$.
But I am unsure what to do from here because the ratio is constant whenever $\sum x_i = \sum y_i$. So does that mean that $\sum x_i$ is a minimally sufficient statistic for both?
The likelihood for $\theta$ given $\lambda$ and a sample $\boldsymbol x = (x_1, \ldots, x_n)$ is $$\mathcal L (\theta \mid \lambda, \boldsymbol x) \propto \lambda^n e^{-\lambda n\bar x} (1 - e^{-\lambda \theta})^{-n} \mathbb 1 (x_{(1)} \ge 0) \mathbb 1 (x_{(n)} \le \theta)$$ where $\bar x$ is the sample mean, $x_{(1)}$ is the first (minimum) order statistic, and $x_{(n)}$ is the last (maximum) order statistic. Then choose $$h(\boldsymbol x) = \lambda^n e^{-\lambda n \bar x} \mathbb 1(x_{(1)} \ge 0),$$ which is the factor that does not depend on $\theta$; choose $$g(T \mid \theta) = (1 - e^{-\lambda \theta})^{-n} \mathbb 1(T \le \theta),$$ which is the factor containing all instances of $\theta$; and choose $$T(\boldsymbol x) = x_{(n)}$$ which by the factorization theorem is sufficient for $\theta$ when $\lambda$ is known. That it is minimal is left to the reader as an exercise.
In the case where $\theta$ is known but $\lambda$ is not, then we have the same functional form of the likelihood as above, but now our choices of $h, g, T$ may be different. This is also left to the reader as an exercise.