# Minimal Sufficient Statistic for $f(x) = e^{-(x-\theta)}, \; \theta < x < \infty, \; x \in \mathbb{R}$

My question comes from Exercise 6.9(b) of Statistical Inference by Casella and Berger:

6.9: Find a minimal sufficient statistic for $$\theta$$
(b) $$f(x|\theta) = e^{-(x-\theta)}, \quad \theta < x < \infty, \quad -\infty < \theta < \infty$$.

This exercise appears to be a straighforward application of the following theorem:

Theorem 6.2.13. Let $$f(\mathbf{x}|\theta)$$ be the pmf or pdf of a sample $$\mathbf{X}$$. Suppose there exists a function $$T(\mathbf{x})$$ such that, for every two sample points $$\mathbf{x}$$ and $$\mathbf{y}$$, the ratio $$f(\mathbf{x}|\theta)/f(\mathbf{y}|\theta)$$ is constant as a function of $$\theta$$ if and only if $$T(\mathbf{x}) = T(\mathbf{y})$$. Then $$T(\mathbf{X})$$ is a minimal sufficient statistic for $$\theta$$.

A little algebra shows that $$\frac{f(\mathbf{x}|\theta)}{f(\mathbf{y}|\theta)} = \exp\left(\sum_{i=1}^{n} (y_i - x_i) \right) \frac{I_{(\theta,\infty)}(\mathbf{x}_{(1)})}{ I_{(\theta,\infty)}(\mathbf{y}_{(1)})}$$

for all $$\mathbf{x},\mathbf{y} \in (\theta,\infty)^n$$. Now it's clear that the desired implication "$$f(\mathbf{x}|\theta)/f(\mathbf{y}|\theta)$$ is constant as a function of $$\theta$$ if and only if $$T(\mathbf{x}) = T(\mathbf{y})$$" depends solely upon the above ratio of indicator functions. If $$\mathbf{x}_{(1)} = \mathbf{y}_{(1)}$$, then $$\frac{I_{(\theta,\infty)}(\mathbf{x}_{(1)})}{ I_{(\theta,\infty)}(\mathbf{y}_{(1)})} = 1$$ for all $$\theta \in (-\infty, \mathbf{y}_{(1)})$$ (and undefined on $$[\mathbf{y_{(1)}},\infty)$$), so the ratio is constant in $$\theta$$ for all $$\theta$$ where it is defined. And if $$\mathbf{x}_{(1)} < \mathbf{y}_{(1)}$$, then \begin{align*} \frac{I_{(\theta,\infty)}(\mathbf{x}_{(1)})}{ I_{(\theta,\infty)}(\mathbf{y}_{(1)})} &= \begin{cases} 1 & \text{ if } \theta \in (-\infty,\mathbf{x}_{(1)}) \\[2pt] 0 & \text{ if } \theta \in [\mathbf{x}_{(1)},\mathbf{y}_{(1)}) \end{cases} \end{align*} (and is undefined for $$\theta \geq \mathbf{y}_{(1)}$$), so in this case the ratio clearly depends on $$\theta$$. But if $$\mathbf{y}_{(1)} < \mathbf{x}_{(1)}$$, then $$\frac{I_{(\theta,\infty)}(\mathbf{x}_{(1)})}{ I_{(\theta,\infty)}(\mathbf{y}_{(1)})} = 1$$ for all $$\theta < \mathbf{y}_{(1)}$$ (and undefined everywhere else). If the ratio is constant as a function of $$\theta$$ if and only if $$\mathbf{x}_{(1)} = \mathbf{y}_{(1)}$$, then we can straightforwardly conclude that $$T(\mathbf{X}_{(1)})$$ is a minimal sufficient statistic. But in the case where $$\mathbf{y}_{(1)} < \mathbf{x}_{(1)}$$, is the ratio of indicator functions considered to be constant as a function of $$\theta$$? Why or why not? Any feedback would be appreciated.

Edit 2/23/22: As @Henry mentioned in the comments, the situation is clarified if we change the part of the theorem that says

the ratio $$f(\mathbf{x}|\theta)/f(\mathbf{y}|\theta)$$ is constant as a function of $$\theta$$

to the new statement

there exists some $$k(\mathbf{x},\mathbf{y}) > 0$$ (a strictly positive function which does not depend on $$\theta$$) such that $$f(\mathbf{x}∣\theta)=k(\mathbf{x},\mathbf{y})f(\mathbf{y}∣\theta)$$ for all $$\mathbf{x}, \mathbf{y}$$ in the sample space and all $$\theta \in \Theta$$.

I would just like some "official" confirmation of this in the form of a theorem in a textbook. Any relevant references would be greatly appreciated.

• On the right track. Delayed exponential dist'n where $\theta$ is the delay. Minimum is sufficient, but because it always at least a bit above $\theta.,$ it is biased. So, the next step using it in practice is to see if you can multiply it by a constant (depending on sample size) to make it unbiased. Maybe google the relevant Wikipedia page. Feb 14, 2022 at 2:03
• The answer to your second question is essentially no. You have the slightly difficult case that your definition suggests division by $0$: to avoid it here you want the similar statement there is a strictly positive value $k$ such that $f(\mathbf{x}\mid \theta) = k f(\mathbf{y}\mid \theta)$ for all $\theta$. That is not the case here when $y_{(1)} < x_{(1)}$ or $y_{(1)} > x_{(1)}$ and $\theta$ is in the gap between them, since you would then need $0$ to be strictly positive Feb 14, 2022 at 3:40
• @Henry: Thanks for your comment! I guess the theorem given by Casella and Berger implicitly assumes that $f(\mathbf{x}|\theta) > 0$ for all $\mathbf{x}$. In your restatement, I take it that $k$ is allowed to be a (strictly positive) function of $\mathbf{x}$ and $\mathbf{y}$, but not $\theta$? Also, do you have a reference containing this version of the theorem? Feb 14, 2022 at 16:43

Theorem 2.3(iii) (Shao) Let $$\mathcal{P}$$ be a family of distributions on $$\mathbb{R}^k$$. Suppose that $$\mathcal{P}$$ contains p.d.f.s $$f_P$$ with respect to a $$\sigma$$-finite measure and that there exists a sufficient statistic $$T(X)$$ such that, for any possible values $$x$$ and $$y$$ of $$X$$, $$f_P(x) = f_P(y)\phi(x,y)$$ for all $$P$$ implies $$T(x)=T(y)$$, where $$\phi$$ is a measurable function. Then $$T(X)$$ is minimal sufficient for $$P\in\mathcal{P}$$.