# Minimal sufficient statistic for location exponential family

Let $$X_1,\dots,X_n$$ iid with pdf $$f(x|\theta)=e^{-(x-\theta)},\,\,\,\theta Part (b) of Problem 6.9 in Casella and Berger asks to find a minimal sufficient statistic for $$\theta$$.

Theorem 6.2.13 in the book states:

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

Now by independence of the sample we have $$f(\textbf{x}|\theta)=e^{-\sum_i (x_i-\theta)}$$. Thus $$\frac{f(\textbf{x}|\theta)}{f(\textbf{y}|\theta)}=e^{-\sum_i (x_i-\theta)+\sum_i (y_i-\theta)}=e^{\sum_i y_i-\sum_i x_i}$$ which is always constant in $$\theta$$. This would mean that the zero function is a minimal sufficient statistic for $$\theta$$ which intuitively means it is impossible to say anything about $$\theta$$ based on the sample, which doesn't make sense. Is there a mistake in my work?

• What happens to the indicators? Jan 17, 2019 at 19:29
• @Xi'an what do you mean by indicator?
– Seth
Jan 17, 2019 at 19:32
• Oh I see, the indicator function of $x > \theta$. I will take that into account and see if I can fix my calculation.
– Seth
Jan 17, 2019 at 19:45
• Ah I see, so there is still dependence on $\theta$ since the function is really piecewise defined, that makes sense. I knew I must have been doing something silly.
– Seth
Jan 17, 2019 at 19:48

I think I'm late trying to write it properly, but here it is: $$\frac{f(\textbf{x}|\theta)}{f(\textbf{y}|\theta)}=\frac{e^{-\sum_i (x_i-\theta)}\prod I(x_i>\theta)}{e^{-\sum_i (y_i-\theta)}\prod I(y_i>\theta)}=\frac{e^{-\sum_i x_i}I(\min(\boldsymbol{x})>\theta)}{e^{-\sum_i y_i}I(\min(\boldsymbol{y})>\theta)}$$.
And, this is independent of $$\theta$$ when $$\min(\boldsymbol{x})=\min(\boldsymbol{y})$$, so $$T(\boldsymbol{x})=\min(\boldsymbol{x})$$.