# Finding 2-dimensional sufficient statistic via Fisher-Neyman factorization when marginal pdf functions for x don't contain x

Let $$X_1,...,X_n$$ be mutually independent with pdfs given by $$f_i(X_i\mid\theta) = 1/(2i\theta)$$ where $$-i(\theta - 1) and $$\theta>0.$$ To find a two-dimensional sufficient statistic, I'm going to use Fisher-Neyman factorization.

The joint pdf of $$X$$ can be written as $$f(x|\theta)=1/((2\theta)^n\sum_i^n i) = (1- \bar x + \bar x)/((2\theta)^n\sum_i^n i)=(1- \bar x)/((2\theta)^n\sum_i^n i) + \bar x/((2\theta)^n\sum_i^n i)$$ over the domain of $$X$$. I need to factor $$f(x|\theta)$$ as $$f(x|\theta)=h(x)g(T(x)|\theta)$$, so letting $$h(x)=1$$, is $$T=(T_1, T_2)=(\bar x,\sum_i^n i)$$ a sufficient statistic? My concern is that $$T_2$$ doesn't involve $$X$$.

• Every distribution function depends on its argument, period: for otherwise you would be describing a distribution comprising either zero or infinite probability. Look more closely at your results and make sure to track the information about the set of values on which the distribution is nonzero (its support).
– whuber
Jul 11, 2018 at 22:22
• stats.stackexchange.com/q/510173/119261 Feb 19, 2021 at 18:56

The support of the distribution depends on the parameter $$\theta$$. So use indicator functions for writing down the pdf correctly and hence get a sufficient statistic for $$\theta$$ using Factorization theorem.

First note that

$$-i\theta+i\max\left(\frac{x_i-i}{i},\frac{i-x_i}{i}\right)=\frac{|{x_i-i}|}{i}$$

So for $$\theta>0$$, joint density of $$X=(X_1,X_2,\ldots,X_n)$$ is

\begin{align} f_{\theta}( x)&=\prod_{i=1}^n\frac{1}{2i\theta}\mathbf1_{-i(\theta-1)\max\left(|x_1-1|,|x_2-2|/2,\ldots,|x_n-n|/n\right)} \end{align}

By the Factorization theorem, you get a sufficient statistic $$T(X)=\max_{1\le i\le n}\{|X_i-i|/i\}$$ for $$\theta$$. If you take any other statistic $$T'$$, then $$(T,T')$$ would also be sufficient for $$\theta$$ as $$T'$$ plays no further role in the amount of data reduction or condensation without losing information about $$\theta$$.

• It may be easier to note that $-\theta<\frac{x_i}{i}-1<\theta\implies |\frac{x_i}{i}-1|<\theta\implies \max\limits_{1\le i\le n}|\frac{x_i}{i}-1|<\theta$. The $T$ here is a minimal sufficient statistic. Jul 2, 2020 at 17:52