Find a two dimensional sufficient statistic for $\theta$

Question

Let $\{X_i\}_{i=1}^n$ be conditional independent given $\theta$ with distribution

$$p_{X_i | \theta} (x |\theta) = \frac{1}{2i\theta}, \ -i\theta<x<i\theta.$$

Find a two dimensional sufficient statistic for $\theta.$

Attempted solution with the factorization theorem:

$$p_{X | \theta} = \prod_{i=1}^{n}\frac{1}{2i\theta}1_{x_i > -i\theta} 1_{x_i < i\theta} = \frac{1}{(2i\theta)^n} 1_{\min(x_i) > -i\theta} 1_{\max(x_i) < i\theta}$$

So we can choose $T(x) = \big(T_1(x), T_2(x)\big) = \big(\min(x_i), \max(x_i)\big) $ and $g\big(\theta, T_1(x), T_2(x)\big) = \frac{1}{(2i\theta)^n} 1_{T_1 > -i\theta} 1_{T_2 < i\theta}$.

By the factorization theorem, $T(x)$ should be a sufficient two dimensional statistic, correct?

By a side note about the dimension of a sufficient statistic, we can always increase the dimension trivially, e.g by saying that $T_n(x) = 1$ and multiplying by it? On the other hand, we can't reduce the dimension to a certain point? In this example, is a two dimensional statistic the lowest we can go?

Answer 1

The reasoning is ok, but there are some mistakes in the execution

$$p_{X \mid \theta} = \prod_{i=1}^{n}\frac{1}{2i\theta}1_{x_i > -i\theta} 1_{x_i < i\theta} = \frac{1}{(2i\theta)^n} 1_{\min(x_i) > -i\theta} 1_{\max(x_i) < i\theta} $$

• The product isn't right (the $i$ should be eliminated and be replaced with some function of $i$, search for 'factorials' to get inspiration)

  $$ \prod_{i=1}^{n}\frac{1}{2i\theta} \neq \frac{1}{(2i\theta)^n}$$

• The indicator functions aren't transformed correctly

  $$\prod_{i=1}^{n} 1_{x_i > -i\theta} 1_{x_i < i\theta} \neq 1_{\min(x_i) > -i\theta} 1_{\max(x_i) < i\theta} ;$$

  this $i$ should be brought inside the $\min$ and $\max$ functions. Expressions like $\max(x_i) < i\theta$ make no sense. On the left side $\max(x_i)$ turns $x_1,x_2,x_3, \dots, x_n$ into a single number. On the right side you still have this $i$, but what is its value?