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

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

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?

• That's not quite right, because the indicator functions correspond to different sets. You want $x_1 \in (-\theta, \theta)$, $x_2 \in (-2\theta, 2\theta)$, etc. Can you see how to make all of them indicators of the same set? Commented Mar 22 at 17:29
• Also, the product of the $i$ terms in the denominator of your likelihood is incorrect, but this doesn't particularly matter (it's a multiplicative constant). Commented Mar 22 at 17:31
• On your side note: that's correct. The sufficient statistic with the lowest possible dimension is said to be minimal sufficient. In this particular example, you can find a one-dimensional sufficient statistic if you look a bit harder. Commented Mar 22 at 18:48
• Yeah, you might say $|x_i| < 2 \theta$, correct? What do you mean with $i$ in the denominator? Commented Mar 23 at 14:53
• The product of the $2 i \theta$ terms in the denominator in the likelihood is not $(2i\theta)^n$. Commented Mar 23 at 18:30

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?