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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?

Oskar
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    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? – Doctor Milt Mar 22 '24 at 17:29
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    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). – Doctor Milt Mar 22 '24 at 17:31
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    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. – Doctor Milt Mar 22 '24 at 18:48
  • Yeah, you might say $|x_i| < 2 \theta$, correct? What do you mean with $i$ in the denominator? – Oskar Mar 23 '24 at 14:53
  • The product of the $2 i \theta$ terms in the denominator in the likelihood is not $(2i\theta)^n$. – Doctor Milt Mar 23 '24 at 18:30
  • Taking the absolute values of $x_i$ is the right idea. Can you see how to modify the indicator functions so they correspond to the set $(-\theta,\theta)$ for all values of $i$? – Doctor Milt Mar 23 '24 at 21:02

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