# Find the sufficient statistic ($T$) that makes a reduction in data.

Suppose that a statistical model is comprised of three distributions $$\{f_{\theta}, \theta \in \{1,2,3\}\}$$ given by the following table

$$\begin{array}{|c|c|c|c|c|} \hline & s = 1 & s= 2 & s = 3 & s = 4 \\ \hline f_{1}(s) & 0.12 & 0.06 & 0.46& 0.36\\ \hline f_{2}(s)& 0.20& 0.10 & 0.10& 0.60\\ \hline f_{3}(s) & 0.16 & 0.08 & 0.28 & 0.48\\ \hline \end{array}$$

(a) Find the sufficient statistic ($$T$$) that makes a reduction in data.

Not sure.

I know how to do it if there was only $$f_1$$ and $$f_2$$ but not with the third.

I'd show

$$L(1 | s = 1) / L(2 | s = 1) = L(1| s = 2) / L(2 | s = 2)$$

not sure...

As a hint,

your $$\dfrac{L(1 \mid s = 1)}{L(2 \mid s = 1)} = \dfrac{L(1 \mid s = 2)}{L(2 \mid s = 2)}$$ is equivalent to $$\dfrac{L(1 \mid s = 1)}{L(1 \mid s = 2)} = \dfrac{L(2 \mid s = 1)}{L(2 \mid s = 2)}$$ and it seems that you can in fact say here that $$\dfrac{L(1 \mid s = 1)}{L(1 \mid s = 2)} = \dfrac{L(2 \mid s = 1)}{L(2 \mid s = 2)} = \dfrac{L(3 \mid s = 1)}{L(3 \mid s = 2)}$$

so what you were thinking about doing if there were only $$f_1$$ and $$f_2$$ may work with $$f_1$$, $$f_2$$ and $$f_3$$

• $\frac{L(1|1)}{L(2 | 1)} = 0.6, \frac{L(1|2)}{L(2 | 2)} = 0.6, \frac{L(1|3)}{L(2 | 3)} = 4.6, \frac{L(1|4)}{L(2 | 4)} = 0.6$ and so a statistic $T : S \to \{1,2\}$ given by $T(1) = T(2) = T(3) = 1, T(3) = 2$ is a sufficient statistic. Would that be right? I'm confused because I didn't use the third row at all =/
– shah
Feb 28, 2019 at 0:49
• @shah Your final answer looks correct but you should also note $\frac{L(1|1)}{L(3 | 1)} = \frac{L(1|2)}{L(3 | 2)} = \frac{L(1|4)}{L(3 | 4)} = 0.75$ and $\frac{L(2|1)}{L(3 | 1)} = \frac{L(2|2)}{L(3 | 2)} = \frac{L(2|4)}{L(3 | 4)} = 1.25$. A simpler but equivalent sufficient statistic would be to count the number of $3$s (and the total sample size) Feb 28, 2019 at 14:30

I will generalise your problem slightly and suppose we have categorical data with $$m$$ categories. (In your problem we have $$m=4$$.) Suppose you observe $$n$$ IID values from this distribution and let $$n_i$$ be the count of the number of observations with $$s=i$$ (so you have $$n_1 + ... + n_m = n$$). Define the count vector $$\mathbb{n} = (n_1,...,n_m)$$ and write the likelihood function for the data as:

\begin{aligned} L_\mathbb{s}(\theta) &= \prod_{i=1}^n f_\theta(s_i) \\[6pt] &= \prod_{k=1}^m f_\theta(k)^{n_k} \\[6pt] &= g_\theta(\mathbb{n}). \\[6pt] \end{aligned}

Using the Fisher-Neyman factorisation theorem we see that the count vector $$\mathbb{n}$$ is a sufficient statistic for $$\theta$$. So, if you only intend to make inferences about $$\theta$$ and are happy to discard other information, you can reduce your data by just keeping the total counts over the four categories.