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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...

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2 Answers 2

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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 $f_3$

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  • $\begingroup$ $\frac{L(1|1)}{L(2 | 1)} = 0.6, \frac{L(1|2)}{L(2 | 2)} = 0.6, \frac{L(1|3)}{L(2 | 3)} = 4.6, \frac{L(1|4)}{L(2 | 4)} = 0.6$ 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 =/ $\endgroup$
    – shah
    Feb 28, 2019 at 0:49
  • $\begingroup$ @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) $\endgroup$
    – Henry
    Feb 28, 2019 at 14:30
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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:

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

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.

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