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I'm trying to work through an example from Richard Dudas Pattern Classification on Recursive Bayes Learning. My main question is why do we choose the $max[D^n] $ in: $$max[D^n] \le \theta \le 10 $$

In this example, we look at data points $ \{4, 7, 2, 8\} $ where after each iteration the LHS of the inequality is the max value of all observed data points. Conceptually, I'm just not 100% on why this it converges to maximum, instead of say, the arithmetic mean? Any explanation or intuition for this would be appreciated!

I have included pictures (from the book) of the example.

Posterior Density Recursion Relation

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  • $\begingroup$ The data come from a uniform$[0, \theta]$ distribution (likelihood) where we assume $\theta$ is between 0 and 10 (prior). So $\theta$ is the upper bound for the data that we expect to see. After observing some data points, we know $\theta$ has to be at least as large as the largest observed point, otherwise it wouldn't be an upper bound for the data anymore. $\endgroup$ – Maurits M Jan 14 at 9:07
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This is standard Bayesian analysis, where

posterior $\propto$ prior x likelihood

In this case the data is made of iid observations from a $\mathcal{U}(0,\theta)$, hence with density$$\frac{1}{\theta}\mathbb{I}_{0\le x\le \theta}$$

The prior $\pi(\theta)=\mathbb{I}_{\theta\le 10}$ thus gets updated as \begin{align} \pi(\theta|x_1) &\propto \mathbb{I}_{\theta\le 10} \times \theta^{-1}\mathbb{I}_{x_1\le \theta}=\theta^{-1}\mathbb{I}_{x_1\le\theta\le 10}\\ \pi(\theta|x_1,x_2) &\propto \theta^{-1}\mathbb{I}_{x_1\le\theta\le 10} \times \theta^{-1}\mathbb{I}_{x_2\le \theta}=\theta^{-2}\mathbb{I}_{\max\{x_1,x_2\}\le\theta\le 10}\\ \pi(\theta|x_1,x_2,x_3) &\propto \theta^{-2}\mathbb{I}_{\max\{x_1,x_2\}\le\theta\le 10} \times \theta^{-1}\mathbb{I}_{x_3\le \theta}=\theta^{-3}\mathbb{I}_{\max\{x_1,x_2.x_3\}\le\theta\le 10}\\ \end{align} and so on. (For the Uniform distribution, $\max\{x_i,i=1,...,n\}$ is a sufficient statistic.)

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