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A dominated posterior on a nonstandard-Borel observation space

Example

Let the data set be a disjoint union $Y = U ⊔ V ⊔ W$ of three uncountable sets, with sigma-algebra consisting of all unions of these three atoms and reference probability $ν (U) = ν (V) = ν (W) = 1 / 3$ . This measurable space is not standard Borel: distinct points of $U$ cannot be separated by a measurable set, whereas every standard-Borel space separates distinct points. Let $Θ = {0, 1}$ have prior probabilities $1 / 2, 1 / 2$ . With respect to $ν$ , use sampling densities $f_{0} (y) = \frac{3}{2} 1_{U \cup V} (y)$ and $f_{1} (y) = 31_{U} (y)$ . The dominated formula gives posterior mass $2 / 3$ on $U$ , $0$ on $V$ , and the prior filler $1 / 2$ on the zero-evidence atom $W$ .

Facts & Assumptions

Given: The measurable data space, reference probability $ν$ , finite parameter space, prior, and sampling densities specified in the example.

[F1]

The proper model is a prior probability together with a parameter-measurable sampling probability kernel. Bayesian model as a prior and sampling kernel

[F2]

The posterior is a conditional law; a data-indexed posterior version must satisfy the rectangle identity. Bayesian posterior as a regular conditional law

[F3]

A proper prior and a sampling kernel with a specified jointly measurable density against a sigma-finite data measure have the displayed posterior kernel on fibres where the marginal density is finite and positive. The formula fills zero or infinite marginal-density fibres with the prior and does not require a standard-Borel data space. Bayes formula for dominated kernels

[F4]

A standard-Borel space is measurably isomorphic to a Polish space. Standard Borel spaces

[F5]

In a Polish space the topology is induced by a metric, so singleton sets are closed and Borel. Polish spaces are separable completely metrizable spaces

[F6]

The product sigma-algebra contains measurable rectangles; on the finite parameter space, simple functions constant on such rectangles are jointly measurable. The product sigma-algebra and its finite iterates

Verification

Given: The assumptions and finite model in the example.

1.1givenF4F5construct

The eight unions of $U, V, W$ form a sigma-algebra. Assigning mass $1 / 3$ to each disjoint atom defines a probability measure $ν$ : a pairwise disjoint family has at most three nonempty members, so countable additivity reduces to finite additivity on the atoms. Distinct $u, u^{'} \in U$ have the same membership in every measurable set. If $(Y, Y)$ were standard Borel, [F4] would give a bimeasurable bijection $h$ to a Polish space. Let $d$ be a compatible metric. For every $z \neq h (u)$ , the ball about $z$ of radius $d (z, h (u)) / 2$ misses $h (u)$ , so ${h (u)}$ is closed and Borel. Its inverse image under $h$ would be a measurable set separating $u$ and $u^{'}$ , a contradiction.

2.1givenF1F3F6step 1.1algebra

The joint density is $ℓ (θ, y) = \frac{3}{2} 1_{{0} \times (U \cup V)} (θ, y) + 31_{{1} \times U} (θ, y)$ , a finite sum of indicators of measurable rectangles by [F6]. Its parameter sections are $f_{0}, f_{1}$ , with integrals $\int f_{0} d ν = \frac{3}{2} \cdot \frac{2}{3} = 1$ and $\int f_{1} d ν = 3 \cdot \frac{1}{3} = 1$ . Thus the sections define sampling probabilities, and the finite reference measure $ν$ is sigma-finite, so [F3] applies.

3.1F3step 2.1algebra

The marginal density is $m (y) = \frac{1}{2} f_{0} (y) + \frac{1}{2} f_{1} (y)$ , hence $m = 9 / 4$ on $U$ , $m = 3 / 4$ on $V$ , and $m = 0$ on $W$ . Thus $J_{Y} (U) = 3 / 4$ , $J_{Y} (V) = 1 / 4$ , and $J_{Y} (W) = 0$ . The normalized formula gives $Π_{y} (1) = 2 / 3$ on $U$ and $0$ on $V$ ; on $W$ the theorem fills the zero-evidence fibre with the prior, so $Π_{y} (1) = 1 / 2$ .

4.1F2step 3.1algebra∎

For $A = {1}$ , the rectangle identities on $U, V, W$ are $(2 / 3) (3 / 4) = 1 / 2 = J ({1} \times U)$ , $0 (1 / 4) = 0 = J ({1} \times V)$ , and $(1 / 2) 0 = 0 = J ({1} \times W)$ . For $A = {0}$ they are $(1 / 3) (3 / 4) = 1 / 4 = J ({0} \times U)$ , $1 (1 / 4) = 1 / 4 = J ({0} \times V)$ , and $(1 / 2) 0 = 0 = J ({0} \times W)$ . Adding over atoms gives the identity for every $A \subseteq Θ$ and $B \in Y$ , so this is a data-indexed posterior kernel despite the nonstandard-Borel data space.

Depends on

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Sources

Alicia A. Johnson, Miles Q. Ott, and Mine Dogucu, Bayes Rules!, Chapter 2 §§2.1.1–2.1.4 (standard reference, not scraped)
Andrew Gelman, John B. Carlin, Hal S. Stern, David B. Dunson, Aki Vehtari, and Donald B. Rubin, Bayesian Data Analysis, 3rd ed., Chapter 1 §1.3 (standard reference, not scraped)