Dirichlet-process priors and finite-partition updates — Examples

The nonempty finite discrete space $\mathsf{X}$ is standard Borel by [F4]. Define $\alpha =2{\delta }_a+{\delta }_b+{\delta }_c$, equivalently $\alpha (B)={\mathbf{21}}_B(a)+{\mathbf{1}}_B(b)+{\mathbf{1}}_B(c)$. By [F4] it is a measure with total mass $4$, hence finite and nonzero, so [F1] and [F2] apply. The prior vector $(G(\{a\}),G(\{b\}),G(\{c\}))$ has law $\mathrm{Dirichlet}(2,1,1)$ by [F1]. The observed counts are $(1,1,0)$, so [F2] gives posterior $\mathrm{Dirichlet}(3,2,1)$, whose parameters sum to $6$.

Applying [F3] after the two observations gives next-draw probabilities $3\mathrm{/}6=1\mathrm{/}2$ for $a$, $2\mathrm{/}6=1\mathrm{/}3$ for $b$, and $1\mathrm{/}6$ for $c$; they sum to one.

Before any draw, [F3] gives $\mathrm{Pr}(X_1=a)=2\mathrm{/}4=1\mathrm{/}2$. After observing $X_1=a$, the same rule gives $\mathrm{Pr}(X_2=a\mid X_1=a)=3\mathrm{/}5$. Therefore $\mathrm{Pr}(X_1=a,X_2=a)=(1\mathrm{/}2)(3\mathrm{/}5)=3\mathrm{/}10$: the first draw increases the next draw's mass for its cell from $2\mathrm{/}4$ to $3\mathrm{/}5$.

The nonempty two-point discrete space $\mathsf{X}$ is standard Borel by [F6], and $\alpha ={\delta }_a$ is a probability measure by [F6], hence a finite nonzero base measure of total mass $1$. Thus [F1] applies: $G(\{b\})=0$ almost surely. Since $\mathrm{Pr}(X_1=b\mid G)=G(\{b\})$, integration over the prior gives $\mathrm{Pr}(X_1=b)=0$, also agreeing with [F3].

The formula in [F2] assigns $K(b)=DP({\delta }_a+{\delta }_b)$ at the null data value. Define another kernel by $K^{\mathrm{\prime }}(a)=K(a)$ and $K^{\mathrm{\prime }}(b)=DP({\delta }_a)$. Both sections are probability measures, and $K^{\mathrm{\prime }}$ is measurable because $\{b\}$ is measurable in the discrete data space. For any measurable event $C$ in $\mathcal{P}(\mathsf{X})$, the disintegration integrals for $K$ and $K^{\mathrm{\prime }}$ differ only at $b$, which has data-marginal mass zero by step 1.1. Thus both are posterior versions; the update formula at $b$ is a version choice, not evidence-supported learning.

In the separate diffuse-base model, [F3] gives data-marginal mass $M(\{y\})=\gamma (\{y\})\mathrm{/}\gamma (\mathsf{Y})=0$. Let $K(z)=DP(\gamma +{\delta }_z)$ be the posterior kernel from [F2] and set $K^{\mathrm{\prime }}(z)=K(z)$ for $z\mathrm{\ne }y$, $K^{\mathrm{\prime }}(y)=DP(\gamma )$. By [F4], $\{y\}$ is measurable. For every measurable $C\subseteq \mathcal{P}(\mathsf{Y})$, $z\mapsto K^{\mathrm{\prime }}(z,C)={\mathbf{1}}_{\mathsf{Y}\setminus \{y\}}(z)K(z,C)+{\mathbf{1}}_{\{y\}}(z)DP(\gamma )(C)$ is measurable; each section is a probability measure, so [F5] makes $K^{\mathrm{\prime }}$ a probability kernel. The disintegration integrals for $K$ and $K^{\mathrm{\prime }}$ agree on every measurable rectangle because the kernels differ only on $\{y\}$ and $M(\{y\})=0$. Thus $K^{\mathrm{\prime }}$ is also a posterior version, proving that the displayed update value at this fixed null fiber is not uniquely determined.