Bayesian models, priors, and posterior laws — Examples

The joint masses are $J(1,0)=1\mathrm{/}12$, $J(0,0)=1\mathrm{/}2$, $J(1,1)=1\mathrm{/}4$, and $J(0,1)=1\mathrm{/}6$, while both masses at data $2$ are zero. Therefore $J_Y(0)=7\mathrm{/}12$, $J_Y(1)=5\mathrm{/}12$, and $J_Y(2)=0$.

Define ${\mathrm{\Pi }}_0(1)=1\mathrm{/}7$, ${\mathrm{\Pi }}_1(1)=3\mathrm{/}5$, and choose any probability law for ${\mathrm{\Pi }}_2$; set ${\mathrm{\Pi }}_y(0)=1-{\mathrm{\Pi }}_y(1)$. Each ${\mathrm{\Pi }}_y$ is a probability, and measurability is automatic on the finite power-set data space. The displayed ratios give ${\mathrm{\Pi }}_0(1)J_Y(0)=J(1,0)$ and ${\mathrm{\Pi }}_1(1)J_Y(1)=J(1,1)$; subtraction from the marginal gives the same identities for parameter value $0$.

For any $A\subseteq \mathrm{\Theta }$ and $B\subseteq Y$, summing the singleton identities in step 2.1 over $y\in B$ gives ${\int }_B{\mathrm{\Pi }}_y(A)J_Y(dy)=J(A\times B)$; the zero-mass term at $y=2$ contributes zero for every choice of ${\mathrm{\Pi }}_2$. Thus every such kernel is a data-indexed posterior version by [F2]. In particular, ${\mathrm{\Pi }}_2={\delta }_0$ and ${\tilde{\mathrm{\Pi }}}_2={\delta }_1$, with the same values at $0,1$, are distinct versions of the same posterior law.

The joint law assigns mass $3\mathrm{/}8$ to each diagonal point $(0,0),(1,1)$ and $1\mathrm{/}8$ to each off-diagonal point; hence $J_Y(0)=J_Y(1)=1\mathrm{/}2$.

For each $y$, ${\mathrm{\Pi }}_y$ has masses $3\mathrm{/}4$ at $y$ and $1\mathrm{/}4$ at $1-y$, so it is a probability, with ${\mathrm{\Pi }}_y(\{y\})J_Y(y)=3\mathrm{/}8=J(y,y)$ and ${\mathrm{\Pi }}_y(\{1-y\})J_Y(y)=1\mathrm{/}8=J(1-y,y)$. Define $L(\omega ,A)={\mathrm{\Pi }}_{Y(\omega )}(A)$. Since the spaces are finite, both $y\mapsto {\mathrm{\Pi }}_y(A)$ and $\omega \mapsto L(\omega ,A)$ are measurable for every $A\subseteq \mathrm{\Theta }$.

Every event $H\in \sigma (Y)$ is $\mathrm{\Theta }\times B$ for some $B\subseteq Y$. For any $A\subseteq \mathrm{\Theta }$, ${\int }_{H}L(\omega ,A)J(d\omega )={\sum }_{y\in B}{\mathrm{\Pi }}_y(A)J_Y(y)={\sum }_{y\in B}J(A\times \{y\})=J(H\cap \{X\in A\})$. Thus $L$ is an RCD of $X$ given $\sigma (Y)$, and by [F2] it is the posterior RCD.

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

The joint density is $\mathrm{\ell }(\theta ,y)=\frac{3}{2}{\mathbf{1}}_{\{0\}\times (U\cup V)}(\theta ,y)+{\mathbf{31}}_{\{1\}\times U}(\theta ,y)$, a finite sum of indicators of measurable rectangles by [F6]. Its parameter sections are $f_0,f_1$, with integrals $\int f_{0}d\nu =\frac{3}{2}\cdot \frac{2}{3}=1$ and $\int f_{1}d\nu =3\cdot \frac{1}{3}=1$. Thus the sections define sampling probabilities, and the finite reference measure $\nu$ is sigma-finite, so [F3] applies.

The marginal density is $m(y)=\frac{1}{2}{f}_0(y)+\frac{1}{2}{f}_1(y)$, hence $m=9\mathrm{/}4$ on $U$, $m=3\mathrm{/}4$ on $V$, and $m=0$ on $W$. Thus $J_Y(U)=3\mathrm{/}4$, $J_Y(V)=1\mathrm{/}4$, and $J_Y(W)=0$. The normalized formula gives ${\mathrm{\Pi }}_y(1)=2\mathrm{/}3$ on $U$ and $0$ on $V$; on $W$ the theorem fills the zero-evidence fibre with the prior, so ${\mathrm{\Pi }}_y(1)=1\mathrm{/}2$.

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

The discrete metric on ${\mathbb{N}}_0$ is complete because every Cauchy sequence is eventually constant; ${\mathbb{N}}_0$ itself is a countable dense subset. Its topology is discrete, so its Borel sigma-algebra is the power set. The finite discrete parameter space has the same properties. Thus both spaces are standard Borel by [F5]–[F6]. By [F4], ${\sum }_n{f}_0(n)=1$ and ${\sum }_n{f}_1(n)=1$, so they define probability sampling laws and the prior in [F1] gives a Bayesian model.

Since both spaces are standard Borel and AC holds, [F3] supplies a data-indexed posterior kernel. For every $n$, both $f_0(n)$ and $f_1(n)$ are positive and finite, so $m(n)>0$ and the displayed values are in $[0,1]$ and sum to one. Every function on the discrete data space is measurable, hence $n\mapsto {\mathrm{\Pi }}_n$ is a probability kernel.

For $i\in \{0,1\}$ and every $B\subseteq {\mathbb{N}}_0$, ${\sum }_{n\in B}{\mathrm{\Pi }}_n(i)m(n)=\frac{1}{2}{\sum }_{n\in B}{f}_i(n)=J(\{i\}\times B)$. Adding the identities for $i=0,1$ proves the rectangle identity for every $A\subseteq \mathrm{\Theta }$ and every data event $B$. Thus the explicit kernel is a data-indexed posterior version; all its data values have positive prior-predictive mass.