Posterior propriety and dominated Bayes formula — Examples

The prior has total mass $1\mathrm{/}3+2\mathrm{/}3=1$, and both conditional probability vectors sum to one. The likelihood at $y=1$ is $L_1(a)=1\mathrm{/}2$ and $L_1(b)=1\mathrm{/}4$, so the evidence is $Z_{\mathrm{\Pi }}(1)=(1\mathrm{/}3)(1\mathrm{/}2)+(2\mathrm{/}3)(1\mathrm{/}4)=1\mathrm{/}3\in (0,+\mathrm{\infty })$.

The unnormalized masses are $(1\mathrm{/}3)(1\mathrm{/}2)=1\mathrm{/}6$ at $a$ and $(2\mathrm{/}3)(1\mathrm{/}4)=1\mathrm{/}6$ at $b$. Dividing each by $Z_{\mathrm{\Pi }}(1)=1\mathrm{/}3$ gives posterior mass $1\mathrm{/}2$ at each point, a probability update by [F2].

The likelihood sections are normalized: $L_0(a)=1$, $L_1(b)=1$, and all other parameter-data pairs have likelihood zero. Therefore $Z(0)=1\mathrm{/}2$, $Z(1)=1\mathrm{/}2$, and $Z(2)=0$.

By [F3], the prior-predictive mass at $2$ is $Z(2)$ times its counting measure, hence zero. At this fiber every numerator ${\int }_A{L}_{2}d\mathrm{\Pi }$ is also zero, so the displayed quotient is $0\mathrm{/}0$ and is not defined; the theorem's chosen filler is $K(2,A)=\mathrm{\Pi }(A)$. This filler is a version value on a prior-predictive null event, not an update inferred from the observation.

By [A1] and [F6], countable choice is available for [F3]. The boxes $E_k=(-k,k]$ have finite measure $2k$ and cover $\mathbb{R}$ by [F3], so $\lambda$ is sigma-finite. Since $E_k\subseteq \mathbb{R}$, [F4] gives $\lambda (\mathbb{R})\ge 2k$ for every $k$, hence $\lambda (\mathbb{R})=+\mathrm{\infty }$ and the flat prior is improper.

Fix $\mu$ and set $T(x)=(x-\mu )\mathrm{/}\sigma$. This is a $C^1$ diffeomorphism of $\mathbb{R}$ with inverse $t\mapsto \mu +\sigma t$ and $\mathrm{\mid }{T}^{\mathrm{\prime }}(x)\mathrm{\mid }=1\mathrm{/}\sigma$. The formula for $q_{\mu }$ is continuous in $x$, hence Borel by [F12] and Lebesgue measurable by [F15]. Since $\varphi$ is also Lebesgue measurable by [F5, F15], [F7] gives ${\int }_{\mathbb{R}}{q}_{\mu }d\lambda ={\int }_{\mathbb{R}}\varphi d\lambda =1$. The set function $P_{\mu }(A)={\int }_A{q}_{\mu }d\lambda$ is a measure by [F16], and its total mass one makes it a probability by [F17]. For a Borel $A$, apply [F7] to $f={\mathbf{1}}_A{q}_{\mu }$ and $S(z)=\mu +\sigma z$: $N(\mu ,{\sigma }^2)(A)=\int {\mathbf{1}}_A(S(z))\varphi (z)d\lambda (z)={\int }_A{q}_{\mu }(x)d\lambda (x),$ since $\sigma q_{\mu }(S(z))=\varphi (z)$ and [F14] defines $N(\mu ,{\sigma }^2)$ as this affine pushforward. Thus each coordinate law has the stated density.

Put $S={\sum }_{i=1}^n(y_i-\bar{y}{)}^2$. Expanding and using ${\sum }_i(y_i-\bar{y})=0$ gives ${\sum }_i(y_i-\mu {)}^2=S+n(\mu -\bar{y}{)}^2$. Hence $L_{\mathbf{y}}(\mu )=(2\pi {\sigma }^2{)}^{-n\mathrm{/}2}{e}^{-S\mathrm{/}(2{\sigma }^2)}{e}^{-n(\mu -\bar{y}{)}^2\mathrm{/}(2{\sigma }^2)}$. This is continuous in $\mu$, hence Borel by [F12] and $\mathcal{L}(\mathbb{R})$-measurable by [F15]; therefore its evidence integral is defined by [F1].

For $k\ge 1$, recursively form the finite product data measure ${\nu }_k={\lambda }^{\otimes k}$ on $\mathcal{L}(\mathbb{R}{)}^{\otimes k}$ using [F8] and [F9]. Starting from the sigma-finite $\lambda$, [F10] ensures that each ${\nu }_k$ exists and is sigma-finite. The coordinate projections are measurable by [F8], so each pullback of $q_{\mu }$ is measurable by step 1.2; finite multiplication is continuous, making $f_k(y_1,\dots ,y_k)={\prod }_{i=1}^k{q}_{\mu }(y_i)$ product-measurable. Step 1.2 gives $\int f_{1}d{\nu }_1=1$. If $\int f_{k}d{\nu }_k=1$, the recursive product definition and Tonelli [F11] give $\int f_{k+1}d{\nu }_{k+1}=\int f_{k}d{\nu }_k\int q_{\mu }d\lambda =1$. Thus $P_{\mu ,n}(E):={\int }_E{f}_{n}d{\nu }_n$ is a measure by [F16] and a probability by [F17]. For measurable rectangles $E=A_1\times \cdots \times A_n$, repeated Tonelli gives $P_{\mu ,n}(E)={\prod }_{i=1}^n{P}_{\mu }(A_i)$. The product measure theorem [F10], iterated over the finite factors, identifies $P_{\mu ,n}$ as the product of the one-coordinate laws. Therefore these coordinates are independent and $L_{\mathbf{y}}(\mu )$ is their joint density.

Set $T(\mu )=\sqrt{n}(\mu -\bar{y})\mathrm{/}\sigma$, a $C^1$ diffeomorphism of $\mathbb{R}$ with inverse $u\mapsto \bar{y}+\sigma u\mathrm{/}\sqrt{n}$ and derivative $\sqrt{n}\mathrm{/}\sigma$. Since $e^{-u^2\mathrm{/}2}=\sqrt{2\pi }\varphi (u)$ is nonnegative and Lebesgue measurable by [F5, F15], [F7] gives ${\int }_{\mathbb{R}}{e}^{-n(\mu -\bar{y}{)}^2\mathrm{/}(2{\sigma }^2)}d\lambda (\mu )=\frac{\sigma }{\sqrt{n}}\sqrt{2\pi }$. Integrating the factorization in step 1.3 against the flat prior gives $Z(\mathbf{y})=(2\pi {\sigma }^2{)}^{-n\mathrm{/}2}{e}^{-S\mathrm{/}(2{\sigma }^2)}\frac{\sigma \sqrt{2\pi }}{\sqrt{n}}$, which is strictly positive and finite for every finite data vector.

Dividing step 1.3 by the finite positive evidence in step 2.2 gives $h(\mu )=\frac{\sqrt{n}}{\sigma }\varphi \text{\hspace{0.17em}⁣}\left(\frac{\sqrt{n}(\mu -\bar{y})}{\sigma }\right)$, the displayed Lebesgue density. Put $s=\sigma \mathrm{/}\sqrt{n}>0$; $S(z)=\bar{y}+sz$ is a $C^1$ diffeomorphism with inverse $\mu \mapsto (\mu -\bar{y})\mathrm{/}s$. For every Borel $A$, [F14] defines $N(\bar{y},s^2)$ as this pushforward; [F7] applied to $f={\mathbf{1}}_{A}h$, which is Lebesgue measurable by [F12, F13, F15], gives ${\int }_{A}h(\mu )d\lambda (\mu )=\int {\mathbf{1}}_A(S(z))h(S(z))sd\lambda (z)=\int {\mathbf{1}}_A(S(z))\varphi (z)d\lambda (z)=N(\bar{y},s^2)(A)\mathrm{.}$ The update on the full Lebesgue sigma-algebra is a probability by [F2]. If $n=0$ were allowed, its empty-product likelihood would be $1$ and the evidence would be $\lambda (\mathbb{R})=+\mathrm{\infty }$ by step 1.1, so no normalized update would exist.

By [A1] and [F7], countable choice is available for [F3], [F4], [F10], and [F11]. Write $f(p)=1\mathrm{/}[p(1-p)]$ on $(0,1)$. For any continuous $g:(0,1)\to \mathbb{R}$ and open $U\subseteq \mathbb{R}$, continuity makes $g^{-1}(U)$ open in the subspace $(0,1)$, so $g^{-1}(U)=O\cap (0,1)$ for some open $O\subseteq \mathbb{R}$. This intersection is open in $\mathbb{R}$, hence Borel by [F9] and Lebesgue measurable by [F10]. Thus $g$ is measurable for the restricted Lebesgue sigma-algebra. This applies to $f$ and to every binomial likelihood section, which is a polynomial. By [F11], Lebesgue measure on $\mathbb{R}$ is a measure; its restriction to the measurable interval $(0,1)$ is a measure as well. Thus [F5] defines the prior measure $\mathrm{\Pi }(A)={\int }_{A}fd\lambda$. For each $p\in (0,1)$, let $\epsilon =\min \{p,1-p,1\mathrm{/}3\}>0$; by [F8] choose $k\ge 1$ with $1\mathrm{/}k<\epsilon$. Then $k\ge 3$ and $p\in E_k=[1\mathrm{/}k,1-1\mathrm{/}k]$, so these sets cover $(0,1)$. On $E_k$, $f(p)\le k^2$. Their Lebesgue lengths are at most one by [F4], so homogeneity and monotonicity [F6] give $\mathrm{\Pi }(E_k)\le k^2<\mathrm{\infty }$; [F3] makes this a sigma-finite prior. It is not finite: on each disjoint interval $I_j=(2^{-j-1},2^{-j}]$, $j\ge 1$, one has $f(p)\ge 1\mathrm{/}p\ge 2^j$ and $\lambda (I_j)=2^{-j-1}$ by [F4]. Homogeneity and monotonicity [F6] give $\mathrm{\Pi }(I_j)={\int }_{I_j}fd\lambda \ge 1\mathrm{/}2$. By [F5], $\mathrm{\Pi }$ is a measure, so the prior mass of the union of the first $m$ such intervals is at least $m\mathrm{/}2$ for every $m$. Hence $\mathrm{\Pi }((0,1))=+\mathrm{\infty }$ and the prior is improper.

For $y=0$, put ${\nu }_0(A):={\int }_A{L}_{0}d\mathrm{\Pi }$; it is a measure by [F5], and $Z(0)={\nu }_0((0,1))$ by [F1]. The likelihood is $L_0(p)=\left(\genfrac{}{}{0px}{}{n}{0}\right)(1-p{)}^n=(1-p{)}^n$, so on each $I_j$ from step 1.1, $p\le 1\mathrm{/}2$ gives $L_0(p)\ge 2^{-n}$. Also $\mathrm{\Pi }(I_j)\ge 1\mathrm{/}2$ by step 1.1. Monotonicity and homogeneity [F6] therefore give ${\nu }_0(I_j)\ge 2^{-n-1}$. The intervals are pairwise disjoint, so measure additivity gives ${\nu }_0({\bigcup }_{j=1}^m{I}_j)\ge m{2}^{-n-1}$; since this union is contained in $(0,1)$, countable additivity gives ${\nu }_0((0,1))\ge m{2}^{-n-1}$ for every $m$. Thus $Z(0)=+\mathrm{\infty }$ and [F1] leaves the quotient undefined.

For $y=n$, put ${\nu }_n(A):={\int }_A{L}_{n}d\mathrm{\Pi }$; it is a measure by [F5], and $Z(n)={\nu }_n((0,1))$ by [F1]. Here $L_n(p)=\left(\genfrac{}{}{0px}{}{n}{n}\right)p^n=p^n$. On each reflected interval $J_j=[1-2^{-j},1-2^{-j-1})$, $j\ge 1$, one has $p\ge 1\mathrm{/}2$, so $L_n(p)\ge 2^{-n}$. Also $f(p)\ge 1\mathrm{/}(1-p)\ge 2^j$ there and $\lambda (J_j)=2^{-j-1}$ by [F4], so $\mathrm{\Pi }(J_j)\ge 1\mathrm{/}2$ by [F6]. Monotonicity and homogeneity [F6] give ${\nu }_n(J_j)\ge 2^{-n-1}$. The intervals are pairwise disjoint, and additivity followed by nonnegativity on their complement gives $Z(n)={\nu }_n((0,1))\ge m{2}^{-n-1}$ for every $m$. Thus $Z(n)=+\mathrm{\infty }$ and [F1] leaves this quotient undefined. By [F2], neither endpoint fiber admits a normalized posterior probability.