Conjugate families and Gaussian conditioning

By [F8], $f(p)=p^{\alpha -1}(1-p{)}^{\beta -1}$ is continuous and hence Borel and Lebesgue measurable on $(0,1)$ with the measure in [F9]. For each integer $m\ge 3$, one has $1\mathrm{/}m<1-1\mathrm{/}m$, so its restriction to $[1\mathrm{/}m,1-1\mathrm{/}m]$ is continuous, hence bounded and Riemann integrable by [F5], and [F4] identifies its Lebesgue integral with its Riemann integral. The indicators of these intervals increase pointwise to one, so the truncated functions increase to $f$.

Fix $s>0$ and put $h_s(t)=t^{s-1}{e}^{-t}$ for $t>0$. For $1\le t\le 2$, [F12] gives $0\le \log t\le \log 2$, so $(s-1)\log t\ge \min \{0,(s-1)\log 2\}$. By [F11] and [F13], $h_s(t)\ge c_s:=\exp (\min \{0,(s-1)\log 2\})\exp (-2)>0$. The function $h_s$ is continuous on $[1,2]$ by [F8] and [F13], hence bounded and Riemann integrable by [F5]. The lower integral bound [F14] therefore gives ${\int }_1^2{h}_s(t)dt\ge c_s(2-1)=c_s>0$.

By [F6], the Lebesgue integral of $f$ is the limit of the truncated integrals from step 1.1. The same truncations define the improper integral in [F1], which converges by [F7]; therefore the limit is $B(\alpha ,\beta )$.

By [F3] and [F10], $\mathrm{\Gamma }(s)$ is finite. The integrand $h_s$ is positive on $(0,\mathrm{\infty })$ by [F11] and [F13]. Every compact truncation integral on either side of $[1,2]$ is nonnegative by [F14], and its finite improper limit is nonnegative as well. Splitting at $1$ and $2$ by [F15] gives $\mathrm{\Gamma }(s)={\int }_0^1{h}_s+{\int }_1^2{h}_s+{\int }_2^{\mathrm{\infty }}{h}_s\ge {\int }_1^2{h}_s\ge c_s>0$. Thus $0<\mathrm{\Gamma }(s)<\mathrm{\infty }$ for every $s>0$, with strict positivity proved here rather than inferred from convergence alone.

Apply step 2.2 with $s=\alpha ,\beta ,\alpha +\beta$. The identity in [F2] expresses $B(\alpha ,\beta )$ as a quotient of finite positive numbers, hence $0<B(\alpha ,\beta )<\mathrm{\infty }$. Together with step 2.1 this proves the displayed Lebesgue integral identity and its endpoint bounds.

For $n\ge 1$, each particular subset $J\subseteq \{1,\dots ,n\}$ of size $y$ describes the disjoint event that $X_i=1$ exactly for $i\in J$. Independence [F3] gives this event probability $p^y(1-p{)}^{n-y}$. There are $\left(\genfrac{}{}{0px}{}{n}{y}\right)$ such subsets by [F2], and their union is $\{S_n=y\}$.

Summing the equal probabilities in step 1.1 gives the formula. If $n=0$, the sum is deterministically zero and the empty product is one, which gives the stated boundary case as well.

By [F7], $g(t)=t^{a-1}{e}^{-t}$ is continuous and hence Borel and Lebesgue measurable on $(0,\mathrm{\infty })$ with the measure in [F8]. For each integer $m\ge 2$, its restriction to $[1\mathrm{/}m,m]$ is continuous, hence bounded Riemann integrable by [F4], and [F3] identifies its Riemann integral with its Lebesgue integral. The truncated functions $g{\mathbf{1}}_{[1\mathrm{/}m,m]}$ increase to $g$.

Monotone convergence [F5] identifies $\int gd{\lambda }_1$ with the limit of those truncated integrals. By the mixed-improper-integral definition [F1] and convergence [F2], this limit is $\mathrm{\Gamma }(a)$, so $g$ has finite positive Lebesgue integral.

Apply [F6] to the diffeomorphism $t=bx$ of $(0,\mathrm{\infty })$. Since $dt=bdx$ and $x=t\mathrm{/}b$, it gives ${\int }_0^{\mathrm{\infty }}{x}^{a-1}{e}^{-bx}dx=b^{-a}{\int }_0^{\mathrm{\infty }}{t}^{a-1}{e}^{-t}dt=b^{-a}\mathrm{\Gamma }(a)\mathrm{.}$ Multiplication by $b^a\mathrm{/}\mathrm{\Gamma }(a)$ is valid by [F2] and yields one.

Every $d_{\mathrm{\infty }}$-ball about $x$ of radius $r>0$ is the product box ${\prod }_{i<n}(x_i-r,x_i+r)$, so it is open in the product topology. Conversely, let $U$ be product-open and $x\in U$. By [F6], choose a rational box $Q={\prod }_{i<n}(a_i,b_i)$ with $x\in Q\subseteq U$. The finite set $S=\{x_i-a_i,b_i-x_i:i<n\}$ is nonempty because $n\ge 1$, and every member is positive. By [F2], $r=\min S$ exists and is positive. If $d_{\mathrm{\infty }}(x,y)<r$, then $\mathrm{\mid }{x}_i-y_i\mathrm{\mid }\le d_{\mathrm{\infty }}(x,y)<r\le x_i-a_i$ and $\mathrm{\mid }{x}_i-y_i\mathrm{\mid }<r\le b_i-x_i$ for every $i<n$, hence $y\in Q\subseteq U$. Thus each product-open set is $d_{\mathrm{\infty }}$-open, and the topologies agree.

Fix $x,y\in {\mathbb{R}}^n$, put $u_i=x_i-y_i$ and $M=d_{\mathrm{\infty }}(x,y)$. By [F1] and [F2], $M=\mathrm{\mid }{u}_j\mathrm{\mid }$ for some $j<n$ and $0\le \mathrm{\mid }{u}_i\mathrm{\mid }\le M$ for every $i<n$. Hence [F3] gives $M^2=u_j^2\le {\sum }_{i<n}{u}_i^2=d_2(x,y{)}^2$, so [F4] yields $M\le d_2(x,y)$. Also $u_i^2=\mathrm{\mid }{u}_i{\mathrm{\mid }}^2\le M^2$ for each $i$, whence $d_2(x,y{)}^2={\sum }_{i<n}{u}_i^2\le {\sum }_{i<n}{M}^2=n{M}^2$ by [F3]. Since $n>0$, [F4] makes $\sqrt{n}M$ nonnegative and $(\sqrt{n}M{)}^2=n{M}^2$; square-root monotonicity now gives $d_2(x,y)\le \sqrt{n}M$. Thus $d_{\mathrm{\infty }}\le d_2\le \sqrt{n}{d}_{\mathrm{\infty }}$ for all pairs.

Let $(x^{(m)})$ be $d_{\mathrm{\infty }}$-Cauchy and fix a positive rational $\epsilon$. By [F10], choose a positive rational $\rho <\epsilon \mathrm{/}(1+\sqrt{n})$. The Cauchy condition [F9] gives a tail on which $d_{\mathrm{\infty }}(x^{(m)},x^{(l)})<\rho$. Step 1.2 then gives $d_2(x^{(m)},x^{(l)})\le \sqrt{n}{d}_{\mathrm{\infty }}(x^{(m)},x^{(l)})<\sqrt{n}\rho <\epsilon$, so the sequence is $d_2$-Cauchy.

By [F5], this sequence has a $d_2$-limit $x\in {\mathbb{R}}^n$. Step 1.2 gives $d_{\mathrm{\infty }}(x^{(m)},x)\le d_2(x^{(m)},x)\to 0$; by [F9] this is convergence in $d_{\mathrm{\infty }}$. Therefore $d_{\mathrm{\infty }}$ is complete.

By [F6], ${\mathbb{Q}}^n$ is at most countable and dense in the product topology. Step 1.1 identifies that topology with the $d_{\mathrm{\infty }}$ topology, so the same subset is dense for $d_{\mathrm{\infty }}$. Thus the product topology is separable by [F7], and step 3.1 supplies its compatible complete metric. Consequently ${\mathbb{R}}^n$ with the product topology is Polish.

The identity from $({\mathbb{R}}^n,\mathcal{B}({\mathbb{R}}^n))$ to the Polish space of step 4.1 is a bijection. The topologies are identical by step 1.1, so they have the same open sets and therefore, by [F8], the same Borel sigma-algebra. The identity is thus a measurable isomorphism as required by the definition of standard Borel.

Apply [F1] and [F2] to expand $({\sum }_j{p}_j{)}^n$; each term is exactly the stated nonnegative mass. Since ${\sum }_j{p}_j=1$, the sum of all masses is one.

When $n=0$ there is one all-zero count vector and its mass is $0!\mathrm{/}(0!{)}^k$ times a product of $k$ zero-th powers, hence one. When some $p_j=0$, terms with $y_j>0$ vanish and terms with $y_j=0$ use $0^0=1$; the same expansion remains valid.

Since $B$ is positive definite, it is invertible and symmetric. For any $u\in {\mathbb{R}}^d$, put $v=-B^{-1}{C}^{T}u$. Positive semidefiniteness of $\mathrm{\Sigma }$ gives $0\le {\left(\begin{array}{c}u\\ v\end{array}\right)}^T\mathrm{\Sigma }\left(\begin{array}{c}u\\ v\end{array}\right)=u^T(A-C{B}^{-1}{C}^T)u=u^{T}Su\mathrm{.}$ Thus $S$ is positive semidefinite. This calculation uses no inverse of $A$.

Put $D=C{B}^{-1}$ and $R=X-a-D(Y-b)$. Every linear projection of $(R,Y)$ is an affine scalar projection of the jointly Gaussian pair $(X,Y)$, hence is normal by the projection characterization in [F1]; therefore $(R,Y)$ is jointly Gaussian. Its mean is $(0,b)$, and block covariance calculation gives $\mathrm{Cov}(R,Y)=C-DB=0,\mathrm{Cov}(R)=A-D{C}^T-C{D}^T+DB{D}^T=S\mathrm{.}$ The block-independence result [F2] therefore makes $R$ independent of $Y$, even if $S$ is singular.

Let $\rho$ be the law of $R$, and define $K(y,E)={\int }_{{\mathbb{R}}^d}{\mathbf{1}}_E\text{\hspace{0.17em}⁣}(a+D(y-b)+r)\rho (dr)\mathrm{.}$ For each fixed $y$, this is the pushforward of the probability $\rho$ under the continuous affine map $r\mapsto a+D(y-b)+r$, so it is a probability measure by [F3]. The map $(y,r)\mapsto a+D(y-b)+r$ is continuous, hence the integrand is product-measurable for Borel $E$. The finite-kernel measurability theorem [F4] makes $y\mapsto K(y,E)$ Borel; therefore $K$ is a probability kernel by [F7]. Since $R\sim N_d(0,S)$ by [F1], its pushforward is $N_d(a+D(y-b),S)$ for every $y$.

By [F2] and [F5], the joint law of $(R,Y)$ is $\rho \otimes {\mathbb{P}}_Y$. The identity $X=a+D(Y-b)+R$ and Tonelli [F6] then give, for Borel $E,F$, $\mathbb{P}(X\in E,Y\in F)={\int }_F{\int }_{{\mathbb{R}}^d}{\mathbf{1}}_E(a+D(y-b)+r)\rho (dr){\mathbb{P}}_Y(dy)={\int }_{F}K(y,E){\mathbb{P}}_Y(dy)\mathrm{.}$ This is the defining conditional-law identity. It holds for every Borel observation set $F$, so the same kernel is a conditional law of $X$ given $Y$. The proof requires $B^{-1}$ only; singularity of $A$ or $S$ causes no change.

For fixed $\eta$, each coordinate pullback of $p_{\eta }$ is measurable on the product sigma-algebra [F2], and the product density $g_{\eta }(y_{1:n})={\prod }_i{p}_{\eta }(y_i)$ is measurable by [F9]. Its integral against ${\mu }^{\otimes n}$ is one by [F2], repeated Tonelli [F3], and $\int p_{\eta }d\mu =1$. Thus $g_{\eta }d{\mu }^{\otimes n}$ is a probability measure by [F4]. On every measurable rectangle it equals ${\prod }_i{P}_{\eta }(A_i)$ by the rectangle formula and Tonelli, so the sigma-finite pi-system uniqueness theorem [F5] identifies it with the IID product law. Expanding $g_{\eta }$ gives the density displayed in the Statement.

Since $\mathrm{\Omega }$ is nonempty, choose ${\eta }_0\in \mathrm{\Omega }$. The density ratio on the common support is $\frac{g_{\eta }(y_{1:n})}{g_{{\eta }_0}(y_{1:n})}=\exp \{(\eta -{\eta }_0{)}^T{T}_n(y_{1:n})-n(A(\eta )-A({\eta }_0))\}\mathrm{.}$ If any $h(y_i)=0$, both product densities vanish, so the same positive Borel function of $T_n$ satisfies $g_{\eta }=r_{\eta }(T_n)g_{{\eta }_0}$ everywhere. Integrating this identity on each measurable set proves that it is a version of the Radon–Nikodym derivative under $P_{{\eta }_0}^{\otimes n}$.

By [F6] the finite sample space is standard Borel. If $k\ge 1$, [F6] also gives that the statistic target ${\mathbb{R}}^k$ is standard Borel. If $k=0$, ${\mathbb{R}}^0$ is the empty product, a singleton with its full two-element sigma-algebra, which is standard Borel by the empty-product case of [F6]. In this case $T_n$ is the unique constant map, the dot products in step 2.1 are empty sums equal to zero, and its likelihood-ratio expression is still a Borel function of this singleton statistic. Thus both target cases meet the standard-Borel hypothesis. The reference $P_{{\eta }_0}^{\otimes n}$ belongs to the family, and step 2.1 makes every likelihood ratio a Borel function of $T_n$. Applying [F7] gives a common conditional-law kernel, which is sufficiency by [F8]. The sole choice-axiom use is inherited from [A1] and [F7].

On the open simplex, each coordinate $p_i$ for $i<k$ and the residual coordinate $p_k=1-{\sum }_{i<k}{p}_i$ is continuous and strictly positive. By [F7], $p\mapsto {\prod }_{j=1}^k{p}_j^{{\alpha }_j-1}$ is therefore continuous, hence Borel measurable there. Since the simplex is open, extending this nonnegative function by zero outside it gives a Borel measurable function on ${\mathbb{R}}^{k-1}$; Borel sets are Lebesgue measurable by [F7], so the integrand has the Lebesgue measurability required by [F2]. Define the stick-breaking map from $(0,1{)}^{k-1}$ to ${\mathrm{\Delta }}_{k-1}$ by $p_i=z_i{\prod }_{j<i}(1-z_j)$ for $i<k$. Its inverse is $z_i=p_i\mathrm{/}(1-{\sum }_{j<i}{p}_j)$, so it is a $C^1$ diffeomorphism. Its derivative matrix is lower triangular with diagonal entries ${\prod }_{j<i}(1-z_j)$; hence its Jacobian is ${\prod }_{i=1}^{k-2}(1-z_i{)}^{k-i-1}$.

Under this change of variables, each factor $p_j^{{\alpha }_j-1}$ contributes $z_j^{{\alpha }_j-1}$ and a factor $(1-z_i{)}^{{\alpha }_j-1}$ for each $i<j$; the Jacobian adds exponent $k-i-1$ to $1-z_i$. Thus the transformed integrand is ${\prod }_{i=1}^{k-1}{z}_i^{{\alpha }_i-1}(1-z_i{)}^{{\sum }_{j>i}{\alpha }_j-1}\mathrm{.}$ This factorization is valid because all coordinates and residuals are strictly positive on the open cube.

Each one-variable factor in step 2.1 is continuous on $(0,1)$, hence Borel measurable by [F7]. Using the product-measure identification in [F3], repeated Tonelli and [F4] now give the integral as ${\prod }_{i=1}^{k-1}B\text{\hspace{0.17em}⁣}({\alpha }_i,{\sum }_{j>i}{\alpha }_j)\mathrm{.}$ Applying [F5] to each factor cancels the intermediate Gamma terms and leaves ${\prod }_{j=1}^k\mathrm{\Gamma }({\alpha }_j)\mathrm{/}\mathrm{\Gamma }({\alpha }_{+})$.

By [F6] the displayed ratio is finite and positive. Multiplying the integrand by the reciprocal constant in [F1] therefore gives integral one, which proves normalization.