Sufficiency, Completeness and Exponential Families

If $k=0$, then ${\mathbb{R}}^0$ is a singleton and the open neighborhood in the hypothesis contains its only point; $L(0)=\nu ({\mathbb{R}}^0)=0$, so $\nu =0$. If $k\ge 1$, set $D=\{z\in {\mathbb{C}}^k:\mathrm{\mid }\mathrm{Re}{z}_j\mathrm{\mid }<\delta \mathrm{/}2\text{ for all }j\}$. Fix $z\in D$ and a coordinate $j$. Choose a neighborhood of $z$ on which every real coordinate has absolute value at most $3\delta \mathrm{/}4$. The moment assumption makes $e^{\delta \mathrm{\parallel }x{\mathrm{\parallel }}_1}$ integrable with respect to the positive measure [F1], and all integrands below belong to $L^1(\mathrm{\mid }\nu \mathrm{\mid })$ by [F2]. On this neighborhood, $\mathrm{\mid }{e}^{z\cdot x}\mathrm{\mid }\le e^{(3\delta \mathrm{/}4)\mathrm{\parallel }x{\mathrm{\parallel }}_1}$, while for small complex $h$ the $j$th difference quotient is bounded by $\mathrm{\mid }{x}_j\mathrm{\mid }{e}^{(3\delta \mathrm{/}4)\mathrm{\parallel }x{\mathrm{\parallel }}_1}\le C{e}^{\delta \mathrm{\parallel }x{\mathrm{\parallel }}_1}$. For every sequence $h\to 0$, [F3] gives convergence of the difference quotients in $L^1(\mathrm{\mid }\nu \mathrm{\mid })$ to $x_j{e}^{z\cdot x}$; the bound in [F2] transfers this convergence to their integrals against $\nu$. Thus $L$ is complex differentiable in coordinate $j$ at $z$, and, since $z$ and $j$ were arbitrary, $L$ is holomorphic in each coordinate on $D$.

Assume $k\ge 1$. The real neighborhood contains a product of real intervals around zero. Fix real values of coordinates $2,\dots ,k$ in those intervals. As a function of $z_1$, $L$ is holomorphic on the connected strip and vanishes on a real interval; [F4] makes it zero for every $z_1$ in that strip. Now fix such a complex $z_1$ and real coordinates $3,\dots ,k$ in their intervals and apply [F4] to coordinate $z_2$. Repeating this finite process for coordinates $1$ through $k$ proves $L(z)=0$ for every $z\in D$.

For $k\ge 1$ and each $\xi \in {\mathbb{R}}^k$, the library Fourier transform is $\widehat{\nu }(\xi )=\int e^{-2\pi i\xi \cdot x}\nu (dx)=L(-2\pi i\xi )=0$, since $-2\pi i\xi \in D$. The Fourier uniqueness result [F5] then gives $\nu =0$.

Choose a countable measurable partition $(E_n{)}_{n\ge 0}$ of $\mathcal{X}$ with $\mu (E_n)<\mathrm{\infty }$ by disjointifying a sigma-finite cover [F2]. Put $c_n=2^{-(n+1)}\mathrm{/}(1+\mu (E_n))$ and $\lambda (A)={\sum }_{n\ge 0}{c}_n\mu (A\cap E_n)$. For disjoint measurable $A_m$, countable additivity of $\mu$ and interchange of nonnegative double series give $\lambda ({\bigcup }_m{A}_m)={\sum }_n{c}_n{\sum }_m\mu (A_m\cap E_n)={\sum }_m\lambda (A_m)$; the interchange is valid because both sides are the supremum of the same finite subsums. Thus $\lambda$ is a measure. Also $\lambda (\mathcal{X})\le {\sum }_{n\ge 0}{2}^{-(n+1)}=1$. Since every $c_n>0$, $\lambda (A)=0$ iff every $\mu (A\cap E_n)=0$, iff $\mu (A)=0$. Thus $\lambda$ is finite and equivalent to $\mu$, so every $P_{\theta }\ll \lambda$.

For each $\theta$, choose the RN density $p_{\theta }=d{P}_{\theta }\mathrm{/}d\lambda$ from [F3], finite outside a $\lambda$-null set, and let $A_{\theta }=\{p_{\theta }>0\}$. Fix ${\theta }_{\ast }$; since $\int p_{{\theta }_{\ast }}d\lambda =1$, $\lambda (A_{{\theta }_{\ast }})>0$. Define $s={\sup }_{F\subseteq \mathrm{\Theta },F\text{ finite}}\lambda ({\bigcup }_{\theta \in F}{A}_{\theta })$, so $0<s\le \lambda (\mathcal{X})<\mathrm{\infty }$. By [A1], choose finite $F_n$ for each $n\ge 0$ with union measure greater than $s-1\mathrm{/}(n+1)$. For all sufficiently large $n$, these unions have positive measure, so the selected index set ${\bigcup }_n{F}_n$ is nonempty. Set $A={\bigcup }_n{\bigcup }_{\theta \in F_n}{A}_{\theta }$. Finite unions from the selected sets have measure at most $s$, so [F4] applied to their indicators gives $\lambda (A)\le s$ by continuity from below; the approximations give $\lambda (A)\ge s$. Hence $\lambda (A)=s$.

Fix $\theta$. If $\lambda (A_{\theta }\setminus A)>0$, the increasing finite unions $({\bigcup }_{n\le N}{\bigcup }_{\varphi \in F_n}{A}_{\varphi })\cup A_{\theta }$ converge to $A\cup A_{\theta }$, whose measure is greater than $s$. [F4] applied to their indicators shows that some finite union then exceeds $s$, contradicting its definition. Therefore $p_{\theta }=0$ $\lambda$-almost everywhere on $A^c$.

Enumerate the nonempty countable collection of selected indices ${\bigcup }_n{F}_n$ as $({\theta }_j)$, allowing repetitions if needed, and choose positive weights summing to one. The mixture $Q={\sum }_j{w}_j{P}_{{\theta }_j}$ has density $q={\sum }_j{w}_j{p}_{{\theta }_j}$ by [F4], so $q>0$ exactly on $A$ up to a $\lambda$-null set. If $Q(C)=0$, then ${\int }_{C}qd\lambda =0$. For each $m\ge 1$, $\lambda (C\cap A\cap \{q\ge 1\mathrm{/}m\})\mathrm{/}m\le {\int }_{C}qd\lambda =0$; these sets cover $C\cap A$ up to the null set where $q$ is not positive on $A$. Hence $\lambda (C\cap A)=0$. Also every $p_{\theta }$ vanishes on $A^c$ by step 3.1, so $P_{\theta }(C)={\int }_C{p}_{\theta }d\lambda =0$. Since $\theta$ was arbitrary, $P_{\theta }\ll Q$ for all $\theta$.

Fix a Borel set $C$ in the target of $U$. The function $q_C(y)=K(y,U^{-1}(C))$ is Borel by [F4]. For every $\theta$, the conditional-law identity [F1] shows $q_C(T)$ is a version of $P_{\theta }(U\in C\mid T)$.

By ancillarity [F3], $p_C=P_{\theta }(U\in C)$ is constant in $\theta$. Hence the bounded function $g_C(y)=q_C(y)-p_C$ satisfies $E_{\theta }{g}_C(T)=0$ for every $\theta$. Completeness [F2] implies $q_C(T)=p_C$ $P_{\theta }$-almost surely for each $\theta$.

For any Borel $D$ in the target of $T$, $P_{\theta }(T\in D,U\in C)=E_{\theta }[1_{\{T\in D\}}{q}_C(T)]=p_C{P}_{\theta }(T\in D)=P_{\theta }(U\in C)P_{\theta }(T\in D)$. This holds for every pair of Borel sets $D,C$, which is the definition of independence.

Suppose $T$ is sufficient, and let $K$ be its common kernel. Fix $\theta$. Disintegrating $P_{\theta }$ and $P_0$ with this same kernel gives $P_{\theta }(A)=\int K(y,A)P_{\theta }^T(dy)$ and $P_0(A)=\int K(y,A)P_0^T(dy)$ for every $A$. Since $P_{\theta }\ll P_0$, also $P_{\theta }^T\ll P_0^T$; let $r_{\theta }=d{P}_{\theta }^T\mathrm{/}d{P}_0^T$. Change of measure [F5] gives $P_{\theta }(A)=\int K(y,A)r_{\theta }(y)P_0^T(dy)$. The disintegration identity under $P_0$ extends to the nonnegative weight $r_{\theta }$ by [F9], making this ${\int }_A{r}_{\theta }(T)d{P}_0$. Uniqueness in [F4] gives $d{P}_{\theta }\mathrm{/}d{P}_0=r_{\theta }(T)$ $P_0$-almost surely.

This proves that the derivative has a $\sigma (T)$-measurable version. Modify that version on the $\sigma (T)$-measurable null set where it is negative or infinite to obtain a finite nonnegative version. By [F6, F7] this version has a Borel factor $g_{\theta }(T)$ on the standard Borel target; reset target values where $g_{\theta }$ is negative or infinite to zero, which preserves the identity $P_0$-almost surely.

Conversely suppose $d{P}_{\theta }\mathrm{/}d{P}_0=g_{\theta }(T)$ $P_0$-almost surely for every $\theta$. By [F3], a regular conditional distribution under $P_0$ factors through $T$ to a kernel $K_0$ on $\mathcal{Y}$. Change of measure [F5] shows $P_{\theta }^T(C)={\int }_C{g}_{\theta }(y)P_0^T(dy)$. For every $A$ and Borel $C$, $P_{\theta }(A\cap T^{-1}C)=\int 1_A{1}_C(T)g_{\theta }(T)d{P}_0={\int }_C{g}_{\theta }(y)K_0(y,A)P_0^T(dy)={\int }_C{K}_0(y,A)P_{\theta }^T(dy)$. The weighted conditional identity follows from [F9]. Thus the same $K_0$ is a conditional law for every parameter, and [F1] gives sufficiency.

Choose ${\eta }_0$ in the open subset and $ϵ>0$ so the closed cube ${\eta }_0+[-ϵ,ϵ{]}^k$ lies in $\mathrm{\Omega }$. Define the finite signed measure $\rho (A)={\int }_{A}g(T(x))P_{{\eta }_0}(dx)$ by [F4], and set $\nu (B)=\rho (T^{-1}B)$ for Borel $B\subseteq {\mathbb{R}}^k$. Since $T$ is measurable, preimages preserve disjoint countable unions, so $\nu$ is a finite signed Borel measure. The preimages of a countable partition form a countable partition, giving $\mathrm{\mid }\nu \mathrm{\mid }\le T_{\mathrm{\#}}\mathrm{\mid }\rho \mathrm{\mid }$ by [F4]. Hence $\mathrm{\mid }\nu \mathrm{\mid }({\mathbb{R}}^k)\le E_{{\eta }_0}\mathrm{\mid }g(T)\mathrm{\mid }<\mathrm{\infty }$.

Choose $0<\delta <ϵ$. Since $e^{\delta \mathrm{\parallel }s{\mathrm{\parallel }}_1}\le {\sum }_{\sigma \in \{-1,1{\}}^k}{e}^{\delta \sigma \cdot s}$ and $\mathrm{\mid }\nu \mathrm{\mid }\le T_{\mathrm{\#}}\mathrm{\mid }\rho \mathrm{\mid }$ by step 1.1, the weighted-integral bound follows first for nonnegative simple functions from that setwise domination and then for $e^{\delta \mathrm{\parallel }s{\mathrm{\parallel }}_1}$ by [F5]. For each sign vector, changing from $P_{{\eta }_0}$ to $P_{{\eta }_0+\delta \sigma }$ contributes the factor $Z({\eta }_0+\delta \sigma )\mathrm{/}Z({\eta }_0)$, so $\int e^{\delta \mathrm{\parallel }s{\mathrm{\parallel }}_1}\mathrm{\mid }\nu \mathrm{\mid }(ds)\le {\sum }_{\sigma }\frac{Z({\eta }_0+\delta \sigma )}{Z({\eta }_0)}{E}_{{\eta }_0+\delta \sigma }\mathrm{\mid }g(T)\mathrm{\mid }<\mathrm{\infty }$, the exponential moment required by [F3].

For every real $t$ with $\mathrm{\parallel }t{\mathrm{\parallel }}_{\mathrm{\infty }}<\delta$, its Laplace transform is $\int e^{t\cdot s}\nu (ds)=\frac{Z({\eta }_0+t)}{Z({\eta }_0)}{E}_{{\eta }_0+t}g(T)=0$ by the mean-zero premise. The transform-uniqueness result [F3], under axiom branch [A1], gives $\nu =0$.

For every Borel $B$, $0=\nu (B)=\rho (T^{-1}B)={\int }_{B}g(s)P_{{\eta }_0}^T(ds)$ by the definition of the pushforward measure and $\rho$. Taking $B=\{g>0\}$ and $B=\{g<0\}$ shows both positive and negative parts have zero integral, so $g(T)=0$ $P_{{\eta }_0}$-almost surely. Every family density is strictly positive on $\{h>0\}$, so all $P_{\eta }$ are equivalent to $P_{{\eta }_0}$ and the same vanishing holds for every $\eta$. By [F2], $T$ is complete.

Suppose $T$ is sufficient and fix a common kernel $K$ from [F1]. For $t\in T(\mathcal{X})$ set $m_{\theta }(t)=P_{\theta }(T=t)$ and $M(t)={\sup }_{\theta }{m}_{\theta }(t)$. The image is countable by countability of $\mathcal{X}$, hence Borel by [F3]; every function on this image extended by zero is Borel. Define $h(x)=M(Tx)K(Tx,\{x\})$ and define $g_{\theta }(t)=m_{\theta }(t)\mathrm{/}M(t)$ when $M(t)>0$, and zero when $M(t)=0$. Then $g_{\theta }$ is Borel and nonnegative.

If $M(Tx)>0$, choose ${\theta }_0$ with $m_{{\theta }_0}(Tx)>0$. The conditional identity in [F1] gives $K(Tx,T^{-1}\{Tx\})=1$ and $p_{\theta }(x)=m_{\theta }(Tx)K(Tx,\{x\})$ for every $\theta$; if $m_{\theta }(Tx)=0$, both sides vanish by the same identity. Thus $p_{\theta }(x)=g_{\theta }(Tx)h(x)$. If $M(Tx)=0$, every law gives that fiber probability zero, so $p_{\theta }(x)=h(x)=g_{\theta }(Tx)=0$. This proves the forward implication pointwise.

Conversely suppose the factorization holds. For each $t\in T(\mathcal{X})$ let $F_t=T^{-1}\{t\}$ and $H(t)={\sum }_{x\in F_t}h(x)$. If some $m_{\theta }(t)>0$, then $g_{\theta }(t)>0$ and $0<H(t)=m_{\theta }(t)\mathrm{/}{g}_{\theta }(t)<\mathrm{\infty }$, since the sum of the probabilities on the fiber is at most one. In that case put $K(t,\{x\})=h(x)\mathrm{/}H(t)$ on $F_t$ and zero off $F_t$. If all $m_{\theta }(t)=0$, choose the first point of $F_t$ in a fixed enumeration and let $K(t,\cdot )$ be its point mass. For $t\mathrm{\notin }T(\mathcal{X})$ choose one fixed $x_0\in \mathcal{X}$ and set $K(t,\cdot )={\delta }_{x_0}$. The countable image and enumeration make every coordinate $t\mapsto K(t,\{x\})$ Borel. For each $A\subseteq \mathcal{X}$, $K(t,A)={\sum }_{x\in A}K(t,\{x\})$ is a countable sum of Borel functions, so it is Borel; the point-mass assignments are countably additive probability measures.

On an active fiber, for every $A\subseteq \mathcal{X}$, $P_{\theta }(A\cap F_t)=g_{\theta }(t){\sum }_{x\in A\cap F_t}h(x)=m_{\theta }(t)K(t,A)$. On an inactive fiber both sides are zero. Summing over the countably many $t\in C\cap T(\mathcal{X})$ gives the conditional-law identity for every Borel $C\subseteq \mathcal{Y}$. Thus [F1] holds and $T$ is sufficient.

Fix any ${\eta }_0\in \mathrm{\Omega }$; nonemptiness is part of the family being considered. The density in [F1] is positive exactly on $\{h>0\}$ for every parameter. For $\eta \in \mathrm{\Omega }$, dividing the densities there gives $d{P}_{\eta }\mathrm{/}d{P}_{{\eta }_0}=\exp \{(\eta -{\eta }_0)\cdot T-(A(\eta )-A({\eta }_0))\}$ $P_{{\eta }_0}$-almost surely. The right side is a Borel function of $T$.

The reference $P_{{\eta }_0}$ is a member of the model and dominates every other member because the densities have the common support $\{h>0\}$. Apply [F3] to the derivative in step 1.1; its criterion yields a common conditional law of $X$ given $T$. By [F2], $T$ is sufficient.

By [F4], choose a positive-weight countable mixture $Q={\sum }_j{w}_j{P}_{{\theta }_j}$ dominating the family, and let $h=dQ\mathrm{/}d\mu$. This RN derivative exists under AC because $Q$ is a probability and $Q\ll \mu$.

Suppose $T$ is sufficient. Let $K$ be the common conditional law from [F1]. For each $A,C$, sum the conditional-law identity under $P_{{\theta }_j}$ with weights $w_j$; [F9] applied to increasing partial sums gives the same identity under $Q$. Thus the same $K$ is common for the augmented family $\{Q\}\cup \{P_{\theta }\}$.

Conversely suppose the factorization holds. By [F4], $P_{\theta }\ll Q$ for every $\theta$. The finite measures $P_{\theta },Q$ and sigma-finite $\mu$ have a common finite exhaustion, so the RN chain rule [F6] identifies $d{P}_{\theta }\mathrm{/}dQ=p_{\theta }\mathrm{/}h$ on the $Q$-full set $\{h>0\}$. There the assumed equality gives $d{P}_{\theta }\mathrm{/}dQ=g_{\theta }(T)$.

In this augmented family $Q$ is a member and dominates all laws. Apply [F5] to obtain Borel $g_{\theta }$ with $d{P}_{\theta }\mathrm{/}dQ=g_{\theta }(T)$ $Q$-almost surely. A sigma-finite cover for $\mu$ is a common finite-measure exhaustion for $P_{\theta },Q,\mu$, since the first two are finite; the chain rule [F6] and density identities [F3, F7] then give $p_{\theta }=h{g}_{\theta }(T)$ $\mu$-almost everywhere. The exceptional set may depend on the fixed $\theta$.

Also $dQ\mathrm{/}dQ=1$ is a function of $T$. In the augmented family $\{Q\}\cup \{P_{\theta }\}$, $Q$ is a member reference and all laws are dominated by it; apply [F5] to these T-measurable derivatives. The criterion gives a common conditional law, so $T$ is sufficient.

Since $P_k\ll Q$, [F5] gives finite-valued nonnegative Borel densities $f_k$ after changing each on its null set. Their exceptional sets have countable union $N$ with $Q(N)=0$; reset all $f_k$ to zero on $N$. By [F6], $[0,\mathrm{\infty }{)}^{\mathbb{N}}$ is standard Borel. The coordinate maps are Borel, hence $T$ is Borel.

By [F7], choose a conditional-law kernel $K$ for $X$ given $T$ under $Q$. Its defining identity extends from indicator weights to nonnegative Borel weights by [F9]. By [F5], $P_k(A\cap T^{-1}C)=\int 1_A{1}_C(T)f_{k}dQ$. Since $T_k=f_k$, applying the weighted conditional identity with weight $1_C(t)t_k$ gives $P_k(A\cap T^{-1}C)={\int }_{C}K(t,A)t_k{Q}^T(dt)$; taking $A=\mathcal{X}$ gives $P_k^T(C)={\int }_C{t}_k{Q}^T(dt)$. Thus $K$ works under every $P_k$, so [F1] gives sufficiency.