Finite-window spatial likelihood

Choose at most countable bases $\mathcal{U}$ for $X$ and $\mathcal{V}$ for $Y$ using [F1]. Their product $\mathcal{U}\times \mathcal{V}$ is at most countable by [F2], and its members $U\times V$ form a basis for $X\times Y$ by [F3]. Each such rectangle lies in $\mathcal{B}(X)\otimes \mathcal{B}(Y)$; every open subset of $X\times Y$ is a union of a subfamily of this countable basis, hence a countable union of those rectangles. By [F4], this proves $\mathcal{B}(X\times Y)\subseteq \mathcal{B}(X)\otimes \mathcal{B}(Y)$.

The coordinate projections ${\pi }_X:X\times Y\to X$ and ${\pi }_Y:X\times Y\to Y$ are continuous by the definition of the product topology [F3]. For Borel $A\subseteq X$ and $C\subseteq Y$, $A\times C={\pi }_X^{-1}(A)\cap {\pi }_Y^{-1}(C)$ is Borel in $X\times Y$. Thus every generator rectangle of the product sigma-algebra [F5] belongs to $\mathcal{B}(X\times Y)$, giving the reverse inclusion.

First suppose $f$ is simple, with value $c_j\ge 0$ on each member $B_j$ of a finite Borel partition of $W$. Then $G_f(\eta )={\prod }_j{c}_j^{\eta (B_j)}$, using $0^0=1$; every exponent is an evaluation map. For each fixed $c_j$, the map $m\mapsto c_j^m$ on ${\mathbb{N}}_0$ is measurable: the inverse image of any Borel set is a union of the countable level sets $\{\eta (B_j)=m\}$. This also covers $c_j=0$ with $0^0=1$. The finite product is measurable by [F3].

For general $f$, choose simple functions $s_m\uparrow f$ by [F2]. If $\eta ={\sum }_{i=1}^n{\delta }_{x_i}$, then $G_{s_m}(\eta )={\prod }_{i=1}^n{s}_m(x_i)\uparrow {\prod }_{i=1}^{n}f(x_i)=G_f(\eta )$; the limit holds because the product has finitely many nonnegative factors, and it also holds for the empty configuration. Thus $G_f$ is a pointwise limit of measurable functions from step 1.1 and is measurable.

For a measurable rectangle $B\times C$, the map $\eta \mapsto (\eta \otimes \eta )(B\times C)=\eta (B)\eta (C)$ is measurable by [F1]. Let $\mathcal{D}:=\{A\in {\mathcal{B}}_W^{\otimes 2}:\eta \mapsto (\eta \otimes \eta )(A)\text{ is measurable}\}\mathrm{.}$ The whole space belongs to $\mathcal{D}$ because its value is $\eta (W{)}^2$: for each integer $k\ge 0$, the level set $\{\eta (W{)}^2\ge k\}$ is a countable union of evaluation level sets $\{\eta (W)=n\}$. The total product mass is finite. For $A\in \mathcal{D}$, finite additivity gives $(\eta \otimes \eta )((W\times W)\setminus A)=\eta (W{)}^2-(\eta \otimes \eta )(A)$; this finite-valued measurable difference is measurable by [F4], so complements belong. Countable disjoint unions belong by countable additivity, since the partial sums are measurable by [F4] and their pointwise limit is measurable by [F5]. Thus $\mathcal{D}$ is a lambda-system containing the rectangles.

Rectangles form a pi-system and generate the product Borel sigma-algebra by [F6] and [F7]. Dynkin's theorem [F3] applied to step 1.1 therefore gives that $\eta \mapsto (\eta \otimes \eta )(A)$ is measurable for every Borel $A\subseteq W\times W$. The set $D_R$ is closed, so its product-measure evaluation is measurable.

If $\eta ={\sum }_{i=1}^n{\delta }_{x_i}$, then $(\eta \otimes \eta )(D_R)={\sum }_{i,j=1}^n{1}_{\{\mathrm{\parallel }{x}_i-x_j{\mathrm{\parallel }}_2\le R\}}$. The $n$ terms with $i=j$ all occur since $R\ge 0$; every remaining unordered pair of distinct occurrences contributes exactly two terms. Hence the displayed formula for $S_R$ is a finite nonnegative integer and is measurable by step 2.1 and [F1].

Each coordinate projection ${\pi }_i:W^n\to W$ is measurable because inverse images of members of ${\mathcal{B}}_W$ are measurable rectangles. Hence each $f\circ {\pi }_i$ is measurable, and [F5] makes $F_n={\prod }_{i=1}^{n}f\circ {\pi }_i$ measurable. Thus $A\mapsto {\int }_A{F}_{n}d{\mu }^{\otimes n}$ is a measure by the same indefinite-integral construction as [F1]. Its total mass is $({\int }_{W}fd\mu {)}^n<\mathrm{\infty }$ by repeated Tonelli [F3].

For a measurable rectangle $B_1\times \cdots \times B_n$, repeated use of [F2] gives ${\nu }^{\otimes n}(B_1\times \cdots \times B_n)={\prod }_i\nu (B_i)={\prod }_i{\int }_{B_i}fd\mu$. Repeated Tonelli [F3] gives the same value for ${\int }_{B_1\times \cdots \times B_n}{F}_{n}d{\mu }^{\otimes n}$.

Rectangles form a generating pi-system containing the whole space $W^n$, and the product formula for $W^n$ in step 1.2 shows that the two finite measures have the same finite total mass. Therefore [F4] applies and gives equality on all of ${\mathcal{B}}_W^{\otimes n}$. For $n=0$ the statement is the definition of the empty product measure.

Put $\mathrm{\Lambda }=\nu ({\mathbb{R}}^d)$. If $\mathrm{\Lambda }=0$, [F2] says the process is the zero measure, which is simple by [F1]. Suppose $\mathrm{\Lambda }>0$ and set $Q=\nu \mathrm{/}\mathrm{\Lambda }$. Atomlessness of $\nu$ implies $Q(\{x\})=0$ for every $x\in {\mathbb{R}}^d$.

For $m\in {\mathbb{N}}_0$ and $k\in {\mathbb{Z}}^d$, put $C_{m,k}=2^{-m}(k+[0,1{)}^d)$. The set $H=\{\xi :\xi ({\mathbb{R}}^d)<\mathrm{\infty }\}\cap {\bigcup }_{m\ge 0}{\bigcap }_{k\in {\mathbb{Z}}^d}\{\xi :\xi (C_{m,k})\le 1\}$ is evaluation-measurable, since all cubes are Borel and the unions and intersections are countable. It is exactly the set of finite simple configurations: a repeated location forces a count at least two in its cube at every scale; conversely, a finite set of distinct locations has positive minimum pairwise distance when it has at least two points, so cubes of sufficiently small diameter contain at most one point. Counts zero and one satisfy the condition at every scale.

Fix $n\ge 2$ and a pair $i<j$ of coordinate indices. The diagonal is product-measurable by the preceding basis argument. Under $Q^{\otimes n}$, the event $x_i=x_j$ has measure $\int Q(\{x\})Q(dx)=0$ by [F3] and step 1.1, since every singleton has zero $Q$-mass by step 1.1. The other $n-2$ coordinates contribute total mass one. There are only finitely many pairs among $n$ coordinates, so with $Q^{\otimes n}$-probability one no two coordinates coincide.

In the mixed-binomial construction [F2], the image configuration belongs to $H$ with probability one on each component: step 2.1 treats $n\ge 2$, and counts $n=0,1$ are simple directly. Summing over the component probabilities gives mass one to $H$. Since $H$ is measurable by step 1.2 and every Poisson process with intensity $\nu$ has this push-forward law by [F2], $\mathbb{P}\{\eta \in H\}=1$. Thus $\eta$ is simple almost surely.

The zero extension of ${\lambda }_{\theta }$ is Borel: for Borel $E\subseteq [0,\mathrm{\infty })$ its inverse image is $W\cap {\lambda }_{\theta }^{-1}(E)$ if $0\mathrm{\notin }E$, and $W^c\cup (W\cap {\lambda }_{\theta }^{-1}(E))$ if $0\in E$; both are Borel because $W$ is Borel and ${\lambda }_{\theta }$ is Borel on $W$. Thus [F2] defines its finite measure of total mass ${\mathrm{\Lambda }}_{\theta }$ on ${\mathbb{R}}^d$, supported on $W$. The unit-rate reference intensity $1_{W}d{\mathrm{\ell }}_d$ has total mass $v$ by [F1]. Thus [F3] applies to both Poisson laws, and their finite-mixture tuple integrals are supported on $W^n$.

The product factor in $g_{\theta }$ is measurable by [F5], and the exponential factor is a finite nonnegative constant. Hence $g_{\theta }$ is a measurable nonnegative function on $\mathcal{C}(W)$.

Let $A\subseteq \mathcal{C}(W)$ be measurable. Expanding [F3] for $P_{\theta }$ on its support $W$ and using [F4] gives $P_{\theta }(A)=e^{-{\mathrm{\Lambda }}_{\theta }}{\sum }_{n=0}^{\mathrm{\infty }}\frac{1}{n!}{\int }_{W^n}{1}_A\text{\hspace{0.17em}⁣}\left({\sum }_{i=1}^n{\delta }_{x_i}\right){\prod }_{i=1}^n{\lambda }_{\theta }(x_i){\mathrm{\ell }}_W^{\otimes n}(d(x_1,\dots ,x_n))\mathrm{.}$ For $P_0$, the same formula holds with ${\mathrm{\Lambda }}_{\theta }$ replaced by $v$ and every product factor replaced by $1$.

Expand ${\int }_A{g}_{\theta }d{P}_0$ by the $P_0$ formula in step 2.1. On the $n$-tuple component, $g_{\theta }({\sum }_i{\delta }_{x_i})=e^{v-{\mathrm{\Lambda }}_{\theta }}{\prod }_i{\lambda }_{\theta }(x_i)$, including $n=0$ by the empty-product convention. Cancellation of $e^{-v}$ with $e^v$ yields exactly the series for $P_{\theta }(A)$ in step 2.1.

Since step 3.1 holds for every measurable $A$, $g_{\theta }$ is a Radon–Nikodym density of $P_{\theta }$ relative to $P_0$, and therefore $P_{\theta }\ll P_0$. The explicitly displayed $g_{\theta }$ supplies the density version for every $\theta$, so the arbitrary selector in [F6] is not used. Taking its logarithm gives the stated expression when all event factors are positive; a zero factor makes the finite product zero and the log likelihood $-\mathrm{\infty }$.

Suppose $n>0$ and put $\widehat{\rho }=n\mathrm{/}v>0$. For $\rho >0$ set $t=\rho \mathrm{/}\widehat{\rho }$. By [F1] and the logarithm quotient law [F2], the log-likelihood difference is $\mathrm{\ell }(\rho )-\mathrm{\ell }(\widehat{\rho })=n(\log t-t+1)$.

Suppose $n=0$. For $\rho >0$, [F1] gives log likelihood $-v\rho$, which is strictly below $0$ and tends to $0$ as $\rho \downarrow 0$. At $\rho =0$, the empty product is $1$ and the density in Finite-window spatial Poisson likelihood relative to unit rate has log likelihood equal to its parameter-independent constant, so the normalized log likelihood value is $0$. Thus zero is the maximizer when admitted and is only a supremum when the parameter space is $(0,\mathrm{\infty })$.

Let $\psi (t)=\log t-t+1$. For each $s>0$, $\frac{\psi (s+h)-\psi (s)}{h}=\frac{\log (s+h)-\log s}{h}-1⟶\frac{1}{s}-1$ as $h\to 0$, by [F2]; hence ${\psi }^{\mathrm{\prime }}(s)=1\mathrm{/}s-1$. If $t>1$, apply [F3] on $[1,t]$: its intermediate point $c>1$ has ${\psi }^{\mathrm{\prime }}(c)<0$, so $\psi (t)<\psi (1)=0$. If $0<t<1$, apply [F3] on $[t,1]$: its intermediate point has ${\psi }^{\mathrm{\prime }}(c)>0$, so again $\psi (t)<\psi (1)=0$. Equality holds at $t=1$. Therefore [F1] and step 1.1 give the unique maximum at $\rho =\widehat{\rho }$ among positive rates; at $\rho =0$ the observed nonempty configuration has likelihood zero.

Steps 1.1 and 2.1 prove the positive-count claim, and step 1.2 proves the zero-count boundary claim.

Extend $g$ and $h$ by zero outside $\mathcal{C}(W)$ and $W\times \mathcal{C}(W)$, respectively; [F5] makes these measurable extensions. For a test function $h$ as in the Statement, define on the domain where $(x,\zeta )$ contains an occurrence at $x$ $k(x,\zeta ):=g(\zeta )h(x,\zeta -{\delta }_x),$ and set $k=0$ off that domain. The domain and deletion map are measurable by [F2]; the extensions, addition operation, and $g,h$ are measurable by [F4, F5], so $k$ is a nonnegative measurable test function for [F1]. Also the addition map is measurable by [F4], and since $g$ is finite and strictly positive, its ratio ${\lambda }_g(u;\eta )=g(\eta +{\delta }_u)\mathrm{/}g(\eta )$ is a measurable function on $W\times \mathcal{C}(W)$.

The left side of [F1] with this $k$ is ${\mathbb{E}}_Q[g(\xi ){\sum }_{x\in \xi }h(x,\xi -{\delta }_x)]={\mathbb{E}}_P{\sum }_{x\in \eta }h(x,\eta -{\delta }_x)\mathrm{.}$ On the right side, $(u,\xi +{\delta }_u)$ is in the deletion domain and $(\xi +{\delta }_u)-{\delta }_u=\xi$, so the integrand is $g(\xi +{\delta }_u)h(u,\xi )=g(\xi ){\lambda }_g(u;\xi )h(u,\xi )$. Consequently [F1] gives, after Tonelli's interchange [F6], ${\mathbb{E}}_P{\sum }_{x\in \eta }h(x,\eta -{\delta }_x)={\mathbb{E}}_P{\int }_{W}h(u,\eta ){\lambda }_g(u;\eta ){\mathrm{\ell }}_W(du)\mathrm{.}$ which is the defining identity [F3].

Step 2.1 proves the claimed Papangelou identity for every admissible nonnegative $h$, hence proves the Statement.