Exact and pseudo likelihoods for finite-window Poisson and Strauss models

Both $N$ and $S_R$ are finite nonnegative integer-valued measurable functions by [F2]. For $m,k\in {\mathbb{N}}_0$, the set $\{N=m,S_R=k\}$ is measurable, and on it $H_{\beta ,\gamma }=-m\log \beta -k\log \gamma$ and $w={\beta }^m{\gamma }^k$. Thus both the energy and weight are measurable by their countable measurable level-set partitions. By [F10], $e^{-H_{\beta ,\gamma }(\eta )}=w(\eta )$. Since $0<\gamma \le 1$, $0<w(\eta )\le {\beta }^{N(\eta )}$. The unit-rate reference count $N$ has law $\mathrm{Pois}(4)$ by [F9], with masses given by [F6], so $Z={\mathbb{E}}_{P_0}w(\eta )\le {\mathbb{E}}_{P_0}{\beta }^N=e^{-4}{\sum }_{n=0}^{\mathrm{\infty }}\frac{(4\beta {)}^n}{n!}=e^{4(\beta -1)}<\mathrm{\infty }\mathrm{.}$ The empty configuration contributes $P_0\{\eta =0\}=e^{-4}$ to $Z$, hence $Z\ge e^{-4}>0$. Thus [F1] defines the stated Gibbs law.

Inserting $u$ adds one point and creates exactly $T_R(u,\eta ):=\eta (\{v\in W:\mathrm{\parallel }v-u{\mathrm{\parallel }}_2\le R\})$ unordered interaction pairs, counting occurrences: if the pair count is written as the number of unordered pairs, the increase is $S_R(\eta +{\delta }_u)-S_R(\eta )=(2T_R(u,\eta )+1-1)\mathrm{/}2=T_R(u,\eta )$; the $1$ in the numerator is the new diagonal term removed in the definition of $S_R$. Therefore $\frac{w(\eta +{\delta }_u)}{w(\eta )}=\beta {\gamma }^{T_R(u,\eta )}\mathrm{.}$ The same ratio holds for the normalized density because $Z$ cancels; [F3] shows that the Strauss Papangelou conditional intensity is $\lambda (u;\eta )=\beta {\gamma }^{T_R(u,\eta )}$.

Every singleton is a degenerate box of Lebesgue measure zero by [F7], so the unit-rate reference intensity is atomless. It is simple almost surely by [F8]. The normalized Strauss law is absolutely continuous with respect to that reference and is therefore also supported on simple configurations. Thus on a probability-one set the occurrence-pair energy is the usual simple-pattern Strauss energy.

When $\gamma =1$, $S_R$ drops out and the normalizer from the earlier calculation gives $Z=e^{4(\beta -1)}$, so the exact density is $e^{4(1-\beta )}{\beta }^{N(\eta )}$, matching the Poisson formula [F5]. The Papangelou intensity is the constant $\beta$, so [F4] gives pseudolikelihood ${\beta }^{N(\eta )}{e}^{-4\beta }$. The ratio of exact to pseudo likelihood is the observation-independent constant $e^4$.

Now suppose $0<\gamma <1$, let $x=(1,1)$, and compare the empty configuration $0$ with ${\delta }_x$. By [F1] and the fact that neither configuration has an interacting pair, their exact Gibbs density ratio is $g({\delta }_x)\mathrm{/}g(0)=\beta$. By [F4] and step 2.1, the pseudolikelihood ratio is $\frac{\mathrm{PL}({\delta }_x)}{\mathrm{PL}(0)}=\beta \exp \{\beta (1-\gamma )a\},a:=\mathrm{\mid }\{u\in W:\mathrm{\parallel }u-x{\mathrm{\parallel }}_2\le R\}\mathrm{\mid }\mathrm{.}$ Indeed the event factor at $x$ is $\beta$, and the integrated intensity changes from $\beta \mathrm{\mid }W\mathrm{\mid }$ to $\beta (\mathrm{\mid }W\mathrm{\mid }-(1-\gamma )a)$.

The square $[4\mathrm{/}5,6\mathrm{/}5{]}^2$ lies inside the radius-$1\mathrm{/}2$ ball about $x$, so $a\ge 4\mathrm{/}25>0$ by [F12]. Hence $\exp \{\beta (1-\gamma )a\}>1$ by [F11], and the pseudo ratio is strictly larger than the exact ratio $\beta$. A constant multiple of the exact likelihood would have equal observation ratios, so the Strauss pseudolikelihood is not the joint likelihood in this case. The example makes no consistency or asymptotic claim.