2.1 The configuration model

Working with the uniform distribution on $d$ -regular $k$ -uniform hypergraphs directly is challenging. Instead, we show Theorem1.1 for occupation problems on so-called configurations. A $d$ -regular $k$ -configuration is a bijection $g:[n]\times [d]\rightarrow [m]\times [k]$ , where the v-edges $(i,h^{\prime})\in [n]\times [d]$ represent pairs of variables $i\in [n]$ and so-called $i$ -edges, i.e. half-edge indices $h^{\prime}\in [d]$ . The image $(a,h)=g(i,h^{\prime})$ is an f-edge, i.e. a pair of a constraint (factor) $a\in [m]$ and an $a$ -edge (or half-edge) $h\in [k]$ , indicating that the $i$ -edge $h^{\prime}$ of the variable $i$ is wired to the $a$ -edge $h$ of $a$ and thereby suggesting that $i$ is connected to $a$ in the corresponding $d$ -regular $k$ -factor graph. Notice that we can represent $g$ by an equivalent, four-partite, graph with (disjoint) vertex sets given by the variables $V=[n]$ , constraints (factors) $F=[m]$ , v-edges $H^{\prime}=[n]\times [d]$ and f-edges $H=[m]\times [k]$ , where each variable $i\in [n]$ connects to all its v-edges $(i,h^{\prime})\in H^{\prime}$ , each constraint $a\in [m]$ to all its f-edges $(a,h)\in H$ and a v-edge $(i,h^{\prime})$ connects to an f-edge $(a,h)$ if $g(i,h^{\prime})=(a,h)$ .

Let $\mathcal{G}$ be the set of all $d$ -regular $k$ -configurations on $n$ variables, and notice that $|\mathcal{G}|=\emptyset$ iff $dn\neq km$ and $|\mathcal{G}|=(dn)!=(km)!$ for $m=dn/k\in \mathbb {Z}$ , which we assume from here on. Further, the occupation problem on factor graphs directly translates to configurations, i.e. an assignment $x\in \{0,1\}^n$ is a solution of $g\in \mathcal{G}$ if for each constraint $a\in [m]$ there exist exactly two distinct $a$ -edges $h$ , $h^{\prime}\in [k]$ such that $x_{i(a,h)}=x_{i(a,h^{\prime})}=1$ , where $i(a,h)=(g^{-1}(a,h))_1$ denotes the $h$ -th neighbour of $a$ . Say, the occupation problem on a configuration corresponding to the example in Figure 1a is shown in Figure 2a.

Let $Z(g)$ be the number of solutions of $g \in \mathcal{G}$ . As before, $Z=0$ almost surely unless $2n\in k\mathbb {Z}$ . Theorem1.1 is a straightforward consequence of the following result.

Theorem2.1 translates to the models in Section 1.1 and Section 1.2 using standard arguments, namely symmetry, the discussion of parallel edges, contiguity, and the fact that both the variable and the factor neighbourhoods are unique with probability tending to one. Hence, in the remainder of this contribution we exclusively consider the occupation problem on configurations.

Figure 2. The figure on the left shows the solution on a configuration corresponding to the solution in Figure 1. We only denoted $a$ -edges (small boxes, filled if they the $a$ -edge takes the value one) and $i$ -edges (small circles, filled if the $i$ -edge takes the value one) instead of f-edges and v-edges for brevity (e.g. $h_{a_1,1}$ instead of $(a_1,h_{a_1,1})$ ).The figure on the right illustrates the corresponding $2$ -in- $3$ vertex cover (given by the filled circles).

2.2 The first moment method

In the first step we apply the first moment method to the occupation problem on configurations, yielding the following result.

Lemma 2.2 (First Moment Method). Let $k\in \mathbb {Z}_{\ge 4}$ , $d\in \mathbb {Z}_{\ge 2}$ . For $n\in \mathcal N$ tending to infinity

\begin{align*} \mathbb {E}[Z]&\sim \sqrt {d}e^{n\phi _1}, \quad \textrm {where} \quad \phi _1=\frac {d}{k}\ln \binom {k}{2}-(d-1)H\left (\frac 2k\right ) \textrm {.} \end{align*}

In particular, $\mathbb {E}[Z]\rightarrow \infty$ for $d\lt d^*$ and $\mathbb {E}[Z]\rightarrow 0$ for $d\gt d^*$ with $d^*$ as in (1). So, Markov’s inequality implies $\mathbb {P}(Z\gt 0)\rightarrow 0$ for $d\gt d^*$ . The map $\phi _1$ is known as annealed free entropy density.

2.3 The second moment method

Let $k\in \mathbb {Z}_{\ge 4}$ and $d\in \mathbb {Z}_{\ge 2}$ . We denote the set of distributions on a finite set $\mathcal S$ by $\mathcal P(\mathcal S)$ and identify $p\in \mathcal P(\mathcal S)$ with its probability mass function, meaning $\mathcal P(\mathcal S)=\{p\in [0,1]^{\mathcal S}:\sum _{x\in \mathcal S}p(s)=1\}$ . Further, let $\mathcal P_{\ell }(\mathcal S)=\{p\in \mathcal P(\mathcal S):\ell p\in \mathbb {Z}^{\mathcal S}\}$ be the empirical distributions over $\ell \in \mathbb {Z}_{\gt 0}$ trials.

In order to apply the second moment method we will consider a (new) CSP with $m$ factors on $n$ variables with the larger domain $\{0,1\}^2$ , and where the constraint $a\in [m]$ is satisfied by an assignment $x\in (\{0,1\}^2)^n$ if $\sum _{i\in v(a)}x_{i,1}=\sum _{i\in v(a)}x_{i,2}=2$ . Here, there are qualitatively three types of satisfying assignments for the constraints, namely with $0$ , $1$ or $2$ overlapping ones. We will analyse the empirical overlap distributions $p\in \mathcal P_m(\{0,1,2\})$ of assignments satisfying all constraints, which determine the empirical distributions $p_{\mathrm {e}}\in \mathcal P_{km}(\{0,1\}^2)$ of the values $\{0,1\}^2$ over the $km$ edges, given by

\begin{align*} p_{\mathrm {e}}(11)=\frac {1}{k}p(1)+\frac {2}{k}p(2) \,\,\textrm { and } \,\, p_{\mathrm {e}}(10)=p_{\mathrm {e}}(01)=\frac {1}{k}p(1)+\frac {2}{k}p(0)\textrm {.} \end{align*}

So, if $p\in \mathcal P_m(\{0,1,2\})$ is an achievable empirical overlap distribution on the $m$ factors, then $p_{\mathrm {e}}$ is necessarily an empirical distribution on the $n$ variables; thus the achievable overlap distributions are contained in $\mathcal P_n=\left \{p\in \mathcal P_m(\{0,1,2\})\,:\,p_{\mathrm {e}}\in \mathcal P_n(\{0,1\}^2)\right \}$ .

In the first – combinatorial – part we establish that the second moment can be written as a sum of all contributions over all achievable overlap distributions.

Lemma 2.3 (Second Moment Combinatorics). For any $n\in \mathcal N$ we have

\begin{align*} \mathbb {E}[Z^2]=\sum _{p \in \mathcal P_n}E(p), \text { where } E(p)=\binom {m}{mp}\prod _{s\in \{0,1,2\}}\binom {k}{s,2-s,2-s,k-4+s}^{mp(s)}\binom {n}{np_{\mathrm {e}}}\binom {dn}{dnp_{\mathrm {e}}}^{-1}\textrm {.} \end{align*}

Here, we use the notation $\binom {m}{mp}$ , $p\in \mathcal P_m(\{0,1,2\})$ , for multinomial coefficients.

To study further the second moment in Lemma 2.3, we identify the maximal contributions. For this purpose, let $p^*\in \mathcal P(\{0,1,2\})$ be the hypergeometric distribution with

(2) \begin{align} p^*(s)=\frac {\binom {2}{s}\binom {k-2}{2-s}}{\binom {k}{2}} \, \textrm { for }\, s\in \{0,1,2\}, \quad \textrm { and }\quad p^*_{\mathrm {e}}(1,1)=\frac {4}{k^2}, \, p^*_{\mathrm {e}}(1,0)=\frac {2(k-2)}{k^2}\textrm {.} \end{align}

The overlap distribution $p^*$ is a natural candidate for maximizing $E(p)$ . Indeed, we obtain $p^*$ when we consider two independent uniformly random assignments in $\{0,1\}^k$ with $2$ ones each, and $p^*_{\mathrm {e}}$ is exactly the marginal probability if we jointly consider two independent uniformly random assignments in $\{0,1\}^n$ to the variables with $2n/k$ ones each. In the next step, we derive the limits of the log-densities $\frac {1}{n}\ln (E(p))$ . Recall that the K(ullback)-L(eibler) divergence $D_{\textrm {KL}}(p\parallel q)$ of two distributions $p$ , $q\in \mathcal P(\mathcal S)$ , such that $p$ is absolutely continuous with respect to $q$ , is

\begin{align*} D_{\textrm {KL}}(p\parallel q)=\sum _{x\in \mathcal S}p(x)\ln \left (\frac {p(x)}{q(x)}\right )\textrm {.} \end{align*}

Lemma 2.4 (Second Moment Asymptotics). For any fully supported $p\in \mathcal P(\{0,1,2\})$ and any sequence $(p_n)_{n\in \mathcal N}\subseteq \mathcal P_n$ with $\lim _{n\rightarrow \infty }p_n=p$ we have $\lim _{n\rightarrow \infty }\frac {1}{n}\ln (E(p_n))=\phi _2(p)$ , where

\begin{align*} \phi _2(p)=2\phi _1-\frac {d}{k}\Delta _d(p) \quad \text {and} \quad \Delta _d(p)=D_{\textrm {KL}}(p\parallel p^*)-\frac {(d-1)k}{d}D_{\textrm {KL}} (p_{\mathrm {e}}\parallel p^*_{\mathrm {e}})\textrm {.} \end{align*}

The following proposition is the main contribution of this work.

With Proposition 2.5, we easily verify that $p^*$ is the unique minimizer of $\Delta _d$ for any $d\lt d^*(k)$ , since the KL divergence is minimized by its unique root and $(d-1)k/d$ is increasing in $d$ . This conclusion then allows us to compute the limit of the scaled second moment using Laplace’s method for sums. More than that, we confirm the conjecture by the authors in [Reference Panagiotou and Pasch36] as an immediate corollary.

Proposition 2.6 (Second Moment Limit). For any $k\in \mathbb {Z}_{\ge 4}$ and $d\lt d^*(k)$

\begin{align*} \frac {\mathbb {E}[Z^2]}{\mathbb {E}[Z]^2}&\sim \sqrt {\frac {k-1}{k-d}},\,\,\textrm { as }n\in \mathcal N\textrm { tends to infinity.} \end{align*}

Proposition 2.6 and the Paley-Zygmund inequality yield $\textrm {lim inf}_{n\rightarrow \infty }\mathbb {P}(Z\gt 0)\ge \sqrt {\frac {k-d}{k-1}}$ . While this bound suggests that a threshold exists, we need to show that the threshold at $d^*$ is sharp.

2.4 Small subgraph conditioning

We complete the proof of Theorem2.1 using the small subgraph conditioning method. For this purpose let $a^{\underline {b}}=\prod _{c=0}^{b-1}(a-c)$ denote the falling factorial.

We will apply Theorem 2.7 to the number $Z$ of solutions from Section 2.1 and the numbers $X_\ell$ of small cycles in the configuration $G$ . In order to understand what a cycle in a configuration is, we recall the representation of a configuration $g$ as a four-partite graph from Section 2.1.

Since we are mostly interested in the factor graph associated with a configuration we divide the lengths of paths by three, e.g. what we call a cycle of length four in the bijection, is actually a cycle of length twelve in its equivalent four-partite graph representation. Figures 1a and 2a show an example of a factor graph and the corresponding configuration in its graph representation. Showing the following statement, which establishes Assumption2.7a), is rather routine.

Lemma 2.8 (Small Cycles). For $\ell \in \mathbb {Z}_{\gt 0}$ let $X_\ell$ be the number of $2\ell$ -cycles in $G$ , and set

\begin{align*} \lambda _\ell =\frac {[(k-1)(d-1)]^\ell }{2\ell }. \end{align*}

Then $X_1,\ldots, X_{L}$ are asymptotically independent and Poisson with $\mathbb {E}[X_\ell ]\sim \lambda _\ell$ for all $L\in \mathbb {Z}_{\gt 0}$ .

We give a self-contained proof of Lemma 2.8 in the appendix, which we build upon to argue that Assumption2.7b) in Theorem2.7 holds. With Lemma 2.8 in place, we consider the base case in Assumption2.7b), i.e. for $\ell \in \mathbb {Z}_{\gt 0}$ we let $r_\ell =1$ and $r_{\ell ^{\prime}}=0$ otherwise, to determine $\delta _\ell =(1-k)^{-\ell }$ . We easily verify that $\sum _{\ell \ge 1}\lambda _\ell \delta _\ell ^2=\frac 12\ln (\frac {k-1}{k-d})$ and thereby establish Assumption2.7c) using Proposition 2.6. Finally, we follow the proof of Lemma 2.8 to complete the verification of Assumption2.7b) and thereby complete the proof of Theorem1.1.

3.1 Notation

We use the notation $[n]=\{1,\ldots, n\}$ and $[n]_0=\{0\}\cup [n]$ for $n\in \mathbb {Z}_{\gt 0}$ , denote the falling factorial with $n^{\underline {k}}$ for $n,k\in \mathbb {Z}_{\ge 0}$ , $k\le n$ , and multinomial coefficients with $\binom {n}{k}$ for $n\in \mathbb {Z}_{\ge 0}$ and $k\in \mathbb {Z}_{\ge 0}^d$ , $d\in \mathbb {Z}_{\gt 1}$ , such that $\sum _{i\in [d]}k_i=n$ . For functions $f$ , $g$ on integers with $\lim _{n\rightarrow \infty }{f(n)}/{g(n)}=1$ we write $f(n)\sim g(n)$ . We make heavy use of Stirling’s formula [Reference Robbins38], i.e.

\begin{align*} \sqrt {2\pi n}\left (\frac {n}{e}\right )^ne^{\frac {1}{12n+1}}\le n!\le \sqrt {2\pi n}\left (\frac {n}{e}\right )^ne^{\frac {1}{12n}}, \quad n\in \mathbb {Z}_{\gt 0}, \end{align*}

and in particular $n!\sim \sqrt {2\pi n}(\frac {n}{e})^n$ . If a random variable $X$ has law $P$ we write $X\sim P$ and use $\textrm {Po}(\lambda )$ to denote the Poisson distribution with parameter $\lambda$ . Distributions $p\in \mathcal P(\mathcal S)$ in the convex polytope $\mathcal P(\mathcal S)$ of distributions with finite support $\mathcal S$ are identified with their probability mass functions $p\in [0,1]^{\mathcal S}$ . Further, $\mathcal P_n(\mathcal S)=\{p\in \mathcal P(\mathcal S)\,:\,np\in [n]_0\}$ denotes the set of empirical distributions obtained from $n\in \mathbb {Z}_{\ge 1}$ trials. Let $v^{\mathrm {t}}$ denote the transpose of a vector $v$ . Finally, we use ‘iff’ for ‘if and only if’.

3.2 Basic observations

We briefly establish the claims in Section 1.1 for the configuration version, and the claim that $d^*$ is not an integer.

Proof. Since $g\in \mathcal{G}$ is a bijection $g:[n]\times [d]\rightarrow [m]\times [k]$ , the set $\mathcal{G}$ is empty for $dn\neq km$ and $|\mathcal{G}|=(dn)!=(km)!$ otherwise, which proves the first assertion. Next, we fix a solution $x\in \{0,1\}^n$ of $g\in \mathcal{G}$ with $n^{\prime}_1$ ones. Then two $a$ -edges $h$ have to take the value one, i.e. $x_{i(a,h)}=1$ , for each $a\in [m]$ and hence $2m$ f-edges $(a,h)\in [m]\times [k]$ in total. On the other hand, there are $dn^{\prime}_1$ v-edges $(i,h)\in [n]\times [d]$ that take the value one. Since $g$ is a bijection, $dn^{\prime}_1=2m$ , so $n_1=n^{\prime}_1\in \mathbb {Z}$ .

For the last assertion, we first focus on the denominator of $d^*$ , i.e.

\begin{align*} kH\left (\frac {2}{k}\right )-\ln \binom {k}{2}=-\ln \left (\binom {k}{2}\left (\frac {2}{k}\right )^2\left (\frac {k-2}{k}\right )^{k-2}\right )\gt 0\textrm {,} \end{align*}

so $d^*\gt 0$ for $k\in \mathbb {Z}_{\ge 3}$ . Next, notice that $d^*$ is a solution of $f(d)=1$ with

\begin{align*} f(d)=e^{(d-1)(kH(2/k)-\ln \binom {k}{2})-\ln \binom {k}{2}}=\frac {2}{k(k-1)}\left (\frac {k^{k-1}}{2(k-2)^{k-2}(k-1)}\right )^{d-1}\textrm {,} \end{align*}

which directly implies that $d^*\gt 1$ and further, since $\gcd (k,k-1)=1$ , that $d^*\in (1,\infty )\setminus \mathbb {Z}$ .

5.1 How to square a constraint satisfaction problem

In order to facilitate the presentation we introduce the squared $d$ -regular $2$ -in- $k$ occupation problem. As before, an instance of this problem is given by a bijection $g\,:\,[n]\times [d]\rightarrow [m]\times [k]$ . Now, for an assignment $x\in (\{0,1\}^2)^n$ let $y_{g,x}=(x_{i(a,h)})_{a\in [m],h\in [k]}$ be the corresponding f-edge assignment under $g$ , where we recall from Section 2.1 that $i(a,h)=(g^{-1}(a,h))_1\in [n]$ is the variable $i(a,h)$ wired to the f-edge $(a,h)$ under $g$ . A constraint $a\in [m]$ is satisfied by a constraint assignment $x\in (\{0,1\}^2)^k$ iff $x\in \mathcal S^{(2)}$ , where

\begin{align*} \mathcal S^{(2)} =\left \{x\in (\{0,1\}^2)^k:\sum _{h\in [k]}x_{h,1}=\sum _{h\in [k]}x_{h,2}=2\right \}\textrm {.} \end{align*}

An f-edge assignment $x\in (\{0,1\}^2)^{m\times k}$ is satisfying if $x_a=(x_{a,h})_{h\in [k]}$ satisfies $a$ for all $a\in [m]$ . Finally, an assignment $x\in (\{0,1\}^2)^n$ is a solution of $g$ if $y_{g,x}$ is satisfying. Notice that the pairs of solutions $x$ , $x^{\prime}\in \{0,1\}^n$ of the standard problem on $g$ are in one to one correspondence with the solutions $y\in (\{0,1\}^2)^n$ of the squared problem on $g$ via $y=(x_i,x^{\prime}_i)_{i\in [n]}$ . So, $z^{(2)}(g)=z(g)^2$ for the number $z^{(2)}(g)$ of solutions of the squared problem, hence $Z^{(2)}=Z^2$ for $Z^{(2)}=z^{(2)}(G)$ and in particular $\mathbb {E}[Z^{(2)}]=\mathbb {E}[Z^2]$ . This equivalence allows us to entirely focus on the squared problem.

5.2 Proof of Lemma 2.3

As before, we can write $\mathbb {E}[Z^{(2)}]=\frac {1}{(dn)!}|\mathcal{E}|$ , where $|\mathcal{E}|$ is the number of pairs $(g,x)\in \mathcal{E}$ such that $x\in (\{0,1\}^2)^n$ solves $g$ . Set

\begin{align*} \mathcal Y=\left \{y\in (\{0,1\}^2)^{m\times k}\,:\,y_a\in \mathcal S^{(2)} \,\textrm {for all}\,a\in [m]\right \}. \end{align*}

For $y\in \mathcal Y$ let the overlap distribution $p_y\in \mathcal P_m(\{0,1,2\})$ be given by

\begin{align*} p_y(s)=\frac {1}{m}|\{a\in [m]\,:\,|y_a^{-1}(1,1)|=s\}|, \quad s\in \{0,1,2\}. \end{align*}

Further, let the edge distribution $q_y\in \mathcal P_{km}(\{0,1\}^2)$ be given by

\begin{align*} q_y(x)=\frac {1}{km}|\{(a,h)\in [m]\times [k]\,:\,y_{a,h}=x\}|=\frac {1}{km}|y^{-1}(x)|,\quad x\in \{0,1\}^2\textrm {.} \end{align*}

Using that $|y_a^{-1}(1,0)|=|y_a^{-1}(0,1)|=2-|y_a^{-1}(1,1)|$ and hence $|y^{-1}(0,0)|=k-4+|y^{-1}(1,1)|$ we directly get

\begin{align*} q_y(1,1)&=\frac {1}{km}\sum _{a\in [m]}|y_a^{-1}(1,1)|=\frac {1}{km}\sum _{s\in \{0,1,2\}}s|\{a\in [m]\,:\,|y_a^{-1}(1,1)|=s\}|=\sum _{s\in \{0,1,2\}}\frac {s}{k}p_y(s)\textrm {,}\\ q_y(1,0)&=q_y(0,1)=\frac {1}{km}\sum _{s\in \{0,1,2\}}(2-s)|\{a\in [m]\,:\,|y_a^{-1}(1,1)|=s\}|=\sum _{s\in \{0,1,2\}}\frac {2-s}{k}p_y(s)\textrm {,}\\ q_y(0,0)&=\sum _{s\in \{0,1,2\}}\frac {k-4+s}{k}p_y(s)\textrm {.} \end{align*}

Hence, let $p_{\mathrm {e}}=Wp\in \mathcal P(\{0,1\}^2)$ denote the edge distribution of any (not necessarily empirical) overlap distribution $p\in \mathcal P(\{0,1,2\})$ , where $W\in [0,1]^{\{0,1\}^2\times \{0,1,2\}}$ is given by

(3) \begin{align} W_{11,s}=\frac {s}{k}\textrm {, } \,\,W_{10,s}=W_{01,s}=\frac {2-s}{k} \,\textrm { and }\,W_{00,s}=\frac {k-4+s}{k}, \quad s\in \{0,1,2\}\textrm {.} \end{align}

Now, notice that for any $(g,x)\in \mathcal{E}$ we have $y_{g,x,a,h}=x_{i(a,h)}$ for all $a\in [m]$ , $h\in [k]$ , hence $g(x^{-1}(z)\times [d])=y_{g,x}^{-1}(z)$ and by that

\begin{align*} q_{y_{g,x}}(z)=\frac {|y_{g,x}^{-1}(z)|}{km}=\frac {d|x^{-1}(z)|}{km}=\frac {1}{n}|x^{-1}(z)|=q_{x}(z)\textrm { for }z\in \{0,1\}^2\textrm {,} \end{align*}

i.e. the relative frequencies of the values in the f-edge assignment $y_{g,x}$ coincide with the relative frequencies $q_x\in \mathcal P_n(\{0,1\}^2)$ of the values in the variable assignment $x$ . In particular, this shows that a satisfying f-edge assignment $y\in \mathcal Y$ is only attainable if $q_y\in \mathcal P_n(\{0,1\}^2)$ , and thereby

\begin{align*} \mathbb {E}[Z^{(2)}]=\frac {1}{(dn)!}\sum _{p\in \mathcal P_n}|\{(g,x)\in \mathcal{E}:p_{y_{g,x}}=p\}|. \end{align*}

Now, fix an attainable satisfying f-edge assignment $y\in \mathcal Y$ and an assignment $x\in (\{0,1\}^2)^n$ with $q_y=q_x$ , i.e. $|x^{-1}(z)\times [d]|=|y^{-1}(z)|$ for all $z\in \{0,1\}^2$ . Any bijection $g$ with $y=y_{g,x}$ needs to respect $g(x^{-1}(z)\times [d])=y_{g,x}^{-1}(z)$ for $z\in \{0,1\}^2$ and can hence be uniquely decomposed into its restrictions $g_z\,:\,x^{-1}(z)\times [d]\rightarrow y_{g,x}^{-1}(z)$ . On the other hand, any choice of such restrictions $g_z$ gives a bijection $g$ with $y=y_{g,x}$ , and so

\begin{align*} |\mathcal{E}_{x,y}|&=\prod _{z\in \{0,1\}^2}(dnq_x(z))!=\prod _{z\in \{0,1\}^2}(dnp_{y,\mathrm {e}}(z))!\,\,\textrm {, where }\, \mathcal{E}_{x,y}=\{(g,x)\in \mathcal{E}:y_{g,x}=y\}\textrm {.} \end{align*}

Notice that $\mathcal{E}_{x,y}\cap \mathcal{E}_{x^{\prime},y^{\prime}}=\emptyset$ for any $(x,y)\neq (x^{\prime},y^{\prime})$ , which is obvious for $x\neq x^{\prime}$ , and also for $y\neq y^{\prime}$ , since $y_{g,x}=y\neq y^{\prime}=y_{g^{\prime},x}$ implies that $g\neq g^{\prime}$ . But since $|\mathcal{E}_{x,y}|$ only depends on $p_{y}$ (actually only on $p_{y,\mathrm {e}}$ ) this completes the proof, because for any fixed attainable overlap distribution $p\in \mathcal P_n$ , we can now independently choose the satisfying f-edge assignment $y$ and variable assignment $x$ , subject to $q_x=p_{\mathrm {e}}$ and $p_y=p$ (which implies $q_y=q_x$ ). So we have $\mathbb {E}[Z^{(2)}]=\sum _{p}E(p)$ with $p\in \mathcal P_n$ and

\begin{align*} E(p)=\frac {1}{(dn)!}\binom {n}{np_{\mathrm {e}}}\binom {m}{mp}\prod _{s\in \{0,1,2\}}\binom {k}{s,2-s,2-s,k-4+s}^{mp(s)}\prod _{x\in \{0,1\}^2}(dnp_{\mathrm {e}}(z))!\textrm {,} \end{align*}

where we choose a variable assignment $x$ with $q_x=p_{\mathrm {e}}$ , an f-edge assignment $y$ with $p_y=p$ by first choosing one of the $\binom {m}{mp}$ options for $(|y_a^{-1}(1,1)|)_{a\in [m]}$ and then independently one of the $\binom {k}{s,2-s,2-s,k-4+s}$ satisfying constraint assignments for each of the $mp(s)$ constraints with overlap $s\in \{0,1,2\}$ , and finally choosing a bijection $g$ with $(g,x)\in \mathcal{E}_{x,y}$ .

5.3 Empirical overlap distributions

This section is dedicated to deriving properties of the set $\mathcal P_n$ for $n\in \mathcal N$ . In the following we will use the canonical ascending order on $\{0,1,2\}$ to denote points in $\mathbb {R}^{\{0,1,2\}}$ and the ascending lexicographical order on $\{0,1\}^2$ to denote points in $\mathbb {R}^{\{0,1\}^2}$ . Recall that $p^{(s)}\in \mathcal P(\{0,1,2\})$ given by $p^{(s)}(s)=1$ for $s\in \{0,1,2\}$ denote the corners of the convex polytope $\mathcal P(\{0,1,2\})$ and further consider the vectors in $\mathbb {R}^{\{0,1,2\}}$

(4) \begin{align} b_1=({-}d,d,0)^{\mathrm {t}}, \quad b_2=(1,-2,1)^{\mathrm {t}}. \end{align}

Finally, let

\begin{align*} \mathcal X=\Big \{x\in \mathbb {R}^2\,:\,(b_1,b_2)x\ge -p^{(0)}\Big \},\quad \mathcal X_n=\mathcal X\cap (m^{-1}\mathbb {Z})^2. \end{align*}

Proof. We use the shorthands $1_{\{0,1,2\}}=(1)_{s\in \{0,1,2\}}$ and

\begin{align*} 1_{\{0,1,2\}}^{\perp }&=\left \{x\in \mathbb {R}^{\{0,1,2\}}:1_{\{0,1,2\}}^{\mathrm {t}}x=0\right \} =\left \{x\in \mathbb {R}^{\{0,1,2\}}:\sum _{s\in \{0,1,2\}}x_s=0\right \}. \end{align*}

Note that $\mathcal P(\{0,1,2\})\subseteq p^{(0)}+1_{\{0,1,2\}}^{\perp } = \{p^{(0)}+x:x\in 1_{\{0,1,2\}}^{\perp }\}$ . On the other hand, $(b_1,b_2)$ is a basis of $1_{\{0,1,2\}}^{\perp }$ , and hence

(5) \begin{equation} \iota \,:\,\mathbb {R}^2\rightarrow p^{(0)}+1_{\{0,1,2\}}^{\perp }, \quad x\mapsto p^{(0)}+(b_1,b_2)x, \end{equation}

is bijective. This gives that $\iota (\mathcal X)=\mathcal P(\{0,1,2\})$ and that $\iota _n$ is the restriction of $\iota$ to $\mathcal X_n$ , so $\iota _n$ is a bijection from $\mathcal X_n$ to $\mathcal P(\{0,1,2\})\cap \iota \left ((m^{-1}\mathbb {Z})^2\right )$ . Consequently, it remains to show that $\mathcal P_n=\mathcal P(\{0,1,2\})\cap \iota \left ((m^{-1}\mathbb {Z})^2\right )$ , where

\begin{align*} \iota \left ((m^{-1}\mathbb {Z})^2\right )=\left \{p^{(0)}+\frac {i_1}{m}b_1+\frac {i_2}{m}b_2:i\in \mathbb {Z}^2\right \} \end{align*}

is a grid anchored at $p^{(0)}$ and spanned by $m^{-1}b_1$ and $m^{-1}b_2$ . Note that

\begin{align*} p^{(0)}_{\mathrm {e}}=\left (\frac {k-4}{k},\frac {2}{k},\frac {2}{k},0\right )^{\mathrm {t}}\textrm {,} \end{align*}

so $np^{(0)}_{\mathrm {e}}\in \mathbb {Z}^{\{0,1\}^2}$ since $n\in \mathcal N$ , and hence $p^{(0)}\in \mathcal P_n$ by the definition of $\mathcal P_n$ . Next, we show that $\mathcal P_n$ is on the grid, i.e. $\mathcal P_n\subseteq \iota \left ((m^{-1}\mathbb {Z})^2\right )$ . For this purpose fix $p\in \mathcal P_n$ and let $x=\iota ^{-1}(p)$ , i.e. $mp\in \mathbb {Z}^{\{0,1,2\}}$ , $n(Wp)\in \mathbb {Z}^{\{0,1\}^2}$ and $p=p^{(0)}+x_1b_1+x_2b_2$ . This directly gives $mx_2=mp(2)\in \mathbb {Z}$ . Further, we notice that $b_2$ is in the kernel of $W$ from Equation (3), i.e. $Wb_2=0_{\{0,1\}^2}$ , and $Wb_1=\frac {d}{k}w$ with $w=(1,-1,-1,1)^{\mathrm {t}}$ . This directly gives $p_{\mathrm {e}}(1,1)=0+\frac {d}{k}x_1+0$ and hence $mx_1=np_{\mathrm {e}}(1,1)\in \mathbb {Z}$ , i.e. $x\in (m^{-1}\mathbb {Z})^2$ and hence $p=\iota (x)\in \iota \left ((m^{-1}\mathbb {Z})^2\right )$ . Conversely, for any $x\in \mathcal X_n$ and with $p=\iota (x)$ we have $p\in \mathcal P(\{0,1,2\})$ since $x\in \mathcal X$ , further $mp=mp^{(0)}+(b_1,b_2)(mx)\in \mathbb {Z}^{\{0,1,2\}}$ since $mx\in \mathbb {Z}^2$ and the other terms on the right-hand side are integer valued by definition, and finally $np_{\mathrm {e}}=np^{(0)}_{\mathrm {e}}+mx_1w\in \mathbb {Z}^{\{0,1\}^2}$ .

Using Lemma 5.1 we have $\mathbb {E}[Z^{(2)}]=\sum _{x\in \mathcal X_n}E(\iota _n(x))$ , where $\mathcal X_n\subseteq \mathbb {R}^2$ may be considered as a normalization of the grid $\mathcal P_n\subseteq p^{(0)}+1_{\{0,1,2\}}^{\mathrm {t}}$ . In order to prepare the upcoming asymptotics of the second moment, we give a complete characterization of the convex polytope $\mathcal X$ and the image of $\mathcal X$ under $W(b_1,b_2)$ , i.e. the image $p_{\mathrm {e}}=Wp$ of $p\in \mathcal P(\{0,1,2\})$ under $W$ from Equation (3). Let $w=(1,-1,-1,1)^{\mathrm {t}}$ from the proof of Lemma 5.1, and set

\begin{align*} \mathcal W = \Big \{p^{(0)}_{\mathrm {e}}+yw\,:\,y\in [0,2/k]\Big \}\subseteq \mathcal P(\{0,1\}^2), \quad \mathcal X_p=\left \{x\in \mathcal X:x_1=\frac {k}{d}p(1,1)\right \} \text { for } p\in \mathcal W. \end{align*}

Moreover, recall the definition of $p^*$ from (2) and the bijection $\iota$ from (5), and let

\begin{align*} x^{(0)}=\begin{pmatrix}0\\ 0\end{pmatrix}\textrm {, } x^{(1)}=d^{-1}\begin{pmatrix}1\\ 0\end{pmatrix}\textrm {, } x^{(2)}=d^{-1}\begin{pmatrix}2\\ d\end{pmatrix}\in \mathbb {R}^2\,\textrm { and }\, x^*=\iota ^{-1}(p^*)\textrm {.} \end{align*}

Proof. Notice that $\iota (x^{(s)})=p^{(s)}$ for $s\in \{0,1,2\}$ , so since $\mathcal P(\{0,1,2\})$ is the convex hull of its corners $p^{(s)}$ , $s\in \{0,1,2\}$ , we have that $\mathcal X$ is the convex hull of $x^{(s)}$ , $s\in \{0,1,2\}$ , i.e. a two-dimensional convex polytope with corners $x^{(s)}$ , since $\iota$ is an affine transformation. In particular this also directly yields that $x^*$ is in the interior of $\mathcal X$ . Further, this shows that for any $x\in \mathcal X$ we have $x_1\ge 0$ with equality iff $x=x^{(0)}$ and further $x_1\le \frac {2}{d}$ with equality iff $x=x^{(2)}$ . Using $Wb_2=0_{\{0,1\}^2}$ and $Wb_1=\frac {d}{k}w$ from the proof of Lemma 5.1 we directly get that

\begin{align*} W(b_1,b_2)x=p^{(0)}_{\mathrm {e}}+\frac {d}{k}x_1w\textrm { with }\frac {d}{k}x_1\in [0,2/k]\textrm {,} \end{align*}

hence the image of $\mathcal X$ under $W(b_1,b_2)$ is a subset of $\mathcal W$ . Conversely, for $y\in [0,2/k]$ and $x=\frac {k}{2}yx^{(2)}\in \mathcal X$ we have $W(b_1,b_2)x=p^{(0)}_{\mathrm {e}}+yw$ , which shows that $\mathcal W$ is the image of $\mathcal X$ under $W(b_1,b_2)$ . This also shows that $\mathcal X_p$ is the preimage of $p\in \mathcal W$ , since for $y\in [0,2/k]$ and $p=p^{(0)}_{\mathrm {e}}+yw$ we have $p(1,1)=y$ . This in turn directly yields that $\mathcal X_{p^{(s)}_{\mathrm {e}}}=\{p^{(s)}\}$ for $s\in \{0,2\}$ . To see that $\mathcal X_p$ contains interior points of $\mathcal X$ otherwise, we can consider non-trivial convex combinations of $x^*$ and $x^{(0)}$ for $\frac {k}{d}p(1,1)\lt x^*_1$ and non-trivial convex combinations of $x^*$ and $x^{(2)}$ for $\frac {k}{d}p(1,1)\gt x^*_1$ , which are points in the interior of $\mathcal X$ .

Notice that in the two-dimensional case at hand, the proof of Lemma 5.2 is overly formal. The set $\mathcal X$ is simply (the convex hull of) the triangle given by $x^{(s)}$ , $s\in \{0,1,2\}$ , with $\mathcal X_p$ given by the vertical lines in $\mathcal X$ with $x_1=\frac {d}{k}p(1,1)$ . Further, the set $\mathcal X_n$ is a canonical discretization of $\mathcal X$ in that it is given by the points of the grid $(m^{-1}\mathbb {Z})^2$ contained in the triangle $\mathcal X$ .

5.4 Proof of Lemma 2.4

We derive Lemma 2.4 from the following stronger assertion.

Lemma 5.3. Let $\mathcal U\subseteq \mathcal P(\{0,1,2\})$ be a subset with non-empty interior and such that the closure of $\mathcal U$ is contained in the interior of $\mathcal P(\{0,1,2\})$ . Then there exists a constant $c=c(\mathcal U)\in \mathbb {R}_{\gt 0}$ such that for all $n\in \mathcal N$ and all $p\in \mathcal P_n\cap \mathcal U$ we have $\tilde E(p)e^{-c/n}\le E(p)\le \tilde E(p)e^{c/n}$ , where

\begin{align*} \tilde E(p)=\sqrt {\frac {d^3}{(2\pi )^2m^2\prod _{s}p(s)}}e^{n\phi _2(p)}\textrm {.} \end{align*}

Proof. Let $\mathcal C$ denote the closure of $\mathcal U$ and $\pi _s\,:\,\mathcal C\rightarrow [0,1]$ , $p\mapsto p(s)$ the projection for $s\in \{0,1,2\}$ . Since $\mathcal C$ is compact, the continuous map $\pi _s$ attains its maximum $p_+(s)$ and its minimum $p_-(s)$ , which directly gives $0\lt p_-(s)\lt p_+(s)\lt 1$ since all $p\in \mathcal C$ are fully supported and the interior of $\mathcal C$ is non-empty (that gives the second inequality). Using Lemma 5.2, the continuous map $\pi \,:\,\mathcal C\rightarrow [0,2/k]$ , $p\mapsto p_{\mathrm {e}}(1,1)$ , and the same reasoning as above we obtain the maximum $p_{\mathrm {e},+}(1,1)$ and minimum $p_{\mathrm {e},-}(1,1)$ of $\pi$ with $0\lt p_{\mathrm {e},-}(1,1)\lt p_{\mathrm {e},+}(1,1)\lt 2/k$ , which directly give the bounds $p_{\mathrm {e},-}(x)$ , $p_{\mathrm {e},+}(x)\gt 0$ for $x\in \{0,1\}^2$ as functions of $p_{\mathrm {e},+}(1,1)$ and $p_{\mathrm {e},-}(1,1)$ . Now, we can use these bounds with the Stirling bound to obtain a constant $c\in \mathbb {R}_{\gt 0}$ such that for all $n\in \mathcal N$ and $p\in \mathcal P_n\cap \mathcal C$ we have $E^{\prime}(p)e^{-c/n}\le E(p)\le E^{\prime}(p)e^{c/n}$ , where

\begin{align*} E^{\prime}(p)&=\sqrt {\frac {2\pi md^3}{\prod _{s}(2\pi mp(s))}}\prod _{s\in \{0,1,2\}}\binom {k}{s,2-s,2-s,k-4+s}^{mp(s)} e^{mH(p)-(d-1)nH(p_{\mathrm {e}})}\\ &=\sqrt {\frac {d^3}{(2\pi )^2m^2\prod _{s}p(s)}}e^{2m\ln \binom {k}{2}-mD_{\textrm {KL}}(p\parallel p^*)-(d-1)nH(p_{\mathrm {e}})}\textrm {.} \end{align*}

To see that $E^{\prime}(p)=\tilde E(p)$ , we observe that $D_{\textrm {KL}}(p_{\mathrm {e}} \parallel p^*_{\mathrm {e}})=2H(2/k)-H(p_{\mathrm {e}})$ , since $H(p^*_{\mathrm {e}})=2H(2/k)$ , $p_{\mathrm {e}}(1,0)=p_{\mathrm {e}}(0,1)$ , and $p_{\mathrm {e}}(1,1)+p_{\mathrm {e}}(1,0)=2/k$ for any $p\in \mathcal P(\{0,1,2\})$ .

Now, Lemma 2.4 is an immediate corollary. To see this, fix a fully supported overlap distribution $p\in \mathcal P(\{0,1,2\})$ and a sequence $(p_n)_{n\in \mathcal N}\subseteq \mathcal P_n$ converging to $p$ , e.g. $p_n=\iota (m^{-1}\lfloor mx_1\rfloor, m^{-1}\lfloor mx_2\rfloor )$ with $\iota (x)=p$ and $n$ sufficiently large. Further, fix a neighbourhood $\mathcal U$ of $p$ as described in Lemma 5.3, which is possible since $p$ is fully supported. Then we have $p_n\in \mathcal U$ for sufficiently large $n$ , hence with the continuity of $\phi _2$ we have

\begin{align*} \lim _{n\rightarrow \infty }\frac {1}{n}\ln (E(p_n)) =\lim _{n\rightarrow \infty }\frac {1}{n}\ln (\tilde E(p_n)) =\lim _{n\rightarrow \infty }\phi _2(p_n)=\phi _2(p)\textrm {.} \end{align*}

5.5 Proof of Proposition 2.6

We postpone the proof of Proposition 2.5 and continue with Laplace’s method for sums using the result. We obtain that

\begin{align*} \frac {\mathbb {E}[Z^2]}{\mathbb {E}[Z]^2}=\sum _pe(p)\textrm {, } \quad e(p)=\frac {E(p)}{\mathbb {E}[Z]^2} =\frac {\binom {2n/k}{np_{\mathrm {e}}(1,1)}\binom {(k-2)n/k}{np_{\mathrm {e}}(0,1)}\binom {dn}{2dn/k}\binom {m}{mp}\prod _sp^*(s)^{mp(s)}} {\binom {2dn/k}{dnp_{\mathrm {e}}(1,1)}\binom {(k-2)dn/k}{dnp_{\mathrm {e}}(0,1)}\binom {n}{2n/k}}\textrm {,} \end{align*}

where the sum is over $p\in \mathcal P_n$ . First, we use Proposition 2.5 to show that Laplace’s method of sums is applicable. While we have already established that $\Delta _{d^*}$ is non-negative, we still need to ensure that $p^*$ is the unique minimizer of $\Delta _d$ for $d\lt d^*$ and that the Hessian at $p^*$ is positive definite. We will need the second order Taylor approximation of the KL divergence. To be specific, let $\mu ^*$ have finite non-trivial support $\mathcal S$ and let $f\,:\,\mathcal P(\mathcal S)\rightarrow \mathbb {R}_{\ge 0}$ , $\mu \mapsto D_{\textrm {KL}}(\mu \parallel \mu ^*)$ , be the corresponding KL divergence. Then

\begin{align*} f^{(2)}\,:\,\mathcal P(\mathcal S)\rightarrow \mathbb {R}_{\ge 0},\quad \mu \mapsto \frac {1}{2}D_{\chi ^2}(\mu \parallel \mu ^*) =\frac 12\sum _s\frac {(\mu (s)-\mu ^*(s))^2}{\mu ^*(s)} =\frac 12(\mu -\mu ^*)^{\mathrm {t}}D_{\mu ^*}^{-1}(\mu -\mu ^*)\textrm {, } \end{align*}

is the second order Taylor approximation of $f$ at $\mu ^*$ , where $D_{\chi ^2}(\mu \parallel \mu ^*)$ denotes Pearson’s $\chi ^2$ divergence, $D_{\mu ^*}=(\delta _{i,j}\mu ^*(i))_{i,j\in \mathcal S}$ the matrix with $\mu ^*$ on the diagonal, and $\delta _{i,j}=1$ if $i=j$ and $0$ otherwise; this can be easily seen by considering the extension of $f$ to $\mathbb {R}_{\ge 0}^{\mathcal S}$ . On the other hand, we would like to consider $\Delta _d$ as a function over the suitable domain $\mathcal X$ from Section 5.3, however relative to the base point $p^*$ . Hence, let $\mathcal X^*=\{x-x^*:x\in \mathcal X\}$ be the triangle $\mathcal X$ centred at $x^*$ instead of $x^{(0)}$ , and $\iota ^*\,:\,\mathcal X^*\rightarrow \mathcal P(\{0,1,2\})$ the bijection given by

\begin{align*} \iota ^*(x)=\iota (x+x^*)=p^{(0)}+(b_1,b_2)x+(b_1,b_2)x^*=\iota (x^*)+(b_1,b_2)x=p^*+(b_1,b_2)x \end{align*}

for $x\in \mathcal X^*$ , with $b_1,b_2$ from Equation (4). Now, let $\gamma _d\,:\,\mathcal X^*\rightarrow \mathbb {R}_{\ge 0}$ , $x\mapsto \Delta _d(\iota ^*(x))$ , denote the corresponding parametrization of $\Delta _d$ . Then, using the chain rule for multivariate calculus as indicated above for both $(b_1,b_2)$ and $W$ from (3), we derive the Hessian

(6) \begin{align} H_d=(b_1,b_2)^{\mathrm {t}}\left (D_{p^*}^{-1}-\frac {(d-1)k}{d}W^{\mathrm {t}}D_{p^*_{\mathrm {e}}}^{-1}W\right )(b_1,b_2) \end{align}

of $\gamma _d$ at $0_{[2]}\in \mathbb {R}^2$ , using the shorthand $D_{\mu ^*}=(\delta _{i,j}\mu ^*(i))_{i,j}$ . The properties of the KL divergence imply that $\gamma _d(0_{[2]})=0$ and $\gamma _d$ has a stationary point at $0_{[2]}$ . Now, the second order Taylor approximation $\gamma ^{(2)}_d\,:\,\mathcal X^*\rightarrow \mathbb {R}$ , $x\mapsto \frac {1}{2}x^{\mathrm {t}}H_dx$ , of $\gamma _d$ at $0_{[2]}$ can be written as $\gamma ^{(2)}_d=\Delta ^{(2)}_d\circ \iota ^*$ with

(7) \begin{align} \Delta ^{(2)}_d(p)&=\frac {1}{2}\left [D_{\chi ^2}(p\parallel p^*)-\frac {(d-1)k}{d}D_{\chi ^2}(p_{\mathrm {e}}\parallel p^*_{\mathrm {e}})\right ]. \end{align}

Further, for any neighbourhood $\mathcal U$ of $0_{[2]}$ such that the closure of $\mathcal U$ is contained in the interior of $\mathcal X^*$ , Taylor’s theorem yields a constant $c\in \mathbb {R}_{\gt 0}$ such that

(8) \begin{align} \gamma ^{(2)}_d(x)-c\|x\|_2^3\le \gamma _d(x)\le \gamma ^{(2)}_d(x)+c\|x\|_2^3 \end{align}

for all $x\in \mathcal U$ . Since $H_d$ is symmetric, let $\lambda _1$ , $\lambda _2\in \mathbb {R}$ with $\lambda _1\le \lambda _2$ denote its eigenvalues and fix a corresponding orthonormal basis of eigenvectors $v_1$ , $v_2\in \mathbb {R}^2$ , i.e. $H_dv_1=\lambda _1v_1$ and $H_dv_2=\lambda _2v_2$ .

Formally and analogously to the KL divergence we will take the liberty to identify $\Delta _d$ and $\Delta ^{(2)}_d$ with their extensions to the maximal domains $\mathcal D\subseteq \mathbb {R}^{\{0,1,2\}}$ and $\mathcal D^{(2)}=\mathbb {R}^{\{0,1,2\}}$ respectively. In particular, Lemma 5.2 shows that for any fully supported $p\in \mathcal P(\{0,1,2\})$ the edge distribution $p_{\mathrm {e}}$ also has full support, hence we can use the Lipschitz continuity of $W$ on $\mathbb {R}^{\{0,1,2\}}$ to find $\varepsilon \in \mathbb {R}_{\gt 0}$ such that both $p^{\prime}\gt 0$ and $Wp^{\prime}\gt 0$ for any $p^{\prime}\in \mathcal B_\varepsilon (p)\subseteq \mathbb {R}_{\gt 0}^{\{0,1,2\}}$ and thereby $\Delta _d$ is well-defined and smooth on $\mathcal B_\varepsilon (p)$ .

Proof. Using Proposition 2.5 we know that $H_{d^*}$ is positive semidefinite since $0_{[2]}$ is a global minimum of $\gamma _{d^*}$ . This in turn yields that $\gamma ^{(2)}_{d^*}\ge 0$ or equivalently $\Delta ^{(2)}_{d^*}\ge 0$ . Now, for any $d\lt d^*$ the unique minimizer of $\Delta _d$ is $p^*$ since $\Delta _d(p^*)=0$ , further $\Delta _d(p)\gt 0$ for any $p\neq p^*$ with $p_{\mathrm {e}}=p^*_{\mathrm {e}}$ and

\begin{align*} \Delta _d(p)=D_{\textrm {KL}}(p\parallel p^*)-\left (1-\frac {1}{d}\right )kD_{\textrm {KL}}(p_{\mathrm {e}}\parallel p^*_{\mathrm {e}}) \gt \Delta _{d^*}(p)\ge 0 \end{align*}

for any $p$ with $p_{\mathrm {e}}\neq p^*_{\mathrm {e}}$ . But the same argumentation shows that $p^*$ is the unique minimizer of $\Delta ^{(2)}_d$ , since $D_{\chi ^2}(\mu \parallel \mu ^*)$ is also minimal with value $0$ iff $\mu =\mu ^*$ . This in turn shows that $\gamma ^{(2)}_d$ is uniquely minimized at $0_{[2]}$ and hence $H_d$ is positive definite.

Let $\eta _{\mathrm {KL}}=\sup _{p\neq p^*}\frac {D_{\textrm {KL}}(p_{\mathrm {e}}\parallel p^*_{\mathrm {e}})}{D_{\textrm {KL}}(p\parallel p^*)}$ denote the contraction coefficient with respect to the KL divergence. Notice that by Proposition 2.5 we have $\frac {d^*}{(d^*-1)k}\ge \frac {D_{\textrm {KL}}(p_{\mathrm {e}}\parallel p^*_{\mathrm {e}})}{D_{\textrm {KL}}(p\parallel p^*)}$ for all $p\neq p^*$ with equality for $p=p^{(2)}$ , hence $\eta _{\mathrm {KL}}=\frac {d^*}{(d^*-1)k}$ (so Proposition 2.5 indeed confirms the conjecture by the authors in [Reference Panagiotou and Pasch36]). Further, let $\eta _{\chi ^2}=\sup _{p\neq p^*}\frac {D_{\chi ^2}(p_{\mathrm {e}}\parallel p^*_{\mathrm {e}})}{D_{\chi ^2}(p\parallel p^*)}$ denote the contraction coefficient with respect to the $\chi ^2$ divergence. The proof of Lemma 5.4 suggests that $\eta _{\chi ^2}\le \eta _{\mathrm {KL}}$ , a result known from the literature.

In the rest of this section we discuss the straightforward (but cumbersome) application of Laplace’s method for sums. For convenience, we first show that the boundaries can be neglected and derive the asymptotics of the sum on the interior using the uniform convergence established in Lemma 5.3.

Lemma 5.5. Let $d\in (0,d^*)$ and let $\mathcal U$ be a neighbourhood of $p^*$ such that its closure is contained in the interior of $\mathcal P(\{0,1,2\})$ . Then

\begin{align*} \frac {\mathbb {E}[Z^2]}{\mathbb {E}[Z]^2}=\sum _{\in \mathcal P_n}e(p)\sim \sum _{p\in \mathcal P_n\cap \mathcal U}\sqrt {\frac {d}{(2\pi )^2m^2\prod _sp(s)}}e^{-m\Delta _d(p)}\textrm {.} \end{align*}

Proof. Let $\Delta _{\min }\gt 0$ denote the global minimum of $\Delta _d$ on $\mathcal P(\{0,1,2\})\setminus \mathcal U$ . Now, we can use the well-known bounds $\frac {1}{a+1}\exp (aH(\frac {b}{a}))\le \binom {a}{b}\le \exp (aH(\frac {b}{a}))$ for binomial coefficients and the corresponding upper bound for multinomial coefficients (using the entropy of the distribution determined by the weights $\frac {b_i}{a}$ ) to derive

\begin{align*} \sum _{p\not \in \mathcal U}e(p) &\le \rho (n)\sum _{p\not \in \mathcal U}e^{-m\Delta _d(p)} \le \rho (n)e^{-m\Delta _{\min }}|\mathcal P_m(\{0,1,2\})| = \rho (n)\binom {m+1}{2}e^{-m\Delta _{\min }}\textrm {, where}\\ \rho (n)&=(n+1)\left (\frac {2dn}{k}+1\right )\left (\frac {(k-2)dn}{k}+1\right )\textrm {.} \end{align*}

Here, we used the form of $e(p)$ introduced at the beginning of this section and further notice that the bounds used are tight for the log-densities, i.e. the exponent is $\Delta _d(p)$ by the computations in Section 5.4. The right-hand side vanishes for $n$ tending to infinity, hence we have

\begin{align*} \frac {\mathbb {E}[Z^2]}{\mathbb {E}[Z]^2}\sim \sum _{p\in \mathcal U}e(p)\textrm {.} \end{align*}

Now, the result directly follows using Lemma 5.3 and Lemma 2.2.

Lemma 5.5 shows that the overlap distributions $p$ with material contributions $e(p)$ to the second moment are concentrated around $p^*$ . Hence, instead of considering a fixed neighbourhood $\mathcal U$ of $p^*$ we consider a sequence $(\mathcal U_n)_{n\in \mathcal N}$ of decreasing neighbourhoods. First, we choose a scaling that improves the assertion of Lemma 5.5 and further allows to simplify the asymptotics of the right-hand side, in the sense that the leading factor collapses to a constant and $\gamma _d(x)=\Delta _d(\iota ^*(x))$ can be replaced by its second order Taylor approximation $\gamma ^{(2)}_d(x)=\Delta ^{(2)}_d(\iota ^*(x))=\frac {1}{2}x^{\mathrm {t}}H_dx$ from above (7). For this purpose let $\mathcal U^*\subseteq \mathcal X^*$ be a sufficiently small neighbourhood of $0_{[2]}$ (in particular bounded away from the boundary of $\mathcal X^*$ ), further

\begin{align*} \mathcal U^*_n=\left \{x\in \mathcal X^*\,:\, \|x\|_2\lt \frac {\ln (m)}{\sqrt {m}}\right \}\textrm { and } \mathcal X^*_n=\left \{x-x^*\,:\,x\in \mathcal X_n\right \}\cap \mathcal U^*_n \textrm { for }n\in \mathcal N\textrm {.} \end{align*}

In the following we restrict to $n\ge n_0$ where $n_0\in \mathcal N$ is such that $\mathcal U^*_{n_0}\subseteq \mathcal U^*$ .

Lemma 5.6. For $d\in (0,d^*)$ we have

\begin{align*} \frac {\mathbb {E}[Z^2]}{\mathbb {E}[Z]^2} \sim \sqrt {\frac {d}{(2\pi )^2m^2\prod _sp^*(s)}}\sum _{x\in \mathcal X^*_n}e^{-\frac {m}{2}x^{\mathrm {t}}H_dx}\textrm {.} \end{align*}

Proof. First, notice that we can apply Lemma 5.5 to $\iota ^*(\mathcal U^*)$ . So, we need to show that the sum over $\mathcal U^*\setminus \mathcal U^*_n$ is negligible. Then we proceed to derive the asymptotics of the sum over $\mathcal U^*_n$ . Obviously, we have $\gamma _{\min }(n)\rightarrow 0$ for $n\rightarrow \infty$ with $\gamma _{\min }(n)=\min _{x\not \in \mathcal U^*_n}\gamma _d(x)\gt 0$ , since $\gamma _d(x)=\Delta _d(\iota ^*(x))$ is continuous and $\gamma _d(0_{[2]})=0$ . The main objective of the proof is to show that $\gamma _{\min }(n)$ converges to zero sufficiently slow. But with $\gamma ^{(2)}_d(x)=\frac {1}{2}x^{\mathrm {t}}H_dx$ from above (7) and for any $x\in \mathbb {R}^2$ we have

\begin{align*} \gamma ^{(2)}_d((v_1,v_2)x)&=\frac {1}{2}(\lambda _1x_1^2+\lambda _2x_2^2) \ge \frac {\lambda _1}{2}\|x\|_2^2=\frac {\lambda _1}{2}\|(v_1,v_2)x\|_2^2 \end{align*}

since $(v_1,v_2)$ is an orthonormal basis, so $\gamma ^{(2)}_d(x)\ge \frac {\lambda _1}{2}\|x\|_2^2$ for all $x\in \mathbb {R}^2$ . Now, for any sufficiently small $\varepsilon \in (0,1)$ let $c\in \mathbb {R}_{\gt 0}$ be the constant for $\mathcal B_{\varepsilon }(0_{[2]})$ from Taylor’s theorem applied to $\gamma _d$ at $0_{[2]}$ , then for any $x\in \mathcal U^*=\mathcal B_{\varepsilon }(0_{[2]})\cap \mathcal B_\delta (0_{[2]})$ , with $\delta =\frac {\varepsilon \lambda _1}{2c}$ , we have $\gamma _d(x)\ge (1-\varepsilon )\gamma ^{(2)}_d(x)$ since

\begin{align*} \gamma _d(x)-(1-\varepsilon )\gamma ^{(2)}_d(x)\ge \varepsilon \gamma ^{(2)}_d(x)-c\|x\|_2^3 \ge \left (\frac {\varepsilon \lambda _1}{2}-c\delta \right )\|x\|_2^2=0\textrm {.} \end{align*}

In combination we have $\gamma _d(x)\ge \frac {(1-\varepsilon )\lambda _1}{2}\|x\|_2^2$ and using $p=\iota ^*(x)$ hence

\begin{align*} \lim _{n\rightarrow \infty }\sum _{x\not \in \mathcal U^*_n}e(p) &=\lim _{n\rightarrow \infty }\sum _{x\in \mathcal U^*\setminus \mathcal U^*_n}\sqrt {\frac {d}{(2\pi )^2m^2\prod _sp(s)}}e^{-m\gamma _d(x)} \le \lim _{n\rightarrow \infty }Cme^{-\frac {(1-\varepsilon )\lambda _1}{2}\ln (m)^2} =0\textrm {,} \end{align*}

by using (the proof of) Lemma 5.5 and some sufficiently large constant $C$ . With this we have

\begin{align*} \frac {\mathbb {E}[Z^2]}{\mathbb {E}[Z]^2} \sim \sum _{x\in \mathcal X^*_n}e(\iota ^*(x)) \sim \sqrt {\frac {d}{(2\pi )^2m^2\prod _sp^*(s)}}\sum _{x\in \mathcal X^*_n}e^{-m\gamma ^{(2)}_d(x)}\textrm {,} \end{align*}

where the last equivalence follows from the fact that the leading factor converges to the respective constant uniformly on $\mathcal U^*_n$ and by (8) on $\mathcal U^*$ .

Lemma 5.6 completes the analytical part of the proof. For the last, measure theoretic, part we recall the bijection $\iota _n$ from Lemma 5.1. The translation of the sum on the right-hand side of Lemma 5.6 into a Riemann sum and further into the integral $\int g_\infty (x)\mathrm {d}x$ , where

\begin{align*} g_{\infty }\,:\,\mathbb {R}^2\rightarrow \mathbb {R}_{\gt 0},\quad y \mapsto \sqrt {\frac {d}{(2\pi )^2\prod _sp^*(s)}}\exp \left ({-}\frac {1}{2}y^{\mathrm {t}}H_dy\right )\textrm {, } \end{align*}

is essentially given by the grid $\mathcal X_n\subseteq (m^{-1}\mathbb {Z})^2\subseteq \mathbb {R}^2$ . We make this rigorous in the following.

Lemma 5.7. We have

\begin{align*} \frac {\mathbb {E}[Z^2]}{\mathbb {E}[Z]^2} \sim \int g_\infty (x)\mathrm {d}x\textrm {.} \end{align*}

Proof. We start with the partition of $\mathbb {R}^2$ into the squares

\begin{align*} \mathcal Q_{n,x}=\left \{x+\alpha _1\begin{pmatrix}1\\ 0\end{pmatrix}+\alpha _2\begin{pmatrix}0\\ 1\end{pmatrix}:\alpha \in \left [-\frac {1}{2m},\frac {1}{2m}\right )^2\right \}\textrm {, }x\in (m^{-1}\mathbb {Z})^2\textrm {.} \end{align*}

Next, we need a suitable selection of squares to cover the disc

\begin{align*} x^*+\mathcal U^*_n=\left \{x^*+x:x\in \mathbb {R}^2,\|x\|_2\lt \frac {\ln (m)}{\sqrt {m}}\right \}\subseteq \mathbb {R}^2 \end{align*}

corresponding to the disc $\mathcal U^*_n$ . For this purpose let $x_{\min }$ , $x_{\max }\in (m^{-1}\mathbb {Z})^2$ be given by

\begin{align*} x_{\min, 1}&=m^{-1}\left \lfloor m\left (x^*_1-\frac {\ln (m)}{\sqrt {m}}\right )\right \rfloor \textrm {, } x_{\min, 2}=m^{-1}\left \lfloor m\left (x^*_2-\frac {\ln (m)}{\sqrt {m}}\right )\right \rfloor \textrm {, }\\ x_{\max, 1}&=m^{-1}\left \lceil m\left (x^*_1+\frac {\ln (m)}{\sqrt {m}}\right )\right \rceil \textrm {, } x_{\max, 2}=m^{-1}\left \lceil m\left (x^*_2+\frac {\ln (m)}{\sqrt {m}}\right )\right \rceil \textrm {.} \end{align*}

Further, let $\mathcal{G}_n=(m^{-1}\mathbb {Z})^2\cap \left ([x_{\min, 1},x_{\max, 1}]\times [x_{\min, 2},x_{\max, 2}]\right )$ . By the definition of $x_{\min }$ and $x_{\max }$ the points on the boundary are not in $x^*+\mathcal U^*_n$ , which ensures that $x^*+\mathcal U^*_n\subseteq \mathcal Q_n$ with $\mathcal Q_n=\bigcup _{x\in \mathcal{G}_n}\mathcal Q_{n,x}$ . Further, we have $\mathcal Q_-\subseteq \mathcal Q_n\subseteq \mathcal Q_+$ with

\begin{align*} \mathcal Q_-=\left \{x\in \mathbb {R}^2:\|x-x^*\|_{\infty }\le \frac {\ln (m)}{\sqrt {m}}\right \} \textrm {, } \mathcal Q_+=\left \{x\in \mathbb {R}^2:\|x-x^*\|_{\infty }\le \frac {\ln (m)}{\sqrt {m}}+\frac {3}{2m}\right \}\textrm {,} \end{align*}

which ensures that $\mathcal Q_n\subseteq \mathcal X$ for $n\in \mathcal N$ sufficiently large. Now, we translate the notions back to $\mathcal X^*$ using the bijection $\tau \,:\,\mathbb {R}^2\rightarrow \mathbb {R}^2$ , $x\mapsto x-x^*$ , i.e. let $\mathcal{G}^*_n=\tau (\mathcal{G}_n)$ , $\mathcal Q^*_{n,x}=\tau (\mathcal Q_{n,\tau ^{-1}(x)})$ for $x\in \mathcal{G}^*_n$ , $\mathcal Q^*_n=\tau (\mathcal Q_n)$ , $\mathcal Q^*_-=\tau (\mathcal Q_-)$ and $\mathcal Q^*_+=\tau (\mathcal Q_+)$ . This directly gives

\begin{align*} \mathcal Q^*_{n,x}&=\left \{x+\alpha _1\begin{pmatrix}1\\ 0\end{pmatrix}+\alpha _2\begin{pmatrix}0\\ 1\end{pmatrix}:\alpha \in \left [-\frac {1}{2m},\frac {1}{2m}\right )^2\right \}\textrm {, }x\in \mathcal{G}^*_n\textrm {, } \mathcal Q^*_n=\bigcup _{x\in \mathcal{G}^*_n}\mathcal Q^*_{n,x}\textrm {,}\\ \mathcal Q^*_-&=\left \{x\in \mathbb {R}^2\,:\,\|x\|_\infty \le \frac {\ln (m)}{\sqrt {m}}\right \}\textrm {, } \mathcal Q^*_+=\left \{x\in \mathbb {R}^2\,:\,\|x\|_\infty \le \frac {\ln (m)}{\sqrt {m}}+\frac {3}{2m}\right \}\textrm {, } \end{align*}

and $\mathcal U^*_n\subseteq \mathcal Q^*_-\subseteq \mathcal Q^*_n\subseteq \mathcal Q^*_+\subseteq \mathcal X^*$ for $n\in \mathcal N$ sufficiently large. Further, with Lemma 5.6 and the definition of $\gamma ^{(2)}_d$ we now have

\begin{align*} \frac {\mathbb {E}[Z^2]}{\mathbb {E}[Z]^2}\sim \sum _{x\in \mathcal{G}^*_n}\sqrt {\frac {d}{(2\pi )^2m^2\prod _sp^*(s)}}\exp \left ({-}\frac {1}{2}mx^{\mathrm {t}}H_dx\right ). \end{align*}

Finally, we need to adjust the scaling to turn the sum on the right-hand side into a Riemann sum. For this purpose let $\sigma \,:\,\mathbb {R}^2\rightarrow \mathbb {R}^2$ , $x\mapsto \sqrt {m}x$ , further $\mathcal{G}^{\prime}_n=\sigma (\mathcal{G}^*_n)$ , $\mathcal Q^{\prime}_{n,x}=\sigma (\mathcal Q^*_{n,\sigma ^{-1}(x)})$ for $x\in \mathcal{G}^{\prime}_n$ , $\mathcal Q^{\prime}_n=\sigma (\mathcal Q^*_n)$ , $\mathcal Q^{\prime}_-=\sigma (\mathcal Q^*_-)$ and $\mathcal Q^{\prime}_+=\sigma (\mathcal Q^*_+)$ . This directly gives

\begin{align*} \mathcal Q^{\prime}_{n,x}&=\left \{x+\alpha _1\begin{pmatrix}1\\ 0\end{pmatrix}+\alpha _2\begin{pmatrix}0\\ 1\end{pmatrix}:\alpha \in \left [-\frac {1}{2\sqrt {m}},\frac {1}{2\sqrt {m}}\right )^2\right \}\textrm {, }x\in \mathcal{G}^{\prime}_n\textrm {, } \mathcal Q^{\prime}_n=\bigcup _{x\in \mathcal{G}^{\prime}_n}\mathcal Q^{\prime}_{n,x}\textrm {,}\\ \mathcal Q^{\prime}_-&=\left \{x\in \mathbb {R}^2:\|x\|_\infty \le \ln (m)\right \}\textrm {, } \mathcal Q^{\prime}_+=\left \{x\in \mathbb {R}^2\,:\,\|x\|_\infty \le \ln (m)+\frac {3}{2\sqrt {m}}\right \}\textrm {, } \end{align*}

and $\mathcal Q^{\prime}_-\subseteq \mathcal Q^{\prime}_n\subseteq \mathcal Q^{\prime}_+$ . Using that $mx^{\mathrm {t}}H_dx=\sigma (x)^{\mathrm {t}}H_d\sigma (x)$ for all $x\in \mathcal{G}^*_n$ and further that the area of $\mathcal Q^{\prime}_{n,x}$ is $m^{-1}$ for all $x\in \mathcal{G}^{\prime}_n$ we have

\begin{align*} \frac {\mathbb {E}[Z^2]}{\mathbb {E}[Z]^2}&\sim \sum _{x\in \mathcal{G}^{\prime}_n}\sqrt {\frac {d}{(2\pi )^2m^2\prod _sp^*(s)}}\exp \left ({-}\frac {1}{2}x^{\mathrm {t}}H_dx\right )=\int g_n(y)\mathrm {d}y\textrm {,}\\ g_n(y)&=\sum _{x\in \mathcal{G}^{\prime}_n}\unicode {x1D7D9}\{y\in \mathcal Q^{\prime}_{n,x}\}\sqrt {\frac {d}{(2\pi )^2\prod _sp^*(s)}}\exp \left ({-}\frac {1}{2}x^{\mathrm {t}}H_dx\right )\textrm {, }y\in \mathbb {R}^2\textrm {.} \end{align*}

In order to show that $\int g_n(y)\mathrm {d}y$ converges to $\int g_{\infty }(y)\mathrm {d}y$ we recall from Lemma 5.4 that $H_d$ is positive definite, which ensures that $\int g_{\infty }(y)\mathrm {d}y$ exists and is finite. Now, using Taylor’s theorem with order $0$ and the Lagrange form of the first order remainder with the fact that the absolutes of the first derivatives of $g_{\infty }$ are bounded from above yields a constant $c\in \mathbb {R}_{\gt 0}$ such that for all $n\in \mathcal N$ and all $y\in \mathcal Q^{\prime}_n$ , with $x\in \mathcal{G}^{\prime}_n$ such that $y\in \mathcal Q^{\prime}_{n,x}$ , we have

\begin{align*} \|g_{\infty }(y)-g_n(y)\|_{\infty }=\|g_{\infty }(y)-g_{\infty }(x)\|_{\infty }\le \frac {c}{\sqrt {m}}\textrm {.} \end{align*}

This bound directly suggests that

\begin{align*} &\left |\int \unicode {x1D7D9}\{y\in \mathcal Q^{\prime}_{n,x}\}g_{\infty }(y)\mathrm {d}y-\int \unicode {x1D7D9}\{y\in \mathcal Q^{\prime}_{n,x}\}g_{n}(y)\mathrm {d}y\right |\le cm^{-\frac {3}{2}}\textrm {, }\\ &\left |\int \unicode {x1D7D9}\{y\in \mathcal Q^{\prime}_{n}\}g_{\infty }(y)\mathrm {d}y-\int \unicode {x1D7D9}\{y\in \mathcal Q^{\prime}_{n}\}g_{n}(y)\mathrm {d}y\right |\le \frac {c}{\sqrt {m}}\left (2\ln (m)+\frac {3}{\sqrt {m}}\right )^2\textrm { and }\\ &\left |\int g_{\infty }(y)\mathrm {d}y-\int g_{n}(y)\mathrm {d}y\right |\le \frac {c}{\sqrt {m}}\left (2\ln (m)+\frac {3}{\sqrt {m}}\right )^2+\int \unicode {x1D7D9}\{y\not \in \mathcal Q^{\prime}_{n}\}g_{\infty }(y)\mathrm {d}y\textrm {.} \end{align*}

In particular the last bound suggests that $\int g_n(y)\mathrm {d}y\rightarrow \int g_{\infty }(y)\mathrm {d}y$ since the error on the right-hand side tends to zero as $n$ tends to infinity.

The only remaining part of the proof is to compute $\int g_\infty (x)\mathrm {d}x$ .

Proof. The Gaussian integral gives

\begin{align*} \int g_\infty (x)\mathrm {d}x =\sqrt {\frac {d}{(2\pi )^2\prod _sp^*(s)}}\sqrt {\frac {(2\pi )^2}{\det (H_d)}} =\sqrt {\frac {d}{\det (H_d)\prod _sp^*(s)}}\textrm {.} \end{align*}

Recall that the Hessian $H_d\in \mathbb {R}^{2\times 2}$ from (6) is a $2\times 2$ matrix and all entries are given explicitly, so a straightforward calculation asserts that $\det (H_d)=(k-d)d/[(k-1)\prod _sp^*(s)]$ .

Finally, combining Lemma 5.7 with Lemma 5.8 completes the proof of Proposition 2.6.

5.6 Proof of Proposition 2.5

This section is dedicated to the identification of the minimizers of $\Delta _d$ . First, we show that the stationary points of $\Delta _d$ including their characterization are in one to one correspondence with the fixed points of a ratio of belief propagation messages for the constraint satisfaction problem defined in Section 5.1, cf. Lemmas 5.9 to 5.12 below. Considering the ratio turns out to be sufficient and allows to consider a one-dimensional problem. For more background on the correspondence of stationary points of $\Delta _d$ and fixed points of belief propagation we refer the reader to [Reference Mézard and Montanari31, Reference Yedidia, Freeman and Weiss41].

The major advantage of this equivalent fixed point problem, cf.Equation (9) below, is that the base function $\iota _{\mathrm {BP}}^*$ is a simple rational function and the target function $\iota _{\mathrm {BP}}$ is obtained from $\iota _{\mathrm {BP}}^*$ by a single exponentiation. Thus, in the second step we discuss $\iota _{\mathrm {BP}}^*$ and introduce a piecewise linear bound on $\iota _{\mathrm {BP}}^*$ on $[1,k]$ given by $g_1^*$ and $g_k^*$ , cf.Lemma 5.13 below. The piecewise function given by $g_1^*$ and $g_k^*$ on $[1,k]$ and $\iota _{\mathrm {BP}}^*$ on $[k,\infty )$ is increasing and piecewise convex, which allows to establish the unique fixed point of $\iota _{\mathrm {BP}}$ on $(1,\infty )$ (and for reasonable choices of $d$ ) in Lemmas 5.16 and 5.17. For symmetry reasons this also completes the discussion for $k=4$ .

In the last step, we establish that $\iota _{\mathrm {BP}}$ has no fixed points on $(0,1)$ , for reasonable $d$ and $k\gt 4$ , cf.Lemma 5.19 below. By combining the results we obtain that $\iota _{\mathrm {BP}}$ has two fixed points (three for $k=4$ ), one at $1$ corresponding to the desired interior minimum of $\Delta _d$ and one in $[k,\infty )$ corresponding to a saddle point of $\Delta _d$ (two for $k=4$ , and for reasonable $d$ ). This allows to identify the two minimizers of $\Delta _{d^*}$ (three for $k=4$ ) and completes the proof.

We begin with characterizing the stationary points of $\Delta _d$ for any $d\in \mathbb {R}_{\gt 0}$ .

In order to pin them down we first determine the stationary points of the restriction of $\Delta _d$ to overlap distributions with the same fixed edge distribution. For this purpose, recall the line $\mathcal W\subseteq \mathcal P(\{0,1\}^2)$ of attainable edge distributions and the lines $\mathcal P_q=\iota (\mathcal X_q)=\{p\in \mathcal P(\{0,1,2\})\,:\,p_{\mathrm {e}}=q\}$ of overlap distributions with fixed edge distribution $q\in \mathcal W$ from Lemma 5.2. Further, let $\Delta _{d,q}\,:\,\mathcal P_q\rightarrow \mathbb {R}$ denote the restriction of $\Delta _d$ to $\mathcal P_q$ . For $x\in \mathbb {R}_{\gt 0}$ let $p_x\in \mathcal P(\{0,1,2\})$ be given by $p_x(s)=p^*(s)x^s/\sum _sp^*(s)x^s$ , $s\in \{0,1,2\}$ , further let $p_{0}=p^{(0)}$ , $p_{\infty }=p^{(2)}$ , and $\mathcal P_{\min }=\{p_x\,:\,x\in [0,\infty ]\}$ . Finally, let $\iota _{\mathrm {rp}}\,:\,[0,\infty ]\rightarrow \mathcal P_{\min }$ , $x\mapsto p_x$ , denote the induced map and $\iota _{\mathrm {pe}}\,:\,\mathcal P_{\min }\rightarrow \mathcal W$ , $p\mapsto p_{\mathrm {e}}$ , the corresponding edge distributions.

Proof. Recall from Lemma 5.2 that $\mathcal P_q$ is one-dimensional for $q\in \mathcal W\setminus \{p^{(0)}_{\mathrm {e}},p^{(2)}_{\mathrm {e}}\}$ . Further, the map $\Delta _{d,q}$ is strictly convex since the KL divergence $D_{\textrm {KL}}(p\parallel p^*)$ (respectively $x\ln (x)$ ) is and further $D_{\textrm {KL}}(p_{\mathrm {e}}\parallel p^*_{\mathrm {e}})=D_{\textrm {KL}} (q\parallel p^*_{\mathrm {e}})$ is constant. Now, fix an interior point $p_\circ \in \mathcal P_q$ and let a boundary point $p_{\mathrm {b}}\in \mathcal P_q$ be given. Then $p_{\mathrm {b}}$ is not fully supported since it is on the boundary of $\mathcal P(\{0,1,2\}$ and hence the derivative of $D_{\textrm {KL}}(\alpha p_{\circ }+(1-\alpha )p_{\mathrm {b}}\parallel p^*)$ tends to $-\infty$ as $\alpha$ tends to $0$ , which shows that $\Delta _{d,q}$ is not minimized on the boundary. Hence, we know that there exists exactly one stationary point $p_q\in \mathcal P_q$ and that $\Delta _{d,q}(p)$ is minimal iff $p=p_q$ . As discussed in Lemma 5.2 we have $\mathcal P_q=\{p^{(s)}\}$ for $q=p^{(s)}_{\mathrm {e}}$ and $s\in \{0,2\}$ , so $p_q=p^{(s)}$ is obviously the unique global minimizer of $\Delta _{d,q}$ in this case and further $\Delta _{d,q}$ has no stationary points (since $\mathcal P_q$ has empty interior). This shows that the map $q\mapsto p_q$ for $q\in \mathcal W$ is a bijection.

Further, for $q$ in the interior of $\mathcal W$ the stationary point $p_q$ is fully supported and the unique root of the first derivative of $\Delta _{d,q}$ in the direction $b_2$ from (4), i.e.

\begin{align*} \ln \left (\frac {p_q(0)}{p^*(0)}\right )+\ln \left (\frac {p_q(2)}{p^*(2)}\right )=2\ln \left (\frac {p_q(1)}{p^*(1)}\right )\textrm { or equivalently } \frac {p_q(2)/p^*(2)}{p_q(1)/p^*(1)}=\frac {p_q(1)/p^*(1)}{p_q(0)/p^*(0)}\textrm {.} \end{align*}

Let $\mathcal P^{\prime}_{\min }$ denote the set of all fully supported $p\in \mathcal P(\{0,1,2\})$ satisfying $\frac {p(2)/p^*(2)}{p(1)/p^*(1)}=\frac {p(1)/p^*(1)}{p(0)/p^*(0)}$ , i.e. our set of candidates for stationary points. Now, for $p\in \mathcal P^{\prime}_{\min }$ let $q=p_{\mathrm {e}}$ , then we obviously have $p\in \mathcal P_q$ and $p$ is a root of the first derivative of $\Delta _{d,q}$ in the direction $b_2$ , so $p$ is the unique root and $p=p_q$ . Hence, the map $\iota _{\mathrm {pe}}^{\prime}\,:\,\mathcal P^{\prime}_{\min }\rightarrow \mathcal W$ , $p\mapsto p_{\mathrm {e}}$ , is a bijection (up to the corners of $\mathcal W$ ) with inverse $q\mapsto p_q$ . Now, let $\iota _{\mathrm {pr}}:\mathcal P^{\prime}_{\min }\rightarrow \mathbb {R}_{\gt 0}$ , $p\mapsto x_p$ , with $x_p=\frac {p(1)p^*(0)}{p^*(1)p(0)}$ . Notice that $\iota _{\mathrm {pr}}$ is surjective since for any $x\in \mathbb {R}_{\gt 0}$ we have

\begin{align*} \frac {p_x(2)/p^*(2)}{p_x(1)/p^*(1)}=\frac {x^2}{x}=x=\frac {p_x(1)/p^*(1)}{p_x(0)/p^*(0)} \end{align*}

and hence $p_x\in \mathcal P^{\prime}_{\min }$ . To show that $\iota _{\mathrm {pr}}$ is injective let $p\in \mathcal P^{\prime}_{\min }$ and $x=x_p$ . Using the definition of $x_p$ and the defining property of $\mathcal P^{\prime}_{\min }$ we get

\begin{align*} p(0)&=p^*(0)\frac {p(0)}{p^*(0)}\textrm {, } p(1)=p^*(1)x\frac {p(0)}{p^*(0)}\textrm {, } p(2)=p^*(2)x\frac {p(1)}{p^*(1)}=p^*(2)x^2\frac {p(0)}{p^*(0)}\textrm {, so}\\ p(s)&=\frac {p(s)}{p(0)+p(1)+p(2)}=\frac {p^*(s)x^s}{\sum _sp^*(s)x^s}=p_x(s)\textrm {, }s\in \{0,1,2\}\textrm {.} \end{align*}

This shows that $\mathcal P_{\min }=\mathcal P^{\prime}_{\min }\cup \{p_0,p_\infty \}$ , that $\iota _{\mathrm {rp}}$ is a bijection with inverse $\iota _{\mathrm {pr}}$ (canonically extended to the endpoints), and finally that $\iota _{\mathrm {pe}}=\iota ^{\prime}_{\mathrm {pe}}$ is a bijection as well.

Lemma 5.9 has a few immediate consequences. For one, the only minimizers of $\Delta _d$ in the direction $b_2$ , from (4), on the boundary are $p^{(0)}$ and $p^{(2)}$ , while all other boundary points are maximizers in the direction $b_2$ , hence if $p$ is a global minimizer of $\Delta _d$ on the boundary, we have $p\in \{p^{(0)},p^{(2)}\}$ . Further, all stationary points of $\Delta _d$ are either local minima or saddle points. Finally, we have $p\in \mathcal P_{\min }$ for any stationary point $p\in \mathcal P(\{0,1,2\})$ of $\Delta _d$ since then also the derivative in the direction of $b_2$ vanishes.

For the upcoming characterization of the stationary points of $\Delta _d$ let

\begin{align*} \iota _{\mathrm {rr}}\,:\,\mathbb {R}_{\gt 0}\rightarrow \mathbb {R}_{\gt 0},\quad \iota _{\mathrm {rr}}(x)=\iota ^*_{\mathrm {rr}}(x)^{\frac {d-1}{d}}\textrm {, } \iota ^*_{\mathrm {rr}}(x)=\frac {p_{x,\mathrm {e}}(1,1)p_{x,\mathrm {e}}(0,0)}{p_{x,\mathrm {e}}(1,0)p_{x,\mathrm {e}}(0,1)} \textrm {.} \end{align*}

Notice that $\iota _{\mathrm {rr}}(x)\in \mathbb {R}_{\gt 0}$ for $x\in \mathbb {R}_{\gt 0}$ since then $p_x$ is fully supported and hence $p_{x,\mathrm {e}}$ is fully supported by Lemma 5.2. Finally, let $\mathcal X_{\mathrm {st}}=\{x\in \mathbb {R}_{\gt 0}\,:\,\iota _{\mathrm {rr}}(x)=x\}$ denote the fixed points of $\iota _{\mathrm {rr}}$ and $\mathcal P_{\mathrm {st}}=\{p_x\,:\,x\in \mathcal X_{\mathrm {st}}\}$ the corresponding distributions. Notice that $p_x=p^*$ for $x=1$ and further $\iota _{\mathrm {rr}}^*(1)=1$ , i.e. $\iota _{\mathrm {rr}}(1)=1$ for all $d\in \mathbb {R}_{\gt 0}$ , hence $1\in \mathcal X_{\mathrm {st}}$ and $p^*\in \mathcal P_{\mathrm {st}}$ for all $d\in \mathbb {R}_{\gt 0}$ .

Proof. Using Lemma 5.9, a fully supported distribution $p\in \mathcal P(\{0,1,2\})$ is a stationary point of $\Delta _d$ iff there exists $x\in \mathbb {R}_{\gt 0}$ such that $p=p_x$ and the derivative of $\Delta _d$ at $p_x$ in the direction $b_1$ vanishes, i.e. $p_x$ is a solution of

\begin{align*} 0=\left (\left (\ln \left (\frac {p_x(s)}{p^*(s)}\right )\right )_{s\in \{0,1,2\}}^{\mathrm {t}}-\frac {(d-1)k}{d}\left (\ln \left (\frac {p_{x,\mathrm {e}}(y)}{p^*_{\mathrm {e}}(y)}\right )\right )_{y\in \{0,1\}^2}^{\mathrm {t}}W\right )b_1\textrm {,} \end{align*}

where we used the chain rule for multivariate calculus, that $W$ is column stochastic and that $b_1\in 1_{\{0,1,2\}}^{\perp }$ . Recall from Section 5.3, e.g. from the proof of Lemma 5.1, that $Wb_1=\frac {d}{k}w$ , hence computing the dot product with $b_1$ gives

\begin{align*} 0=d\ln (x)-(d-1)\ln \left (\frac {p_{x,\mathrm {e}}(1,1)p_{x,\mathrm {e}}(0,0)}{p_{x,\mathrm {e}}(1,0)p_{x,\mathrm {e}}(0,1)}\right )\textrm {.} \end{align*}

Obviously, equality holds if and only if $x\in \mathcal X_{\mathrm {st}}$ , hence $p$ is a stationary point of $\Delta _d$ iff $p\in \mathcal P_{\mathrm {st}}$ .

Lemma 5.10 does not only allow to translate the stationary points of $\Delta _d$ into fixed points of $\iota _{\mathrm {rr}}$ , it also allows to translate the types as follows.

Proof. Fix $x\in \mathbb {R}_{\gt 0}$ . The proof of Lemma 5.10 directly suggests that the first derivative of $\Delta _d$ at $p_x$ in the direction $b_1$ is strictly positive iff

\begin{align*} 0\lt \ln (x)-\frac {d-1}{d}\ln (\iota ^*_{\mathrm {rr}}(x))\textrm {,} \end{align*}

which holds iff $\iota _{\mathrm {rr}}(x)\lt x$ . We’re left to establish that the direction of $\iota _{\mathrm {rp}}$ is consistent with $b_1$ . Intuitively, using Lemma 5.2 and Lemma 5.9 we can argue that $x\mapsto p_{x,\mathrm {e}}(1,1)$ is a bijection and hence either increasing or decreasing. Taking the limits $x\rightarrow 0$ and $x\rightarrow \infty$ suggests that it is increasing, hence with $c\in \mathbb {R}^2$ given by $\iota ^{\prime}_{\mathrm {rp}}(x)=(b_1,b_2)c$ , we know that $c_1\ge 0$ .

Formally, we quantify the direction of $\iota _{\mathrm {rp}}$ . For this purpose we compute the derivative of $\iota _{\mathrm {rp}}$ at $x\in \mathbb {R}_{\gt 0}$ , given by

\begin{align*} \iota ^{\prime}_{\mathrm {rp}}(x)=\left (\frac {sp^*(s)x^{s-1}\sum _{s^{\prime}\in \{0,1,2\}}p^*(s^{\prime})x^{s^{\prime}}-p^*(s)x^{s}\sum _{s^{\prime}\in \{0,1,2\}}s^{\prime}p^*(s^{\prime})x^{s^{\prime}-1}}{\left (\sum _{s^{\prime}\in \{0,1,2\}}p^*(s^{\prime})x^{s^{\prime}}\right )^2}\right )_{s\in \{0,1,2\}}\textrm {.} \end{align*}

Notice that $v=\iota ^{\prime}_{\mathrm {rp}}(x)\in 1_{\{0,1,2\}}^{\perp }$ , since $\iota _{\mathrm {rp}}(\mathbb {R}_{\gt 0})\subseteq \mathcal P(\{0,1,2\})$ or by computing $\sum _sv_s=0$ directly. Now, let $c\in \mathbb {R}^2$ be given by $v=(b_1,b_2)c$ . This directly gives $c_2=v_2$ and hence $c_1=d^{-1}(v_1+2v_2)=d^{-1}\sum _ssv_s$ . Now, notice that

\begin{align*} S&=dxc_1 =\sum _{s,s^{\prime}\in \{0,1,2\}}p_x(s)p_x(s^{\prime})s(s-s^{\prime})\\ &=\sum _{s\gt s^{\prime}}p_x(s)p_x(s^{\prime})s(s-s^{\prime})-\sum _{s\gt s^{\prime}}p_x(s)p_x(s^{\prime})s^{\prime}(s-s^{\prime}) =\sum _{s\gt s^{\prime}}p_x(s)p_x(s^{\prime})(s-s^{\prime})^2\gt 0\textrm {,} \end{align*}

which directly gives $c_1=\frac {S}{dx}\in \mathbb {R}_{\gt 0}$ . Now, with $\nabla =\left (\frac {\partial \Delta _d}{\partial p(s)}(p_x)\right )_{s\in \{0,1,2\}}\in \mathbb {R}^{\{0,1,2\}}$ denoting the partial derivatives of $\Delta _d$ at $p_x$ and using the chain rule we have

\begin{align*} (\Delta _d\circ \iota _{\mathrm {rp}})^{\prime}(x) &=\nabla ^{\mathrm {t}}\iota ^{\prime}_{\mathrm {rp}}(x) =c_1\nabla ^{\mathrm {t}}b_1 +c_2\nabla ^{\mathrm {t}}b_2 =c_1\nabla ^{\mathrm {t}}b_1\textrm {,} \end{align*}

since the derivative $\nabla ^{\mathrm {t}}b_2$ of $\Delta _d$ at $p_x$ in the direction $b_2$ is zero, hence we have $(\Delta _d\circ \iota _{\mathrm {rp}})^{\prime}(x)\gt 0$ iff the derivative $\nabla ^{\mathrm {t}}b_1$ of $\Delta _d$ at $p_x$ in the direction $b_1$ is strictly positive, which is the case iff $\iota _{\mathrm {rr}}(x)\lt x$ .

Lemma 5.11 with Lemma 5.10 shows that control over $\iota _{\mathrm {rr}}$ gives complete control over the location and characterization of the stationary points of $\Delta _d$ . However, instead of solving the fixed point equation given by $\iota _{\mathrm {rr}}$ directly, we use a slight modification inspired by the belief propagation algorithm applied to the constraint satisfaction discussed in Section 5.1 and initialized with uniform messages.

For this purpose let $N\in \mathbb {Z}_{\ge 0}$ , further $N_1$ , $N_2\in [N]_0$ and the hypergeometric distribution $p_{N,N_1,N_2}\in \mathcal P(\mathbb {Z})$ be given by

\begin{align*} p_{N,N_1,N_2}(s)=\frac {\binom {N_1}{s}\binom {N-N_1}{N_2-s}}{\binom {N}{N_2}}=\frac {\binom {N}{N-N_1-N_2+s,N_1-s,N_2-s,s}}{\binom {N}{N_1}\binom {N}{N_2}}\textrm { for }s\in \mathbb {Z}\textrm {.} \end{align*}

The latter form directly shows that $p_{N,N_1,N_2}=p_{N,N_2,N_1}$ . Now, for $y\in \{0,1\}^2$ let $p^*_y\in \mathcal P(\{0,1,2\})$ be given by $p^*_{(1,1)}(s)=p_{k-1,1,1}(s-1)$ , $p^*_{(1,0)}(s)=p_{k-1,1,2}(s)$ , $p^*_{(0,1)}(s)=p_{k-1,2,1}(s)$ , $p^*_{(0,0)}(s)=p_{k-1,2,2}(s)$ for $s\in \{0,1,2\}$ . On the other hand, for $y\in \{0,1\}^2$ let $p^{\prime}_y\in \mathcal P(\{0,1,2\})$ be given by $p^{\prime}_{(1,1)}(s)=p_{k-1,1,1}(s)$ for $s\in \{0,1,2\}$ and further $p^{\prime}_y=p^*_y$ for $y\in \{0,1\}^2\setminus \{(1,1)\}$ . Finally, for a distribution $p\in \mathcal P(\{0,1,2\})$ let $f_p\,:\,\mathbb {R}_{\ge 0}\rightarrow \mathbb {R}_{\ge 0}$ , $x\mapsto \sum _{s\in \{0,1,2\}}p(s)x^s$ , be its probability generating function, and further $\iota _{\mathrm {BP}}\,:\,\mathbb {R}_{\gt 0}\rightarrow \mathbb {R}_{\gt 0}$ given by

(9) \begin{align} \iota _{\mathrm {BP}}(x)=\iota ^*_{\mathrm {BP}}(x)^{d-1}\textrm {, } \iota ^*_{\mathrm {BP}}(x)=\frac {f_{p^{\prime}_{(1,1)}}(x)f_{p^{\prime}_{(0,0)}}(x)}{f_{p^{\prime}_{(1,0)}}(x)f_{p^{\prime}_{(0,1)}}(x)}\textrm {, for }x\in \mathbb {R}_{\gt 0}\textrm {.} \end{align}

Proof. First, notice that the normalization constant of $p_x$ cancels out in $\iota ^*_{\mathrm {rr}}$ , as does the normalization constant $\binom {k}{2}^2$ of $p^*$ . Further, with $v=(k-4+s,2-s,2-s,s)^{\mathrm {t}}\in \mathbb {R}^{\{0,1\}^2}$ we have $W_{y,s}\binom {k}{v}=\frac {v_y}{k}\binom {k}{v}=\binom {k-1}{v-(\delta _{y,z})_z}$ for $y\in \{0,1\}^2$ , $s\in \{0,1,2\}$ , and thereby

\begin{align*} \iota ^*_{\mathrm {rr}}(x)=\frac {\left (\sum _s\binom {k-1}{k-4+s,2-s,2-s,s-1}x^s\right )\left (\sum _s\binom {k-1}{k-5+s,2-s,2-s,s}x^s\right )}{\left (\sum _s\binom {k-1}{k-4+s,2-s,1-s,s}x^s\right )\left (\sum _s\binom {k-1}{k-4+s,1-s,2-s,s}x^s\right )} \textrm { for }x\in \mathbb {R}_{\gt 0}. \end{align*}

Now, since the normalization constants cancel out in total, this directly gives

\begin{align*} \iota ^*_{\mathrm {rr}}(x)=\frac {f_{p^*_{(1,1)}}(x)f_{p^*_{(0,0)}}(x)}{f_{p^*_{(1,0)}}(x)f_{p^*_{(0,1)}}(x)} =x\iota ^*_{\mathrm {BP}}(x) \textrm { for }x\in \mathbb {R}_{\gt 0}\textrm {,} \end{align*}

using that $p^*_{(1,1)}(s)=p^{\prime}_{(1,1)}(s-1)$ for $s\in \{0,1,2\}$ , hence $f_{p^*_{(1,1)}}(x)=xf_{p^{\prime}_{(1,1)}}(x)$ , and $p^*_y(s)=p^{\prime}_y(s)$ for $y\neq (1,1)$ . Now, we have $x=\iota _{\mathrm {rr}}(x)$ iff $x=x^{\frac {d-1}{d}}\iota _{\mathrm {BP}}(x)^{\frac {1}{d}}$ , which holds iff $x^{\frac {1}{d}}=\iota _{\mathrm {BP}}(x)^{\frac {1}{d}}$ , which then again is equivalent to $x=\iota _{\mathrm {BP}}(x)$ . Equivalence of the inequalities follows analogously.

The following part is dedicated to the identification of the fixed points of $\iota _{\mathrm {BP}}$ . We start with a discussion of $\iota ^*_{\mathrm {BP}}$ . For this purpose let $g^*_1$ , $g^*_k\,:\,\mathbb {R}_{\ge 1}\rightarrow \mathbb {R}_{\ge 1}$ be given by

\begin{align*} g^*_1(x)=\frac {1}{k-1}(x-1)+1\textrm { and } g^*_k(x)=\frac {13k-12}{27(k-1)(k-2)}(x-k)+\frac {2(7k-12)}{9(k-2)}\textrm {, }x\in \mathbb {R}_{\ge 1}\textrm {.} \end{align*}

Lemma 5.13. For any $k\in \mathbb {Z}_{\ge 4}$ we have

\begin{align*} \lim _{x\rightarrow 0}\iota ^*_{\mathrm {BP}}(x)=\frac {k-4}{k-3}\textrm {, } \quad \iota ^*_{\mathrm {BP}}(1)=1=g^*_1(1)\quad \textrm {and} \quad \iota ^*_{\mathrm {BP}}(k)=g^*_k(k)\textrm {.} \end{align*}

For the first derivative we have

\begin{align*} {\iota ^*}^{\prime}_{\mathrm {BP}}(1)={g^*}^{\prime}_1(1)\textrm {, } \quad {\iota ^*}^{\prime}_{\mathrm {BP}}(k)={g^*}^{\prime}_k(k)\quad \textrm {and} \quad \lim _{x\rightarrow \infty }{\iota ^*}^{\prime}_{\mathrm {BP}}(x)=\frac {1}{2(k-2)}\textrm {.} \end{align*}

Moreover, for the second derivative we have

\begin{align*} {\iota ^*}^{\prime\prime}_{\mathrm {BP}}(x)\lt 0\textrm { for }x\in (0,k)\textrm {, } \quad {\iota ^*}^{\prime\prime}_{\mathrm {BP}}(k)=0\textrm {,} \quad {\iota ^*}^{\prime\prime}_{\mathrm {BP}}(x)\gt 0\textrm { for }x\in \mathbb {R}_{\gt k}\textrm {.} \end{align*}

For $k=4$ and $x\in \mathbb {R}_{\gt 0}$ we have $\iota ^*_{\mathrm {BP}}(x^{-1})=\left (\iota ^*_{\mathrm {BP}}(x)\right )^{-1}$ .

Proof. Using $f_p(1)=1$ for the moment generating function of any finitely supported law $p$ , we have $\iota ^*_{\mathrm {BP}}(1)=1$ . Further, using that the first moment of a hypergeometric distribution $p_{N,N_1,N_2}$ is $\frac {N_1N_2}{N}$ and that $f^{\prime}_p(1)$ is the first moment of $p$ , we have ${\iota ^*}^{\prime}_{\mathrm {BP}}(1)=\frac {1}{k-1}+\frac {4}{k-1}-2\cdot \frac {2}{k-1}=\frac {1}{k-1}$ . The symmetry of $\iota ^*_{\mathrm {BP}}$ for the special case $k=4$ can be seen as follows. First, recall that $p_{N,N_1,N_2}(s)=p_{N,N-N_1,N_2}(N_2-s)$ for any hypergeometric distribution $p_{N,N_1,N_2}$ . For $s\in \{0,1,2\}$ this gives

\begin{align*} p^{\prime}_{(0,0)}(s)&=p_{3,2,2}(s)=p_{3,1,2}(2-s)=p^{\prime}_{(1,0)}(2-s)\\ &=p^{\prime}_{(0,1)}(2-s)=p_{3,2,1}(2-s)=p_{3,1,1}(s-1)=p^{\prime}_{(1,1)}(s-1)\textrm {.} \end{align*}

These relations can be directly translated to the moment generating functions, i.e.

\begin{align*} f_{p^{\prime}_{(0,0)}}(x)&=x^2f_{p^{\prime}_{(1,0)}}(x^{-1})=x^2f_{p^{\prime}_{(0,1)}}(x^{-1})=xf_{p^{\prime}_{(1,1)}}(x) \end{align*}

for $x\in \mathbb {R}_{\gt 0}$ . Using these transformations we have

\begin{align*} \iota ^*_{\mathrm {BP}}(x^{-1}) =\frac {f_{p^{\prime}_{(1,1)}}(x^{-1})f_{p^{\prime}_{(0,0)}}(x^{-1})}{f_{p^{\prime}_{(1,0)}}(x^{-1})f_{p^{\prime}_{(0,1)}}(x^{-1})} =\frac {x^{-1}f_{p^{\prime}_{(1,0)}}(x)x^{-2}f_{p^{\prime}_{(0,1)}}(x)}{x^{-1}f_{p^{\prime}_{(1,1)}}(x)x^{-2}f_{p^{\prime}_{(0,0)}}(x)} =\left (\iota ^*_{\mathrm {BP}}(x)\right )^{-1}\textrm {.} \end{align*}

For $k\in \mathbb {Z}_{\ge 4}$ and $x\in \mathbb {R}_{\gt 0}$ direct computation gives

\begin{align*} \iota ^*_{\mathrm {BP}}(x)&=\frac {1}{2(k-2)}x+\frac {2k-5}{2(k-2)}+\frac {(k-1)(k-3)(x-1)}{2(k-2)(2x+k-3)^2}\textrm {,}\\ {\iota ^*}^{\prime}_{\mathrm {BP}}(x)&=\frac {1}{2(k-2)}+\frac {(k-1)(k-3)({-}2x+k+1)}{2(k-2)(2x+k-3)^3}\textrm {,}\\ {\iota ^*}^{\prime\prime}_{\mathrm {BP}}(x)&=\frac {4(k-1)(k-3)(x-k)}{(k-2)(2x+k-3)^4}\textrm {.} \end{align*}

The remaining assertions follow immediately or with routine computations.

Lemma 5.13 has the following immediate consequences.