Mean Field at Distance One

Abstract

To be able to understand how infectious diseases spread on networks, it is important to understand the network structure itself in the absence of infection. In this text we consider dynamic network models that are inspired by the (static) configuration network. The networks are described by population-level averages such as the fraction of the population with k partners, k = 0, 1, 2, … This means that the bookkeeping contains information about individuals and their partners, but no information about partners of partners. Can we average over the population to obtain information about partners of partners? The answer is ‘it depends’, and this is where the mean field at distance one assumption comes into play. In this text we explain that, yes, we may average over the population (in the right way) in the static network. Moreover, we provide evidence in support of a positive answer for the network model that is dynamic due to partnership changes. If, however, we additionally allow for demographic changes, dependencies between partners arise. In earlier work we used the slogan ‘mean field at distance one’ as a justification of simply ignoring the dependencies. Here we discuss the subtleties that come with the mean field at distance one assumption, especially when demography is involved. Particular attention is given to the accuracy of the approximation in the setting with demography. Next, the mean field at distance one assumption is discussed in the context of an infection superimposed on the network. We end with the conjecture that an extension of the bookkeeping leads to an exact description of the network structure.

Appendices

Appendix 1: Relationship Between $\boldsymbol{\pi _{1}(a)}$ and $\boldsymbol{H}$

The probability density function π ₁(a) for the age of an occupied binding site (see Eq. (5.13)) is related to the probability density function H(a, α) for the ages of two partners u and v in a randomly chosen partnership. This relation is formulated in Eq. (5.20).

Lemma A.1.

$$\displaystyle{ \int _{\alpha =0}^{\infty }H(a,\alpha )d\alpha =\pi _{ 1}(a). }$$

(5.20)

Proof.

First note that we can rewrite H(a, α) as

$$\displaystyle{ H(a,\alpha ) = \frac{\mu e^{-\mu a}} {1 - F}\int _{\xi =0}^{\min (a,\alpha )}\rho F\varphi (a-\xi )\pi _{ 0}(\alpha -\xi )e^{-(\sigma +\mu )\xi }d\xi. }$$

Next, one finds that

$$\displaystyle{ \int _{\alpha =0}^{\infty }\int _{ \xi =0}^{\min (a,\alpha )}\rho F\varphi (a-\xi )\pi _{ 0}(\alpha -\xi )e^{-(\sigma +\mu )\xi }d\xi d\alpha = 1 -\varphi (a), }$$

(5.21)

by direct calculations using Eqs. (5.11) and (5.13) for φ and π ₀, and we conclude that Eq. (5.20) holds. One can also reason as follows for Eq. (5.21): ρFφ(a −ξ)π ₀(α −ξ)e ^−(σ+μ)ξ is the probability that a binding site with age a has a partner with age α and partnership duration ξ given that the owner of the binding site under consideration does not die. By integrating over all possible partnership durations 0 ≤ ξ ≤ min(a, α), we obtain the probability that a binding site with age a has a partner with age α (given that the owner does not die): ∫ ₀ ^min(a, α) ρFφ(a −ξ)π ₀(α −ξ)e ^−(σ+μ)ξ dξ. Then finally, by integrating over all possible α ≥ 0 we obtain the probability 1 −φ(a) that a binding site with age a is occupied.

Appendix 2: The Marginal Degree Distribution $\boldsymbol{Q_{k} =\sum _{l}P(k,l)}$

We obtain the probability Q _k that an individual involved in a randomly chosen partnership has degree k from the joint probability distribution P(k, l), cf. (5.17), by summing over all l = 1, …, n, i.e. Q _k = ∑ _{l = 1} ⁿ P(k, l). On the other hand, in previous work [7] we have derived an expression for Q _k from the stable degree distribution (P _k ) _k in the population: Q _k = kP _k ∕n(1 − F). We show that both ways of arriving at Q _k yield the same expression, i.e. 

$$\displaystyle{ Q_{k} =\sum _{ l=1}^{n}P(k,l) = kP_{ k}/n(1 - F). }$$

(5.22)

First, we work out the right-hand side. P _k is expressed in terms of the probability φ(a) as follows:

$$\displaystyle{ P_{k} = \binom{n}{k}\int _{0}^{\infty }\mu e^{-\mu a}\varphi (a)^{n-k}(1 -\varphi (a))^{k}da. }$$

On the other hand, we can simplify ∑ _{l = 1} ⁿ P(k, l). First of all, note that since q _l (a) is a probability distribution, ∑ _{l = 1} ⁿ q _l (α) = 1. Therefore

$$\displaystyle\begin{array}{rcl} & & \sum _{l=1}^{n}P(k,l) \\ & & =\sum _{ l=1}^{n}\int _{ a=0}^{\infty }\int _{ \alpha =0}^{\infty }\int _{ 0}^{\min (a,\alpha )}q_{ k}(a)q_{l}(\alpha )\frac{\rho F^{2}\pi _{0}(a-\xi )\pi _{0}(\alpha -\xi )e^{-(\sigma +2\mu )\xi }} {1 - F} d\xi d\alpha da \\ & & =\int _{ a=0}^{\infty }\int _{ \alpha =0}^{\infty }\int _{ 0}^{\min (a,\alpha )}q_{ k}(a)\frac{\rho F^{2}\pi _{0}(a-\xi )\pi _{0}(\alpha -\xi )e^{-(\sigma +2\mu )\xi }} {1 - F} d\xi d\alpha da. {}\end{array}$$

(5.23)

Next, note that we can simplify Eq. (5.23) as follows:

$$\displaystyle\begin{array}{rcl} \sum _{l=1}^{n}P(k,l)& & {}\\ & =& \frac{k\binom{n}{k}} {n(1 - F)}\int _{a=0}^{\infty }\mu e^{-\mu a}\varphi (a)^{n-k}(1 -\varphi (a))^{k}(1 -\varphi (a))^{-1} {}\\ & & \int _{\alpha =0}^{\infty }\int _{ \xi =0}^{\min (a,\alpha )}\rho F^{2}e^{\mu \xi }\frac{\varphi (a-\xi )} {F} \frac{\mu e^{-\mu (\alpha -\xi )}\varphi (\alpha -\xi )} {F} e^{-(\sigma +2\mu )\xi }d\xi d\alpha da {}\\ & =& \frac{k\binom{n}{k}} {n(1 - F)}\int _{a=0}^{\infty }\mu e^{-\mu a}\varphi (a)^{n-k}(1 -\varphi (a))^{k}(1 -\varphi (a))^{-1} {}\\ & & \int _{\alpha =0}^{\infty }\int _{ 0}^{\min (a,\alpha )}\rho F\varphi (a-\xi )\pi _{ 0}(\alpha -\xi )e^{-(\sigma +\mu )\xi }d\xi d\alpha da {}\\ & =& \frac{k\binom{n}{k}} {n(1 - F)}\int _{a=0}^{\infty }\mu e^{-\mu a}\varphi (a)^{n-k}(1 -\varphi (a))^{k}(1 -\varphi (a))^{-1}(1 -\varphi (a))da. {}\\ \end{array}$$

Here we used Eq. (5.20) in the third equality. So we find that Eq. (5.22) indeed holds.

Appendix 3: The Correlation Coefficient as a Function of Model Parameters

We work out Eq. (5.18) by computing A, B, and C, defined by Eq. (5.19).

$$\displaystyle\begin{array}{rcl} A& =& \sum _{k=1}^{n}\sum _{ l=1}^{n}klP(k,l) =\int _{ a=0}^{\infty }\int _{ \alpha =0}^{\infty }\sum _{ k=1}^{n}kq_{ k}(a)\sum _{l=1}^{n}lq_{ l}(\alpha )H(a,\alpha )dad\alpha \\ & =& \int _{a=0}^{\infty }\int _{ \alpha =0}^{\infty }\big(n(1 -\varphi (a)) +\varphi (a)\big)\big(n(1 -\varphi (\alpha )) +\varphi (\alpha )\big)H(a,\alpha )dad\alpha \\ & =& \frac{1} {(1 - F)(\rho F +\sigma +2\mu )^{2}(2\rho F + 3\sigma + 4\mu )(2(\rho F +\sigma +\mu )+\mu )^{2}} \\ & & \phantom{=\ }\Big\{\rho (\mu ^{2}\left (\rho ^{2}F^{2}(n(33n + 46) + 1) + 2\rho F\sigma (92n + 33) + 179\sigma ^{2})\right. \\ & & \phantom{=\ } \quad \left.+4\mu (\rho Fn+\sigma )(4\rho ^{2}F^{2}n + 10\rho F\sigma (n + 1) + 19\sigma ^{2}\right ) \\ & & \phantom{=\ } \quad + 6\mu ^{3}(2\rho F(8n + 3) + 31\sigma ) + 4\sigma (2\rho F + 3\sigma )(\rho Fn+\sigma )^{2} + 72\mu ^{4})\Big\}, {}\end{array}$$

(5.24)

where the last equality is calculated using Mathematica.

We already calculated the mean B = ∑ _{k = 1} ⁿ kQ _k in [7, eq. (23)]. For completeness, we work it out using the probability distribution (P(k, l)).

$$\displaystyle\begin{array}{rcl} B& =& \sum _{k=1}^{n}kQ_{ k} =\sum _{ k=1}^{n}k\sum _{ l=1}^{n}P(k,l) \\ & =& \int _{a=0}^{\infty }\sum _{ k=1}^{n}kq_{ k}(a)\int _{\alpha =0}^{\infty }\sum _{ l=1}^{n}q_{ l}(\alpha )H(a,\alpha )d\alpha da \\ & =& \int _{a=0}^{\infty }\big(n(1 -\varphi (a)) +\varphi (a)\big)\int _{\alpha =0}^{\infty }H(a,\alpha )d\alpha da \\ & =& \int _{a=0}^{\infty }\big(n(1 -\varphi (a)) +\varphi (a)\big)\pi _{ 1}(a)da \\ & =& 1 + \frac{2\rho F(n - 1)} {2(\rho F +\sigma +\mu )+\mu }. {}\end{array}$$

(5.25)

In the first equality we used that (Q _k ) is the marginal distribution of (P(k, l)), and in the fifth equality we used identity (5.20).

Finally, we consider the second moment C = ∑ _{k = 1} ⁿ k ² Q _k :

$$\displaystyle\begin{array}{rcl} C& =& \sum _{k=1}^{n}k^{2}Q_{ k} =\sum _{ k=1}^{n}k^{2}\sum _{ l=1}^{n}P(k,l) \\ & =& \int _{a=0}^{\infty }\sum _{ k=1}^{n}k^{2}q_{ k}(a)\int _{\alpha =0}^{\infty }\sum _{ l=1}^{n}q_{ l}(\alpha )H(a,\alpha )d\alpha da \\ & =& \int _{a=0}^{\infty }\big(n^{2}(1 -\varphi (a))^{2} +\varphi (a)((3n - 1)(1 -\varphi (a)) +\varphi (a))\big)\int _{\alpha =0}^{\infty }H(a,\alpha )d\alpha da \\ & =& \int _{a=0}^{\infty }\big(n^{2}(1 -\varphi (a))^{2} +\varphi (a)((3n - 1)(1 -\varphi (a)) +\varphi (a))\pi _{ 1}(a)da \\ & =& \frac{\rho F(\mu (12\mu +\rho F(24n - 7) + 17\sigma ) + 6(\rho ^{2}F^{2}n^{2} +\rho F(3n - 1)\sigma +\sigma ^{2}))} {(1 - F)(\rho F +\sigma +2\mu )(2(\rho F +\sigma +\mu )+\mu )(3(\rho F +\sigma +\mu )+\mu )}. {}\end{array}$$

(5.26)

Inserting Eqs. (5.24), (5.25), and (5.26) together in Eq. (5.18), we find an explicit expression for the correlation coefficient corr. Note that the variance of a random variable is always nonnegative (and nonzero if the random variable is not equal to a constant). Therefore, we find that the sign of Cov(D _u , D _v ) = A − B ² determines the sign of the correlation coefficient corr in Eq. (5.18). Note that identity (5.9) for F allows us to express σ in terms of the other parameters: σ = ρF ²∕(1 − F) − 2μ. For clarity, we use this identity for σ in the numerator (but not in the denominator) in the simplification of Cov(D _u , D _v ). We find that

$$\displaystyle{ A - B^{2} = \frac{\mu ^{2}\rho ^{3}F^{2}(1 - F)(n - 1)^{2}} {(1 - F)^{2}(\rho F +\sigma +2\mu )^{2}(2\rho F + 3\sigma + 4\mu )(2\rho F + 2\sigma + 3\mu )^{2}}. }$$

(5.27)

In particular, the covariance (and therefore the correlation coefficient corr) is strictly larger than zero if ρ > 0, σ > 0, μ > 0, and n > 1.