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.