Measures of Assortativity

Abstract

This paper discusses alternative measures of assortative matching and relates them to Sewall Wright’s F-statistic. It also explores applications of measures of assortativity to evolutionary dynamics. We generalize Wright’s statistic to allow the possibility that some types match more assortatively than others, and explore the possibility of identifying parameters of this more general model from the observed distribution of matches by the partners’ types.

Appendix

Proof of Theorem 1

Proof

To motivate Wright’s F-statistic as a correlation, let us construct two random variables I _A and I _B as follows. If one randomly selects one matched pair and then randomly chooses one individual from that pair, let I _A be the random variable that takes on the value 1 or 0 depending on whether this individual is a type 1 or a type 2. Let I _B be the random variable that takes the value 1 or 0 depending on whether the remaining member of the selected pair is of type 1 or type 2. Wright’s F is the correlation coefficient between the random variable I _A and I _B . This correlation coefficient is, by definition,

$$ \rho = \frac{E(I_{A}I_{B})-E(I_{A})E(I_{B})}{\sigma(I_{A})\sigma(I_{B})} $$

(32)

Now E(I _A ) = E(I _B ) = p, and $\sigma(I_{A})=\sigma(I_{B})=\sqrt{p(1-p)}.$ Also $E(I_{A}I_{B})=\pi_{11}=p\pi(1\vert 1).$ Therefore Eq. 32 can be written as

$$ \rho=\frac{p\pi(1\vert 1)-p^{2}}{p(1-p)}=\frac{\pi(1\vert 1)-p}{1-p} $$

(33)

This establishes Eq. 1.$\square$

Since $\pi(1\vert 1)=1-\pi(2\vert 1),$ Eq. 33 can be written as

$$ \rho=\frac{1-\pi(2\vert 1)-p}{1-p}=1-\frac{\pi(2\vert 1)}{1-p} $$

(34)

Then, since $\pi(1,2)=p\pi(2\vert 1),$ it follows that

$$ \rho=1-\frac{\pi_{12}}{p(1-p)} $$

(35)

This establishes Eq. 2.

Since $\pi_{12}=p\pi(2\vert 1)=(1-p)\pi(1\vert 2),$ it must be that

$$ \pi(1\vert 2)=\frac{p}{1-p}\pi(2\vert 1)=\frac{p}{1-p}\left(1-\pi (1\vert 1)\right). $$

(36)

From Eq. 1 it follows that $\pi(1\vert 1)=(1-p)F(p) +p.$ Therefore Eq. 36 simplifies to

$$ \pi(1\vert 2)=p\left(1-F(p)\right). $$

(37)

Therefore we have

$$ \begin{aligned} \pi(1\vert 1)-\pi(1\vert 2)&=(1-p)F(p)+p-p\left(1-F(p)\right) &=F(p) \end{aligned} $$

(38)

This establishes Eq. 3.

Proof of Theorem 3

Proof

The mapping from the G _i ’s and p _i ’s to the probabilities π _ij is immediate from Eqs. 10 and 11.$\square$

To find the inverse mapping from the π _ij ’s to the G _i ’s and p _i ’s, we proceed as follows. From Eq. 10, it follows that

$$ \frac{2\left(p_{1}p_{2}G_{1}G_{2}\right)\left(p_{1}p_{3}G_{1}G_{3}\right)} {p_{2}p_{3} G_{2}G_{3}}= \frac{\pi_{12}\pi_{13}}{\pi_{23}} \left(\sum_{k=1}^{3}p_{k}G_{k}\right) $$

(39)

Simplifying and rearranging Eq. 39, we have

$$ p_{1}G_{1}=\frac{1}{\sqrt{2}}\frac{\sqrt{\pi_{12}\pi_{13}}}{ \sqrt{\pi_{23}}}\sqrt{\sum_{k}p_{k}G_{k}}= \frac{1}{\sqrt{2}}\frac{\sqrt{\pi_{12}\pi_{23}\pi_{13}}}{\pi_{23}} \sqrt{\sum_{k}p_{k}G_{k}}. $$

(40)

Symmetric reasoning shows that also

$$ p_{2}G_{2} = \frac{1}{\sqrt{2}}\frac{\sqrt{\pi_{12}\pi_{23}\pi_{13}}} {\pi_{13}} \sqrt{\sum_{k}p_{k}G_{k}} $$

(41)

$$ p_{3}G_{3} = \frac{1}{\sqrt{2}}\frac{\sqrt{\pi_{12} \pi_{23}\pi_{13}}}{\pi_{12}} \sqrt{\sum_{k}p_{k}G_{k}} $$

(42)

Summing the terms in Eqs. 40–42, we have

$$ \sum_{k}p_{k}G_{k}=\frac{1}{\sqrt{2}}\sqrt{\pi_{12}\pi_{23}\pi_{13}} \sqrt{\sum_{k}p_{k}G_{k}} \left(\frac{1}{\sqrt{\pi_{23}}+\sqrt{\pi_{13}}+\sqrt{\pi_{12}}}\right) $$

(43)

which in turn implies

$$ \sqrt{\sum_{k}p_{k}G_{k}}=\frac{1}{\sqrt{2}}\sqrt{\pi_{12}\pi_{23}\pi_{13}} \left(\frac{1}{\sqrt{\pi_{23}}+\sqrt{\pi_{13}}+\sqrt{\pi_{12}}}\right) $$

(44)

From Eqs. 40–42 it then follows that

$$ p_{1}G_{1}=\frac{1}{2}\pi_{12}\pi_{13}\left(\frac{1}{\sqrt{\pi_{23}} +\sqrt{\pi_{13}}+\sqrt{\pi_{12}}}\right) $$

(45)

$$ p_{2}G_{2}=\frac{1}{2}\pi_{12}\pi_{23}\left(\frac{1}{\sqrt{\pi_{23}} +\sqrt{\pi_{13}}+\sqrt{\pi_{12}}}\right) $$

(46)

$$ p_{3}G_{3}=\frac{1}{2}\pi_{13}\pi_{23}\left(\frac{1}{\sqrt{\pi_{23}} +\sqrt{\pi_{13}}+\sqrt{\pi_{12}}}\right) $$

(47)

Since

$$p_{i}=\pi_{ii}+\frac{1}{2}\sum_{j\neq i}\pi_{ij} $$

(48)

the p _i ’s are uniquely determined by the π _ij ’s. Given that the p _i ’s are uniquely determined, it follows from Eqs. 40–42 that the G _i ’s are also uniquely determined by the π _ij ’s. This proves Theorem 3.

Proof of Theorem 6

Proof

The function $F(\cdot)$ is continuous and strictly decreasing for p in the interval [1/2, 1], with $F(1/2)=\frac{s-m}{s+m}$ and F(1) = 0. Therefore $F^{-1}(\cdot)$ is a continuous, decreasing function from the interval $[0,\frac{s-m}{s+m}]$ onto [0,1/2]. Our assumptions imply that function $\rho(\cdot)$ is a continuous, increasing function from (0,x*] onto the interval $(\rho_{0},\frac{s-m}{s+m}].$ Therefore there is a well-defined function p(x) = F ⁻¹(ρ(x)) mapping the non-empty interval (0, x*] onto [1/2,1). Since $\rho(\cdot)$ is an increasing function and $F^{-1}(\cdot)$ is a decreasing function, it must be that $p(\cdot)$ is a decreasing function of x.$\square$

Let $x\in (0,x^{*}]$ and suppose that the fraction p(x) of the population seeking matches are of the type that contributes x and the fraction 1 − p(x) are of the type that contributes 0. Then the expected payoff of a type x is $\pi(x\vert x)b(x)-c(x)$ and the expected payoff of a type 0 is $\pi(x\vert 0)b(x).$ The difference between the expected payoffs of the two types is F(p(x)b((x) − c(x). From the definition of p(x), it follows that $F\left(p(x)\right)=\rho(x)$ and hence $b(x)F\left(p(x)\right)-c(x)=0.$ Therefore, when the fraction p(x) are of type x and fraction 1 − p(x) are of type 0, the expected fitnesses of the two types are equal. A mutant individual of another type, who contributes a non-zero amount x′ ≠ x will almost always meet either a type x or a type 0. The probability that a type x′ matches with a type x is the same as the probability that a type 0 matches with a type x. Therefore the fitness of a type x′ is $\pi(x\vert 0)b(x)-c(x')<\pi(x\vert 0,$ which is the expected payoff of the two incumbent types x and 0.