Assortativity and correlation coefficients

This is a technical follow-up to a previous post on assortative mixing in networks. In a footnote, I claimed that Newman’s (2003) assortativity coefficient equals the Pearson correlation coefficient when there are two possible node types. This post proves that claim.

Notation

Consider an undirected network $N$ in which each node has a type belonging to a (finite) set $T$ . The assortativity coefficient is defined as $r = \frac{\sum_{t \in T} x_{t t} - \sum_{t \in T} y_{t}^{2}}{1 - \sum_{t \in T} y_{t}^{2}},$ where $x_{s t}$ is the proportion of edges joining nodes of type $s$ to nodes of type $t$ , and where $y_{t} = \sum_{s \in T} x_{s t}$ is the proportion of edges incident with nodes of type $t$ . The Pearson correlation of adjacent nodes’ types is given by $ρ = \frac{Cov (t_{i}, t_{j})}{\sqrt{Var (t_{i}) Var (t_{j})}},$ where $t_{i} \in T$ and $t_{j} \in T$ are the types of nodes $i$ and $j$ , and where (co)variances are computed with respect to the frequency at which nodes of type $t_{i}$ and $t_{j}$ are adjacent in $N$ .

Proof

Let $T = {a, b} \subset R$ with $a \neq b$ . I show that the correlation coefficient $ρ$ and assortativity coefficient $r$ can be expressed as the same function of $y_{a}$ and $x_{a b}$ , implying $ρ = r$ .

Consider $ρ$ . It can be understood by presenting the mixing matrix $X = (x_{s t})$ in tabular form:

$t_{i}$	$t_{j}$	$x_{t_{i} t_{j}}$
$a$	$a$	$x_{a a}$
$a$	$b$	$x_{a b}$
$b$	$a$	$x_{b a}$
$b$	$b$	$x_{b b}$

The first two columns enumerate the possible type pairs $(t_{i}, t_{j})$ and the third column stores the proportion of adjacent node pairs $(i, j)$ with each type pair. This third column defines the joint distribution of types across adjacent nodes. Thus $ρ$ equals the correlation of the first two columns, weighted by the third column. (Here $x_{a b} = x_{b a}$ since $N$ is undirected.) Now $t_{i}$ has mean $\begin{aligned} E [t_{i}] & = x_{a a} a + x_{a b} a + x_{b a} b + x_{b b} b \\ = y_{a} a + y_{b} b \end{aligned}$ and second moment $\begin{aligned} E [t_{i}^{2}] & = x_{a a} a^{2} + x_{a b} a^{2} + x_{b a} b^{2} + x_{b b} b^{2} \\ = y_{a} a^{2} + y_{b} b^{2}, \end{aligned}$ and similar calculations reveal $E [t_{j}] = E [t_{i}]$ and $E [t_{j}^{2}] = E [t_{i}^{2}]$ . Thus $t_{i}$ has variance $\begin{aligned} Var (t_{i}) & = E [t_{i}^{2}] - E [t_{i}]^{2} \\ = y_{a} a^{2} + y_{b} b^{2} - (y_{a} a + y_{b} b)^{2} \\ = y_{a} (1 - y_{a}) a^{2} + y_{b} (1 - y_{b}) b^{2} - 2 y_{a} y_{b} a b \end{aligned}$ and similarly $Var (t_{j}) = Var (t_{i})$ . We can simplify this expression for the variance by noticing that $x_{a a} + x_{a b} + x_{b a} + x_{b b} = 1,$ which implies $\begin{aligned} y_{b} & = x_{a b} + x_{b b} \\ = 1 - x_{a a} - x_{b a} \\ = 1 - y_{a} \end{aligned}$ and therefore $\begin{aligned} Var (t_{i}) & = y_{a} (1 - y_{a}) a^{2} + (1 - y_{a}) y_{a} b^{2} - 2 y_{a} (1 - y_{a}) a b \\ = y_{a} (1 - y_{a}) (a - b)^{2} . \end{aligned}$ We next express the covariance $Cov (t_{i}, t_{j}) = E [t_{i} t_{j}] - E [t_{i}] E [t_{j}]$ in terms of $y_{a}$ and $x_{a b}$ . Now $\begin{aligned} E [t_{i} t_{j}] & = x_{a a} a^{2} + x_{a b} a b + x_{b a} a b + x_{b b} b^{2} \\ = (y_{a} - x_{a b}) a^{2} + 2 x_{a b} a b + (y_{b} - x_{a b}) b^{2} \\ = y_{a} a^{2} + y_{b} b^{2} - x_{a b} (a - b)^{2} \end{aligned}$ because $x_{a b} = x_{b a}$ . It follows that $\begin{aligned} Cov (t_{i}, t_{j}) & = y_{a} a^{2} + y_{b} b^{2} - x_{a b} (a - b)^{2} - (y_{a} a + y_{b} b)^{2} \\ = y_{a} (1 - y_{a}) a^{2} + y_{b} (1 - y_{b}) b^{2} - 2 y_{a} y_{b} a b - x_{a b} (a - b)^{2} \\ = y_{a} (1 - y_{a}) (a - b)^{2} - x_{a b} (a - b)^{2}, \end{aligned}$ where the last line uses the fact that $y_{b} = 1 - y_{a}$ . Putting everything together, we have $\begin{aligned} ρ & = \frac{Cov (t_{i}, t_{j})}{\sqrt{Var (t_{i}) Var (t_{j})}} \\ = \frac{y_{a} (1 - y_{a}) - x_{a b}}{y_{a} (1 - y_{a})}, \end{aligned}$ a function of $y_{a}$ and $x_{a b}$ .

Now consider $r$ . Its numerator equals $\begin{aligned} \sum_{t \in T} x_{t t} - \sum_{t \in T} y_{t}^{2} & = x_{a a} + x_{b b} - y_{a}^{2} - y_{b}^{2} \\ = (y_{a} - x_{a b}) + (y_{b} - x_{a b}) - y_{a}^{2} - y_{b}^{2} \\ = y_{a} (1 - y_{a}) + y_{b} (1 - y_{b}) - 2 x_{a b} \\ \overset{⋆}{=} 2 y_{a} (1 - y_{a}) - 2 x_{a b} \end{aligned}$ and its denominator equals $\begin{aligned} 1 - \sum_{t \in T} y_{t}^{2} & = 1 - y_{a}^{2} - y_{b}^{2} \\ \overset{⋆ ⋆}{=} 1 - y_{a}^{2} - (1 - y_{a})^{2} \\ = 2 y_{a} (1 - y_{a}), \end{aligned}$ where $⋆$ and $⋆ ⋆$ both use the fact that $y_{b} = 1 - y_{a}$ . Thus $r = \frac{y_{a} (1 - y_{a}) - x_{a b}}{y_{a} (1 - y_{a})},$ the same function of $y_{a}$ and $x_{a b}$ , and so $ρ = r$ as claimed.

Writing $ρ = r$ in terms of $y_{a}$ and $x_{a b}$ makes it easy to check the boundary cases: if there are no within-type edges then $y_{a} = x_{a b} = 1 / 2$ and so $ρ = r = - 1$ ; if there are no between-type edges then $x_{a b} = 0$ and so $ρ = r = 1$ .

Appendix: Constructing the mixing matrix

The proof relies on noticing that $x_{a b} = x_{b a}$ , which comes from undirectedness of the network $N$ and from how the mixing matrix $X$ is constructed. I often forget this construction, so here’s a simple algorithm: Consider some type pair $(s, t)$ . Look at the edges beginning at type $s$ nodes and count how many end at type $t$ nodes. Call this count $m_{s t}$ . Do the same for all type pairs to obtain a matrix $M = (m_{s t})$ of edge counts. Divide the entries in $M$ by their sum to obtain $X$ .