Suppose I have data $$(a_i,b_i)_{i=1}^n$$ on two random variables $$A$$ and $$B$$. I store my data as vectors a and b, and compute their correlation using the cor function in R:

cor(a, b)

##  0.4326075


Now suppose I append a mirrored version of my data by defining the vectors

alpha = c(a, b)
beta = c(b, a)


so that alpha is a concatenation of the $$a_i$$ and $$b_i$$ values, and beta is a concatenation of the $$b_i$$ and $$a_i$$ values. I compute the correlation of alpha and before as before:

cor(alpha, beta)

##  0.4288428


Notice that cor(a, b) and cor(alpha, beta) are not equal. This surprised me. How can appending a copy of the same data change the correlation within those data?

The answer is that the concatenated data $$(\alpha_i,\beta_i)_{i=1}^{2n}$$ have different marginal distributions than the original data $$(a_i,b_i)_{i=1}^n$$. Indeed one can show that \DeclareMathOperator{\Cor}{Cor} \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \begin{align} \E[\alpha]=\E[\beta]=\frac{\E[a]+\E[b]}{2} \end{align} and \begin{align} \E[\alpha^2]=\E[\beta^2]=\frac{\E[a^2]+\E[b^2]}{2}, \end{align} where $$\E[\alpha]\equiv\frac{1}{2n}\sum_{i=1}^n\alpha_i$$ is the empirical mean of the $$\alpha_i$$ values, and where $$\E[\beta]$$, $$\E[a]$$, and $$\E[b]$$ are defined similarly. It turns out that $$\E[\alpha\beta]=\E[ab]$$, but since the marginal distributions are different the empirical correlations are different. In fact $$\Cor(\alpha,\beta)=\frac{\Cov(a,b)-0.25\left(\E[a]+\E[b]\right)^2}{0.5\Var(a)+0.5\Var(b)+0.25\left(\E[a]-\E[b]\right)^2},$$ where $$\Cor$$, $$\Cov$$, and $$\Var$$ are the empirical correlation, covariance, and variance operators. This expression implies that cor(alpha, beta) and cor(a, b) will be equal if the $$a_i$$ and $$b_i$$ values have the same means and variances. We can achieve this by scaling a and b before computing their correlation:

cor(scale(a), scale(b))

##  0.4326075


The scale function de-means its argument and scales it to have unit variance. These operations don’t change the correlation of a and b. But they do change the correlation of alpha and beta:

alpha = c(scale(a), scale(b))
beta = c(scale(b), scale(a))

cor(alpha, beta)

##  0.4326075


Now the two correlations agree!

I came across this phenomenon while writing my previous post, in which I discuss the degree assortativity among nodes in Zachary’s (1977) karate club network. One way to measure this assortativity is to use the degree_assortativity function in igraph:

library(igraph)

G = graph.famous('Zachary')

assortativity_degree(G)

##  -0.4756131


This function returns the correlation of the degrees of adjacent nodes in G. Another way to compute this correlation is to

1. construct a matrix el in which rows correspond to edges and columns list incident nodes;
2. define the vectors d1 and d2 of degrees among the nodes listed in el;
3. compute the correlation of d1 and d2 using cor.

Here’s what I get when I take those three steps:

el = as_edgelist(G)

d = degree(G)
d1 = d[el[, 1]]  # Ego degrees
d2 = d[el[, 2]]  # Alter degrees

cor(d1, d2)

##  -0.4769563


Notice that cor(d1, d2) disagrees with the value of assortativity_degree(G) computed above. This is because the vectors d1 and d2 have different means and variances:

c(mean(d1), mean(d2))

##  7.487179 8.051282

c(var(d1), var(d2))

##  25.94139 32.23110


These differences come from el listing each edge only once: it includes a row c(i, j) for the edge between nodes $$i$$ and $$j\not=i$$, but not a row c(j, i). Whereas assortativity_degree accounts for edges being undirected by adding the row c(j, i) before computing the correlation. This is analogous to the “append the mirrored data” step I took to create $$(\alpha_i,\beta_i)_{i=1}^{2n}$$ above. Appending the mirror of el to itself before computing cor(d1, d2) returns the same value as assortativity_degree(G):

el = rbind(
el,
matrix(c(el[, 2], el[, 1]), ncol = 2)  # el's mirror
)

d1 = d[el[, 1]]
d2 = d[el[, 2]]

c(assortativity_degree(G), cor(d1, d2))

##  -0.4756131 -0.4756131