**Freakonometrics - Tag - R-english**, and kindly contributed to R-bloggers)

Following the course, in order to define assocation measures (from Kruskal (1958)) or concordance measures (from Scarsini (1984)), define a *concordance function* as follows: let be a random pair with copula , and with copula . Then define

the so-called *concordance function*. Thus

As proved last week in class,

Based on that function, several concordance measures can be derived. A popular measure is Kendall’s tau, from Kendall (1938), defined as i.e.

which is simply . Here, computation can be tricky. Consider the following sample,

> set.seed(1) > n=40 > library(mnormt) > X=rmnorm(n,c(0,0), + matrix(c(1,.4,.4,1),2,2)) > U=cbind(rank(X[,1]),rank(X[,2]))/(n+1)

Then, using R function, we can obtain Kendall’s tau easily,

> cor(X,method="kendall")[1,2] [1] 0.3794872

To get our own code (and to understand a bit more how to get that coefficient), we can use

> i=rep(1:(n-1),(n-1):1) > j=2:n > for(k in 3:n){j=c(j,k:n)} > M=cbind(X[i,],X[j,]) > concordant=sum((M[,1]-M[,3])*(M[,2]-M[,4])>0) > discordant=sum((M[,1]-M[,3])*(M[,2]-M[,4])<0) > total=n*(n-1)/2 > (K=(concordant-discordant)/total) [1] 0.3794872

or the following (we’ll use random variable quite frequently),

> i=rep(1:n,each=n) > j=rep(1:n,n) > Z=((X[i,1]>X[j,1])&(X[i,2]>X[j,2])) > (K=4*mean(Z)*n/(n-1)-1) [1] 0.3794872

Another measure is Spearman’s rank correlation, from Spearman (1904),

where has distribution .

Here, which leads to the following expressions

Numerically, we have the following

> cor(X,method="spearman")[1,2] [1] 0.5388368 > cor(rank(X[,1]),rank(X[,2])) [1] 0.5388368

Note that it is also possible to write

Another measure is the cograduation index, from Gini (1914), obtained by sybstituting an L1 norm instead of a L2 one in the previous expression,

Note that this index can also be defined as . Here,

> Rx=rank(X[,1]);Ry=rank(X[,2]); > (G=2/(n^2) *(sum(abs(Rx+Ry-n-1))- + sum(abs(Rx-Ry)))) [1] 0.41

Finally, another measure is the one from Blomqvist (1950). Let denote the median of , i.e.

Then define

or equivalently

> Mx=median(X[,1]);My=median(X[,2]) > (B=4*sum((X[,1]<=Mx)*((X[,2]<=My)))/n-1) [1] 0.4

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