Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

632 Nonparametric and Robust Statistics

bivariate distribution (discrete or continuous). We say these pairs areconcordantif
sgn{(X 1 −X 2 )(Y 1 −Y 2 )}=1andarediscordantif sgn{(X 1 −X 2 )(Y 1 −Y 2 )}=−1.
The variablesX andY have an increasing relationship if the pairs tend to be
concordant and a decreasing relationship if the pairs tend to be discordant. A
measure of this is given byKendall’sτ,

τ=P[sgn{(X 1 −X 2 )(Y 1 −Y 2 )}=1]−P[sgn{(X 1 −X 2 )(Y 1 −Y 2 )}=−1].(10.8.1)

As Exercise 10.8.1 shows,− 1 ≤τ ≤1. Positive values ofτ indicate increasing
monotonicity, negative values indicate decreasing monotonicity, andτ= 0 reflects
neither. Furthermore, as the following theorem shows, ifXandYare independent,
thenτ=0.

Theorem 10.8.1.Let(X 1 ,Y 1 )and(X 2 ,Y 2 )be independent pairs of observations of
(X, Y), which has a continuous bivariate distribution. IfXandYare independent,
thenτ=0.

Proof:Let (X 1 ,Y 1 )and(X 2 ,Y 2 ) be independent pairs of observations with the same
continuous bivariate distribution as (X, Y). Because the cdf is continuous, the sign
function is either−1 or 1. By independence, we have

P[sgn(X 1 −X 2 )(Y 1 −Y 2 )=1] = P[{X 1 >X 2 }∩{Y 1 >Y 2 }]

+P[{X 1 <X 2 }∩{Y 1 <Y 2 }]

= P[X 1 >X 2 ]P[Y 1 >Y 2 ]

+P[X 1 <X 2 ]P[Y 1 <Y 2 ]

=

(

1

2

) 2

+

(

1

2

) 2

=

1

2

.

Likewise,P[sgn(X 1 −X 2 )(Y 1 −Y 2 )=−1] =^12 ; hence,τ=0.

Relative to Kendall’sτas the measure of association, the two-sided hypotheses
of interest here are
H 0 : τ=0versusH 1 :τ =0. (10.8.2)
As Exercise 10.8.1 shows, the converse of Theorem 10.8.1 is false. However, the
contrapositive is true; i.e.,τ = 0 implies thatXandYare dependent. As with the
correlation coefficient, in rejectingH 0 , we conclude thatXandYare dependent.
Kendall’sτhas a simple unbiased estimator. Let (X 1 ,Y 1 ),(X 2 ,Y 2 ),...,(Xn,Yn)
be a random sample of the cdfF(x, y). Define the statistic

K=

(

n

2

)− (^1) ∑
i<j
sgn{(Xi−Xj)(Yi−Yj)}. (10.8.3)
Note that for alli =j, the pairs (Xi,Yi)and(Xj,Yj) are identically distributed.
ThusE(K)=
(n
2
)− 1 (n
2
)
E[sgn{(X 1 −X 2 )(Y 1 −Y 2 )}]=τ.
In order to useKas a test statistic of the hypotheses (10.8.2), we need its
distribution under the null hypothesis. UnderH 0 ,τ =0,soEH 0 (K)=0. The