Why Dependence Measures Can Disagree
Two risk variables can move together in one sense and diverge in another. Their losses might follow a consistent ordering without a straight-line relationship; a few extreme observations might drive a high linear correlation; or ordinary observations might look weakly associated while the largest losses often occur together. A single dependence statistic cannot describe all these features.
This tutorial diagnoses paired observations using four complementary summaries: Pearson correlation, Spearman rank correlation, Kendall’s tau, and finite-threshold tail co-exceedance. Each summarizes a different aspect of the observed data. None, by itself, identifies a complete joint probability model or proves how dependence behaves beyond the observed range.
What Each Measure Summarizes
For \(n\) paired observations without ties, let \(R_j\) and \(S_j\) be the ranks of \(x_j\) and \(y_j\). Pearson’s sample correlation is
It is sensitive to distances between observations and can be strongly affected by extreme values. Spearman’s rank correlation, denoted here by \(\rho_S\), measures association between the two orderings. With no tied ranks, it can be calculated from the rank differences \(d_j=R_j-S_j\):
For Kendall’s tau, a pair of observations is concordant if the observation with the larger \(x\) also has the larger \(y\). It is discordant if the observation with the larger \(x\) has the smaller \(y\). Write \(C\) and \(D\) for the numbers of concordant and discordant pairs. With no ties, the total number of pairs is \(\binom{n}{2}\), and
For an upper-tail diagnostic, choose thresholds \(q_X\) and \(q_Y\), and count observations strictly above each threshold. The empirical conditional co-exceedance proportion is the number above both thresholds divided by the number above the \(X\) threshold. State the thresholds and denominator: changing either can change the result. This finite-sample proportion is not, by itself, an asymptotic tail-dependence coefficient.
A Short Justification for the Rank Formula
The formula for Spearman correlation follows from applying the usual Pearson correlation calculation to the ranks. When there are no ties, each rank list is a permutation of \(1,\ldots,n\), so both rank means are \((n+1)/2\), and each centered rank sum of squares is \(n(n^2-1)/12\). Also,
Because the two rank lists have the same sum of squares, this identity gives the covariance numerator in terms of \(\sum d_j^2\). Dividing by the rank standard-deviation denominator \(n(n^2-1)/12\) yields \(1-6\sum d_j^2/[n(n^2-1)]\). Ties require an appropriate rank-based calculation; the simplified difference formula should not be used as though all ranks were distinct.
Worked Diagnostics
Worked Example: A Perfect Ordering With a Different Linear Summary
A hypothetical insurer compares annual losses from two business segments across five matched exposure periods. The losses, in thousands of dollars, are \((1,1),(2,2),(3,3),(4,4),(5,100)\). Calculate the three dependence measures and diagnose their difference.
Method. Calculate Pearson correlation on the original values, then calculate rank measures. There are no ties. The key diagnostic is whether the extreme \(Y\) value changes the linear summary even though the ordering is unchanged.
The means are \(\bar{x}=15/5=3\) and \(\bar{y}=110/5=22\). The centered \(x\) values are \((-2,-1,0,1,2)\); the centered \(y\) values are \((-21,-20,-19,-18,78)\). Thus the cross-product sum is \(42+20+0-18+156=200\). The centered sums of squares are \(10\) for \(X\) and \(441+400+361+324+6084=7610\) for \(Y\). Therefore,
Both rank lists are \((1,2,3,4,5)\). Every rank difference is zero, so \(\sum d_j^2=0\) and \(\rho_S=1-6(0)/[5(25-1)]=1\). All \(\binom{5}{2}=10\) pairs are concordant and none discordant, giving \(\tau=(10-0)/10=1\).
Diagnosis and check. The data have a perfect common ordering, but the original-scale observations do not have perfect linear association: the last \(Y\) value is much larger than the preceding values. The rank measures correctly report the unchanged ordering. The Pearson result is within its allowable range \([-1,1]\), and its lower value is not a contradiction. A common mistake is to infer that a rank correlation of 1 requires Pearson correlation of 1; that is true only for a positive exact linear relationship, not for every perfectly increasing relationship.
Worked Example: Tail Co-Exceedance With Modest Overall Association
A hypothetical risk team has eight paired quarterly loss indicators, recorded as ranks from 1 (lowest) to 8 (highest) in each portfolio. The pairs are \((1,8),(2,1),(3,2),(4,3),(5,4),(6,5),(7,6),(8,7)\). Calculate Pearson, Spearman, Kendall, and the empirical conditional co-exceedance proportion for observations above 6 in both portfolios.
Method. Since each variable is already a rank permutation, Pearson correlation of these values equals Spearman correlation. Then count pair inversions for Kendall’s tau and count threshold exceedances directly.
The rank differences \(d_j=R_j-S_j\) are \((-7,1,1,1,1,1,1,1)\), so \(\sum d_j^2=49+7=56\). Hence
For Pearson correlation, both means are \(4.5\), and both centered sums of squares are \(42\). The centered cross-products sum to \(14\), so \(r_{XY}=14/\sqrt{42(42)}=14/42=1/3\), as expected for the same rank data.
There are \(\binom{8}{2}=28\) pairs. The first \(Y\) rank, 8, is greater than each of the seven later \(Y\) ranks even though its \(X\) rank is the smallest; these are seven discordant pairs. The remaining 21 pairs are concordant. Thus \(\tau=(21-7)/28=14/28=0.5000\).
For the tail calculation, \(X>6\) occurs twice, at \(X=7,8\), and \(Y>6\) occurs twice, at \(Y=7,8\). Only the pair \((8,7)\) exceeds both thresholds. The conditional proportion is therefore \(1/2=0.5000\); the joint proportion of all eight observations is \(1/8=0.1250\).
Diagnosis and check. Overall rank association is modest, but one of the two \(X\) exceedances also exceeds the \(Y\) threshold. This is a threshold-specific observation, not proof of persistent extreme-tail dependence. The common Kendall trap is to report \(D/\binom{n}{2}=7/28\) as tau; that is the discordant-pair proportion, not tau. The correction is to use \((C-D)/\binom{n}{2}=(21-7)/28=0.5000\).
Worked Example: A Paired Extreme Observation and Different Diagnostics
A hypothetical reinsurer reviews six paired annual loss indices from two portfolios: \((1,1),(2,4),(3,2),(4,5),(5,3),(20,20)\). Calculate Pearson correlation, both rank measures, and the empirical conditional co-exceedance proportion when each portfolio’s threshold is its second-largest observed value.
Method. Use centered sums for Pearson, ranks for Spearman, inversion counting for Kendall, and a direct threshold count for co-exceedance. This separates the influence of the extreme paired observation from the full-sample ordering.
Both totals are 35, so \(\bar{x}=\bar{y}=35/6\). The product sum is \(\sum xy=1+8+6+20+15+400=450\). The sums of squares are \(\sum x^2=\sum y^2=455\). The centered cross-product sum is \(450-35^2/6=1475/6\); each centered sum of squares is \(455-35^2/6=1505/6\). Therefore,
The \(X\) ranks are \((1,2,3,4,5,6)\), while the \(Y\) ranks are \((1,4,2,5,3,6)\). The rank differences are \((0,-2,1,-1,2,0)\), with squared differences summing to \(10\). Thus \(\rho_S=1-6(10)/[6(36-1)]=1-60/210=5/7\approx0.7143\).
In the \(Y\)-rank ordering, rank 4 precedes ranks 2 and 3, and rank 5 precedes rank 3. These are three inversions, so \(D=3\). There are \(\binom{6}{2}=15\) pairs and \(C=12\), giving \(\tau=(12-3)/15=0.6000\).
The second-largest value in each portfolio is 5, so the thresholds are \(X>5\) and \(Y>5\). Only the paired observation \((20,20)\) exceeds both thresholds. Two observations exceed the \(X\) threshold, so the conditional co-exceedance proportion is \(1/2=0.5000\).
Diagnosis and check. Pearson correlation is very high because the large observation is extreme in both variables. The rank measures show that the overall ordering is positive but not nearly perfect; the finite-threshold proportion shows one joint exceedance among two \(X\) exceedances. Each result answers a different question. A common error is to treat the high Pearson value as sufficient evidence that most pairs are similarly ordered or that extreme co-exceedances are frequent. The ranks and the small tail denominator show why those conclusions do not follow.
How to Diagnose a Disagreement
Interpret the measures as complementary diagnostics, not as competing estimates of one universal quantity. A structured comparison helps identify what the data support:
Confirm that each row is a matched pair, both variables use consistent units and periods, and any threshold and tie treatment are explicit.
A large gap between Pearson and the rank measures can indicate curvature, influential magnitudes, or both. Inspect the paired observations before assigning a cause.
Spearman summarizes rank distances, whereas Kendall summarizes concordance across pairs. They can differ because they weight ordering patterns differently; neither is a simple rescaling of the other for all data.
Report the chosen thresholds, exceedance counts, joint count, and denominator. A high conditional proportion based on one or two exceedances is uncertain and should not be presented as stable tail behavior.
Several patterns are especially informative. High rank measures with a lower Pearson value are consistent with strong ordering but a nonlinear scale relationship or influential magnitudes. High Pearson correlation with weaker rank measures can occur when a few large paired observations dominate the cross-products. Weak full-sample measures can coexist with a sizable finite-threshold co-exceedance proportion, but the tail estimate may be highly variable when few observations are available.
Common Traps and Practical Limits
- Confusing pair proportions with Kendall’s tau. Kendall’s tau is \((C-D)/\binom{n}{2}\), not \(C/\binom{n}{2}\) and not \(D/\binom{n}{2}\).
- Assuming Pearson and rank correlations must agree. Pearson uses original distances; rank measures use order. Agreement is not required.
- Changing the tail denominator silently. The conditional proportion \(P(Y>q_Y\mid X>q_X)\) in the sample divides by the number of \(X\) exceedances. The joint proportion divides by the full sample size; they are different quantities.
- Overinterpreting selected thresholds. A finite sample at one threshold does not establish asymptotic tail dependence or guarantee future co-exceedance rates.
- Ignoring ties or observation design. Repeated losses, rounded values, and dependent time observations can affect calculations and uncertainty. Use a method appropriate to the data and state its assumptions.
For risk diagnosis, preserve the distinctions among linear association, ordering, and tail clustering. Report the data window, sample size, threshold definition, and tail counts alongside the summaries. This makes a disagreement interpretable and helps avoid treating one convenient coefficient as a complete description of dependence.
Check Your Understanding
For each problem, calculate the requested summaries and interpret what they indicate about the paired observations.
- A hypothetical pair of loss ranks has observations \((1,1),(2,2),(3,4),(4,3),(5,5)\). Calculate Spearman correlation and Kendall’s tau. (Answer: Spearman \(=0.9000\); Kendall’s \(\tau=0.8000\).)
- In eight paired observations, three exceed the \(X\) threshold, four exceed the \(Y\) threshold, and two exceed both. Find the empirical conditional co-exceedance proportion given \(X\) exceeds its threshold, and the joint proportion of all observations. (Answer: Conditional \(=2/3\approx0.6667\); joint \(=2/8=0.2500\).)
- For five paired observations without ties, suppose \(C=8\) and \(D=2\). Calculate Kendall’s tau, and state why the discordant-pair proportion alone is not the answer. (Answer: \(\tau=(8-2)/10=0.6000\); discordant-pair proportion \(=2/10=0.2000\).)
- Two rank lists of length six are \(R=(1,2,3,4,5,6)\) and \(S=(2,1,4,3,6,5)\). Calculate Spearman correlation and Kendall’s tau. (Answer: Spearman \(=29/35\approx0.8286\); Kendall’s \(\tau=0.6000\).)
- In a sample of ten paired loss observations, the chosen upper thresholds leave two \(X\) exceedances and three \(Y\) exceedances; one observation exceeds both. Calculate the conditional co-exceedance proportion given \(X\) exceeds, and identify one limitation of the result. (Answer: \(1/2=0.5000\); it is based on only two \(X\) exceedances and does not establish asymptotic tail dependence.)