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Wikipedia tells us: (link)

Tau-c (also called Stuart-Kendall Tau-c) is more suitable than Tau-b for the analysis of data based on non-square (i.e. rectangular) contingency tables.

What does $\tau_c$ try to solve? More specifically, is there a problem with applying Kendall's Tau-b on non-square contingency tables? When we compute $\tau_b(x,y)$, do we implictly assume that the number of levels of $x$ is the same as the number of levels of $y$? Or is it an issue with how Tau-b omits ties?

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Largely it boils down to this: $\tau_b$ can't attain the values $\pm 1$ in a non-square table but $\tau_c$ can.

You can verify this with examples.

Consider the following tables

    I    II   III
A   30    0    0
B    0   30    0
C    0    0   30

    Ia   Ib   IIa  IIb   IIIa IIIb
A   15   15     0    0     0    0
B    0    0    15   15     0    0
C    0    0     0    0    15   15

Try computing the two measures on each. They're both 1 on the first table, and $\tau_c$ is 1 on both tables, but $\tau_b$ is only around 0.9 on the second table.

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  • $\begingroup$ So is $\tau_c$ a more correct/general version of $\tau_b$? I wonder why $\frac{n_c-n_d}{n_c+n_d}$ isn't used instead of these formulas when we want to consider only non-tied pairs. For this case, it always scales correctly (i.e., attains $\pm1$). $\endgroup$
    – Trisoloriansunscreen
    Commented Jun 9, 2019 at 17:45
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    $\begingroup$ Indeed, I often do just the calculation you suggest, since it usually achieves my intent in each situation I want a Kendall-like measure of association. However, if I remember right, $\tau_c$ is not always identical to $\tau_b$ on square tables so it's not simply "a more correct version" of the same thing (however I think their p-values will be the same on square tables) $\endgroup$
    – Glen_b
    Commented Jun 10, 2019 at 0:39

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