4.4 The Efficiency of the Wilcoxon Test
Combining Theorems 4.2.2 and 4.3.1 gives the central result of the chapter.
Corollary 4.4.1. \[\mathrm {ARE}(W,t) = 12\,\sigma ^{2} \left (\int ^{\infty }_{-\infty }f^{2}(x)\,dx\right )^{2} .\]
Everything now depends on the single functional \(\sigma ^{2}\left (\int f^{2}\right )^{2}\), which can be evaluated for any density of interest.
Example 4.4.2 (Normal data). Let \(f\) be the \(N(0,\sigma ^{2})\) density. Then \[\int ^{\infty }_{-\infty }f^{2}(x)\,dx = \frac {1}{2\sigma \sqrt {\pi }},\] so by Corollary 4.4.1 \[\mathrm {ARE}(W,t) = 12\,\sigma ^{2}\cdot \frac {1}{4\sigma ^{2}\pi } = \frac {3}{\pi } \approx 0.955 .\]
This is the number the subject is built on, and it deserves to be stated in words. On perfectly normal data — the case in which the \(t\)-test is exactly optimal, and the case most favourable to it — the Wilcoxon test, which throws away the numerical values entirely and keeps only their order, is \(95.5\%\) as efficient. Discarding the data’s magnitudes costs about one observation in twenty-two.
Example 4.4.3 (Heavy tails). For the double exponential (Laplace) density \(f(x)=\tfrac {1}{2b}e^{-|x|/b}\), which has \(\sigma ^{2}=2b^{2}\) and \(\int f^{2} = 1/(4b)\), \[\mathrm {ARE}(W,t) = 12\cdot 2b^{2}\cdot \frac {1}{16b^{2}} = \frac {3}{2}.\] The Wilcoxon test needs two thirds of the observations the \(t\)-test requires.
| Distribution | \(\mathrm {ARE}(W,t)\) | \(\mathrm {ARE}(S,t)\) |
| Uniform | \(1.000\) | \(0.333\) |
| Normal | \(3/\pi \approx 0.955\) | \(2/\pi \approx 0.637\) |
| Logistic | \(\pi ^{2}/9 \approx 1.097\) | \(\pi ^{2}/12\approx 0.822\) |
| Double exponential | \(1.500\) | \(2.000\) |
| Cauchy | \(\infty \) | \(\infty \) |
Two features of Table 4.1 deserve comment. The Wilcoxon test is never much worse than the \(t\)-test and is frequently better; already at the logistic — a distribution barely distinguishable from the normal by eye — it has overtaken it. And under the Cauchy the entry is not large but infinite, because the \(t\)-test has no finite variance to work with and fails entirely, while the rank test is unaffected.
The sign test tells a different story: it is markedly inefficient on well-behaved data, at roughly two thirds of the \(t\)-test under normality, but it is the best of the three under the double exponential. Nothing dominates everywhere.
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