Kendall's tau of the Plackett family: a one-dimensional integral #
Starting from the rational representation kendallTau_plackett_eq_rational
(Copula.Families.Plackett.Kendall), the inner integral is elementary: for 0 < v < 1 write
x = (θ-1)u + 1 - (θ+1)v (PlackettKendall.shift) and s = 2√θ √(v(1-v))
(PlackettKendall.arcScale); then disc = x² + s² and the numerator is
1 + (θ-1)(u+v-2uv) = (1-2v) x + 2(θ+1) v(1-v), so the u-integrand has the primitive
((1-2v)/2 · log disc + (θ+1)/√θ · √(v(1-v)) · arctan(x/s)) / (θ-1)
(PlackettKendall.hasDerivAt_ratPrimitive). The boundary terms are symmetric under
v ↦ 1 - v, and the logarithmic part integrates in closed form. The result is
(kendallTau_plackett_eq_arctan, θ ≠ 1)
τ(C_θ) = (θ+1)/(θ-1) - 2θ(θ² - 1 - 2θ log θ)/(θ-1)⁴ + 4(θ+1)√θ/(θ-1)² · J(θ),
J(θ) = ∫₀¹ √(v(1-v)) arctan((1 - (θ+1)v) / (2√θ √(v(1-v)))) dv (plackettTauIntegral),
equivalently, with Mardia's ρ(C_θ) = (θ² - 1 - 2θ log θ)/(θ-1)² (spearmanRho_plackett),
τ(C_θ) = (θ+1)/(θ-1) - 2θ ρ(C_θ)/(θ-1)² + 4(θ+1)√θ/(θ-1)² · J(θ)
(kendallTau_plackett_eq_spearmanRho_arctan).
There is no elementary closed form for J; the formula reduces the computation of τ(C_θ)
to one bounded one-dimensional integral (checked numerically, e.g. τ(C_2) ≈ 0.15305,
τ(C_5) ≈ 0.34550, τ(C_{20}) ≈ 0.59166).
The primitive in u of the rational part (1 + (θ-1)(u+v-2uv))/disc.
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The arctangent part √(v(1-v)) arctan((1 - (θ+1)v)/(2√θ√(v(1-v)))).
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- ProbabilityTheory.Copula.PlackettKendall.arcPart θ v = √(v * (1 - v)) * Real.arctan ((1 - (θ + 1) * v) / ProbabilityTheory.Copula.PlackettKendall.arcScale θ v)
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Half of the boundary term: ∫₀¹ ratPart(u,v) du = half(v) + half(1-v).
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- One or more equations did not get rendered due to their size.
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A primitive of the logarithmic part.
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The double integral of the rational part as twice the integral of half.
The arctangent integral
J(θ) = ∫₀¹ √(v(1-v)) arctan((1 - (θ+1)v)/(2√θ √(v(1-v)))) dv appearing in Kendall's tau of
the Plackett copula.
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Kendall's tau of the Plackett copula as a one-dimensional integral (θ ≠ 1):
τ(C_θ) = (θ+1)/(θ-1) - 2θ(θ² - 1 - 2θ log θ)/(θ-1)⁴ + 4(θ+1)√θ/(θ-1)² · J(θ).
Kendall's tau of the Plackett copula in terms of Spearman's rho and J(θ) (θ ≠ 1):
τ(C_θ) = (θ+1)/(θ-1) - 2θ ρ(C_θ)/(θ-1)² + 4(θ+1)√θ/(θ-1)² · J(θ).