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Modelling Dependence with Copulas - IFOR

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5 Archimedean <strong>Copulas</strong><br />

Proof. Let U and V be uniform (0, 1) random variables <strong>with</strong> joint distribution<br />

function C, and let K C denote the distribution function of C(U, V ). Then from<br />

(3.3.6) we have<br />

τ C = 4E(C(U, V )) − 1,<br />

= 4<br />

∫ 0<br />

1<br />

t dK C (t) − 1,<br />

= 4 ( [tK C (t)] 1 0 −<br />

= 3−<br />

∫ 1<br />

0<br />

∫ 1<br />

0<br />

K C (t)dt.<br />

K C (t)dt ) − 1,,<br />

From Theorem 5.3 and Corollary 5.1 it follows that the distribution function K C<br />

of C(U, V )is<br />

and hence<br />

τ C =3− 4<br />

∫ 1<br />

0<br />

K C (t) =t −<br />

(<br />

t −<br />

ϕ(t)<br />

ϕ ′ (t + )<br />

ϕ(t)<br />

ϕ ′ (t + ) ,<br />

)<br />

dt =1+4<br />

∫ 1<br />

0<br />

ϕ(t)<br />

ϕ ′ (t) dt,<br />

where ϕ ′ (t + ) is replaced by ϕ ′ (t) in the denominator of the integral because concave<br />

functions are differentiable almost everywhere.<br />

Example 5.6. Consider the Gumbel family <strong>with</strong> generator ϕ(t) =(− ln t) θ ,for<br />

θ ≥ 1. Then<br />

ϕ(t)<br />

ϕ ′ (t) = t ln t<br />

θ .<br />

Using Theorem 5.4 we can calculate Kendall’s tau for the Gumbel family.<br />

∫ 1<br />

t ln t<br />

τ θ = 1+4 dt<br />

0 θ<br />

= 1+ 4 ( [t 2 ] 1<br />

∫ 1<br />

)<br />

θ 2 ln t t<br />

−<br />

0 0 2 dt<br />

= 1+ 4 (0 − 1/4) = 1 − 1/θ.<br />

θ<br />

Example 5.7. Consider the Clayton family <strong>with</strong> generator ϕ(t) =(t −θ − 1)/θ, for<br />

θ ∈ [−1, ∞)\{0}. Then<br />

ϕ(t)<br />

ϕ ′ (t) = tθ+1 − t<br />

.<br />

θ<br />

Using Theorem 5.4 we can calculate Kendall’s tau for the Clayton family.<br />

∫ 1<br />

t θ+1 − t<br />

τ θ = 1+4<br />

dt<br />

0 θ<br />

= 1+ 4 ( 1<br />

θ θ +2 − 1 )<br />

2<br />

= 1+ 4 −θ<br />

θ 2(θ +2) = θ<br />

θ +2 .<br />

34

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