Solution: To show ϕ(t) = (− ln t)^θ, θ ≥ 1ϕt=−lntθ,θ≥1 is the generating function of Archimedean copulas, we need to show that it is a continuous strict decreasing function.

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   1. Let f(t) =  ln tft=lnt. It is obvious that f(t)ft is a strictly increasing function of t and thus −f(t) =  −  ln t−ft=−lnt is a strict decreasing function of t with − ln (0)→ ∞ ,  −  ln (1) = 0.−ln0→∞,−ln1=0. Given θ ≥ 1θ≥1, we have (− ln (0))^θ = ∞^θ =  ∞ ; (− ln (1))^θ = 0^∞ = 0−ln0θ=∞θ=∞;−ln1θ=0∞=0. Now we show ϕ(t) = (− ln t)^θ, θ ≥ 1ϕt=−lntθ,θ≥1 satisfies the generating function ϕ(t) = (− ln t)^θ, θ ≥ 1ϕt=−lntθ,θ≥1 is a continuous strictly decreasing from [0, 1]→[0, ∞)01→0∞.

   
   
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   2. The inverse of function ϕ(t)ϕt can be given as follows:

   
   

   Let u = (− ln t)^θ.u=−lntθ. Then we have ϕ−1t=exp(−u1θ). Now we need to show that ϕ⁻¹(t)ϕ−1t is nonincreasing. And, it is obvious that the exponential function above is continuous and a nonincreasing function.

Applying Equation (4.1) for d = 2 or 3, we have the following:

ϕ(u1)+ϕ(u2)=(−lnu1)θ+(−lnu2)θC(u1,u2)=ϕ[−1](ϕ(u1)+ϕ(u2))=exp(−((−lnu1)θ+(−lnu2)θ)1θ)(4.6a)

and

ϕ(u1)+ϕ(u2)+ϕ(u3)=(−lnu1)θ+(−lnu2)θ+(−lnu3)θC(u1,u2,u3)=ϕ[−1](ϕ(u1)+ϕ(u2)+ϕ(u3))=exp(−((−lnu1)θ+(−lnu2)θ+(−lnu3)θ)1θ)(4.6b)

To illustrate the copula with the random variables in a real domain, let random variables {X₁, X₂, X₃}X1X2X3 be positively dependent, and they may be modeled with the symmetric Archimedean copula in Equation (4.6b). In addition, X₁, X₂, X₃X1,X2,X3 follow the marginal distributions, respectively, of

We then have u1=F1x1=1−exp−2×1,u2=F2x2=11+exp−x2−42,u3=Φx3−32. Now we have the following copula functions:

Cu1u2θ=exp−−lnu1θ+−lnu2θ1θ=exp−−ln1−exp−2×1θ+−ln11+expx2−42θ1θ(4.7a)

Cu1u2u3θ=exp−−lnu1θ+−lnu2θ+−lnu3θ1θ=exp−−ln1−exp−2×1θ+−ln11+expx2−42θ+−lnΦx3−32θ1θ(4.7b)

Equations (4.7a) and (4.7b) illustrate how to construct symmetric Archimedean copulas from the correlated random variables with following different marginal distributions. It may be worth noting here again that all cumulative marginal distribution u_i~uniform (0, 1)ui~uniform01.

Solution: Directly from Equation (4.1),

Suppose the symmetric Gumbel–Hougaard copula with parameter θ can be applied to study a given trivariate analysis. Show that the copula is associative, as follows:

Solution: The trivariate symmetric Gumbel–Hougaard copula can be expressed as follows:

Cu1u2u3=ϕ−1∑k=13ϕuk=exp−−lnu1θ+−lnu2θ+−lnu3θ1θ

in which ϕ(u) = (− ln u)^θϕu=−lnuθ.

Now let’s prove the associative property of the symmetric copula using C(u₁, C(u₂, u₃))Cu1Cu2u3 as an example. The inner copula function C(u₂, u₃)Cu2u3 is the bivariate Gumbel–Hougaard copula with the same parameter θ and can be written as follows:

Cu2u3=exp−−lnu2θ+−lnu3θ1θ

Then C(u₁, C(u₂, u₃))Cu1Cu2u3 is also the bivariate Gumbel–Hougaard copula and can be written as follows:

ϕCu2u3=−lnexp−−lnu2θ+−lnu3θ1θθ=−lnu2θ+−lnu3θ

Finally, we have the following:

C(u1,C(u2,u3))=ϕ−1(ϕ(u1)+ϕ(C(u2,u3)))=exp(−((−lnu1)θ+(−lnu2)θ+(−lnu3)θ)1θ)=C(u1,u2,u3)

Similarly, we can prove that C(u₁, u₂, u₃) = C(C(u₁, u₂), u₃).Cu1u2u3=CCu1u2u3.

Solution: The bivariate Farlie–Gumbel–Morgenstern copula can be expressed as follows:

With u=12,v=14,w=16,andθ=12, we have

Cuv=1214+1212141−121−14=0.1484

CCu1u2u3=0.148416+0.148412161−0.14841−16=0.0335

and

Cvw=1416+1214161−141−16=0.0547

Cu1Cu2u3=0.054712+0.054712121−121−0.0547=0.0338

Now we can reach the conclusion: C(u₁, C(u₂, u₃))≠C(C(u₁, u₂), u₃).Cu1Cu2u3≠CCu1u2u3.

Solution: The generating function and the corresponding Gumbel–Hougaard copula can be written as follows:

ϕt=−lntθ,ϕ−1t=exp−t1θand

Cu1u2=exp−−lnu1θ+−lnu2θ1θ

in which θθ is the copula parameter.

For any given α, α>0α,α>0 and let ψ(t) = αϕ(t) = α(− ln t)^θψt=αϕt=α−lntθ, we have the following:

ψ−1t=exp−tα1θ

Rearranging the preceding Gumbel–Hougaard copula function, we have the following:

C(u1,u2)=exp(−((−lnu1)θ+(−lnu2)θ)1θ)=exp(−(a(−lnu1)θ+a(−lnu2)θa)1θ)=ψ−1(ψ(u1)+ψ(u2))(4.10)

Now, we show that αϕαϕ is also a generator of the Archimedean copula C, if α>0α>0.

Solution: Taking the first derivative of ϕ(t)ϕt,ϕ’t=−θ−lntθ−1t,ϕtϕ’t=−lntθ−θ−lntθ−1t=tlntθ; Kendall’s ττ for the Gumbel–Hougaard copula is as follows:

τ=1+4∫01tlntθdt=1+4θ∫01tlntdt=1+4θt2lnt201−∫01t2dt=1+4θ0−14=1−1θ(4.13)

Furthermore, in Equation (4.13) τ = 0τ=0 if θ = 1θ=1 (i.e., the bivariate random variable is independent), and the dependence increases with the increase of copula parameter θθ.

Solution: Taking the first derivative of ϕ(t)ϕt, we have the following:

ϕ’t=−t−θ−1andϕtϕ’t=t−θ−1/θ−t−θ−1=tθ+1−tθ

Kendall’s ττ for the Clayton copula can then be computed as follows:

τ=1+4∫21tθ+1−tθdt=1+4θtθ+2θ+201−t2201=θθ+2(4.14)

In Equation (4.14), τ =  − 1τ=−1 when θ =  − 1θ=−1 (i.e., perfectly negatively dependent). And τ→1τ→1 when θ→∞θ→∞ (i.e., perfectly positively dependent). Similar to Example 4.6, the dependence of the bivariate random variable increases with the increase of parameter θθ.

Solution: It is known that if the Clayton copula can be extended to higher dimensions, i.e., d ≥ 3d≥3, we need to satisfy the theorem (Theorem 4.6.2, Nelsen, 2006) discussed previously. The generating function for the Clayton copula can be written as follows:

ϕt=1θt−θ−1⇒ϕ−1t=θt+1−1θ

According to Nelsen (2006), we know that for θ ≥ 0θ≥0, the generating function is strictly decreasing from I to (0, ∞)0∞. Applying Equation (4.15), we have the following:

−11dϕ−1tdt=θt+1−1+θθ≥0

−12d2ϕ−1tdt=1+θθt+11+2θθ≥0

…

−1kdkϕ−1tdt=−12k∏j=2k1+j−1θθt+1−1+kθθ;k≥2

Now, we reach the conclusion that the bivariate Clayton copula can be extended to multivariate symmetric Clayton copula, as follows:

Cθdu=∑i=1dui−θ−d+1−1θ;θ≥0

Note that the multivariate symmetric Clayton copula (i.e., d ≥ 3d≥3) may only model the positive dependent/independent multivariate random variables. The reason is that if θ < 0θ<0, Equation (4.15) cannot be guaranteed to be fully satisfied.

Solution: Following Nelsen (2006), it is known that the generating function of the Ali–Mikhail–Haq copula is strictly decreasing from I to (0, ∞)0∞. The generating function and its inverse function can be written as follows:

ϕt=ln1−θ1−tt;ϕ−1t=θ−1θ−expt

Rather than directly applying the theorem as in Example 4.9, here we use the inequality proposed by Widder (1941) for function ϕ⁻¹ϕ−1 to be completely monotonic, as follows:

The first and second derivative of the inverse function can be written as follows:

ϕ−1′=dϕ−1dt=θ−1exptθ−expt2;ϕ−1′′=d2ϕtdt2=θ−1exptθ−expt2+2θ−1exp2tθ−expt3

Substituting the first and second derivatives of the inverse function into Equation (4.17), we have the following:

ϕ−1ϕ−1′′−ϕ−1′2=θ−1θ−exptθ−1exptθ−expt2+2θ−1exp2tθ−expt3−θ−1exptθ−expt22=θ−12expt+θθ−expt4