Solution: The bivariate Gaussian copula is a distribution over the unit square [0, 1]²012, which is constructed from the bivariate normal distribution through the probability integral transform.

For a given correlation matrix, RR, the bivariate Gaussian copula can be given as follows:

CRGAUu=ΦRΦ−1u1Φ−1u2,u=u1u2(3.21)

where Φ⁻¹Φ−1 denotes the inverse cumulative distribution function of standard normal distribution; and Φ_RΦR denotes the joint cumulative distribution function of bivariate normal distribution with mean vector of zero and covariance matrix of RR.

The density function of bivariate Gaussian copula can be given as follows:

cRGAUu=12π1−ρ2exp−x∗2−2ρx∗y∗+y∗221−ρ2(3.22)

where x^∗, y^∗x∗,y∗ are the transformed variables as x^∗ = Φ⁻¹(u₁), y^∗ = Φ⁻¹(u₂)x∗=Φ−1u1,y∗=Φ−1u2; and ρρ denotes the correlation coefficient of the bivariate random variable that may be expressed through the Kendall correlation coefficient as follows:

ρ=sinπτ2(3.22a)

It is worth noting that the Gaussian copula may also be called the meta-Gaussian distribution with no constraints on the type of marginal distributions. In what follows, we will further illustrate the bivariate Gaussian copula with two different marginal distributions:

Let u₁ = F_X(x) = N(x; μ, σ²)u1=FXx=Nxμσ2 and u₂ = F_Y(y) = 1 −  exp (−λy)u2=FYy=1−exp−λy. We have

x∗=Φ−1u1=Φ−1Nxμσ2,y∗=Φ−1u2=Φ−11−exp−λy,ρ=sinπτXY2.

Finally, we obtain the bivariate Gaussian copula and its density function as follows:

CGAUu=ΦΦ−1Nxμσ2Φ−11−exp−λysinπτXY2cGAUu=12π1−sinπτXY22exp−Φ−1Nxμσ22−2Φ−1Nxμσ2Φ−11−exp−λy+Φ−11−exp−λy221−sinπτXY22

Consider a simple numerical example with the random sample values x = 2.5 and y = 4x=2.5andy=4 drawn from the probability distributions of X~N(0, 2²); Y~ exp (0.5).X~N022;Y~exp0.5. The rank based Kendall correlation coefficient of X, YX,Y is τ_XY = 0.7τXY=0.7.

Applying Equation (3.22a), we may compute the Pearson correlation coefficient as follows: ρ=sinπτ2=sin0.7π2=0.891.

From the parent normal and exponential distributions, we can compute the transformed variables:

X~N022⇒FX2.5=N2.5022=0.894⇒x∗=Φ−1FX2.501=Φ−10.894401=1.25

Y~exp0.5⇒FY4=1−exp−0.54=0.8647⇒y∗=Φ−1FY401=Φ−10.864701=1.1015

Substituting x^∗ = 1.25, y^∗ = 1.1015, ρ = 0.891x∗=1.25,y∗=1.1015,ρ=0.891 into the bivariate Gaussian copula and the corresponding density function, we have the following:

Solution: The trivariate Gaussian copula is a distribution over the unit cube [0, 1]³013 which is constructed from the trivariate normal distribution through the probability integral transform.

For a given correlation matrix, RR, the trivariate Gaussian copula can be given as follows:

CRGAUu=ΦRΦ−1u1Φ−1u2Φ−1u3,u=u1u2u3(3.32)

where Φ⁻¹Φ−1 denotes the inverse cumulative distribution function of the standard normal distribution; and Φ_RΦR denotes the joint cumulative distribution function of trivariate normal distribution with a mean vector of zero and a covariance matrix of RR.

The density function of trivariate Gaussian copula can be given as follows:

cRGAUu=1∣R∣exp−12Φ−1u1Φ−1u2Φ−1u3TR−1−IΦ−1u1Φ−1u2Φ−1u3(3.33)

where the mean vector is [0,0,0], R denotes the covariance matrix of the random variables, and I is the three-by-three identity matrix.

Similar to the bivariate Gaussian copula example (i.e., Example 3.6), there is no restriction in regard to the marginal distribution that the random variables may follow. More examples will be given in the chapter focused on meta-elliptical copulas.

Solution: Suppose that random variables X₁, X₂ each follow the Gumbel distribution as follows: X₁ ~ Gumbel (a₁, b₁), and X₂ ~ Gumbel (a₂, b₂). Their joint distribution follows the Gumbel mixed distribution. In this example, the univariate Gumbel distribution can be expressed as follows:

Fx=exp−exp−x−ba(3.35)

and the bivariate Gumbel mixed distribution can be expressed as follows:

Fx1x2=F1x1F2x2exp−α1lnF1x1+1lnF2x2−1;α∈01(3.36)

Again, let u = F₁(x₁), v = F₂(x₂)u=F1x1,v=F2x2 with F₁(x₁)F1x1 and F₂(x₂)F2x2 each following the Gumbel distribution given by Equation (3.35). Then, we have

where αα is the parameter of the copula.

The copula function derived as Equation (3.37) is actually the Gumbel-mixed model. Thus, it should be noted that Equations (3.37) can be successfully constructed to represent the joint distribution if and only if the random variables are positively correlated with the correlation coefficient not exceeding 2/3. This may be explained from the properties of the Gumbel-mixed model. Given by Oliveria (1982), the parameter of Gumbel-mixed model is related to the Pearson correlation coefficient as follows:

α=21−cosπρ6;α=0⇒ρ=0;α=1⇒ρ=23(3.38)

Solution: Suppose that random variables X₁, X₂ with X₁~ exp (θ₁), X₂~ exp (θ₂)X1~expθ1,X2~expθ2, the joint distribution of X₁ and X₂, F(x₁, x₂),Fx1x2, follows the bivariate exponential distribution presented by Singh and Singh (1991) as follows:

Fx1x2=1−e−x1θ11−e−x2θ21+δe−x1θ1−x2θ2(3.39)

Let u=F1x1=1−e−x1θ1,v=F2x2=1−e−x2θ2; then we have

where δδ is the parameter of the copula in Equation (3.39).

In Equations (3.39) and (3.40), |δ| ≤ 1. In the case of the bivariate exponential distribution in this example, the correlation of bivariate random variables is in the range of [–0.25, 0.25] to guarantee that the bivariate distribution so derived is valid. In addition, the FGM copula is also expressed as Equation (3.40). In the case of the FGM copula, the correlation of bivariate random variables needs to be in the range of [–1/3, –1/3] (Schucany et al., 1978).

Let (α, β)αβ be a point in I2 such that α>0, β>0, and α + β < 1α>0,β>0,andα+β<1. Suppose that the probability mass α is uniformly distributed on the line segment joining (α,β) and (0, 1), the probability mass β is uniformly distributed on the line segment joining (α,β) and (1, 0), and the probability mass 1-α-β is uniformly distributed on the line segment joining (α,β) and (1, 1). Determine the copula function with these supports.

Solution: Based on the description of the problem statements, Figure 3.2(a) graphs the prescribed support (depicted by the solid line). It is seen from Figure 3.2(a) that (u, v)uv may be reside either in the upper triangle (i.e., Figure 3.2(b)) or in the lower triangle (i.e., Figure 3.2(c)). In addition, we will also check what may happen if (u, v)uv fall out of the prescribed support (i.e., beneath the two triangles). Now, to determine the copula function with the corresponding prescribed support, we will look at three different cases: (a) (u,v) is in the upper triangular support region (Figure 3.2(b)); (b) (u,v) is in the lower triangular support region (Figure 3.2(c)); and (c) (u,v) does not fall into either support region individually.

1. 
   

   1. If (u,v) falls into the region bounded by the upper triangular region with vertices (α,β), (0, 1), and (1, 1), as shown in Figure 3.2(b), then according to the definition of the copula, Figure 3.2(b) clearly shows the following:

   
   
   VC0u×v1=VC0α1−v1−β×v1(3.41)
   
   
   
   

   V_C([0, u] × [v, 1]) = C(u, 1) − C(u, v) − C(0, 1) + C(0, v) = u − C(u, v)

   VC0u×v1=Cu1−Cuv−C01+C0v=u−Cuv(3.42)
   
   
   VC0α1−v1−β×v1=Cα1−v1−β1−Cα1−v1−βv−C01+C0v=α1−v1−β−Cα1−v1−βv(3.43)
   Equating Equation (3.42) to Equation (3.43), we get the following:
   
   Cuv=u−α1−v1−β(3.44a)
   In order to determine the copula function in this region, we can also look at the rectangle α1−v1−βu×v1. This rectangle is not intercepting any support line segment, thus we know the C-volume is zero, as follows:
   
   VCα1−v1−βu×v1=Cu1−Cuv−Cα1−v1−β1+Cα1−v1−βv=0⇒Cuv=u−α1−v1−β(3.44b)
   
   
   
   
2. 
   

   2. Similarly, If (u,v) falls into the region bounded by the lower triangular region with vertices (α,β), (1, 0), and (1, 1), as shown in Figure 3.2(c), then we can use the same approach to find the following:

   
   
   Cuv=v−β1−α1−u(3.45)
   
   
   
   
3. 
   

   3. If (u,v) is not falling into the two triangles bounded by the support segment, then we immediately know that the C-volume is zero and C(u, v)Cuv can be found as follows:

   
   
   
   

   V_C([0, u] × [0, v]) = C(u, v) − C(0, v) − C(u, 0) − C(0, 0) = 0 ⇒ C(u, v) = 0

   VC0u×0v=Cuv−C0v−Cu0−C00=0⇒Cuv=0(3.46)
   
   

Note the following for the limiting cases:

1. 
   

   1. If α = β = 0α=β=0, the support line segment is the main diagonal on I². Nelsen (2006) proved that in this case, C(u, v)Cuv is the Fréchet–Hoeffding upper bound, i.e., C(u, v) = M(u, v) =  min (u, v)Cuv=Muv=minuv.

   
   
2. 
   

   2. If β = 1 − αβ=1−α, Equation (3.44a) and Equation (3.45) reduce to the following:

   
   
   Cuv=u−β1−α1−v=u+v−1(3.47a)
   
   
   
   

   and

   
   
   Cuv=v−α1−β1−u=u+v−1(3.47b)
   
   
   
   

   Equation (3.47) represents the Fréchet–Hoeffding lower bound, i.e.,

   
   
   
   

   C(u, v) = W(u, v) =  max (u + v − 1, 0)

   Cuv=Wuv=maxu+v−10.(3.47)
   
   

Figure 3.2 Schematic of singular copulas with prescribed support.

Show for each of the following choices of the ΨΨ function, the function C given as

is a copula:

1. 
   

   a. Ψv=θπsinπv;θ∈−11

   
   
2. 
   

   b. Ψ(v) = θ[ζ(v) + ζ(1 − v)], θ ∈ [−1, 1]; ζΨv=θζv+ζ1−v,θ∈−11;ζ is the piecewise linear function with the graph connecting [0, 0] to (1/4, 1/4) to (1/2, 0) to (1, 0).

Solution: According to Nelsen (2006), if Equation (3.48) is a copula, it is a copula with quadratic sections in u.

1. 
   

   a. Ψv=θπsinπv,θ∈−11

Corollary 3.2.5 (from Nelsen, 2006) can be applied to prove that the C function with the ΨΨ function so defined is a copula. Corollary 3.2.5 states the necessary and sufficient conditions for the C function to be a copula:

1. 
   

   1. Ψ(v)Ψv is absolutely continuous on I.

   
   
2. 
   

   2. |Ψ^‘(v)| ≤ 1Ψ’v≤1 almost everywhere on I.

   
   
3. 
   

   3. |Ψ(v)| ≤  min (v, 1 − v)Ψv≤minv1−v.

Based on corollary 3.2.5, we conclude the following:

1. 
   

   1. It is easy to see that Ψ(v)Ψv is absolutely continuous on I with sine function being an absolutely continuous function.

   
   
2. 
   

   2. It is seen that for θ ∈ [−1, 1], |θ/π| < 1θ∈−11,θ/π<1, so we have the following: ∣Ψ’v∣=∣θπcosπv∣<1.

   
   
3. 
   

   3. For Ψv=θπsinπv,v∈I, we have the following:

   
   
   0≤πv≤π,sinπv≤πv⇒Ψv=θπsinπv≤θππv=θv≤v(3.49)
   
   
   
   

   Similarly,

   
   
   sinπv=sinπ−πv=sinπ1−v≤π1−v⇒Ψv=θπsinπ1−v≤θ1−v≤1−v(3.50)
   
   
   
   

   From Equations (3.49) and (3.50), we have |Ψ(v)| ≤  min (v, 1 − v)Ψv≤minv1−v for v ∈ Iv∈I.

Now, all the conditions are satisfied and function C with Ψ(v)Ψv defined in a. is a copula.

1. 
   

   b. Ψ(v) = θ[ζ(v) + ζ(1 − v)], θ ∈ [−1, 1]Ψv=θζv+ζ1−v,θ∈−11; ζζ is the piecewise linear function with the graph connecting {[0, 0] to (1/4, 1/4)} to {(1/2, 0) to (1, 0)}.

Theorem 3.2.4 in Nelsen (2006) can be applied to prove that function C is a copula. Theorem 3.2.4 states the necessary and sufficient conditions for C to be a copula as follows:

1. 
   

   1. Ψ(0) = Ψ(1) = 0Ψ0=Ψ1=0

   
   
2. 
   

   2. Ψ(v)Ψv satisfies the Lipschitz condition: |Ψ(v₂) − Ψ(v₁)| ≤ |v₂ − v₁|; v₁, v₂ ∈ IΨv2−Ψv1≤v2−v1;v1,v2∈I

   
   
3. 
   

   3. C is absolutely continuous.

The schematic plot for the piecewise linear function is given in Figure 3.3(a).

Figure 3.3 Plots of functions ζ(v)ζv and Ψ(v)Ψv.

The Ψ(v)Ψv function can be written as follows:

Ψv=θv;v∈014θ12−v;v∈1434θv−1;v∈341(3.51)

1. 
   

   1. For θ ∈ [−1, 1]θ∈−11, we have the following:

   
   
   
   

   Ψ(0) = 0; Ψ(1) = θ(1 − 1) = 0

   Ψ0=0;Ψ1=θ1−1=0
   
   
   
   
2. 
   

   2. Prove the Lipschitz condition with v₁ ≤ v₂v1≤v2 and θ ∈ [−1, 1]θ∈−11.

   
   
   
   
   1. 
      

      i. If v1∈014, we have the following:

   
   
   
   

   |Ψ(v₂) − Ψ(v₂)|

   Ψv2−Ψv2
   
   
   =θv2−v1≤v2−v1;v2∈014θ12−v2−θv1=θ12−v1+v2<θv2−v1≤v2−v1;v2∈1434θv2−1−θv1<θv2−v1−1<v2−v1;v2∈341(3.52)
   
   
   
   
   1. 
      

      ii. If v1∈1434, we have the following:

   
   
   Ψv2−Ψv1=θ12−v2−θ12−v1=θv2−v1≤v2−v1;v2∈1434θv2−1−θ12−v1=∣θ(v2+v1−32∣≤θv2−34≤θv2−v1≤v2−v1;v2∈341(3.53)
   
   
   
   
   1. 
      

      iii. Similarly, it can be easily shown that the Lipschitz condition is also satisfied for v1∈341.

   
   
   
   
3. 
   

   3. Following Nelsen (2006), to prove the absolute continuity of C follows the absolute continuity of Ψ(v)Ψv with the second condition. Figure 3.3(b) plots the Ψ(v)Ψv function with θ = 0.8θ=0.8; as an example, it is shown that there is no discontinuity in domain I:

   
   
   Ψ’v=θ,v∈014−θ,v∈1434θ,v∈341(3.54)
   
   
   
   

   with θ ∈ [−1, 1]θ∈−11, we have proved that |Ψ^‘(v)| ≤ 1Ψ’v≤1 in domain I.

Now all the conditions are satisfied and the C function with the Ψ(v)Ψv function defined in (b) is a copula.

It is worth noting that the copula defined as Equation (3.48) is a copula with quadratic sections in u. The reader can refer to Nelsen (2006) for more complete details of the geometric method and other types of geometric support to construct copulas.

The Ali–Mikhail–Haq copula (Ali et al., 1978) can be expressed as follows:

Cuv=uv1−θ1−u1−v(3.57)

Construct the copula by using the algebraic method.

Solution: Ali et al. (1978) proposed that Ali–Mikhail–Haq copula belongs to the bivariate logistic distribution family with the standard bivariate logistic distribution and standard logistic marginals.

The standard bivariate logistic distribution can be given as follows:

The standard logistic marginal can be given as follows:

The survival ratio of Equation (3.58a) is

1−Fx1x2Fx1x2=1−1+e−x1+e−x2−11+e−x1+e−x2−1=e−x1+e−x2(3.59)

From Equation (3.59), it is seen that the survival ratio of the standard bivariate logistic distribution can be rewritten as follows:

1−Fx1x2Fx1x2=e−x1+e−x2=1−1+e−x1−11+e−x1−1+1−1+e−x2−11+e−x2−1(3.60a)

Substituting Equation (3.58b) into Equation (3.60a), we have the following:

1−Fx1x2Fx1x2=1−F1x1F1x1+1−F2x2F2x2(3.60b)

In Ali et al. (1978), the Ali–Mikhail–Haq copula was considered a bivariate distribution satisfying the survival ratio as follows:

1−Fx1x2Fx1x2=1−F1x1F1x1+1−F2x2F2x2+1−θ1−F1x1F1x11−F2x2F2x2(3.61)

It is concluded from Equation (3.61) that θ = 1 implies that the joint distribution F(x, y) of random variables X₁X1 and X₂X2 follows the standard biviariate logistic distribution; and θ = 0 implies that X and Y are independent with the proof given in example 3.19 in Nelsen (2006).

Applying Sklar’s theorem to Equation (3.59) and letting

Equation (3.61) can be rewritten as follows:

1−CuvCuv=1−uu+1−vv+1−θ1−uu1−vv(3.62)

With simple algebra, we have

Cuv=uv1−θ1−u1−v(3.63)

where θ is the parameter of the Ali–Mikhail–Haq copula.

Solution: To determine the empirical copula, we will first need to rank the flood volume and peak discharge variables in the increasing order. Then we can use Equation (3.65) to compute the empirical copula. Here we will use C1n1n as an illustration example. For the flood data in Table 3.1, Table 3.2 lists the order statistics of flood volume and peak discharge individually.

To apply the empirical copula, using x₍₁₎ = 3360, y₍₁₎ = 121x1=3360,y1=121 as an example, we have C1n1n represent #(x_i ≤ x₍₁₎ & y_i ≤ y₍₁₎, i = 1, 2, …, 54)/54#(xi≤x1&yi≤y1i=12…54)/54. Looking up Table 3.1, we find that there is only one pair, i.e., pair 24 (3360, 121), that satisfies the condition x₍₁₎ = 3360x1=3360 and y₍₁₎ = 121y1=121. Thus, we have C154154=154. With this in mind, we can easily compute the empirical copula for the rest of the values, as shown in Figure 3.4.

Figure 3.4 Empirical copula for peak discharge and flood volume.

Solution: The rank of the dataset is computed as in Table 3.4 and Figure 3.5.

Figure 3.5 Scatter plot of ranked pairs (Ri, Si), chi-plot, and K-plot to detect the dependence for the dataset in Table 3.3. (a) Scatter plot of (Ri, Si); (b) chi-plot with control limit of P = 0.9, 0.95; and (c) K-plot with independent and perfectly positively dependent curves.

Using Equation (3.70), we have the following:

∑i=1nRiSi=56+44+31+65+22+13=83

ρn=12nn+1n−1∑i=1nRiSi−3n+1n−1=12×8666+16−1−36+16−1≈0.54

Solution: To calculate sample Kendall’s τ_nτn, we will use Equation (3.73). To illustrate the calculation procedure, we will use the first pair (x₁ = 15.237, y₁ = 19.2x1=15.237,y1=19.2) as an example shown in Table 3.5.

Similarly, we can compute the sum for the remaining pairs as follows:

• 
  

  Pair (x₂, y₂)x2y2 compared to {(x_i, y_i) : i = 3, …, 6}xiyi:i=3…6, sum = 2;

  
  
• 
  

  Pair (x₃, y₃)x3y3 compared to {(x_i, y_i) : i = 4, …, 6}xiyi:i=4…6, sum = 3;

  
  
• 
  

  Pair (x₄, y₄)x4y4 compared to {(x_i, y_i) : i = 5, 6}xiyi:i=56, sum = –2;

  
  
• 
  

  Pair (x₅, y₅)x5y5 compared to (x₆, y₆)x6y6, sum = –1;

  
  
• 
  

  Finally, using Equation (3.73), we have the following:

τn=266−1∑i=15∑j=26signxi−xjyi−yj=266−13+0+1−2−1≈0.47

Solution: Using Equations (3.71) and (3.72), we have the pairs (λ_i, χ_i),λiχi, as shown in Table 3.6 and Figure 3.5.

Table 3.6. Coordinates of points, displayed on the chi-plot, for the data of Table 3.3.

ii	1	2	3	4	5	6
HiHi	0.2	0.6	0	0.4	0.2	0.8
FiFi	0.2	0.8	0	0.4	0.6	1
GiGi	0.4	0.6	0	1	0.2	0.8
F˜i	–0.3	0.3	–0.5	–0.1	0.1	0.5
G˜i	–0.1	0.1	–0.5	0.5	–0.3	0.3
χiχi	0.61	0.61	—	—	0.41	—
λiλi	0.36	0.36	1	–1	–0.36	1

From Table 3.6, it is seen that only three pairs satisfy the condition, and the control limit for p = 0.9, 0.95, 0.99p=0.9,0.95,0.99 will be CL =  ± 0.63,  ± 0.73,  ± 0.89,CL=±0.63,±0.73,±0.89, respectively.

Solution: Let f(x) = xk₀(w){K₀(x)}^i − 1{1 − K₀(x)}^n − 1; n = 6fx=xk0wK0xi−11−K0xn−1;n=6. We can obtain W_i : nWi:n with the numerical integration. The results are given in Table 3.7 and Figure 3.5.

Solution: Table 3.8 lists the rank of flood volume (VV) and peak discharge (QQ). The computation procedure is exactly the same as that in Examples 3.14–3.17.

1. 
   

   1. Calculate the sample Spearman’s ρ_n and Kendall’s τ_n:

Using Equation (3.70) and the same procedure as in Example 3.14, we can easily compute sample ρ_n = 0.7577ρn=0.7577 using Table 3.8 with sample size of 54, as follows:

ρn=125454+154−12847.5+1328+314+…+2936.5−354+154−1.

Using Equation (3.73) and the same procedure as in Example 3.15, we can compute sample Kendall’s τ_n as τ_n = 0.5695τn=0.5695.

Given the double summation for the computation of sample Kendall’s tau, here we will show the first intersummation (i.e., i = 1, j = 2:54), or in other words, comparing (V, Q) = (8704, 371)VQ=8704371 to the rest of pairs: