[...]... )] + [a1 − min(b2 , a1 )]} ≥ 0 Since b1 ≥ min(b1 , a2 ), and a1 ≥ min(a1 , b2 ) Then C(u, v, w) is a 3 -copula The 1-dimensional margins of C are the 1-copulas 9 C1 (u) = C(u, 1, 1) = u, C2 (v) = C(1, v, 1) = v , and C3 (w) = C(1, 1.w) = w And the 2-margins of C are 2-copulas C1,2 (u, v) = C(u, v, 1) = min(u, v), C2,3 (v, w) = C(1, v, w) = vw, and C1,3 (u, w) = C(u, 1, w) = uw The following theorem... Then a quasi-inverse of F is any function F ( 1) with domain I such that: ˆ if t is in RanF, then F ( 1) (t) is any number x in R such that F −1 (x) = t, i.e, for all t in RanF, F (F ( 1) (t)) = t (1 .9) F ( 1) (t) = inf{x|F (x) ≥ t} = sup{x|F (x) ≤ t} (1 .10) ˆ if t is not in RanF, then If F is strictly increasing, then it has but a single quasi-inverse, which is of course the ordinary inverse, for... > Y } = C(u, v)dC(u, v) [0,1]2 Similarly, P {X < X, Y < Y } = P {X > x, Y > y}dC(F (x), G(y)) R2 {1 − F (x) − G(y) + C(F (x), G(y))}dC(F (x), G(y)) = R2 {1 − u − v + C(u, v)}dC(u, v) = [0,1]2 But since C is the joint distribution function of a vector (U, V )T of U (0 , 1) random variables, E(U ) = E(V ) = 1/2, and hence P {X < X, Y < Y } = 1 − 1 1 − + 2 2 C(u, v)dC(u, v) = [0,1]2 C(u, v)dC(u, v) [0,1]2... bivariate Uniform(0, 1) distribution We can also use the equation (1 .12) to derive an expression for h as a function of x and y instead: h(F −1 (u), G−1 (v)) = f (F −1 (u)).g(G−1 (v)).c(u, v) (1 .13) h(x, y) = f (x).g(y).c(F (x), G(y)) Equation (1 .13) is the density version of Sklar's (1 959) theorem: the joint density, h, can be decomposed into product of the marginal densities, f and g , and the copula density,... w) = w min(u, v) It is easy to see that C satises conditions of Denition 1.7, and the H -volume of the 3-boxes and B = [a1 , b1 ] × [a2 , b2 ] × [a3 , b3 ] is b VC (B) = ∆a3 ∆b2 ∆bn C(u, v, w) a2 a1 3 b = (b3 − a3 )∆a2 ∆bn C(u, v, w) a1 2 = (b3 − a3 )[min(b2 , b1 ) − min(b2 , a1 ) − min(a2 , b1 ) + min(a2 , a1 )] = (b3 − a3 )(b1 + a1 − min(b2 , a1 ) − min(a2 , b1 )) = (b3 − a3 ){[b1 − min(a2 , b1... and (X, Y )T , respectively, so that H(x, y) = C(F (x), G(y)) and H = C(F (x), F (y)) Let Q denotes the dierence between the probability of concordance and discordance of 15 (X, Y )T and (X, Y )T , i.e let Q = P {(X − X)(Y − Y ) > 0} − P {(X − X)(Y − Y ) < 0} Then C(u, v)dC(u, v) − 1 Q = Q(C, C) = 4 [0,1]1 Proof Since the random variables are all continuous, P {(X − X)(Y − Y ) < 0} = 1 − P {(X − X)(Y... (X) )−1 = f (X)−1 and ∂V = ( ∂V )−1 = ∂U ∂X ∂Y ∂G(Y ) −1 ∂X ∂Y −1 ( ∂Y ) = g(Y ) Note that ∂V = ∂U = 0 Then, c(u, v) = h(X(u), Y (v)) ∂X ∂U ∂Y ∂U ∂X ∂V ∂Y ∂V (1 .12) ∂X ∂Y ∂U ∂V h(F −1 (u), G−1 (v)) = f (F −1 (u)).g(G−1 (v)) = h(F −1 (u), G−1 (v)) Equation (1 .12) show that the copula density of X and Y is equal to the ratio of the joint density, h, to the product of marginal densities, f and g From... analysis, and is obtained quite easily, provided that F and G are dierentiable, and H and G are twice dierentiable ∂ 2 H(x, y | ) ∂x∂y 2 ∂ C(F (x | ), G(y | ) | ) ∂F (x | ) ∂G(y | ) = ∂(F (x | ))∂(G(y | )) ∂x ∂y 2 ∂ C(u, v | ) f (x | ).g(y | ), = ∂u∂v h(x, y | ) = c(u, v | ).f (x | ).g(y | ) ∀x, y ∈ R h(x, y | ) ≡ (1 .15) where u ≡ F (x | ), and v ≡ G(y | ) We can also obtain a corollary to Theorem... denote the linear correlation coecient Then ρ(X, Y ) = E(XY ) − E(X)E(Y ) Var(X)Var(Y) = E(XY ) − 1, where ∞ ∞ xydH(x, y) E(XY ) = 0 ∞ 0 ∞ xy (( 1 + θ)e−x−y − 2θe−2x−y − 2θe−x−2y + 4θe−2x−2y )dxdy = 0 0 θ =1+ 4 Hence ρ(X, Y ) = θ/4 But (1 − e−X , 1 − e−Y ) = ρS (X, Y ) C(u, v)dudv − 3 = 12 [0,1]2 (uv + (1 − u )(1 − v))dudv − 3 = 12 [0,1]2 19 1 θ = 1 2( + ) − 3 4 36 = θ/3 Hence ρ(X, Y ) is not invariant... quasi-inverses are F ( 1) (u) = 2u − 1 and G(−1) (v) = − ln(1 − v), u, v ∈ I ( 1) Corollary 1.13 Let H, C, F1 , F2 , , Fn be as in Theorem 1.9 and F 1( 1) , F 2( 1) , , Fn be quasi-inverses of F1 , F2 , , Fn , respectively Then, for any u in I n ( 1) C(u1 , u2 , , un ) = H(F1 ( 1) (u1 ), F2 ( 1) (u2 ), , Fn (un )) (1 .11) We can use the result of this corollary to nd the copulas in Example 1.4 Example