Does Tail Dependence Make A Difference In the ... - Boston College

Where the first component ∂Φρ(X,Y )
∂X
can be further derived as:
∂Φ ρ (X, Y )
∂X
=
=
=
=
=
∫
1
Y
2π √ 1 − ρ 2
−∞
∫
1
Y
2π √ 1 − ρ 2
−∞
∫
1
Y
2π √ 1 − ρ 2
−∞
∫
1
Y
2π √ 1 − ρ 2
−∞
1
√
2π
exp{− x2
2 } 1
√
2π
√
1 − ρ
2
= φ(x) ∗
= φ(x) ∗
1
√
2π
√
1 − ρ
2
∫
1 Y
√
2π
−∞
= φ(x) ∗ Φ( Y − ρX √
1 − ρ
2 )
exp{− (x2 − 2ρxt + t 2 )
}dt
2(1 − ρ 2 )
exp{− (x2 − ρ 2 x 2 + ρ 2 x 2 − 2ρxt + t 2 )
}dt
2(1 − ρ 2 )
exp{− [x2 (1 − ρ 2 ) + (t − ρx) 2 ]
}dt
2(1 − ρ 2 )
exp{− x2 (t − ρx)2
} ∗ exp{−
2 2(1 − ρ 2 ) }dt
∫ Y
−∞
∫ Y
−∞
(t − ρx)2
exp{−
2(1 − ρ 2 ) }dt
(t − ρx)2
exp{−
2(1 − ρ 2 ) }dt
(t − ρx)2 (t − ρx)
exp{− }d √
2(1 − ρ 2 ) 1 − ρ
2
= φ(x) ∗ Φ( Φ−1 (v) − ρΦ −1 (u)
√ )
1 − ρ
2
Therefore
∂C(u, v; ρ)
∂u
= ∂Φ ρ(X, Y )
∂X
1
φ(x)
= Φ( Φ−1 (v) − ρΦ −1 (u)
√ )
1 − ρ
2
(3)
Let τ = ∂C(u,v;ρ) , we can solve v from the above equation(4) as:
∂u
[
v = Φ ρΦ −1 (u) + √ ]
1 − ρ 2 Φ −1 (τ)
(4)
Note that v = F Y (y) and u = F X (x).
We have assumed that both X and Y follow Gaussian marginal distribution
Y ∼ N(µ y , σ y ) X ∼ N(µ x , σ x )
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