“…whereψ(x, y) = x − y, u x − y, v 1{ x − y ≤ r}1{ x ≤ R − r} . (Xi, Xj)ψ(X k , X l )] − E i =jψ(Xi, Xj) ψ(x, y)ψ(z, w)ρ4(x, y, z, w)dxdydzdw + ψ(x, y)ψ(y, w)ρ3(x, y, w)dxdydw + ψ(x, y)ψ(x, w)ρ3(x, y, w)dxdydw+ ψ(x, y)ψ(z, y)ρ3(x, y, z)dxdydz + ψ(x, y)ψ(z, x)ρ3(x, y, z)dxdydz + ψ(x, y) 2 ρ2(x, y)dxdy − ψ(x,y)ρ2(x, y)dxdyBy symmetry of K (and thus of ρ3), all triple integrals are the same, equal to ψ(x, y)ψ(x, z)ρ3(x, y, z)dxdydz .Moreover, since 4-point correlation ρ4 is given byρ4(x1, x2, x3, x4) = det K(x1, x1) K(x1, x2) K(x1, x3) K(x1, x4) K(x2, x1) K(x2, x2) K(x2, x3) K(x2, x4) K(x3, x1) K(x3, x2) K(x3, x3) K(x3, x4) K(x4, x1) K(x4, x2) K(x4, x3) K(x4, x4) the symmetry of K and Fischer's inequality thatρ4(x1, x2, x3, x4) ≤ det K(x1, x1) K(x1, x2) K(x2, x1) K(x2, x2) • det K(x3, x3) K(x3, x4) K(x4, x3) K(x4, x4) = ρ2(x1, x2)ρ2(x3, x4) [12]. Therefore, ψ(x, y)ψ(z, w)ρ4(x, y, z, w)dxdydzdw − ψ(x, y)ρ2(x, y)dxdy 2…”