conditioning with copulas n.
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Conditioning with Copulas. Let C 1 (u,v) denote the first partial derivative of C(u,v). F(x,y) = C(F X (x),F Y (y)), distribution of Y|X=x is given by: F Y|X (y) = C 1 (F X (x),F Y (y)) C(u,v) = uv, the conditional distribution of V given that U=u is C 1 (u,v) = v = Pr(V<v|U=u).

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conditioning with copulas
Conditioningwith Copulas
  • Let C1(u,v) denote the first partial derivative of C(u,v). F(x,y) = C(FX(x),FY(y)), distribution of Y|X=x is given by:
  • FY|X(y) = C1(FX(x),FY(y))
  • C(u,v) = uv, the conditional distribution of V given that U=u is C1(u,v) = v = Pr(V<v|U=u).
  • If C1 is simple enough to invert algebraically, then the simulation of joint probabilities can be done using the derived conditional distribution. That is, first simulate a value of U, then simulate a value of V from C1.
kendall correlation
Kendall correlation
  • t is a constant of the copula
  • t = 4E[C(u,v)] – 1
  • t = 2dE[C(u1, . . .,ud)] – 1

2d – 1 – 1

frank s copula
Frank’s Copula
  • Define gz = e-az – 1
  • Frank’s copula with parameter a  0 can be expressed as:
  • C(u,v) = -a-1ln[1 + gugv/g1]
  • C1(u,v) = [gugv+gv]/[gugv+g1]
  • c(u,v) = -ag1(1+gu+v)/(gugv+g1)2
  • t(a) = 1 – 4/a + 4/a20a t/(et-1) dt
  • For a<0 this will give negative values of t.
  • v = C1-1(p|u) = -a-1ln{1+pg1/[1+gu(1–p)]}
gumbel copula
Gumbel Copula
  • C(u,v) = exp{- [(- ln u)a + (- ln v)a]1/a}, a  1.
  • C1(u,v) = C(u,v)[(- ln u)a + (- ln v)a]-1+1/a(-ln u)a-1/u
  • c(u,v) = C(u,v)u-1v-1[(-ln u)a +(-ln v)a]-2+2/a[(ln u)(ln v)]a-1 {1+(a-1)[(-ln u)a +(-ln v)a]-1/a}
  • t(a) = 1 – 1/a
  • Simulate two independent uniform deviates u and v
  • Solve numerically for s>0 with ues = 1 + as
  • The pair [exp(-sva), exp(-s(1-v)a)] will have the Gumbel copula distribution
heavy right tail copula
Heavy Right Tail Copula
  • C(u,v) = u + v – 1 + [(1 – u)-1/a + (1 – v)-1/a – 1]-a a>0
  • C1(u,v) = 1 – [(1 – u)-1/a + (1 – v)-1/a – 1] -a-1(1 – u)-1-1/a
  • c(u,v) = (1+1/a)[(1–u)-1/a +(1– v)-1/a –1] -a-2[(1–u)(1–v)]-1-1/a
  • t(a) = 1/(2a + 1)
  • Can solve conditional distribution for v
joint burr
Joint Burr
  • F(x) = 1 – (1 + (x/b)p)-a and G(y) = 1 – (1 + (y/d)q)-a
  • F(x,y) = 1 – (1 + (x/b)p)-a – (1 + (y/d)q)-a + [1 + (x/b)p + (y/d)q]-a
  • The conditional distribution of y|X=x is also Burr:
  • FY|X(y|x) = 1 – [1 + (y/dx)q]-(a+1), where dx =d[1 + (x/b)p/q]
partial perfect correlation copula generator
Partial Perfect Correlation Copula Generator
  • Assume logical values 0 and 1 are arithmetic also
  • h : unit square  unit interval
  • H(x) = 0xh(t)dt
  • C(u,v) = uv – H(u)H(v) + H(1)H(min(u,v))
  • C1(u,v) = v – h(u)H(v) + H(1)h(u)(v>u)
  • c(u,v) = 1 – h(u)h(v) + H(1)h(u)(u=v)
h u u a
h(u) = (u>a)
  • H(u) = (u – a)(u>a)
  • t(a) = (1 – a)4
h u u a1
h(u) = ua
  • H(u) = ua+1/(a+1)
  • t(a) = 1/[3(a+1)4] + 8/[(a+1)(a+2)2(a+3)]
the normal copula
The Normal Copula
  • N(x) = N(x;0,1)
  • B(x,y;a) = bivariate normal distribution function,  = a
  • Let p(u) be the percentile function for the standard normal:
  • N(p(u)) = u, dN(p(u))/du = N’(p(u))p’(u) = 1
  • C(u,v) = B(p(u),p(v);a)
  • C1(u,v) = N(p(v);ap(u),1-a2)
  • c(u,v) = 1/{(1-a2)0.5exp([a2p(u)2-2ap(u)p(v)+a2p(v)2]/[2(1-a2)])}
  • t(a) = 2arcsin(a)/p
  • a: 0.15643 0.38268 0.70711 0.92388 0.98769
  • t: 0.10000 0.25000 0.50000 0.75000 0.90000
tail concentration functions
Tail Concentration Functions
  • L(z) = Pr(U<z,V<z)/z2
  • R(z) = Pr(U>z,V>z)/(1 – z)2
  • L(z) = C(z,z)/z2
  • 1 - Pr(U>z,V>z) = Pr(U<z) + Pr(V<z) - Pr(U<z,V<z)
  • = z + z – C(z,z).
  • Then R(z) = [1 – 2z +C(z,z)]/(1 – z)2
  • Generalizes to multi-variate case
cumulative tau
Cumulative Tau
  • t = –1+40101 C(u,v)c(u,v)dvdu
  • J(z) = –1+40z0z C(u,v)c(u,v)dvdu/C(z,z)2
  • Generalizes to multi-variate case
cumulative conditional mean
Cumulative Conditional Mean
  • M(z) = E(V|U<z) = z-10z01 vc(u,v)dvdu
  • M(1) = ½
  • A pairwise concept

Copula Distribution Function

  • K(z) = Pr(C(u,v)<z)
  • Generalizes to multi-variate case
slide22

HRT Gumbel Frank Normal

Parameter 0.968 1.67 4.92 0.624

Ln Likelihood 124 157 183 176

Tau 0.34 0.40 0.45 0.43