Copula functions Advanced Methods of Risk Management Umberto Cherubini

Copula functions Copula functions are based on the principle of integral probability transformation. Given a random variable X with probability distribution F X (X). Then u = F X (X) is uniformly distributed in [0,1]. Likewise, we have v = F Y (Y) uniformly distributed. The joint distribution of X and Y can be written H(X,Y) = H(F X –1 (u), F Y –1 (v)) = C(u,v) Which properties must the function C(u,v) have in order to represent the joint function H(X,Y).

Copula function Mathematics A copula function z = C(u,v) is defined as 1. z, u and v in the unit interval 2. C(0,v) = C(u,0) = 0, C(1,v) = v and C(u,1) = u 3. For every u 1 > u 2 and v 1 > v 2 we have V C (u,v)  C(u 1,v 1 ) – C (u 1,v 2 ) – C (u 2,v 1 ) + C(u 2,v 2 )  0 V C (u,v) is called the volume of copula C

Copula functions: Statistics Sklar theorem: each joint distribution H(X,Y) can be written as a copula function C(F X,F Y ) taking the marginal distributions as arguments, and vice versa, every copula function taking univariate distributions as arguments yields a joint distribution.

Copula function and dependence structure Copula functions are linked to non-parametric dependence statistics, as in example Kendall’s  or Spearman’s  S Notice that differently from non-parametric estimators, the linear correlation  depends on the marginal distributions and may not cover the whole range from – 1 to + 1, making the assessment of the relative degree of dependence involved.

Dualities among copulas Consider a copula corresponding to the probability of the event A and B, Pr(A,B) = C(H a,H b ). Define the marginal probability of the complements A c, B c as H a =1 – H a and H b =1 – H b. The following duality relationships hold among copulas Pr(A,B) = C(H a,H b ) Pr(A c,B) = H b – C(H a,H b ) = C a (H a, H b ) Pr(A,B c ) = H a – C(H a,H b ) = C b (H a,H b ) Pr(A c,B c ) =1 – H a – H b + C(H a,H b ) = C(H a, H b ) = Survival copula Notice. This property of copulas is paramount to ensure put-call parity relationships in option pricing applications.

Radial symmetry Take a copula function C(u,v) and its survival version C(1 – u, 1 – v) = 1 – v – u + C( u, v) A copula is said to be endowed with the radial symmetry (reflection symmetry) property if C(u,v) = C(u, v)

AND/OR operators Copula theory also features more tools, which are seldom mentioned in financial applications. Example: Co-copula = 1 – C(u,v) Dual of a Copula = u + v – C(u,v) Meaning: while copula functions represent the AND operator, the functions above correspond to the OR operator.

Conditional probability I The dualities above may be used to recover the conditional probability of the events.

Conditional probability II The conditional probability of X given Y = y can be expressed using the partial derivative of a copula function.

Tail dependence in crashes… Copula functions may be used to compute an index of tail dependence assessing the evidence of simultaneous booms and crashes on different markets In the case of crashes…

…and in booms In the case of booms, we have instead It is easy to check that C(u,v) = uv leads to lower and upper tail dependence equal to zero. C(u,v) = min(u,v) yields instead tail indexes equal to 1.

The Fréchet family C(x,y) =  C min +(1 –  –  )C ind +  C max, ,   [0,1] C min = max (x + y – 1,0), C ind = xy, C max = min(x,y) The parameters ,  are linked to non-parametric dependence measures by particularly simple analytical formulas. For example  S =  Mixture copulas (Li, 2000) are a particular case in which copula is a linear combination of C max and C ind for positive dependent risks (  >0,  C min and C ind for the negative dependent (  >0, 

Elliptical copulas Ellictal multivariate distributions, such as multivariate normal or Student t, can be used as copula functions. Normal copulas are obtained C(u 1,… u n ) = = N(N – 1 (u 1 ), N – 1 (u 2 ), …, N – 1 (u N );  ) and extreme events are indipendent. For Student t copula functions with v degrees of freedom C (u 1,… u n ) = = T(T – 1 (u 1 ), T – 1 (u 2 ), …, T – 1 (u N ); , v) extreme events are dependent, and the tail dependence index is a function of v.

Archimedean copulas Archimedean copulas are build from a suitable generating function  from which we compute C(u 1,…, u n ) =  – 1 [  (u 1 )+…+  (u n )] The function  (x) must have precise properties. Obviously, it must be  (1) = 0. Furthermore, it must be decreasing and convex. As for  (0), if it is infinite the generator is said strict. In n dimension a simple rule is to select the inverse of the generator as a completely monotone function (infinitely differentiable and with derivatives alternate in sign). This identifies the class of Laplace transform.

Example: Clayton copula Take  (t) = [t –  – 1]/  such that the inverse is  – 1 (s) =(1 –  s) – 1/  the Laplace transform of the gamma distribution. Then, the copula function C(u 1,…, u n ) =  – 1 [  (u 1 )+…+  (u n )] is called Clayton copula. It is not symmetric and has lower tail dependence (no upper tail dependence).

Example: Gumbel copula Take  (t) = (–log t)  such that the inverse is  – 1 (s) =exp(– s – 1/  ) the Laplace transform of the positive stable distribution. Then, the copula function C(u 1,…, u n ) =  – 1 [  (u 1 )+…+  (u n )] is called Gumbel copula. It is not symmetric and has upper tail dependence (no lower tail dependence).

Radial symmetry: example Take u = v = 20%. Take the gaussian copula and compute N(u,v; 0,3) = 0,06614 Verify that: N(1 – u, 1 – v; 0,3) = 0,66614 = = 1 – u – v + N(u,v; 0,3) Try now the Clayton copula and compute Clayton(u, v; 0,2792) = 0,06614 and verify that Clayton(1 – u, 1 – v; 0,2792) = 0,6484  0,66614

Kendall function For the class of Archimedean copulas, there is a multivariate version of the probability integral transfomation theorem. The probability t = C(u,v) is distributed according to the distribution K C (t) = t –  (t)/  ’(t) where  ’(t) is the derivative of the generating function. There exist extensions of the Kendall function to n dimensions. Constructing the empirical version of the Kendall function enables to test the goodness of fit of a copula function (Genest and McKay, 1986).

Kendall function: Clayton copula

Copula product The product of a copula has been defined (Darsow, Nguyen and Olsen, 1992) as A*B(u,v)  and it may be proved that it is also a copula.

Markov processes and copulas Darsow, Nguyen and Olsen, 1992 prove that 1st order Markov processes (see Ibragimov, 2005 for extensions to k order processes) can be represented by the  operator (similar to the product) A (u 1, u 2,…, u n )  B(u n,u n+1,…, u n+k–1 )  i

Properties of  products Say A, B and C are copulas, for simplicity bivariate, A survival copula of A, B survival copula of B, set M = min(u,v) and  = u v (A  B)  C = A  ( B  C) (Darsow et al. 1992) A  M = A, B  M = B (Darsow et al. 1992) A   = B   =  (Darsow et al. 1992) A  B =A  B (Cherubini Romagnoli, 2010)

Symmetric Markov processes Definition. A Markov process is symmetric if 1.Marginal distributions are symmetric 2.The  product T 1,2 (u 1, u 2 )  T 2,3 (u 2,u 3 )…  T j – 1,j (u j –1, u j ) is radially symmetric Theorem. A  B is radially simmetric if either i) A and B are radially symmetric, or ii) A  B = A  A with A exchangeable and A survival copula of A.

Example: Brownian Copula Among other examples, Darsow, Nguyen and Olsen give the brownian copula If the marginal distributions are standard normal this yields a standard browian motion. We can however use different marginals preserving brownian dynamics.

Basic credit risk applications Guarantee: assume a client has default probability 20% and he asks for guarantee from a guarantor with default probability of 1%. What is the default probability of the loan? First to default: you buy protection on a first to default on a basket of “names”. What is the price? Are you long or short correlation? Last to default: you buy protection on the last default in a basket of “names”. What is the price? Are you long or short correlation?

An example: guarantee on credit Assume a credit exposure with probability of default of H a = 20% in a year. Say the credit exposure is guaranteed by another party with default probability equal to H a = 1%. The probability of default on the exposure is now the joint probability DP = C(H a, H b ) The worst case is perfect dependence between default of the two counterparties leading to DP = min(H a, H b )

“First-to-default” derivatives Consider a credit derivative, that is a contract providing “protection” the first time that an element in the basket of obligsations defaults. Assume the protection is extended up to time T. The value of the derivative is FTD = LGD v(t,T)(1 – Q(0)) Q(0) is the survival probability of all the names in the basket: Q(0)  Q(  1 > T,  2 > T…)