Copula functions Advanced Methods of Risk Management Umberto

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 FX(X). Then u = FX(X) is uniformly distributed in [0, 1]. Likewise, we have v = FY(Y) uniformly distributed. • The joint distribution of X and Y can be written H(X, Y) = H(FX – 1(u), FY – 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 VC(u, v) C(u 1, v 1) – C (u 1, v 2) – C (u 2, v 1) + C(u 2, v 2) 0 • VC(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(FX, FY) 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(Ha, Hb). Define the marginal probability of the complements Ac, Bc as Ha=1 – Ha and Hb=1 – Hb. • The following duality relationships hold among copulas Pr(A, B) = C(Ha, Hb) Pr(Ac, B) = Hb – C(Ha, Hb) = Ca(Ha, Hb) Pr(A, Bc) = Ha – C(Ha, Hb) = Cb(Ha, Hb) Pr(Ac, Bc) =1 – Ha – Hb + C(Ha, Hb) = 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) =b. Cmin +(1 – a – b)Cind + a. Cmax , a, b [0, 1] Cmin= max (x + y – 1, 0), Cind= xy, Cmax= min(x, y) • The parameters a, b are linked to non-parametric dependence measures by particularly simple analytical formulas. For example S = a - b • Mixture copulas (Li, 2000) are a particular case in which copula is a linear combination of Cmax and Cind for positive dependent risks (a>0, b =0), Cmin and Cind for the negative dependent (b>0, a =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, … un ) = = 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, … un ) = = 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, …, un) = – 1 [ (u 1)+…+ (un)] • 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, …, un) = – 1 [ (u 1)+…+ (un)] 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, …, un) = – 1 [ (u 1)+…+ (un)] 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 KC (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 Mc. Kay, 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 1 st 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, …, un) B(un, un+1, …, un+k– 1)

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)… Tj – 1, j(uj – 1 , uj) 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 Ha = 20% in a year. • Say the credit exposure is guaranteed by another party with default probability equal to Ha = 1%. • The probability of default on the exposure is now the joint probability DP = C(Ha , Hb) • The worst case is perfect dependence between default of the two counterparties leading to DP = min(Ha , Hb )

“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…)