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Gary G Venter Tails of Copulas. 2 Guy Carpenter Correlation Issues Correlation is stronger for large events Can model by copula methods Quantifying correlation.

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Presentation on theme: "Gary G Venter Tails of Copulas. 2 Guy Carpenter Correlation Issues Correlation is stronger for large events Can model by copula methods Quantifying correlation."— Presentation transcript:

1 Gary G Venter Tails of Copulas

2 2 Guy Carpenter Correlation Issues Correlation is stronger for large events Can model by copula methods Quantifying correlation – Degree of correlation – Part of spectrum correlated

3 3 Guy Carpenter Modeling via Copulas Correlate on probabilities Inverse map probabilities to correlate losses Can specify where correlation takes place in the probability range Conditional distribution easily expressed Simulation readily available

4 4 Guy Carpenter What is a copula? A way of specifying joint distributions A way to specify what parts of the marginal distributions are correlated Works by correlating the probabilities, then applying inverse distributions to get the correlated marginal distributions Formally they are joint distributions of unit uniform variates, as probabilities are uniform on [0,1]

5 5 Guy Carpenter Formal Rules F(x,y) = C(F X (x),F Y (y)) – Joint distribution is copula evaluated at the marginal distributions – Expresses joint distribution as inter-dependency applied to the individual distributions C(u,v) = F(F X -1 (u),F Y -1 (v)) – u and v are unit uniforms, F maps R 2 to [0,1] F Y|X (y) = C 1 (F X (x),F Y (y)) – Derivative of the copula is the conditional distribution E.g., C(u,v) = uv, C 1 (u,v) = v = Pr(V<v|U=u) – So independence copula

6 6 Guy Carpenter Correlation Kendall tau and rank correlation depend only on copula, not marginals Not true for linear correlation rho Tau may be defined as: –1+4E[C(u,v)]

7 7 Guy Carpenter Example C(u,v) Functions Frank: -a -1 ln[1 + g u g v /g 1 ], with g z = e -az – 1 –  (a) = 1 – 4/a + 4/a 2  0 a t/(e t -1) dt Gumbel: exp{- [(- ln u) a + (- ln v) a ] 1/a }, a  1 –  (a) = 1 – 1/a HRT: u + v – 1+[(1 – u) -1/a + (1 – v) -1/a – 1] -a –  (a) = 1/(2a + 1) Normal: C(u,v) = B(p(u),p(v);a) i.e., bivariate normal applied to normal percentiles of u and v, correlation a –  (a) = 2arcsin(a)/ 

8 8 Guy Carpenter Copulas Differ in Tail Effects Light Tailed Copulas Joint Lognormal

9 9 Guy Carpenter Copulas Differ in Tail Effects Heavy Tailed Copulas Joint Lognormal

10 10 Guy Carpenter Partial Perfect Correlation Copulas of Kreps Each simulated probability pair is either identical or independent depending on symmetric function h(u,v), often =h(u)h(v) h(u,v) –> [0,1], e.g., h(u,v) = (uv) 3/5 Draw u,v,w from [0,1] If h(u,v)>w, drop v and set v=u Simulate from u and v, which might be u

11 11 Guy Carpenter Simulated Pareto (1,4) h(u)=u 0.3 (Partial Power Copula)

12 12 Guy Carpenter Partial Cutoff Copula h(u)=(u>k)

13 13 Guy Carpenter Partial Perfect Copula Formulas For case h(u,v)=h(u)h(v) H’(u)=h(u) C(u,v) = uv – H(u)H(v) + H(1)H(min(u,v)) C 1 (u,v) = v – h(u)H(v) + H(1)h(u)(v>u)

14 14 Guy Carpenter Tau’s h(u)=u a,  (a)= (a+1) -4 /3 +8/[(a+1)(a+2) 2 (a+3)] h(u)=(u>k),  (k) = (1 – k) 4 h(u)=h 0.5,  (h) = (h 2 +2h)/3 h(u)= h 0.5 u a ( u>k),  (h,a,k) = h 2 (1-k a+1 ) 4 (a+1) -4 /3 +8h[(a+2) 2 (1-k a+3 )(1-k a+1 )–(a+1)(a+3)(1-k a+2 ) 2 ]/d where d = (a+1)(a+2) 2 (a+3)

15 15 Guy Carpenter Quantifying Tail Concentration L(z) = Pr(U<z|V<z) R(z) = Pr(U>z|V>z) L(z) = C(z,z)/z R(z) = [1 – 2z +C(z,z)]/(1 – z) L(1) = 1 = R(0) Action is in R(z) near 1 and L(z) near 0 lim R(z), z->1 is R, and lim L(z), z->0 is L

16 16 Guy Carpenter LR Function (L below ½, R above)

17 17 Guy Carpenter R usually above tau

18 18 Guy Carpenter Example: ISO Loss and LAE Freez and Valdez find Gumbel fits best, but only assume Paretos Klugman and Parsa assume Frank, but find better fitting distributions than Pareto All moments less than tail parameter converge

19 19 Guy Carpenter Can Try Joint Burr, from HRT F(x,y) = 1–(1+(x/b) p ) -a –(1+(y/d) q ) -a +[1+(x/b) p +(y/d) q ] -a E.g. F(x,y)=1–[1+x/14150] -1.11 –[1+(y/6450) 1.5 ] -1.11 +[1+x/14150 +(y/6450) 1.5 ] -1.11 Given loss x, conditional distribution is Burr: F Y|X (y|x) = 1–[1+(y/d x ) 1.5 ] –2.11 with d x = 6450 +11x 2/3

20 20 Guy Carpenter Example: 2 States’ Hurricanes

21 21 Guy Carpenter L and R Functions, Tau =.45 R looks about.25, which is >0, <tau, so none of our copulas match

22 22 Guy Carpenter Fits

23 23 Guy Carpenter Auto and Fire Claims in French Windstorms

24 24 Guy Carpenter MLE Estimates of Copulas

25 25 Guy Carpenter Modified Tail Concentration Functions Both MLE and R function show that HRT fits best

26 26 Guy Carpenter Conclusions Copulas allow correlation of different parts of distributions Tail functions help describe and fit

27 27 Guy Carpenter finis


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