By Roger B. Nelsen

The learn of copulas and their function in facts is a brand new yet vigorously turning out to be box. during this publication the scholar or practitioner of facts and likelihood will locate discussions of the basic houses of copulas and a few in their fundamental purposes. The purposes comprise the research of dependence and measures of organization, and the development of households of bivariate distributions. This ebook is acceptable as a textual content or for self-study.

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**Additional info for An Introduction to Copulas (Springer Series in Statistics)**

**Example text**

The random variables max(X,Y) and min(X,Y) are the order statistics for X and Y. Prove that the distribution functions of the order statistics are given by P[max( X ,Y ) £ t] = C ( F ( t),G ( t)) and P[min( X ,Y ) £ t] = F ( t) + G ( t) – C ( F ( t),G ( t)) , so that when F = G, P[max( X ,Y ) £ t ] = d C ( F ( t)) and P[min( X ,Y ) £ t] = 2 F ( t) – d C ( F ( t)) . (b) Show that bounds on the distribution functions of the order statistics are given by max( F ( t) + G ( t) - 1,0) £ P[max( X ,Y ) £ t] £ min( F ( t),G ( t)) and max( F ( t),G ( t)) £ P[min( X ,Y ) £ t] £ min( F ( t) + G ( t),1) .

Let (a,b) be any point in R 2 , and consider the following distribution function H: Ï0, x < a or y < b, H ( x, y) = Ì Ó1, x ≥ a and y ≥ b. The margins of H are the unit step functions e a and e b . 4 yields the subcopula C ¢ with domain {0,1}¥{0,1} such that C ¢ (0,0) = C ¢ (0,1) = C ¢ (1,0) = 0 and C ¢ (1,1) = 1. , C(u,v) = uv. Notice however, that every copula agrees with C ¢ on its domain, and thus is an extension of this C ¢ . ■ We are now ready to prove Sklar’s theorem, which we restate here for convenience.

Consider the set of points v where each v k is 0, 1, or tk = = ( v1 , v 2 ,L , v n ) min{( n - 1) u k ( u1 + u 2 + L + u n ) ,1} . Define an n-place function C ¢ on these points by C ¢( v ) = W n ( v ) . 5 and hence is an nsubcopula. 2). Then for each x in the n-box [0,t], t = ( t1 , t2 ,L , tn ) (which includes u), C ( x ) = W n ( x ) = 0. 2. Suppose n – 1 < u1 + u 2 + L + u n < n, and consider the set of points v = ( v1 , v 2 ,L , v n ) where now each v k is 0, 1, or sk = 1 – (1 - u k ) [ n - ( u i + u 2 + L + u n )] .