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MissWawrzyniak
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which Archimedean copula is independent?

February 6th, 2006, 5:28 pm

Is there a Independent copula in the family of Archimedean copulas? I know that the product copula: Pi(u,v)=u*v is an independent, I was asked if there is any Archimedean independent copula. I dont know, cant find the answer. And what about the Frechet upper and lower bounds, do they correspond to some Archimedean copula?Thanks
 
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meteor
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which Archimedean copula is independent?

February 7th, 2006, 1:51 pm

The independant case correspond to a paramter value in many archimedean copulas.By example, in the Clayton and Frank when the generator parameter is equal to zero, in the Gumpel when the parameter is one, in the Marshall Olkin when one parameter is equal to zero,....In the bivariate case the frechet-hoeffdings bounds are copula. In the multivariate case only the upper frechet bounds is a copula while the lower bounds is generally not a copula. So that by example in the bivariate case the upper frecht bounds correspond to the clayton copula when its parameter is equal to +\infty While the lower bounds correspond to the clayton copulas when its parameter is -1). Check the Nelsen book on copulas for more info (in fact, the second edition has just been published)
Last edited by meteor on February 6th, 2006, 11:00 pm, edited 1 time in total.
 
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genkideska
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which Archimedean copula is independent?

February 8th, 2006, 7:02 am

the independence copula IS a archimedian copula. just use the generator phi=Log(x).
 
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actuaryalfred
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which Archimedean copula is independent?

February 9th, 2006, 1:08 am

Don't we use copula for modeling dependence? I mean, if the multivariate distribution is independent, why bother using copula at all?
 
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MissWawrzyniak
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which Archimedean copula is independent?

February 12th, 2006, 12:58 pm

Thx all !good question actuaryalfred !genkideska, yeah thats what I needed. Is there a genrator phi which generates Frechet bounds ?