[R-SIG-Finance] extract parameters from fitCopula outputs !!

spencerg spencer.graves at prodsyse.com
Sun May 17 23:01:30 CEST 2009


      Your example is incomplete, because I do not have your "u". 

      However, the examples in the "fitCopula" help page should suffice 
to answer your question:  Consider in particular the multivariate example: 

    normal.cop <- normalCopula(c(0.6,0.36, 0.6),dim=3,dispstr="un")
     x <- rcopula(normal.cop, n)     ## true observations
     u <- apply(x, 2, rank) / (n + 1)  ## pseudo-observations
     ## inverting Kendall's tau
     fit.tau <- fitCopula(normal.cop, u, method="itau")

     
      Compare str(fit.tau) with print(fit.tau).  This suggests we can 
answer your question with "fit.tau at estimate". 


      Hope this helps. 
      Spencer Graves
   
tawfiq just wrote:
> hello
>
> i have the following output from fitting a normal copula (copula package)
>
>  normal.cop <- normalCopula(c(0,0,0,0,0,0,0,0,0,0),dim=5,dispstr="un")
> fit.tau <- fitCopula(normal.cop, u, method="itau")
>
> The estimation method is  Inversion of Kendall's Tau  based on  52
> observations.
>         Estimate Std. Error   z value Pr(>|z|)
> rho.1  0.7609442 0.08183654  9.298343        0
> rho.2  0.9123854 0.04060923 22.467440        0
> rho.3  0.8854560 0.04804054 18.431432        0
> rho.4  0.8730453 0.05525219 15.801100        0
> rho.5  0.8296553 0.05762548 14.397370        0
> rho.6  0.8022804 0.06063024 13.232347        0
> rho.7  0.8120697 0.06072519 13.372865        0
> rho.8  0.9470253 0.02154186 43.962098        0
> rho.9  0.9333255 0.02565925 36.373847        0
> rho.10 0.9591890 0.01818588 52.743616        0
>
> the question is how to extract the first column from the output, as a vector
> (0.7609442, 0.9123854, .....,0.9591890)
>
> thank you very much
>
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