[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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