2009/5/17 tawfiq just <just.tawfiq@gmail.com>

>
>
> 2009/5/17 spencerg <spencer.graves@prodsyse.com>
>
>     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@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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>>>
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>>
>> thak you spencer for answer, that's what i was looking for
>
> to be more clair, what i wanted to do is to simulate a multivariate normal
> copula using CML method
> suppose i have a matrix of returns 50 * 4 dimension
> x <- matrix(rnorm(200),50,4)
> u <- apply(x, 2, rank) / (n + 1)  ## empirical distribution of x
>   ## inverting Kendall's tau
> normal.cop1 <- normalCopula(c(0,0,0,0,0,0),dim=4,dispstr="un")
>
> fit.tau <- fitCopula(normal.cop1, G, method="itau")  # fit data to copula
> to evaluate rho
>
> normal.cop <- normalCopula(  fit.tau@estimate,dim=4,dispstr="un")
>
> u <- rcopula(normal.cop, 10)     # simulate normal copula that have same
> caracteristics than data
>
>
> is it correct what i have done?
>
> thank you again
>

Error in the instruction
fit.tau <- fitCopula(normal.cop1, G, method="itau")  # fit data to copula to
evaluate rho

u insteade of G

fit.tau <- fitCopula(normal.cop1, u, method="itau")  # fit data to copula to
evaluate rho

sorry

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