Question

我正在尝试使用的方法将Copula拟合成两个变量的样本 'copula'和'VineCopula'套餐，但我有两个问题。

首先，我的数据在[0,4]中有值，但是copulas需要数据在[0,1]的区间内，转换它们的程序是什么而没有松散它们的特征？

它给出错误的代码的最后一行，如下所示：

> fit <- fitCopula(cop_model, m, method = 'ml') 
Error in .local(u, copula, log, ...) : unused argument (checkPar = FALSE)

你有什么想法我做错了吗？

# z <- my data - I know that is very small!!


    # the output of dput(z)
    structure(c(3.5448, 3.3863, 3.1097, 2.8126, 2.5005, 2.2661, 2.1308, 
    2.0771, 2.0377, 2.0769, 2.0608, 2.3478, 2.8888, 3.2404, 3.4623, 
    3.6302, 3.4086, 3.0667, 2.8401, 2.5854, 2.2845, 2.0605), .Dim = c(11L, 
    2L))
    # or 

    > z
            [,1]   [,2]
     [1,] 3.5448 2.3478
     [2,] 3.3863 2.8888
     [3,] 3.1097 3.2404
     [4,] 2.8126 3.4623
     [5,] 2.5005 3.6302
     [6,] 2.2661 3.4086
     [7,] 2.1308 3.0667
     [8,] 2.0771 2.8401
     [9,] 2.0377 2.5854
    [10,] 2.0769 2.2845
    [11,] 2.0608 2.0605



    m <- pobs(as.matrix(cbind(z[,1],z[,2])))
            selectedCopula <- BiCopSelect(m[,1],m[,2],familyset=NA)
            selectedCopula

    # Bivariate copula: Survival Joe (par = 2.11, tau=0.38) 
    # Pseudo observations are the observations in the [0,1] interval. 
    # But my data aren't in the [0,1]

    cop_model <- surJoeBiCopula(selectedCopula$par) 
    cop_model

    # method = mpl or ml
    fit <- fitCopula(cop_model, m, method = 'ml') 

       fit

fitCopula用于R中的双变量数据

0 个答案: