我正在处理一个数据文件,里面的观察值是随机值.在这种情况下,我不知道x的分布(我的观察).我使用函数密度来估计密度,因为必须应用内核估计.
I'm working with a data file, the observations inside are random values. In this case I don't know the distribution of x (my observations). I'm using the function density in order to estimate the density, because I must apply a kernel estimation.
T=density(datafile[,1],bw=sj,kernel="epanechnikov")在此之后,我必须将其整合,因为我正在寻找分位数(类似于95%的VaR). 为此,我有2个选择:
After this I must integrate this because I'm looking for a quantile (similar to VaR, 95%). For this I have 2 options:
ecdf() quantile()现在我有分位数95的值,但这是内核估计的数据.
Now I have the value of the quantile 95, but this is the data estimated by kernel.
是否可以使用一个函数来了解原始数据的分位数95的值?
我指出这是一个未知的分布,为此,我想将一种非参数方法想象为Newton,就像SAS solve()
I remark that this is a distribution unknown, for this I would like to imagine a non parametric method as Newton, like the one that is in SAS solve()
推荐答案您可以使用 quantile() .这是一个使用随机数据的示例:
You can use quantile() for this. Here is an example using random data:
> data<-runif(1000) > q<-quantile(data, .95) > q 95% 0.9450324这里,数据均匀地分布在0和1之间,因此第95个百分位数接近0.95.
Here, the data is uniformly distributed between 0 and 1, so the 95th percentile is close to 0.95.
要执行逆变换:
> ecdf(data)(q) [1] 0.95更多推荐
未知累积函数的反函数
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