将滚动窗口回归应用于 R 中的 XTS 系列 [英] Applying a rolling window regression to an XTS series in R

问题描述

对于 5 个货币对，我有 1033 个每日回报点的 xts，我想在这些货币对上运行滚动窗口回归，但是 rollapply 不适用于我定义的使用 lm() 的函数.这是我的数据:

I have an xts of 1033 daily returns points for 5 currency pairs on which I want to run a rolling window regression, but rollapply is not working for my defined function which uses lm(). Here is my data:

> head(fxr)
                 USDZAR        USDEUR       USDGBP        USDCHF        USDCAD
2007-10-18 -0.005028709 -0.0064079963 -0.003878743 -0.0099537170 -0.0006153215
2007-10-19 -0.001544470  0.0014275520 -0.001842564  0.0023058211 -0.0111410271
2007-10-22  0.010878027  0.0086642116  0.010599365  0.0051899551  0.0173792230
2007-10-23 -0.022783987 -0.0075236355 -0.010804304 -0.0041668499 -0.0144788687
2007-10-24 -0.006561223  0.0008545792  0.001024275 -0.0004261666  0.0049525483
2007-10-25 -0.014788901 -0.0048523001 -0.001434280 -0.0050425302 -0.0046422944

> tail(fxr)
                 USDZAR       USDEUR       USDGBP       USDCHF        USDCAD
2012-02-10  0.018619309  0.007548205  0.005526184  0.006348533  0.0067151342
2012-02-13 -0.006449463 -0.001055966 -0.002206810 -0.001638002 -0.0016995755
2012-02-14  0.006320364  0.006843933  0.006605875  0.005992935  0.0007001751
2012-02-15 -0.001666872  0.004319096 -0.001568874  0.003686840 -0.0015009759
2012-02-16  0.006419616 -0.003401364 -0.005194817 -0.002709588 -0.0019044761
2012-02-17 -0.004339687 -0.003675992 -0.003319899 -0.003043481  0.0000000000

我可以轻松地对整个数据集运行 lm 以针对其他货币对对 USDZAR 进行建模:

I can easily run an lm on it for the whole data set to model USDZAR against the other pairs:

> lm(USDZAR ~ ., data = fxr)$coefficients
  (Intercept)        USDEUR        USDGBP        USDCHF        USDCAD 
-1.309268e-05  5.575627e-01  1.664283e-01 -1.657206e-01  6.350490e-01 

但是我想运行一个滚动的 62 天窗口来获得这些系数随时间的演变，因此我创建了一个函数 dolm 来执行此操作:

However I want to run a rolling 62 day window to get the evolution of these coefficients over time, so I create a function dolm which does this:

> dolm
function(x) {
  return(lm(USDZAR ~ ., data = x)$coefficients)
}

但是，当我对此运行 rollapply 时，我得到以下信息:

However when I run rollapply on this I get the following:

> rollapply(fxr, 62, FUN = dolm)
Error in terms.formula(formula, data = data) : 
  '.' in formula and no 'data' argument

即使 dolm(fxr) 本身工作正常:

that is even though dolm(fxr) on its own works fine:

> dolm(fxr)
  (Intercept)        USDEUR        USDGBP        USDCHF        USDCAD 
-1.309268e-05  5.575627e-01  1.664283e-01 -1.657206e-01  6.350490e-01 

这是怎么回事?如果 dolm 是一个更简单的函数，它似乎工作正常，例如意思:

What's going on here? It seems to work fine if dolm is a simpler function for example mean:

> dolm <- edit(dolm)
> dolm
function(x) {
  return(mean(x))
}
> rollapply(fxr, 62, FUN = dolm)
                  USDZAR        USDEUR        USDGBP        USDCHF        USDCAD
2007-11-29 -1.766901e-04 -6.899297e-04  6.252596e-04 -1.155952e-03  7.021468e-04
2007-11-30 -1.266130e-04 -6.512204e-04  7.067767e-04 -1.098413e-03  7.247315e-04
2007-12-03  8.949942e-05 -6.406932e-04  6.637066e-04 -1.154806e-03  8.727564e-04
2007-12-04  2.042046e-04 -5.758493e-04  5.497422e-04 -1.116308e-03  7.124593e-04
2007-12-05  7.343586e-04 -4.899982e-04  6.161819e-04 -1.057904e-03  9.915495e-04

非常感谢任何帮助.基本上我想要的是在滚动的 62 天窗口中获得 USDZAR ~ USDEUR + USDGBP + USDCHF + USDCAD 回归的权重.

Any help much appreciated. Essentially what I want is to get the weightings for the regression of USDZAR ~ USDEUR + USDGBP + USDCHF + USDCAD over a rolling 62-day window.

推荐答案

这里有几个问题:

• rollapply 传递一个矩阵，但 lm 需要一个 data.frame.
• rollapply 将函数分别应用于每一列，除非我们指定 by.column=FALSE.
• 您可能希望也可能不希望结果是与日期右对齐，但如果您使用 rollapplyr :

• rollapply passes a matrix but lm requires a data.frame.
• rollapply applies the function to each column separately unless we specify by.column=FALSE.
• you may or may not want the result to be right aligned with the dates but if you do use rollapplyr :

1) 结合以上我们有:

dolm <- function(x) coef(lm(USDZAR ~ ., data = as.data.frame(x))))
rollapplyr(fxr, 62, dolm, by.column = FALSE)

2) 上面 dolm 中 lm 的替代方法是使用 lm.fit适用于矩阵并且速度更快:

2) An alternative to the lm in the dolm above is to use lm.fit which directly works with matrices and is also faster:

dolm <- function(x) coef(lm.fit(cbind(Intercept = 1, x[,-1]), x[,1]))

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