滚动窗口估计协方差矩阵

Question

我有4年的时间序列资产回报的，我想，以估计为6个月校准周期方差 - 协方差矩阵进行滚动窗口。总的来说，我应该得到40点的协方差矩阵。我试图运行下面写的代码，但它是错误的。我怎么能修改此R代码里面？

data 

window.size <- 180 #set the size of the window equal to 6 months
windows <- embed(1:nrow(data), window.size)
forApproach <- function(data, windows) {
  l <- vector(mode="list", length=nrow(windows))
  for (i in 1:nrow(data)) {
    l[[i]] <- cov(data[windows[i, ], ])
  }
}

考虑作为数据集的矩阵，其中包括20天以上5个资产回报

data <- matrix(rnorm(100), 20, 5) #data represents the returns of 5 assets over 20 days

我要校准超过5天返回的协方差矩阵，所以考虑到天1，2，3，4，5。然后我要校准另一个协方差矩阵考虑到天6，7，8，9，10。依此类推，使用滚动窗口（我一直在使用循环的此次尝试）。

window.size <- 5

但设置窗口大小等于5，将码认为，对于第一矩阵，天1，2，3，4，5，而对于所述第二矩阵的代码认为天2，3，4，5，6（未6 ，7，8，9，10，我想）。这是我的问题。我不知道如何修改代码，以便该从第2天“分裂”到6一天。

Answer 1

我觉得有一个关于“滚动窗口”一词的误解;通常滚动窗口的方法是指按照一定的窗口内跨行一些指标，你行由行“摇啊摇”。因此，在你，每一个行对应于每天5天的窗口的情况下，行1,2,3,4,5之后将2,3,4,5,6，其次是行3,4,5,6,7等行。

如果我理解正确的话，你反而要计算数据的非重叠排块的协方差矩阵。

鉴于样本数据，你可以这样做：

# Sample data
set.seed(2017);
df <- matrix(rnorm(100), 20, 5)

# Split into groups of 5 corresponding to 5 days and calculate
# covariance matrix
idx <- rep(1:(nrow(df) / 5), each = 5)
lapply(split(as.data.frame(df), idx), cov)
#$`1`
#            V1          V2          V3          V4         V5
#V1  1.42311854  1.12594509 -0.01635956 -0.02680876 -0.9996623
#V2  1.12594509  1.91104181  0.01600511 -0.50270431 -0.4910714
#V3 -0.01635956  0.01600511  0.21584984  0.04264861  0.5356313
#V4 -0.02680876 -0.50270431  0.04264861  0.80241761 -0.3501894
#V5 -0.99966230 -0.49107141  0.53563126 -0.35018940  2.2617564
#
#$`2`
#           V1          V2          V3          V4          V5
#V1  1.6361650  0.28858744  0.55629684 -0.10309928 -0.56784302
#V2  0.2885874  0.32030225  0.09751046 -0.03968577  0.10521384
#V3  0.5562968  0.09751046  0.21460406  0.06921578 -0.20474838
#V4 -0.1030993 -0.03968577  0.06921578  0.44061198 -0.02624344
#V5 -0.5678430  0.10521384 -0.20474838 -0.02624344  0.35858727
#
#$`3`
#            V1         V2          V3          V4         V5
#V1  1.32188749 -0.2504449  0.02865553 -0.83709045  0.7402660
#V2 -0.25044493  0.4449060 -0.45165482  0.18724720 -0.1684300
#V3  0.02865553 -0.4516548  1.59804827 -0.05257944 -0.2588460
#V4 -0.83709045  0.1872472 -0.05257944  2.08276888  0.1345800
#V5  0.74026604 -0.1684300 -0.25884602  0.13457998  0.7381084
#
#$`4`
#          V1         V2         V3         V4         V5
#V1 1.3825793  1.8348434  0.1367480  0.7553666  0.1722815
#V2 1.8348434  3.0679884 -0.7141430  1.9419513  0.4139003
#V3 0.1367480 -0.7141430  1.3646673 -1.3689109 -0.3962832
#V4 0.7553666  1.9419513 -1.3689109  2.1242897  0.7087351
#V5 0.1722815  0.4139003 -0.3962832  0.7087351  0.4589429

---

更新

为了从您的评论解决方案，这里是一个可能性：

# Calculate rows by which to calculate the covariance matrix. 
idx <- lapply(seq(1, nrow(df) - 5, by = 3), function(i) seq(i, i + 4));
idx;
#[[1]]
#[1] 1 2 3 4 5
#
#[[2]]
#[1] 4 5 6 7 8
#
#[[3]]
#[1]  7  8  9 10 11
#
#[[4]]
#[1] 10 11 12 13 14
#
#[[5]]
#[1] 13 14 15 16 17

# Calculate covariance matrix
lapply(idx, function(i) cov(df[i, ]))
[[1]]
            [,1]        [,2]        [,3]        [,4]       [,5]
[1,]  1.42311854  1.12594509 -0.01635956 -0.02680876 -0.9996623
[2,]  1.12594509  1.91104181  0.01600511 -0.50270431 -0.4910714
[3,] -0.01635956  0.01600511  0.21584984  0.04264861  0.5356313
[4,] -0.02680876 -0.50270431  0.04264861  0.80241761 -0.3501894
[5,] -0.99966230 -0.49107141  0.53563126 -0.35018940  2.2617564

[[2]]
           [,1]      [,2]       [,3]        [,4]       [,5]
[1,]  1.2276633 0.8120994 0.68757421  0.43389428 -0.2034626
[2,]  0.8120994 0.9467878 0.54971586  0.32442138  0.0417013
[3,]  0.6875742 0.5497159 0.81237637  0.04317779  0.1016797
[4,]  0.4338943 0.3244214 0.04317779  0.28202885 -0.1328829
[5,] -0.2034626 0.0417013 0.10167967 -0.13288293  0.1941425

[[3]]
             [,1]        [,2]        [,3]         [,4]       [,5]
[1,]  1.611594316  0.20860309  0.57449605 -0.009977472 -0.5998735
[2,]  0.208603088  0.37400181  0.02228603 -0.184461638  0.1758137
[3,]  0.574496047  0.02228603  0.25869591  0.192013428 -0.2558926
[4,] -0.009977472 -0.18446164  0.19201343  0.726477219 -0.1542378
[5,] -0.599873476  0.17581368 -0.25589263 -0.154237772  0.4141949

[[4]]
             [,1]         [,2]       [,3]        [,4]       [,5]
[1,]  0.959758045 -0.002570837 -0.2490718 -0.11965574  0.7669619
[2,] -0.002570837  0.413593056 -0.2238722 -0.05783551 -0.1231235
[3,] -0.249071754 -0.223872167  1.3953139  0.56463838 -0.2210563
[4,] -0.119655741 -0.057835506  0.5646384  1.09879770  0.1947360
[5,]  0.766961857 -0.123123489 -0.2210563  0.19473603  0.7653579

[[5]]
            [,1]       [,2]        [,3]       [,4]       [,5]
[1,]  1.02217247  0.8925311 -0.01480308  0.4282321  0.5941764
[2,]  0.89253109  2.8366577 -1.20242470  2.7991809  0.6818609
[3,] -0.01480308 -1.2024247  1.48751111 -1.7348326 -0.1196483
[4,]  0.42823208  2.7991809 -1.73483255  3.8382883  1.0043009
[5,]  0.59417636  0.6818609 -0.11964826  1.0043009  0.8229246