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我想最大化funcToOpt代码中的。

数据说明：

  1. wbX1是 T x N 矩阵，以及一个Nt长度为 N 的向量
  2. 每行之和wb为1
  3. X1横截面标准化（平均值=1 sd=0）

标准化意味着 的横截面平均值theta*X1为零，这意味着它rowSums(wb + (theta1*X1) / Nt)始终为 1。 这发生在我使用没有 NAs 值的 X1 矩阵时，是否有人知道我如何使行总和wi等于1 当使用带有 NA 的矩阵时？

最大化代码：

X1 <- t(scale(t((X1))))    # X1 STANDARDIZATION
N <- dim(X1)[2]
T <- dim(X1)[1]

M <- wb + r + X1                  # Nt, counting the NAs values
Nt <- rowSums(!is.na(M)) 

funcToOpt <- function (theta1)
{
  inner <- rowSums((wb + (theta1*X1) / Nt) * r,na.rm = TRUE)  

  return (sum(inner, na.rm=TRUE) / T)
}

max<-optimize(funcToOpt, c(-50, 50), maximum = TRUE)

theta1=max[[1]]                                 

wi=wb+(theta1*X1) / Nt     
rowSums(wi,na.rm=T)

矩阵 5X5 无 NA

r=matrix(c(0.1,0.3,0.4,0.3,0.4,0.5,0.3,0.3,0.2,0.1,0.1,0.1,0.1,0.1,0.2,0.9,0.3,0.2,0.1,0.2,0.2,0.4,0.4,0.5,0.1),ncol = 5)
wb=matrix(c(0.3,0.3,-0.4,-0.3,0.4,-0.5,-0.3,0.3,0.2,0.1,0.1,0.1,0.1,0.1,0.2,0.9,0.3,0.6,0.5,0.2,0.2,0.6,0.4,0.5,0.1),ncol = 5)
X1=matrix(c(1:20,0,4,5,0,3),ncol = 5)

MATRIX 5x5 不适用

r=matrix(c(0.1,0.3,0.4,0.3,0.4,0.5,0.3,0.3,0.2,0.1,0.1,0.1,0.1,0.1,0.2,0.9,0.3,0.2,0.1,0.2,0.2,0.4,0.4,0.5,0.1),ncol = 5)
wb=matrix(c(0.3,0.3,-0.4,-0.3,0.4,-0.5,-0.3,0.3,0.2,0.1,0.1,0.1,0.1,0.1,0.2,0.9,0.3,0.6,0.5,0.2,0.2,0.6,0.4,0.5,0.1),ncol = 5)
X1=matrix(c(1:20,NA,4,5,NA,3),ncol = 5)
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