Question

sMean = function(x) {
  sum = 0;
  for (i in 1:length(x)) {
    sum = sum + x[i]
  }
  return(sum/length(x))
}

covariance = function(x,y) {
  #Formula is E((x-E[x])(y-E[y])) => E[xy] - E[x]E[y]
  meanX = sMean(x)
  meanY = sMean(y)
  cov = 0;
  for (i in 1:length(x)) {
    cov = cov + ((x[i] - meanX) * (y[i] - meanY))
  }
  cov = cov/(length(x)-1)
  return (cov)
}

covariance2 = function(x,y) {
  #Formula is E((x-E[x])(y-E[y])) => E[xy] - E[x]E[y]
  meanX = (sMean(x) * length(x)) /(length(x)-1)
  meanY = (sMean(y) * length(x)) /(length(x)-1)
  meanXY = (sMean(x*y) * length(x)) /(length(x)-1)
  return (meanXY - meanX*meanY)
}

#Output
  #> cov(arr,arr2)
  #[1] 16.75
  #> covariance(arr,arr2)
  #[1] 16.75
  #> covariance2(arr,arr2) #Why this function give wrong output?
  #[1] -9.5

为什么covariance2输出错误？根据{{3}}输出应该是相同的。

Answer 1

在covariance2()中，您需要除length(x)而不是length(x) - 1。

covariance2 = function(x,y) {
    meanX <- sMean(x)
    meanY <- sMean(y)
    meanXY <- sMean(x*y)
    return((meanXY - meanX * meanY) * length(x) / (length(x) - 1))
}

这背后的数学是通过E [x] = sum（x）/ N来表示长度N向量x的期望（即平均值），您可以显示E [xy] -E [x] E [y ] = E [（xE（x））（yE（y））]。这是一个普遍的事实，来自期望运算符的线性。您的covariance()函数返回此等式的右侧乘以N /（N-1）（即样本协方差）。因此，在covariance2()中计算的左侧也需要乘以相同的因子。

为什么covariance2（）给出了错误的答案？

1 个答案: