在 1972 年的书中找到了解决方案（George R. Price, Ann. Hum. Genet., Lond, pp485-490, Extension of covariance selection math, 1972）。

有偏加权样本协方差：

$\Sigma=\frac{1}{\sum_{i=1}^{N}w_i}\sum_{i=1}^N w_i \left(x_i - \mu^*\right)^T\left(x_i - \mu^*\right)$

以及通过应用贝塞尔校正给出的无偏加权样本协方差：

$\Sigma=\frac{1}{\sum_{i=1}^{N}w_i - 1}\sum_{i=1}^N w_i \left(x_i - \mu^*\right)^T\left(x_i - \mu^*\right)$

其中是（无偏的）加权样本均值：$\mu^*$

$\mathbf{\mu^*}=\frac{\sum_{i=1}^N w_i \mathbf{x}_i}{\sum_{i=1}^N w_i}$

重要提示：仅当权重是“重复”型权重时才有效，这意味着每个权重代表一个观察的出现次数，并且其中表示实际样本量（实际样本总数，占权重）。$\sum_{i=1}^N w_i=N^*$$N^*$

我更新了 Wikipedia 上的文章，您还可以在其中找到无偏加权样本方差的方程：

https://en.wikipedia.org/wiki/Weighted_arithmetic_mean#Weighted_sample_covariance

和逐列相乘进行矩阵乘法以包装事物并自动执行求和。例如在 Python Pandas/Numpy 代码中：$w_i$$\left(x_i - \mu^*\right)$$\left(x_i - \mu^*\right)$

import pandas as pd
import numpy as np
# X is the dataset, as a Pandas' DataFrame
mean = mean = np.ma.average(X, axis=0, weights=weights) # Computing the weighted sample mean (fast, efficient and precise)
mean = pd.Series(mean, index=list(X.keys())) # Convert to a Pandas' Series (it's just aesthetic and more ergonomic, no differenc in computed values)
xm = X-mean # xm = X diff to mean
xm = xm.fillna(0) # fill NaN with 0 (because anyway a variance of 0 is just void, but at least it keeps the other covariance's values computed correctly))
sigma2 = 1./(w.sum()-1) * xm.mul(w, axis=0).T.dot(xm); # Compute the unbiased weighted sample covariance

使用非加权数据集和等效加权数据集进行了一些健全性检查，它工作正常。

有关无偏方差/协方差理论的更多详细信息，请参阅这篇文章。