# Combining 2 covariance matrices

Given 2 square-symmetric covariance matrices whose sample sizes are not equal, can the following equation be used to compute the combined covariance ? I have been reading this article on wiki https://en.wikipedia.org/wiki/Algorithms_for_calculating_variance#Covariance but was wondering why the equation below is not right (if it is not right).

$C_x = \frac{C_aN_a + C_bN_b}{N_a+N_b}$

where C denotes the covariance and N the sample size.

Can anyone help me on this ?

Assuming your covariance matrices are computed from the samples $\{ \textbf{a}_i \}_{i=1}^{N_a}$ and $\{ \textbf{b}_j \}_{j=1}^{N_b}$, the usual definition for the sample covariance matrix is $$\textbf{C}_a = \frac{1}{N_a - 1} \sum_{i=1}^{N_a} ( \textbf{a}_i - \bar{\textbf{a}}) ( \textbf{a}_i - \bar{\textbf{a}})^T,$$ and similarly for $\textbf{C}_b$. Note that the denominator $N_a - 1$ makes the sample covariance matrix unbiased: $E[\textbf{C}_a] = Cov(\textbf{a}_i )$.
With this is mind, if you now compute the expected value of your proposed combined covariance you get: $$E[\textbf{C}_x] = \frac{N_a}{N_a + N_b} Cov(\textbf{a}_i) + \frac{N_b}{N_a + N_b} Cov(\textbf{b}_j).$$ By itself this is of little use but if we furthermore assume that the two samples come from populations with equal covariance matrices (as is often done, see e.g. Hotelling's $T^2$ test), that is, $Cov(\textbf{a}_i) = Cov(\textbf{b}_j) = \boldsymbol{\Sigma}$, we then have $$E[\textbf{C}_x] = \boldsymbol{\Sigma}.$$ Thus now $\textbf{C}_x$ is unbiased for the common population covariance and what you proposed is indeed the ''correct'' way of combining the two estimators.