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Assume that $X_{i,j} \sim [\mu_i, \Sigma], i=1,...n; j=1...m$ and we have realisations $x_{i,j}=X_{i,j}(\omega)$. Is the formula: $\frac{1}{(n-1)m}\sum_{i=1}^n\sum_{j=1}^m[x_{i,j}-\overline{x_{i}}][x_{i,j}-\overline{x_{i}}]^T$ then an unbiased pooled estimate for the covariance matrix $\Sigma$? I have read almost the same formula on wikipedia: https://en.wikipedia.org/wiki/Hotelling%27s_T-squared_distribution#Pooled_covariance_matrix but there they assumed that the mean-vectors are equal, but if I had to guess I would say that it should also work without the equal mean assumption.

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Yes, what you say is correct. What seems strange is the wikipedia page assuming the two populations have equal parameters! Then it would really be only one population, and there would be no need of pooling separate estimates, you would just pool the individual data.

Neither would there be need of calculating some test statistic ... For this pooling to make sense the needed assuption is of a common covariance matrix.

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