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Assume that $\mu_1, \mu_2$ are vectors ($1\times n$) and $\Sigma_1, \Sigma_2$ are symmetric square matrixes ($n\times n$).

Having $\Sigma$, I want to compute $\mu$ such that : $$ \mu^T \Sigma^{-1} = \mu_1^T \Sigma_1^{-1}+ \mu_2^T \Sigma_2^{-1} $$

How can I do that?

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  • $\begingroup$ No, $\mu_1, \mu_2, \Sigma_1, \Sigma_2$ are fixed. What we want to calculate are $\mu$ and $\Sigma$ $\endgroup$ – AliHadian Jun 1 '14 at 10:44
  • $\begingroup$ Computing the vector $\mu^T\Sigma^{-1}$ is obvious, but then the factorization is undetermined, isn't it ? $\endgroup$ – Yves Daoust Jun 1 '14 at 13:18
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You can't compute $\Sigma$, from this and you should have some other equations for calculation of $\Sigma$, but given $\Sigma$, you can easily calculate $\mu$:

$$\mu^T \Sigma^{-1} = \mu_1^T \Sigma_1^{-1}+ \mu_2^T \Sigma_2^{-1}$$

$$\Rightarrow \mu^T = (\mu_1^T \Sigma_1^{-1}+ \mu_2^T \Sigma_2^{-1})\Sigma$$ $$\Rightarrow \mu = \left((\mu_1^T \Sigma_1^{-1}+ \mu_2^T \Sigma_2^{-1})\Sigma\right)^T$$

$$\Rightarrow \mu = \Sigma^T(\mu_1^T \Sigma_1^{-1}+ \mu_2^T \Sigma_2^{-1})^T$$

$$\Rightarrow \mu = \Sigma^T( \Sigma_1^{-T}\mu_1+ \Sigma_2^{-T}\mu_2)$$

Since, both $\Sigma_1$ and $\Sigma_2$ are symmetric, their inverses are also symmetric:

$$\Rightarrow \mu = \Sigma^T( \Sigma_1^{-1}\mu_1+ \Sigma_2^{-1}\mu_2)$$

Now, if you know that $\Sigma$ is also symmetric, you will have:

$$\mu = \Sigma( \Sigma_1^{-1}\mu_1+ \Sigma_2^{-1}\mu_2)$$

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    $\begingroup$ Thanks for guessing the solution and giving the answer! Yes, I found that $\Sigma$ can be computed using another constraint in my problem. $\endgroup$ – AliHadian Jun 2 '14 at 6:16

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