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?

  • $\begingroup$ 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

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:

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

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.