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Suppose, that we have a random vector $\mathbf{x} \sim \mathcal{N}(\mu,\Sigma)$. What is the distribution of $(a\cdot x)$, where $a$ is a real vector?

It is known, that for a nonsingular real matrix $B$, it would be the normal distribution with

$$\mathbb{E}(\mathbf{x}) = B\mu, \text{Var}(\mathbf{x}) = B\Sigma B^T.$$

For a diagonal matrix $\Sigma$ it is also can be shown to have normal distribution. However, the general case is not that obvious.

Any tips are welcome.

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It is a one-dimensional normal distribution.

https://en.wikipedia.org/wiki/Cram%C3%A9r%E2%80%93Wold_theorem

This follows from an alternative definition of multi-dimensional Gaussian distribution.

EDIT: see the Wikipedia article (for future reference/anyone new to the question)

https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Definition

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  • $\begingroup$ Thank you, did not know about different definitions of normal distribution. Then is it true that the variance of the resulted distribution is just $a\Sigma a^T$? Or I'm wrong? $\endgroup$ – Andrei Kulunchakov Apr 30 '16 at 17:17
  • $\begingroup$ It looks like you are correct. Remember that a multivariate normal distribution is completely determined by its covariance matrix, even to the extent that the components are independent if and only if they are uncorrelated, so the result isn't all too surprising from that perspective. $\endgroup$ – Chill2Macht Apr 30 '16 at 17:24
  • $\begingroup$ An alternative in the spirit of your diagonalization; using $\Sigma=P\Delta P^{-1}=P\Delta P^{T}$ (we assume $\Sigma$ diagonalizable), you have $(P^Tx)^T\Delta(Px)^T$ with $Px$ a Gaussian vector. $\endgroup$ – Jean Marie Apr 30 '16 at 18:58
  • $\begingroup$ @JeanMarie, great approach! $\endgroup$ – Andrei Kulunchakov Apr 30 '16 at 20:53

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