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If $a$ is a multivariate normal random variable, and $x$ is a plain old vector (of the same shape as $a$), then the inner product $x \cdot a$ is a random variable. This post on math exchange suggests that the product will have a univariate normal distribution, but I haven't been able to frame the problem in a way that leads me to calculating $\mu$ and $\sigma$.

Edit: For posterity, the answer to this question was made much clearer to me by understanding a proof of the affine property.

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The dot product is equivalent to linear combination in your situation. $x\cdot a = x_1a_1 + x_2a_2+...+x_ka_k$

Suppose $a \sim N(\mu, \Sigma)$, then scale $x\cdot a$ follows univariate normal distribution: $x\cdot a\sim N(x\cdot \mu, x\Sigma x') $

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  • $\begingroup$ Perhaps it is worth emphasizing that $x\cdot \mu$ is a scalar too, as is $x\Sigma x^\prime$ and thus $x\cdot a \sim N(\ldots)$ is referring to a univariate normal distribution $\endgroup$ – Dilip Sarwate Nov 2 '18 at 22:42
  • $\begingroup$ Thanks. This is axactly what I asked for :) What would be the general approach for answering this for some other multivariate distribution? $\endgroup$ – grge Nov 4 '18 at 0:43
  • $\begingroup$ Let $Y$ be random vector with mean $\mu$ and variance matrix $\Sigma$, $AY$ has the mean vector $A\mu$ and variance matrix $A\Sigma A'$ is true always. But the distribution of $AY$ depends on the distribution of $Y$. $\endgroup$ – user158565 Nov 4 '18 at 0:49

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