5
$\begingroup$

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.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
6
$\begingroup$

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') $

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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, 2018 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, 2018 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, 2018 at 0:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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