# Variance of $a^TX$ for MVN X

How do you show that the variance of $a^TX$ for multivariate normal X is $a^T\Sigma a$?

I have $V(a'X)=E(a'X-E[a'X])^2$, but it seems like the dimensions get messed up or something after that. So I'm not sure what I ought to be doing instead.

• You don't, because no matter whether you are thinking of $a$, $X$, or both as random variables, $a^\prime X a$ is not a variance: it's still a random variable. I believe you intend $X$ to be the MVN variable, for $a$ to be a constant vector, and you probably want something other than "$X$" to appear in "$a^\prime X a$". – whuber Jan 26 '16 at 23:27
• Following up on whuber , I believe what you want is $a^T cov(X) a$, which indeed is a scalar (variance) if $a$ is a compatibly-dimensioned deterministic column vector. – Mark L. Stone Jan 26 '16 at 23:56
• $Cov(a^T X) = E([(a^T X - E(a^T X)] [(a^T X - E(a^T X)]')$. Hint: The order of $a^T$ and $E$ can be interchanged due to linearity. Expand and simplify. – Mark L. Stone Jan 27 '16 at 0:30
• Typo fixed, replacing x with sigma – Hatshepsut Jan 27 '16 at 0:32
• This is really the same question as stats.stackexchange.com/questions/38721 but with different notation. – whuber Jan 27 '16 at 1:16

where $\Sigma=E(XX^T)-E(X)E(X^T)$
$$V(X)=E(X^2)-(E(X))^2$$ and if you multiplied that by a scalar $a$ then you would have $$V(aX)=a^2(E(X^2)-(E(X))^2)$$