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.
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$\begingroup$ 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 "$a^\prime X a$". $\endgroup$– whuber ♦Commented Jan 26, 2016 at 23:27
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$\begingroup$ 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. $\endgroup$– Mark L. StoneCommented Jan 26, 2016 at 23:56
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2$\begingroup$ $Cov(a^T X) = E([(a^T X - E(a^T X)] [(a^T X - E(a^T X)]')$. Hint: The order of $a^T$ and $E$ can be interchanged due to linearity. Expand and simplify. $\endgroup$– Mark L. StoneCommented Jan 27, 2016 at 0:30
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$\begingroup$ Typo fixed, replacing x with sigma $\endgroup$– HatshepsutCommented Jan 27, 2016 at 0:32
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1$\begingroup$ This is really the same question as stats.stackexchange.com/questions/38721 but with different notation. $\endgroup$– whuber ♦Commented Jan 27, 2016 at 1:16
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I imagine it goes something like this (apologies for any typos or steps missed):
\begin{align} Cov(a^TX) &= E([a^TX-E(a^TX)][a^TX-E(a^TX)]^T)\\ &=E([a^TX-a^TE(X)][a^TX-a^TE(X)]^T)\\ &=E(a^TXX^Ta-2a^TE(X) X^Ta+a^TE(X)E(X^T)a)\\ &=a^TE(XX^T)a-2a^TE(X)E(X^T)a+a^TE(X)E(X^T)a\\ &=a^TE(XX^T)a-a^TE(X)E(X^T) a\\ &=a^T[E(XX^T)a-E(X)E(X^T)a]\\ &=a^T[E(XX^T)-E(X)E(X^T)]a\\ &=a^T\Sigma a \end{align}
where $\Sigma=E(XX^T)-E(X)E(X^T)$
which should make sense since it looks familiar to the typical variance formula we are used to, namely:
$$V(X)=E(X^2)-(E(X))^2$$ and if you multiplied that by a scalar $a$ then you would have $$V(aX)=a^2(E(X^2)-(E(X))^2)$$
answered Jan 27, 2016 at 1:03