Given $x$ (n x 1 vector), knowing that $x$ ~ $N(0,C)$, I would like to compute $Q = E[(x'A)(x'A)']$.
Now, $Q$ is clearly a scalar. However, when I try to find a solution in terms of $C$ and $A$, I get something that is not a scalar; therefore, I am making some mistakes somewhere. Can anyone point my mistakes out, and ideally suggest an alternative solution?
Here is how I proceed: $Q = E[(x'A - E[x'A]) (x'A - E[x'A])']$, because $E[x'A]=E[x']A = 0*A = 0$.
Therefore, $Q=cov(x'A)=A'CA$, which would lead to an (n x n) solution..
Thanks for your help!