6
$\begingroup$

Here $X,Y$ are vector of length $n$ and $A$ is a $n\times n$ matrix. Suppose the covariance matrix $D(X)$ is known?

$\endgroup$
14
$\begingroup$

This (linear transform) is typically listed as a property of covariance, but easy to show as well: $$\begin{align}\operatorname{cov}(AX)&=\mathbb E[AXX^TA^T]-\mathbb E[AX]\mathbb E[X^TA^T]\\&=A\mathbb E[XX^T]A^T-A\mathbb E[X]\mathbb E[X^T]A^T\\&=A(\mathbb E[XX^T]-\mathbb E[X]\mathbb E[X^T])A^T\\&=ADA^T\end{align}$$

$\endgroup$
1
  • 1
    $\begingroup$ I think in the general case, for complex-valued variables, those super-T should be super-H $\endgroup$
    – Luis Mendo
    May 11 '20 at 23:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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