It is given that a random vector $\mathbf{y} \sim N(\mathbf{0}, \boldsymbol{\Theta})$, and $\mathbf{A}$ is a constant matrix. Assuming that the dimensions are compatible, what would $$E(\mathbf{A} \mathbf{y} \mathbf{y}^T)$$ equal?

I understand that $E(\mathbf{A} \mathbf{y}) = \mathbf{A} E(\mathbf{y})$ and that $E(\mathbf{y} \mathbf{y}^T)$ is simply the covariance-variance matrix of the random vector. But I am unable to evaluate the above expression. Any help would be much appreciated; thanks for reading!

  • $\begingroup$ I understand that $E(\mathbf{A} \mathbf{y}) = \mathbf{A} E(\mathbf{y}) \mathbf{A}^T \quad\quad$No, : $Var(Ay) = A Var(y) A^T$. Your answer below is correct. $\endgroup$
    – krkeane
    Jul 3 at 13:57
  • $\begingroup$ You're right, thanks. $\endgroup$ Jul 3 at 15:10

Turns out that the expression evaluates to $$\mathbf{A} E(\mathbf{y} \mathbf{y}^T),$$ which is just $\mathbf{A} \mathbf{\Theta}$.

  • $\begingroup$ This is an example of linearity of the expectation operator. $\endgroup$
    – whuber
    Jul 3 at 15:17

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.