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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!

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

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

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