Given a random variable $Y=Ab+DX$ where $X$ is a random variable with mean $\mu$ and covariance matrix $\Sigma$. For a fixed $b$ and $D,A$. How do you compute the covariance of $Y$?
I know that $\mu_{Y} = Ab + D\mu$ and $Cov(Y) = E\{(Y-\mu_{Y})(Y-\mu_{Y})^{T}\}$.
In my notes I have that $(Y-\mu_{Y}) = D(X-\mu)$ which I don't understand. Shoudln't it be $(Y-\mu_{Y}) = (Y-D(X-\mu))$? Or is it supposed to be $E\{(Y-\mu_{Y})\} = D(X-\mu)$?
I also don't understand why $XX^{T} = \Sigma$.
We have that $$ (Y-\mu_Y)=Ab+DX-Ab-D\mu=D(X-\mu). $$ The covariance matrix is given by \begin{align*} \operatorname E[(Y-\mu_Y)(Y-\mu_Y)^{\mathrm T}] &=\operatorname E[D(X-\mu)(D(X-\mu))^{\mathrm T}]\\ &=\operatorname E[D(X-\mu)(X-\mu)^{\mathrm T}D^{\mathrm T}]\\ &=D\operatorname E[(X-\mu)(X-\mu)^{\mathrm T}]D^{\mathrm T}\\ &=D\Sigma D^{\mathrm T}. \end{align*} $XX^{\mathrm T}$ is random and $\Sigma$ is just a matrix. It does not make sense to write $XX^{\mathrm T}=\Sigma$.