Expected value Matrix properties

Question

Let $x\in \mathbb R^3$ be a Gaussian vector with $(\mu_x,\Sigma_y)$, and let $y=Ux$ where $U\in \mathbb R^{3\times 3}$. Find $y$'s expected value and variance as a function of $U$ using $(\mu_x,\Sigma_y)$.

I started by: $$\mu_y=E[y]=E[Ux]=UE[x]=U\mu_x$$

can I simply throw out the matrix Given that $U\in \mathbb R^3$ or do I have to expand the summation? Is this at all correct?

For the variance part, Is this correct:

$$\Sigma_y=Var(y)=E[(y-E[y])^2]=E[(Ux-U\mu_x)^2]$$

And I don't know where to go from here or if it is at all the correct direction.

Answer 1

No, this is not correct. The (co?)variance matrix of $y$ is the $3\times 3$ matrix \begin{align}\Sigma_y&=E[(y-E[y])(y-E[y])^t]=E[yy^t]-E[E[y]y^t]-E[yE[y]^t]+E[y]E[y]^t=\\&=E[yy^t]-E[y]E[y]^t-E[y]E[y]^t+E[y]E[y]^t=\boxed{E[yy^t]-E[y]E[y]^t}=\\&=E[Uxx^tU^t]-UE[x]E[x]^tU^t=U\Sigma_xU^t\end{align}

To my knowledge, when $v$ is a $n\times 1$ vector, the notation $v^2$ is sometimes used to indicate the real number $\langle v,v\rangle=\lVert v\rVert^2=v^tv$. Using it for $vv^t$ would be unusual and, in my opinion, annoyingly confusing.

Answer 2

The next step in the Variance is to expand the squared term, i.e.

$$Var(y)=E[(y-E(y))^2]=E[(Ux-U\mu_x)^2]$$

$=E[(U^2x^2-2U^2x\mu_x+U^2\mu^2_x)]=E(U^2x^2)-E(2U^2x\mu_x)+E(U^2\mu^2_x)=U^2[E(x^2)-2\mu_xE(x)+\mu^2_x]=U^2[E(x^2)-2\mu_x^2+\mu_x^2]$

$=U^2[E(x^2)-\mu_x^2]=U^2Var(x)$

Therefore $Var(Y)=U^2Var(x)$

This can be represeted alternatively

$\Sigma_y=U^T\Sigma_x U$

Hope this helps