Let $X=\begin{bmatrix} x_{1} \\ x_{2} \\ \end{bmatrix}$ and $Y=\begin{bmatrix} y_{1} \\ y_{2} \\ \end{bmatrix}$ be two random variables.
Also let $B=\begin{bmatrix} ax_{1}+by_{1}\\ ax_{2}+by_{2} \end{bmatrix}$ where $a$ and $b$ are constants.
I tried to find the $Cov(X,B)$ and the following is the thing that I got.
$Cov(X,B)=E([X-E(X)][B-E(B)]^T)$
$=E\Bigg[\Bigg(\begin{bmatrix} x_{1} \\ x_{2} \\ \end{bmatrix}-E\begin{bmatrix} x_{1} \\ x_{2} \\ \end{bmatrix}\Bigg)\Bigg(\begin{bmatrix} ax_{1}+by_{1}\\ ax_{2}+by_{2} \end{bmatrix}-E\begin{bmatrix} ax_{1}+by_{1}\\ ax_{2}+by_{2} \end{bmatrix}\Bigg)^T\Bigg]$ .....$(1)$
Since $B=a[X]+b[Y]$, $(1)$ can be written as
$=E\Big[\Big(X-E(X)\Big)-\Big[\Big(a(X)+b(Y)\Big)-E\Big(a(X)+b(Y)\Big)\Big]^T\Big]$
$=E\Big[\Big(X-E(X)\Big)-\Big[\Big(a(X)-aE(X)\Big)+\Big(b(Y)-bE(Y)\Big)\Big]^T\Big]$
$=E\Big[\Big(X-E(X)\Big)-\Big[a\big((X)-E(X)\big)+b\big((Y)-E(Y)\big)\Big]^T\Big]$
I need to further simplify this expression. But I'm having a difficulty in proceeding further at this place.Can someone please give me some hints to do this?
Cheers,
Notice that you've added a minus sign $$E\Big[\Big(X-E(X)\Big)\color{red}-\Big[\Big(a(X)+b(Y)\Big)-E\Big(a(X)+b(Y)\Big)\Big]^T\Big]$$ and instead you should have $$E\Big[\Big(X-E(X)\Big)\Big[\Big(a(X)+b(Y)\Big)-E\Big(a(X)+b(Y)\Big)\Big]^T\Big]$$ $$=E\Bigg[\Big(X-E(X)\Big)\Big[a\big((X)-E(X)\big)^T+b\big((Y)-E(Y)\big)^T\Big]\Bigg]$$ $$=E\Bigg[a\big(X-E(X)\big)\big((X)-E(X)\big)^T+b\big(X-E(X)\big)\big((Y)-E(Y)\big)^T\Big]\Bigg]$$
or using some properties of Cov: $$\text{Cov}[X,aX+bY]=\text{Cov}[X,aX]+\text{Cov}[X,bY]=a\text{Cov}[X,X]+b\text{Cov}[X,Y].$$