I am given a problem for which part of the solution relies on finding $Cov(X,X+cY)$ where c is some positive constant. I am told that this $= Cov(X,X)+c \cdot Cov(X,Y)$
I do know the formula for finding $Cov(aX+bY+c,dZ+eW+q)$, but I don't see how "$ Cov(X,X)+c \cdot Cov(X,Y)$" results from this. To be more specific, I guess I don't understand how to utilize the formula for $Cov(aX+bY+c,dZ+eW+q)$, when instead of one of the random variables, there is a 0...Or is this the wrong way to think about this?
The Bilinearoty of Covariance says:
$$\begin{align}{\mathsf{Cov}(aX+bY+c,dZ+eW+f)} &= {{ad~\mathsf{Cov}(X,Z)}+{ae~\mathsf{Cov}(X,W)}\\+{db~\mathsf{Cov}(Y,Z)}+{be~\mathsf{Cov}(Y,W)}} \end{align}$$
So we have: $$\begin{align} \mathsf{Cov}(X,X+cY) &= \mathsf {Cov}(X+0Y+0,X+cY+0) \\[1ex] &= {{\mathsf{Cov}(X,X)}+{c~\mathsf{Cov}(X,Y)}+{0~\mathsf{Cov}(Y,X)}+{0c~\mathsf{Cov}(Y,Y)}} \\[1ex] &= \mathsf{Var}(X)+c~\mathsf{Cov}(X,Y) \end{align}$$