# Proof that Cov(W+Y, Y-V) = 0 given that W, Y, and V are uncorrelated but not independent

$$\newcommand{\Cov}{\operatorname{Cov}}$$ I'm trying to prove the following statement: $$\Cov(W+Y , Y-V) = 0$$, given the following constraints:

1. $$W$$,$$Y$$, and $$V$$ are Uncorrelated but not independent
2. $$E(W)=E(Y)=E(V)=\mu$$
3. $$V(W)=V(Y)=V(V)=\sigma^2.$$

Can someone help me out with the sketch/proof ?

$$\newcommand{\Cov}{\operatorname{Cov}}$$ The claim is false, if we make the assumption that $$\sigma^2\not=0.$$ "Uncorrelated" occurs if and only if the covariances are zero. We know that \begin{align*} \Cov(W+Y,Y-V) &=\Cov(W,Y)-\Cov(W,V)+V(Y)-\Cov(Y,V)\\ &=0-0+\sigma^2-0\\ &=\sigma^2\\ &\not=0. \end{align*} The independence of the variables is irrelevant, as are the expected values.
\begin{align} Cov(W+Y,Y-V)&=Cov(W,Y)-Cov(W,V)+Cov(Y,Y)-Cov(Y,V)\\ &=0-0+\sigma^2-0\\ &=\sigma^2 \end{align} given that $$Cov(aW+bY,cY+dV)=ac Cov(W,Y)+ad Cov(W,V)+bc Cov(Y,Y)+bd Cov(Y,V)$$ $$a=1,b=1,c=1,d=-1$$ in this case, and $$Cov(Y,Y)=Var(Y)$$.