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$\newcommand{\Cov}{\operatorname{Cov}}$ 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 ?

Thanks in advance.

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2 Answers 2

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$\newcommand{\Cov}{\operatorname{Cov}}$ 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.

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  • $\begingroup$ is it gonna be 0 if W,X, and Y are independent? $\endgroup$
    – user292024
    Jul 22, 2020 at 17:40
  • $\begingroup$ No, because the same calculation is valid whether the variables are independent or not. $\endgroup$ Jul 22, 2020 at 17:43
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\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 $$Cov(Y,Y)=Var(Y)$$.

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