$X$ and $Y$ are two independent variables, with variances $\sigma^2_x$ and $\sigma^2_y$ respectively.

Two other variables $W$ and $V$ are defined by $W=X+Y$ and $V=X-Y$.

Find $Cov(X,V)$ and $Cov(W,V)$


closed as off-topic by Em., GoodDeeds, Silvia Ghinassi, user228113, Michael Albanese Mar 7 '16 at 5:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Em., GoodDeeds, Silvia Ghinassi, Community, Michael Albanese
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Hint: $\operatorname{Cov}(\bullet,\bullet)$ is bilinear. $\endgroup$ – user228113 Mar 7 '16 at 4:37

Hint: Recall that covariance satisfies bilinearity,

$$\text{Cov}(aX+Y, Z) = a\text{Cov}(X,Z)+\text{Cov}(Y,Z)$$ and $$\text{Cov}(X,X) = \text{Var}(X).$$

  • $\begingroup$ So Cov(X,V)=Cov(X,X-Y)=Cov(2X)+Cov(X-Y)? $\endgroup$ – ItsImpulse Mar 7 '16 at 3:32
  • $\begingroup$ No, try again.. $\endgroup$ – Em. Mar 7 '16 at 3:33
  • $\begingroup$ Sorry should the last term be -Cov(X,Y)? $\endgroup$ – ItsImpulse Mar 7 '16 at 3:38
  • $\begingroup$ No, try again.. $\endgroup$ – Em. Mar 7 '16 at 3:39
  • $\begingroup$ Cov(X,X) - Cov(X,Y)? $\endgroup$ – ItsImpulse Mar 7 '16 at 3:41

Not the answer you're looking for? Browse other questions tagged or ask your own question.