Well, the definition of covariance is

Cov(X,Y) = E((X-E(X)) * (Y-E(Y)) )

(which happens to be equal to E(XY)-E(X)E(Y) the definition you may have seen). But in any case, from the definition you can check

Cov(X+Z,Y) = Cov(X,Y) + Cov(Z,Y)

and that's why you can 'expand brackets' (and similarly in the second 'slot'). It's also clear that covariance is 'symmetric': Cov(X,Y)=Cov(Y,X) and that Cov(X,X)=Var(X). These are all the properties of covariance that I used.

The upshot is that you need to work out Var(X) and Var(Y) and then you are sorted, as the answer is their difference.

Reply 4

s_abbott

Thank you very much!
Also, I have to calculate Var(2x+y) but I have worked out that it is 4Var(X)+Var(Y)+4Cov(X,Y)
Is this right??

Reply 5

SPMars

s_abbott

Thank you very much!
Also, I have to calculate Var(2x+y) but I have worked out that it is 4Var(X)+Var(Y)+4Cov(X,Y)
Is this right??

Yes that is right! (For the same reasons I mention in my previous post.)

Reply 6

akbarfaizan2000

Original post by s_abbott

How do I calculate Cov(X+Y,X-Y)?

We Already Know that Cov(X, Y)=E[X, Y]-E[X]E[Y]
Since their fore
Cov(X+Y, X-Y)=E[(X+Y),(X-Y)]-E[(X+Y)]E[(X-Y)]
Cov(X+Y, X-Y) = E[X^2-XY+YX-Y^2]-[{E(X)+E(Y)}{E(X)-E(Y)}]
= E[X^2-XY+XY-Y^2]-[{E(X)}^2-E(X)E(Y)]+E(X)E(Y)-{E(Y)^2}
= E[X^2-Y^2]-[{E(X)}^2-{E(Y)}^2]
= E(X^2)-E(Y^2)-{E(X)}^2+{E(Y)}^2
= E(X^2)-{E(X)}^2-[E(Y^2)-{E(Y)}^2]
= Var(X)-Var(Y)

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