# proof for uncorrelated variables X and Y

• Oct 16th 2012, 05:33 PM
shaunoflobe
proof for uncorrelated variables X and Y
Hello :D.

Quick question, how can I show that, for uncorrelated variables X and Y, p(X+Y, X-Y) = (Var(X)-Var(Y)/(Var(X)+Var(Y)) ?
I know how to show that Cov(X+Y, X-Y) = Var(X) - Var(Y).
Thanks.
• Oct 16th 2012, 05:38 PM
chiro
Re: proof for uncorrelated variables X and Y
Hey shaunoflobe.

Recall that Covariance is distributive in that Cov(A+B,C+D) = Cov(A,C) + Cov(A,D) + Cov(B,C) + Cov(B,D)

Also recall that Cov(X,Y) = E[XY] - E[X]E[Y]. If two random variables are independent, then P(A = a, B = b) = P(A = a)*P(B = b). If you can show Cov(X,Y) = 0 for independent random variables (Hint: What will E[XY] be for this to happen), then you can use Cov(X,X) = Var[X] and collect things together.
• Oct 16th 2012, 06:04 PM
shaunoflobe
Re: proof for uncorrelated variables X and Y
Hi Chiro,

Thanks for the reply,

Yes I know how to show that Cov(X,Y) = 0:
As you showed above, if two variables are independent then E(XY) = E(X)E(Y). Therefore, Cov(X,Y) = E(XY) - E(X)(EY) = E(X)E(Y) - E(X)E(Y) = 0.

and Cov(X,X) = E(X^2) - E(X)^2 = Var(X). Just not sure how I can "collect things together", i.e state the proof. thanks.
• Oct 16th 2012, 08:05 PM
chiro
Re: proof for uncorrelated variables X and Y
Cov(X+Y,X-Y) = Cov(X,X) - Cov(X,Y) + Cov(X,Y) - Cov(Y,Y) = Cov(X,X) - Cov(Y,Y).

I didn't do the expansion and assumed you may get a Cov(X,Y) term but since you don't, you don't have to worry about it.