Why is the variance of X plus or minus Y equal to the variance of X plus the variance of Y?

Question

I want to know why the variance of X-Y is equal to the variance of X+Y. And A source would be really helpful Thanx!

Answer 1

Note that this is true when X and Y are independent, but may not be true if they are not. For example, if X=Y, then Var(X+Y) = 4*Var(X), but Var(X-Y) = 0.

If however, X and Y are independent, which tells you that E[XY] = E[X]*E[Y]. So:

Var[X+Y]
(by definition of variance)
= E[(X+Y)^2] - E[X+Y]^2
(by linearity of expectation)
= E[X^2] + 2E[XY] + E[Y^2] - E[X]^2 - 2E[X]E[Y] - E[Y]^2
(by independence of X and Y)
= E[X^2] - E[X]^2 + E[Y^2] - E[Y]^2
(by definition of variance)
Var[X] + Var[Y]

Near the same thing happens for Var[X-Y]:

Var[X-Y]
(by definition of variance)
= E[(X-Y)^2] - E[X-Y]^2
(by linearity of expectation)
= E[X^2] - 2E[XY] + E[Y^2] - E[X]^2 + 2E[X]E[Y] - E[Y]^2
(by independence of X and Y)
= E[X^2] - E[X]^2 + E[Y^2] - E[Y]^2
(by definition of variance)
Var[X] + Var[Y]

I do not know a source for this, though I would guess Wikipedia is a decent source for related things.

Hope that helps!

Answer 2

FOR ANY ONE WHO CARES 8 YEARS LATER
They are "generally" not. I think the question you are doing had the condition that X and Y are independent. This works because the general formula for doing the variance of two variables is:
Var(A*X+B*Y)= A^2*Var(X)+B^2*Var(Y)+2*A*B*Cov(X,Y)
In your first equation A =1 and B = 1, In the second equation your A= 1 and B= -1.
If the variables are independent the cov = 0.
Additionally, when we plug the values for A and B into the equation they equal out.

Answer 3

They would only be equal if the Covariance of X and Y equals zero.
Var(X+Y)= Var(X)+Var(Y)+2Cov(X,Y)
Var(X-Y)= Var(X)+Var(Y)-2Cov(X,Y)

you see.

As a source you can use any probability or statistics text book.