# Getting variance from known correlation

• MHB
das1
Suppose X and Y have the same distribution, and Corr(X,Y) = -0.5. Find V[X+Y].

I know that Corr(X,Y)*SD[X]SD[Y] = Cov(X,Y)

and also V[X+Y] = V[X] + V[Y] + 2Cov(X,Y)

So there must be a missing link, maybe an identity, that I'm not realizing. I think the fact that the 2 variables have the same distribution is probably important, I'm just not sure how.

Thanks!

## Answers and Replies

Homework Helper
MHB
Suppose X and Y have the same distribution, and Corr(X,Y) = -0.5. Find V[X+Y].

I know that Corr(X,Y)*SD[X]SD[Y] = Cov(X,Y)

and also V[X+Y] = V[X] + V[Y] + 2Cov(X,Y)

So there must be a missing link, maybe an identity, that I'm not realizing. I think the fact that the 2 variables have the same distribution is probably important, I'm just not sure how.

Thanks!

Hey das! (Smile)

Since X and Y have the same distribution, we have that V[Y]=V[X] and SD[Y]=SD[X].
What if you substitute that and combine the equations you already have?

das1
Hi thank you!

So substituting that we have something like $$-0.5 = \frac{Cov(X,Y)}{SD[X]^2}$$ and $$V[X+Y] = 2V[X] + 2Cov[X+Y]$$ Wouldn't we still need to know more info, like what Cov(X,Y) is, before solving? And we can't get those without knowing what SD[X] or V[X] are (or SD[Y] or V[Y]). Don't we need one more constant here?

Last edited:
das1
OK, did it all out and got
$$\frac{-var(X)}{2} = cov(X,Y)$$
then $$-var(X) = 2cov(X,Y)$$
then $$V[X+Y] = V[X] = V[Y]$$
I was thinking I needed a constant but not sure if that's possible here...is there a way I can actually make this a constant or is this as simple as it can get?

Homework Helper
MHB
Yep. That's it and it is as simple as it can get. (Nod)

Note that if X and Y were uncorrelated, we would have $V[X+Y] = 2V[X]^2$.
Now that X and Y are negative correlated by a factor of $-\frac 1 2$ that factor of $2$ effectively disappears.