MHB Getting variance from known correlation

AI Thread Summary
In the discussion, participants explore the variance of the sum of two variables, X and Y, that have the same distribution and a correlation of -0.5. They derive that the covariance can be expressed in terms of the variance of X, leading to the conclusion that V[X+Y] equals V[X] when substituting the known values. The importance of the negative correlation is highlighted, indicating that it alters the expected variance outcome compared to uncorrelated variables. Ultimately, the conversation concludes that the calculations confirm the relationship between variance and correlation, emphasizing that the scenario is simpler than initially perceived. The final takeaway is that the negative correlation significantly impacts the variance of the sum of the two variables.
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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!
 
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das said:
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?
 
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?
 
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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?
 
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.
 
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