Var[X+Y]

[jetoso]

My simulation professor says the following equality is true:
Let X and Y be two random variables, then
Var[X + Y] = Var[X] + Var[Y] - 2Cov[X,Y], where Cov[X,Y] = E[XY]-E[X]E[Y].

I solved this equality and I am still having the following result:
Var[X+Y] = Var[X]+Var[Y]+2Cov[X,Y].

I sent my work to my professor, but he says my work is not correct. Could somebody tell me why? I mean, for which case it is not true?

Var[X+Y] = E[((X+Y) - (E[X]+E[Y]))^2], since Var[Z] = E[(Z-E[Z])^2] is the definition of the variance for a random variable Z.

[mathman]

You're right and the professor is wrong, unless he meant Var[X-Y].

[jetoso]

You're right and the professor is wrong, unless he meant Var[X-Y].

Thanks, but I am afraid that my professor clearly stated Var[X+Y], otherwise we would have the minus sign for Cov[X,Y].

[juvenal]

You can find the result in many authoritative sources. Here's one:

http://mathworld.wolfram.com/Covariance.html

Plus any basic statistics textbook will have this result.

[Bayou Tiger]

My simulation professor says the following equality is true:
Let X and Y be two random variables, then
Var[X + Y] = Var[X] + Var[Y] - 2Cov[X,Y], where Cov[X,Y] = E[XY]-E[X]E[Y].

I solved this equality and I am still having the following result:
Var[X+Y] = Var[X]+Var[Y]+2Cov[X,Y].

I sent my work to my professor, but he says my work is not correct. Could somebody tell me why? I mean, for which case it is not true?

Intuitively, a negative covariance between the two variables seems like it would result in a sum with less variance than if the variables were independent (zero covariance). Because the variables are "moving in different directions" it is more challenging for the observations of the sum to stray from the combined mean (i.e. less total variance with negative covariance), which agrees with your solution. If you're wrong, then I am stumped also.

This reminds me of portfolio optimization from Investments class, but unfortunately my notes are at work.

[mathman]

A simple example is X=Y. Var(X+Y)=Var(2X)=4Var(X). The professor's answers would end up as Var(X+Y)=0, since Cov(X,Y)=Var(X) in this case.

[jetoso]

At the end of the day, our professor found his error, just because a sign. Thank you all.