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Var[X+Y] equality true

  1. Jun 13, 2005 #1
    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.
  2. jcsd
  3. Jun 13, 2005 #2


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    You're right and the professor is wrong, unless he meant Var[X-Y].
  4. Jun 13, 2005 #3

    Thanks, but I am afraid that my professor clearly stated Var[X+Y], otherwise we would have the minus sign for Cov[X,Y].
  5. Jun 13, 2005 #4
  6. Jun 14, 2005 #5
    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.
    Last edited: Jun 14, 2005
  7. Jun 14, 2005 #6


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    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.
  8. Jun 23, 2005 #7
    At the end of the day, our professor found his error, just because a sign. Thank you all.
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