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Adding variances

  1. Jul 20, 2012 #1
    Hi, I have a quick question about variance.

    If you are trying to compute Var[3X-Y] where Var[X]=1 and Var[Y]=2,
    the correct formula is 32*Var[X] + Var[Y],
    but why do you add the variances of x and y?

    If you were computing Var[3X+Y] would it also be 32*Var[X] + Var[Y] ?

    Thanks, Erik
     
  2. jcsd
  3. Jul 20, 2012 #2

    Mark44

    Staff: Mentor

    Your textbook probably has a theorem about Var[X + Y] and Var[kX]
    Yes.
     
  4. Jul 20, 2012 #3

    chiro

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    Science Advisor

    Hey waealu.

    For the proof, the easiest way is to use VAR[X] = E[X^2] - {E[X]}^2 and then prove the identity. You do it for the continuous and discrete cases and you'll get the same result for both.

    Intuitively though, the variance is a squared positive quantity which means that the variance has to always increase (and recall that variances are always positive for a random variable).
     
  5. Jul 21, 2012 #4

    Ray Vickson

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    Homework Helper

    These are only true if X and Y are independent (or, at least, uncorrelated) random variables. For independent (or uncorrelated) X and Y and constants a, b we have
    [itex] \text{Var}[aX + bY] =a^2 \text{Var}(X) + b^2 \text{Var}(Y).[/itex]

    RGV
     
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