Why do you add variances when computing Var[3X-Y]?

[waealu]

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

 

[Mark44]

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?

Your textbook probably has a theorem about Var[X + Y] and Var[kX]

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

Yes.

 

[chiro]

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).

 

[Ray Vickson]

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

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
$\text{Var}[aX + bY] =a^2 \text{Var}(X) + b^2 \text{Var}(Y).$

RGV