Variance sum

patter2809

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Mar 29, 2013
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Need to show Var(X+Y) = Var(X) + Var(Y) given that X,Y are independent RVs.

A. Var(X+Y) = E(X+Y)2 + E2(X+Y)
= E(X2+2XY+Y2) + E(X+Y)E(X+Y)
= E(X2) + 2E(XY) + E(Y2) + E2(X) + 2E(X)E(Y) + E2(Y)
= [E(X2) + E2(X)] + [E(Y2) + E2(Y)] + 2[E(XY) + E(X)E(Y)]
= Var(X) + Var(Y) + 2[E(XY) + E(X)E(Y)]

Is it wrong to say E(X2+2XY+Y2) = E(X2) + 2E(XY) + E(Y2) given that X2 and 2XY are not independent? Or have I gone wrong in some other way?

Thank you for your time.
 
Need to show Var(X+Y) = Var(X) + Var(Y) given that X,Y are independent RVs.

A. Var(X+Y) = E(X+Y)2 + E2(X+Y) OOPS!
= E(X2+2XY+Y2) + E(X+Y)E(X+Y)
= E(X2) + 2E(XY) + E(Y2) + E2(X) + 2E(X)E(Y) + E2(Y)
= [E(X2) + E2(X)] + [E(Y2) + E2(Y)] + 2[E(XY) + E(X)E(Y)]
= Var(X) + Var(Y) + 2[E(XY) + E(X)E(Y)]

Is it wrong to say E(X2+2XY+Y2) = E(X2) + 2E(XY) + E(Y2) given that X2 and 2XY are not independent? Or have I gone wrong in some other way?

Thank you for your time.
Unfortunately, your very first equation for variance is wrong! The Variance is the mean of the square minus the square of the mean:

V[X+Y] = E[(X + Y)^2] - E[X + Y]^2

Perhaps if you change the sign . . .

Since x and y are independent, it is true that E[xy] = E[x]E[y]
 
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