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4.2.4 Expectation and Variance of a Weighted Sum

Let Z = X + Y so:

E [Z ] = μz

by definition, but it can broken into its components X and Y .

∑ ∑ ∑ E [Z ] = E [X + Y] = pi(xi + yi) = pi(xi) + pi(yi) = μx + μy


Recall:

(a + b)2 = a(a + b) + b(a + b) = a2 + 2ab + b2


We will need this to understand V ar(c1X + c2Y )
Again let Z = X + Y

V ar(Z) = E [(Z - μ )2] = σ2 z z


but you may only be able to solve for Z from the X and Y .

2 2 Var (Z) = E [(Z - μz) ] = E[((X + Y) - (μx + μy)) ] =

Think of (X + Y ) as a and (μx + μy) as b.

2 2 E [(X + Y ) - 2(X + Y)(μx + μy ) + (μx + μy) ] =

Again using (a + b)2 = a2 + 2ab + b2

E [X2 + 2XY + Y 2 - 2X μx - 2X μy - 2Y μx - 2Y μy + μ2 + 2 μxμy + μ2] = x y

Yes, a big mess but the terms can be regrouped and come up with the following:

[ ] = E (X2 - 2X μx + μ2x) + (Y2 - 2Y μy + μ2y) + 2(XY - X μy - Y μx + μxμy) [ 2 2 ] = E [(X - μx) + (Y - μy ) + 2(X - μx)(Y - μy) ] = E (X - μx)2] + E[(Y - μy )2] + 2E [(X - μx)(Y - μy ) = V ar(X ) + V ar(Y ) + 2Cov (X, Y) = V ar(Z )

From these concepts:

V ar(c1X + c2Y ) = Var (c1X ) + V ar(c2Y ) + 2Cov (c1X, c2Y ) 2 2 = c1Var (X ) + c2V ar(Y ) + 2c1c2Cov (X, Y )

If we added a constant, c3 this not change the variance, see next formula:

Var (c1X + c2Y + c3) = V ar(c1X ) + V ar(c2Y ) + 2Cov (c1X,c2Y ) = c2V ar(X ) + c2V ar(Y ) + 2c cCov (X, Y ) 1 2 1 2

Note: Cov(X,X) = V ar(X) and correlation, ρ = Covσ(Xσ,Y) x y

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