Page 111 -

110 Chapter 2 Review of Probability

To derive Equation (2.31), use the definition of the variance to write

var(aX + bY ) = E 5 3(aX + bY ) - (amX + bmY)426
= E 5 3a(X - mX) + b(Y - mY)42 6

= E3a2(X - mX)24 + 2E3ab(X - mX)(Y - mY)4
+ E3b2(Y - mY)24

= a2var(X ) + 2ab cov(X, Y ) + b2 var(Y )
= a2sX2 + 2absXY + b2s2Y, (2.49)

where the second equality follows by collecting terms, the third equality follows by expanding
the quadratic, and the fourth equality follows by the definition of the variance and covariance.

To derive Equation (2.32), write E(Y2) = E5 3(Y - mY) + mY]26 = E[(Y - mY)24 +
2 mYE(Y - mY) + mY2 = sY2 + mY2 because E(Y - mY) = 0.

To derive Equation (2.33), use the definition of the covariance to write

cov(a + bX + cV, Y) = E5[a + bX + cV - E(a + bX + cV)43Y - mY]6
= E5[b(X - mX) + c(V - mV)4 3Y - mY]6
= E5 3b(X - mX)4 3Y - mY]6 + E5 3c(V - mV)4 3Y - mY4 6

= bsXY + csVY, (2.50)

which is Equation (2.33).
To derive Equation (2.34), write E(XY ) = E5 3(X - mX) + mX4 3(Y - mY) + mY]6 =

E3(X - mX)(Y - mY)4 + mXE(Y - mY) + mYE(X - mX) + mX mY = sXY + mX mY.

We now prove the correlation inequality in Equation (2.35); that is, 0 corr (X, Y ) 0 … 1.

Let a = - sXY>sX2 and b = 1. Applying Equation (2.31), we have that

var(aX + Y ) = a2s2X + sY2 + 2asXY
= ( - sXY>sX2 )2 s2X + sY2 + 2( - sXY>s2X)sXY
= s2Y - sX2 Y > sX2 . (2.51)

Because var(aX + Y) is a variance, it cannot be negative, so from the final line of Equa-
tion (2.51), it must be that sY2 - s2XY > s2X Ú 0. Rearranging this inequality yields

sX2 Y … s2XsY2 (covariance inequality). (2.52)

The covariance inequality implies that sX2 Y > (sX2 sY2 ) … 1 or, equivalently,

0 sXY > (sX sY) 0 … 1, which (using the definition of the correlation) proves the correlation
inequality, 0 corr (X Y ) 0 … 1.