# 3 2 3 4 3 the variance of x 1 x 2 is v x 1 v x 2 2 c

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3 2 = 3 4 . 3. The variance of x 1 + x 2 is V ( x 1 ) + V ( x 2 ) + 2 C ( x 1 , x 2 ). Extend this result to find the variance of x 1 + x 2 + x 3 . Answer. V ( x 1 + x 2 + x 3 ) = V ' ( x 1 + x 2 ) + x 3 = V ( x 1 + x 2 ) + V ( x 3 ) + 2 C ( x 1 + x 2 , x 3 ) = V ( x 1 ) + V ( x 2 ) + V ( x 3 ) + 2 C ( x 1 + x 2 , x 3 ) + 2 C ( x 1 , x 2 ) . But C ( x 1 + x 2 , x 3 ) = E h ' ( x 1 + x 2 ) - E ( x 1 + x 2 ) “' x 3 - E ( x 3 ) i = E h ' x 1 - E ( x 1 ) + ' x 2 - E ( x 2 ) ih x 3 - E ( x 3 ) i = E h ' x 1 - E ( x 1 ) “' x 3 - E ( x 3 ) i + E h ' x 2 - E ( x 2 ) “' x 3 - E ( x 3 ) i = C ( x 1 , x 3 ) + C ( x 2 , x 3 ) Substituting this result into the expression for V ( x 1 + x 2 + x 3 ) gives V ( x 1 + x 2 + x 3 ) = V ( x 1 )+ V ( x 2 )+ V ( x 3 )+2 C ( x 1 , x 2 )+2 C ( x 1 , x 3 )+2 C ( x 2 , x 3 ) . 4. Find V ( x 1 - x 2 ). Prove that C ( x 1 , x 2 ) p V ( x 1 ) V ( x 2 ) 1 2 ' V ( x 1 ) + V ( x 2 ) . Answer. For the first part, V ( x 1 - x 2 ) = E ' ( x 1 - x 2 ) - E ( x 1 - x 2 ) 2 = E h ' x 1 - E ( x 1 ) - ' x 2 - E ( x 2 ) i 2 = E ' x 1 - E ( x 1 ) 2 - 2 E h ' x 1 - E ( x 1 ) “' x 2 - E ( x 2 ) i + E ' x 2 - E ( x 2 ) 2 = V ( x 1 ) + V ( x 2 ) - 2 C ( x 1 , x 2 ) 2

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SERIES A No.6, ANSWERS For the second part, we may assume that - 1 ρ = C ( x 1 , x 2 ) p V ( x 1 ) V ( x 2 ) 1 . It follows that C ( x 1 , x 2 ) p V ( x 1 ) + V ( x 2 ) Also we have 0 ( p V ( x 1 ) - p V ( x 2 ) ) 2 = V ( x 1 ) - 2 p V ( x 1 ) p V ( x 2 ) + V ( x 2 ) whence 1 2 ' V ( x 1 ) + V ( x 2 ) p V ( x 1 ) + V ( x 2 ) and the result follows from combining the inequalities. 3
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• Spring '12
• D.S.G.Pollock
• Variance, X2 X1, series A

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