Unformatted text preview: Second Edition 421 We need to prove that log( n ) ≥ ∑ n i =1 a i log( 1 a i ). Using Jensen inequality we have that E log( 1 a ) = ∑ n i =1 a i log( 1 a i ) ≤ log(E 1 a ) = log( ∑ n i =1 a i 1 a i ) = log( n ) which establish the result. 4.59 Assume that E X = 0, E Y = 0, and E Z = 0. This can be done without loss of generality because we could work with the quantities X E X , etc. By iterating the expectation we have Cov( X,Y ) = E XY = E[E( XY  Z )] . Adding and subtracting E( X  Z )E( Y  Z ) gives Cov( X,Y ) = E[E( XY  Z ) E( X  Z )E( Y  Z )] + E[E( X  Z )E( Y  Z )] . Since E[E( X  Z )] = E X = 0, the second term above is Cov[E( X  Z )E( Y  Z )]. For the first term write E[E( XY  Z ) E( X  Z )E( Y  Z )] = E [E { XY E( X  Z )E( Y  Z )  Z } ] where we have brought E( X  Z ) and E( Y  Z ) inside the conditional expectation. This can now be recognized as ECov( X,Y  Z ), establishing the identity....
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 Spring '12
 Dr.Hackney
 Statistics, Conditional expectation, X1, Jensen's inequality, XY Z

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