Dr. Hackney STA Solutions pg 65

Dr. Hackney STA Solutions pg 65 - Second Edition 4-21 We...

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Unformatted text preview: Second Edition 4-21 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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