# zz-23 - 7 cov aX b cY d = accov X Y cov X Y = E X E X Y E Y...

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Unformatted text preview: 7. cov ( aX + b, cY + d ) = accov ( X, Y ) cov ( X, Y ) = E [( X- E [ X ])( Y- E [ Y ])] = E [(( aX- b )- E ( aX + b ))(( cY + d )- E ( cY + d ))] E ( aX + bY ) = aE [ X ] + bE [ Y ] cov ( aX + b, cY + d ) = ( aX- b- aE [ X ] + b )( cY + d- cE [ Y ] + d ) E [( aX- aμ x )( cY- cμ y )] = E [ a ( x- μ x ) c ( Y- μ y )] cov ( aX + b, cY + d ) = ac E [( X- μ x )( Y- μ y )] | {z } cov ( X,Y ) = accov ( X, Y ) Q.E.D 8. X ∼ N (0 , sigma 2 ) Φ X ( μ ) = E [ e iμx ] = Z ∞-∞ e iμx f x ( x ) dx f x ( x ) = 1 √ 2 π 2 σ x e- ( x- μx ) 2 2 σ 2 x Φ x ( μ ) = e iμu- 1 2 u 2 σ 2 = e- 1 2 u 2 σ 2 E [ X k ] = i- k d k du k Φ x ( u ) | u =0 = i- k d k du k e- 1 2 u 2 σ 2 | u =0 = d k du k i- k (1- x + x 2 2!- x 3 3! ... ) | u =0 = d k du k i- k (1- σ 2 u 2 2 1 2! + σ 4 u 4 2 2 2!- σ 6 u 6 2 3 3! + σ 8 u 8 2 4 4! ... ) | u =0 E [ x k ] = when k is odd σ k k ! 2 k 2 ( k 2 )! when k is even 1 1.4.4 Solution: The final calculation of 2 3 refers not to a single draw of one ball from an urn containing three, but rather to a composite experiment comprising more than one stage. While it is true that { two black, one white } is the only fixed collection of balls for which a random choice is black with probability 2 3 , the composition of the urn is not determined...
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zz-23 - 7 cov aX b cY d = accov X Y cov X Y = E X E X Y E Y...

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