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Unformatted text preview: = 0Knowledge of X gives us full knowledge of Y , since Y is the square of X . However, their covariance is 0, because X 2 is a nonlinear function of X Question 2 Assume E[u]=0. Show that E [ u  x ] = 0 ⇒ Cov ( x,u ) = 0 Cov ( x,u ) = E [( xE [ x ])( uE [ u ])] 1 by E [ u ] = 0 then Cov ( x,u ) = E [ xu ] by the law of iterated expectations: = E [ E [ xu  x ]] Since x is now ﬁxed, = E [ xE [ u  x ]] = E [ x 0] = 0 2...
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 Fall '06
 LUCARELLI
 Normal Distribution, Standard Deviation, Variance, Probability theory, Cov, Multivariate Analysis Fall

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