Suggested_solutions_HW1 - = 0-Knowledge of X gives us full...

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PAM 305 - Introduction to Multivariate Analysis Fall 2006 Suggested Solutions Homework 1 Question 1: Covariance as a measure of linear association If X N (0 , 1) and define Y = X 2 , prove that Cov ( X,Y ) = 0 Cov ( X,Y ) = Cov ( X,X 2 ) = E [( X - E [ X ])( X 2 - E [ X 2 ])] But E [ X ] = 0 by our assumption of X N (0 , 1) and E [ X 2 ] = 1 because the square of a standard normally distributed random variable is distributed χ 2 1 (chi-square with 1 degree of freedom). Therefore, our equation is now: = E [( X - 0)( X 2 - 1)] = E [ X 3 - X ] = E [ X 3 ] - E [ X ] But E [ X 3 ] = 0 due to symmetry of the normal distribution (this moment is called skewness). Therefore,
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Unformatted text preview: = 0-Knowledge 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 non-linear function of X Question 2 Assume E[u]=0. Show that E [ u | x ] = 0 Cov ( x,u ) = 0 Cov ( x,u ) = E [( x-E [ x ])( u-E [ 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 xed, = E [ xE [ u | x ]] = E [ x 0] = 0 2...
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Suggested_solutions_HW1 - = 0-Knowledge of X gives us full...

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