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Unformatted text preview: CHAPTER 5 SOLUTIONS TO PROBLEMS 5.1 Write y = β + 1 β x 1 + u , and take the expected value: E( y ) = β + 1 β E( x 1 ) + E( u ), or μ y = β + 1 β μ x since E( u ) = 0, where μ y = E( y ) and μ x = E( x 1 ). We can rewrite this as β = μ y 1 β μ x . Now, ˆ β = y − 1 ˆ β 1 x . Taking the plim of this we have plim( ˆ β ) = plim( y − 1 ˆ β 1 x ) = plim( y ) – plim( 1 ˆ β ) ⋅ plim( 1 x ) = μ y − 1 β μ x , where we use the fact that plim( y ) = μ y and plim( 1 x ) = μ x by the law of large numbers, and plim( 1 ˆ β ) = 1 β . We have also used the parts of Property PLIM.2 from Appendix C. 5.3 The variable cigs has nothing close to a normal distribution in the population. Most people do not smoke, so cigs = 0 for over half of the population. A normally distributed random variable takes on no particular value with positive probability. Further, the distribution of cigs is skewed, whereas a normal random variable must be symmetric about its mean....
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 Spring '08
 BREMAN
 Econometrics, Normal Distribution, Standard Deviation, probability density function

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