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PS6Soln

# PS6Soln - Problem Set 6 ECON 231W Spring 2010 Suggested...

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Problem Set 6 ECON 231W Spring 2010 Suggested Solutions 1. a. Since Y i is a binary variable, we know E ( Y i | X i ) = 1 × Pr ( Y i = 1 | X i ) + 0 × Pr ( Y i = 0 | X i ) = Pr ( Y i = 1 | X i ) = β 0 + β 1 X i . Thus: E ( u i | X i ) = E [ Y i - ( β 0 + β 1 X i ) | X i ] = E ( Y i | X i ) - ( β 0 + β 1 X i ) = 0 b. Using Equation (2.7) in the book, we have: V AR ( Y i | X i ) = Pr ( Y i = 1 | X i )[1 - Pr ( Y i = 1 | X i )] = ( β 0 + β 1 X i )[1 - ( β 0 + β 1 X i )] Thus: V AR ( u i | X i ) = V AR [ Y i - ( β 0 + β 1 X i ) | X i ] = V AR ( Y i | X i ) = ( β 0 + β 1 X i )[1 - ( β 0 + β 1 X i )] c. V AR ( u i | X i ) depends on the value of X i , so u i is heteroskedastic. 2. a. Let n 1 = #( Y = 1), the number of observations on the random variable Y which equals 1; and n 2 = #( Y = 2). Then #( Y = 3) = n - n 1 - n 2 . The joint probability distribution of Y 1 , ..., Y n is: Pr ( Y 1 = y 1 , ..., Y n = y n ) = n Y i =1 Pr ( Y i = y i ) = p n 1 q n 2 (1 - p - q ) n - n 1 - n 2 The likelihood function is the above joint probability distribution treated as a function of the un- known coefficients ( p and q ). b. The MLEs of p and q maximize the likelihood function. Let’s use the log-likelihood function: L = ln[ Pr ( Y 1 = y 1 , ..., Y n = y n )] = n 1 ln p + n 2 ln q + ( n - n 1 - n 2 ) ln(1 - p - q ) Using calculus, the partial derivatives of L are: L ∂p = n 1 p - n - n 1 - n 2 1 - p - q L ∂q = n 2 q - n - n 1 - n 2 1 - p - q Setting these two equations equal to zero and solving for p and q yield the MLE of p and q : b p = n 1 n , b q = n 2 n 1

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PS6Soln - Problem Set 6 ECON 231W Spring 2010 Suggested...

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