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Introduction MLE LDV Problems EC220 Revision Lectures Lecture3 Bonsoo Koo May 4, 2010 B KOO EC220 Revision Lectures
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Introduction MLE LDV Problems Today’s Lecture Today: MLE & Limited Dependent Variables Maximum Likelihood Estimation (Chapter 10) Limited Dependent Variables (Chapter 10) B KOO EC220 Revision Lectures
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Introduction MLE LDV Problems Least Squares Criterion: Ordinary Least Squares Estimation Suppose we have a regression model: Y i = β 1 + β 2 X i + ε i where ε is an iid random variable and all the Gauss Markov assumptions are satisfied. Then the OLS estimator for β is b = argmax β N X i = 1 ( Y i - β 1 - β 2 X i ) 2 B KOO EC220 Revision Lectures
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Introduction MLE LDV Problems Alternative approach: Maximum Likelihood Estimation. We have 3 pieces of information: 1 A sample of observations - ( Y 1 , X 1 ) , ( Y 2 , X 2 ) , · · · , ( Y N , X N ) 2 The model - Y i = β 1 + β 2 X i + ε i 3 A specified pdf for the error - ε i N ( 0 , σ 2 ) What is the density of ε i ? f [ ε i ] = 1 σ 2 π e - 1 2 ( ε i σ ) 2 From this normal density function, f [ Y i | β 1 , β 2 , σ ] = 1 σ 2 π e - 1 2 Y i - β 1 - β 2 X i σ 2 Given independence, Joint density function is f [ Y | β 1 , β 2 , σ ] = n Y i = 1 f [ Y i | β 1 , β 2 , σ ] B KOO EC220 Revision Lectures
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Introduction MLE LDV Problems Alternative approach: Maximum Likelihood Estimation. We have 3 pieces of information: 1 A sample of observations - ( Y 1 , X 1 ) , ( Y 2 , X 2 ) , · · · , ( Y N , X N ) 2 The model - Y i = β 1 + β 2 X i + ε i 3 A specified pdf for the error - ε i N ( 0 , σ 2 ) What is the density of ε i ? f [ ε i ] = 1 σ 2 π e - 1 2 ( ε i σ ) 2 From this normal density function, f [ Y i | β 1 , β 2 , σ ] = 1 σ 2 π e - 1 2 Y i - β 1 - β 2 X i σ 2 Given independence, Joint density function is f [ Y | β 1 , β 2 , σ ] = n Y i = 1 f [ Y i | β 1 , β 2 , σ ] B KOO EC220 Revision Lectures
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Introduction MLE LDV Problems Alternative approach: Maximum Likelihood Estimation. We have 3 pieces of information: 1 A sample of observations - ( Y 1 , X 1 ) , ( Y 2 , X 2 ) , · · · , ( Y N , X N ) 2 The model - Y i = β 1 + β 2 X i + ε i 3 A specified pdf for the error - ε i N ( 0 , σ 2 ) What is the density of ε i ? f [ ε i ] = 1 σ 2 π e - 1 2 ( ε i σ ) 2 From this normal density function, f [ Y i | β 1 , β 2 , σ ] = 1 σ 2 π e - 1 2 Y i - β 1 - β 2 X i σ 2 Given independence, Joint density function is f [ Y | β 1 , β 2 , σ ] = n Y i = 1 f [ Y i | β 1 , β 2 , σ ] B KOO EC220 Revision Lectures
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Introduction MLE LDV Problems Alternative approach: Maximum Likelihood Estimation. We have 3 pieces of information: 1 A sample of observations - ( Y 1 , X 1 ) , ( Y 2 , X 2 ) , · · · , ( Y N , X N ) 2 The model - Y i = β 1 + β 2 X i + ε i 3 A specified pdf for the error - ε i N ( 0 , σ 2 ) What is the density of ε i ? f [ ε i ] = 1 σ 2 π e - 1 2 ( ε i σ ) 2 From this normal density function, f [ Y i | β 1 , β 2 , σ ] = 1 σ 2 π e - 1 2 Y i - β 1 - β 2 X i σ 2 Given independence, Joint density function is f [ Y | β 1 , β 2 , σ ] = n Y i = 1 f [ Y i | β 1 , β 2 , σ ] B KOO EC220 Revision Lectures
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Introduction MLE LDV Problems For each ( Y i , X i ) L [ β 1 , β 2 , σ | Y i , X i ] = 1 σ 2 π e - 1 2 Y i - β 1 - β 2 X i σ 2 Therefore, we can get the following likelihood function of jointly obtaining all the observations.
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