Lecture33-2010 - Generalized Method of Moments Estimator...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Generalized Method of Moments Estimator: Lecture XXXIII Charles B. Moss November 30, 2010 I. Basic Derivation of the Linear Estimator A. Starting with the basic linear model y t = x t θ 0 + u t (1) where y t is the dependent variable, x t is the vector of independent variables, θ 0 is the parameter vector, and u t is the residual. In addition to these variables we will introduce the notion of a vector of instrumental variables denoted z t . 1. Reworking the original formulation slightly, we can express the residual as a function of the parameter vector u t ( θ 0 ) = y t x t θ 0 (2) 2. Based on this expression, estimation follows from the popu- lation moment condition E [ z t u t ( θ 0 )] = 0 (3) Or more specifically, we select the vector of parameters so that the residuals are orthogonal to the set of instruments. a) Note the similarity between these conditions and the or- thogonality conditions implied by the linear projection space P c = X ( X X ) 1 X (4) 1
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
AEB 6571 Econometric Methods I Professor Charles B. Moss Lecture XXXIII Fall 2010 3. Further developing the orthogonality condition, note that if a single θ 0 solves the orthogonality conditions, or that θ 0 is unique that E [ z t u t ( θ )] = 0 if and only if θ = θ 0 (5) Alternatively E [ z t u t ( θ )] = 0 if θ = θ 0 (6) a) Going back to the original formulation E [ z t u t ( θ )] = E [ z t ( y t x t θ )] (7) b) Taking the first-order Taylor series expansion E [ z t ( y t x t θ )] = E [
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern