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Econometrics-I-5 - Applied Econometrics William Greene...

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Applied Econometrics William Greene Department of Economics Stern School of Business
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Applied Econometrics 5. Regression Algebra and a Fit Measure
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The Sum of Squared Residuals       b  minimizes  e e  = ( y  -  Xb ) ( y  -  Xb ).        Algebraic equivalences, at the solution         b  = ( X X ) -1 X y       e’e   =    y e  (why?   e’  =  y’  –  b’X’ )       e  =   y y  -  y’Xb   =   y y  -  b X y               =   e y  as  e 0          (This is the F.O.C. for least squares.)
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Minimizing e e Any other coefficient vector has a larger sum of squares.  A  quick proof:      d  = the vector, not  b      u  =  y  -  Xd .    Then,  u u  = ( y  -  Xd ) ( y - Xd )                = [ y  -  Xb  -  X ( d  -  b )] [ y  -  Xb  -  X ( d  -  b )]               = [ e   -  X ( d  -  b )]  [ e   -  X ( d  -  b )] Expand to find  u u  =  e e  + ( d - b ) X X ( d - b )  >    e e  
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Dropping a Variable An important special case.  Suppose [ b ,c]=the regression coefficients in a  regression of  y  on [ X , z ]and  d  is the same, but computed to force the  coefficient on  z  to be 0.  This removes  z  from the regression.  (We’ll  discuss how this is done shortly.)  So, we are comparing the results that  we get with and without the variable  z  in the equation.   Results which we  can show: Dropping a variable(s) cannot improve the fit - that is, reduce the sum of  squares. Adding a variable(s) cannot degrade the fit - that is, increase the sum of  squares. The algebraic result is on text page 38.  Where  u  = the residual in the  regression of  y  on [ X,z ] and  e  = the residual in the regression of  y  on  X   alone,          u’u  =  e e  –  c 2   ( z * z *)      e e  where  z * =  M X z . This result forms the basis of the Neyman-Pearson class of tests of the  regression model.
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The Fit of the Regression “Variation:” In the context of the “model” we  speak of  variation  of a variable as movement of  the variable, usually associated with (not  necessarily caused by) movement of another  variable. Total variation =              =  y M 0 y . M 0  =  I – i(i’i) -1 i’  = the  M  matrix for  X  = a column  of ones. n 2 i i=1 (y - y)
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Decomposing the Variation of y Decomposition:         y  =  Xb  +  e  so        M 0 y  =  M 0 Xb  +  M 0 e  =  M 0 Xb  +  e .         (Deviations from means. Why is  M 0 e  =  e ? )        y M 0 y  =  b ( X’ M 0 )( M 0 X ) b  +  e e                  =  b X M 0 Xb  +  e e.  
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