1 Ordinary Least Squares Regression: The Multivariate
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1 1 Ordinary Least Squares Regression The Multivariate Case Let Y Xβ+ where the dimensions of the matrices are as pointed out in the previous lecture. The Least Square regression method involves minimizing the sum of squaresofresiduals. Forany β, Y X β Intermsofthevector ˆ,squaredsumofresiduals as where standsfortranspose. ( Y Xβ ( Y Xβ ( ( Y β X. Y Xβ ( n Y Y Y Xβ β X Y + β i1 2 i can be written LetusexplorewhatthedimensionofY Xβis? Thedimensionofthismatrix is(1xnx(nxkx(kx1(1x1.thereforethisisamatrixwithjustone elementandthereforeequaltoitstransposeβ X Y.Sowecanwrite Y Y 2β X Y + β (1 The Ordinary Least Square Estimator β OLS ( minimizes the squared sum of residual. Note that β OLS OLS is the vector β1, β OLS 2,..., β OLS. So k first order conditions(foc have to be true β 1 β 2 ˆ β k OLS. Thisiswritten,inmatrixnotation,as 1 β. k
2 Now, note that ( β1 βx Y, β 2,..., β k X Y. NowwhatisthedimensionofX Y (kxnx(nx1(kx1.soitsavector withkrows. Letsdenotethisvectorby(a 1,a 2,...,a k.then ( a 1 β1 βx Y, β 2,..., β k a 2 β 1 a 1 + β 2 a β k a k i1 a k Therefore, X Y β i1 β 1 i1 β 2 i1 a 1 a 2 a k X Y β k Ingeneral A A.Similarly, Vβ (V +V β.ifv issymmetric V V ;hence Vβ 2Vβ. So β 2X X β DifferentiationwithrespecttoβandsettingtheFOCat β OLS β 2X Y +2 OLS (2 Equation(2 represents k equations. These are called Normal Equations. SinceX hasfullcolumnrankandx X issquare, X X hasaninverse. To seethis, letusdenotecolumnsofx bythevectorsx 1,x 2,...,x k. Eachvector x i containsthenrowsofdataforvariablei.therefore X ( x 1 x 2... x k Note that linear independence of X (full column rank means that there does notexistscalarsa i i1,..,ksuchthata 1 x 1 +a 2 x 2 +..a k x k 2
3 Nowsupposetothecontrary,X X wasnotinvertiblesinceasquarematix is invertible if it has linearly independent columns, it must be that the columns ofx X arenotlinearlyindependent. WhatarethecolumnsofX X? X X x 1 x 2 x k ( x1 x 2... x k x 1x 1 x 1x 2... x 1x k x 2x 1 x 2x 2... x 2x k x k x 1 x k x 2... x k x k Ifthecolumnsarelinearlydependent,itmustbethat x 1x 1 x 1x 2 x 1x k c 1 x 2x 1 +c x 2x c x 2x k k x k x 1 x k x 2 x k x k Now consider the first row Collecting terms Sincex 1isnotzero,thisimplies c 1 x 1 x 1+c 2 x 1 x c k x 1 x k x 1(c 1 x 1 +c 2 x c k x k (c 1 x 1 +c 2 x c k x k butthatisacontradictionofthelinearindependenceofx.therefore(x X 1 existsandthe linearly idependence of X (assumption2of full columnrank k guarantees that. Asanasidenotethat x 2 i1 x i1 x i2... x i1 x ik i1 i1 i1 n X X x i2 x i1 x 2 i2... x i2 x ik i1 i1 i1 n x ik x i1 x ik x i2... i1 i1 Equation(2 can be rearranged as OLS X Y i1 x 2 ik 3
4 andsince(x X 1 exists OLS β (X X 1 X Y 1.1 Algebric Properties of the OLS estimator Notation1 1. X To see this, recall 2X Y +2 OLS which implies But So by construction of OLS that is or X (Y X βols Y Xβ OLS ˆ X x 1 x 2 x k x k ˆ x 1 x 2 i1 i1 i1 x i1 i x i2 i x ik i This would imply that the covariance between each variable and the residuals iszerobyconstruction. Thisisnotthesameastheorthogonalityofthexand the error term. 4
5 2 i ifoneofthevariablesis1(thatis,ifyourunaregressionwith i1 an intercept term Notice that if you run a regression with a intercept(also sometimes called theconstanttermterm,thisimpliesx i1 1forallobservationsi.Since therefore x i1i i1 i (3 i1 Equation(3isONLYtrueinwhentheregressionhasaninterceptterm! 3 The regression hyperplane passes through the mean value of all the variable. Since then, i i1 (y i β 1 x i2β2... x ikβk i1 y i nβ 1 + β n 2 x i β n k i1 i1 i1 x ik y β1 + β 2 x β k xk 5
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