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Econometrics-I-3

Is regressed on x p mx 0 this result is fundamental

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 is regressed on  X p MX  =  0  (This result is fundamental!)       How do we interpret this result in terms of residuals?       When a column of X is regressed on X, we get a   perfect fit and zero residuals. p (Therefore)   My   =   MXb  +  Me  =  Me   =         (You should be able to prove this. p y = Py + My, P = X ( X’X )-1 X’ = (I - M).              PM = MP = 0. p Py  is the projection of  y  into the column space of  X .   ™    18/26
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Part 3: Least Squares Algebra The M Matrix p M = I- X(X’X)-1X’  is an nxn matrix p M  is symmetric –  M  =  M p M  is idempotent –  M * M  =  M     (just multiply it out) p M  is singular –  M -1 does not exist.     (We will prove this later as a side result in  another derivation.) ™    19/26
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Part 3: Least Squares Algebra Results when X Contains a Constant Term p X  = [ 1 , x 2,…, x K] p The first column of  X  is a column of ones p Since  X’e  =  0 x1’e  = 0 – the residuals sum to zero. ™    20/26 = = = = = = + n i i=1 Define  [1,1,...,1] '  a column of n ones  =   y ny implies (after dividing by n) y    (the regression line passes through the means) These do not apply if the model has no y Xb e i i'y i'y i'Xb + i'e = i'Xb x b  constant term.
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Part 3: Least Squares Algebra Least Squares Algebra ™    21/26
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Part 3: Least Squares Algebra Least Squares ™    22/26
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Part 3: Least Squares Algebra Residuals ™    23/26
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Part 3: Least Squares Algebra Least Squares Residuals ™    24/26
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Part 3: Least Squares Algebra Least Squares Algebra-3 M is nxn potentially huge ™    25/26 I X X X X M X e
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Part 3: Least Squares Algebra Least Squares Algebra-4 MX   =     26/26
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