Why does this expression simplify as such?General linear hypothesis test statistic: equivalence of two expressionsSlope Derivation for the variance of a least square problem via Matrix notationCovariance of OLS estimator and residual = 0. Where is the mistake?A doubt on SUR modelWhy trace of $I−X(X′X)^-1X′$ is $n-p$ in least square regression when the parameter vector $beta$ is of p dimensions?Getting the posterior for Bayesian linear regression with a flat priorDistribution of coefficients in linear regressionFitted values and residuals: are they random vectors?Proving that Covariance of residuals and errors is zeroWhat is the relationship of long and short regression when we have an intercept?

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Why does this expression simplify as such?


General linear hypothesis test statistic: equivalence of two expressionsSlope Derivation for the variance of a least square problem via Matrix notationCovariance of OLS estimator and residual = 0. Where is the mistake?A doubt on SUR modelWhy trace of $I−X(X′X)^-1X′$ is $n-p$ in least square regression when the parameter vector $beta$ is of p dimensions?Getting the posterior for Bayesian linear regression with a flat priorDistribution of coefficients in linear regressionFitted values and residuals: are they random vectors?Proving that Covariance of residuals and errors is zeroWhat is the relationship of long and short regression when we have an intercept?













3












$begingroup$


I'm reading through my professor's lecture notes on the multiple linear regression model and at one point he writes the following:



$$E[(b-beta)e']=E[(X'X)^-1epsilonepsilon'M_[X]]. $$



In the above equation, $b$, $beta$, $e$, and $epsilon$ are all vectors, $X$ is a regressor matrix and $M$ is the residual maker matrix. In general, I have no idea why these expressions are equivalent, and I'm particularly confused at how the $e$ vector disappears and the $epsilon$ vector appears.










share|cite|improve this question











$endgroup$
















    3












    $begingroup$


    I'm reading through my professor's lecture notes on the multiple linear regression model and at one point he writes the following:



    $$E[(b-beta)e']=E[(X'X)^-1epsilonepsilon'M_[X]]. $$



    In the above equation, $b$, $beta$, $e$, and $epsilon$ are all vectors, $X$ is a regressor matrix and $M$ is the residual maker matrix. In general, I have no idea why these expressions are equivalent, and I'm particularly confused at how the $e$ vector disappears and the $epsilon$ vector appears.










    share|cite|improve this question











    $endgroup$














      3












      3








      3





      $begingroup$


      I'm reading through my professor's lecture notes on the multiple linear regression model and at one point he writes the following:



      $$E[(b-beta)e']=E[(X'X)^-1epsilonepsilon'M_[X]]. $$



      In the above equation, $b$, $beta$, $e$, and $epsilon$ are all vectors, $X$ is a regressor matrix and $M$ is the residual maker matrix. In general, I have no idea why these expressions are equivalent, and I'm particularly confused at how the $e$ vector disappears and the $epsilon$ vector appears.










      share|cite|improve this question











      $endgroup$




      I'm reading through my professor's lecture notes on the multiple linear regression model and at one point he writes the following:



      $$E[(b-beta)e']=E[(X'X)^-1epsilonepsilon'M_[X]]. $$



      In the above equation, $b$, $beta$, $e$, and $epsilon$ are all vectors, $X$ is a regressor matrix and $M$ is the residual maker matrix. In general, I have no idea why these expressions are equivalent, and I'm particularly confused at how the $e$ vector disappears and the $epsilon$ vector appears.







      regression multiple-regression linear-model residuals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 4 hours ago









      Benjamin Christoffersen

      1,264519




      1,264519










      asked 5 hours ago









      DavidDavid

      24311




      24311




















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          I am assuming $b$ is the OLS estimate of $beta$ and $e$ is the corresponding estimate of $epsilon$. Also I believe you have a typo above in your expression, as there should be $X'$ in front of $epsilon epsilon'$ and behind $(X'X)^-1$.



          Start with the definition of $b$:
          $$b=(X'X)^-1X'Y.$$
          Replacing $Y$ with $Xbeta+epsilon$ in our expression above, we get
          $$b=(X'X)^-1X'(Xbeta+epsilon)=beta+(X'X)^-1X'epsilon.$$
          It follows that
          $$b-beta = (X'X)^-1X'epsilon$$



          Now turn to the defintion of $e$:
          $$e=Y-hatY=Y-Xb=Y-X(X'X)^-1X'Y.$$



          Notice $X(X'X)^-1X'$ is the projection matrix for $X$, which we will denote with $P_[X]$.
          Replacing this in our expression for $e,$ we get
          $$e=(I-P_[X])Y=M_[X]Y.$$
          Replacing $Y$ in the expression above with $Xbeta+epsilon$, we get
          $$e=M_[X](Xbeta+epsilon)=M_[X]epsilon,$$
          since $M_[X]X$ is a matrix of zeros.



          Post-multiplying $b-beta$ with $e'$, we get
          $$(b-beta)e'=(X'X)^-1X'epsilon epsilon' M_[X],$$
          since $e'=epsilon'M_[X].$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Ah. The key thing I was missing was what you wrote in the last line.
            $endgroup$
            – David
            3 hours ago


















          3












          $begingroup$

          Assuming that the coefficient estimator $b$ is calculated by OLS estimation, you have:



          $$beginequation beginaligned
          b-beta
          &= (X'X)^-1 X'y - beta \[6pt]
          &= (X'X)^-1 X'(X beta + epsilon)- beta \[6pt]
          &= (X'X)^-1 (X'X) beta + (X'X)^-1 X' epsilon - beta \[6pt]
          &= beta + (X'X)^-1 X' epsilon - beta \[6pt]
          &= (X'X)^-1 X' epsilon. \[6pt]
          endaligned endequation$$



          Presumably $e$ is the residual vector (different to the error vector $epsilon$) so we have $e = M_[X] Y = M_[X] epsilon$. Substituting this vector gives:



          $$beginequation beginaligned
          (b-beta) e'
          &= (X'X)^-1 X' epsilon (M_[X] epsilon)' \[6pt]
          &= (X'X)^-1 X' epsilon epsilon' M_[X]' \[6pt]
          &= (X'X)^-1 X' epsilon epsilon' M_[X]. \[6pt]
          endaligned endequation$$



          (The last step follows from the fact that $M_[X]$ is a symmetric matrix.) So the expression given by your professor is missing the $X'$ term.






          share|cite|improve this answer









          $endgroup$








          • 2




            $begingroup$
            Nice, I think we both must have been typing our answers at the same time. I'm glad you also found the mistake.
            $endgroup$
            – dlnB
            3 hours ago






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            3 hours ago










          Your Answer





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          2 Answers
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          2 Answers
          2






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          active

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          3












          $begingroup$

          I am assuming $b$ is the OLS estimate of $beta$ and $e$ is the corresponding estimate of $epsilon$. Also I believe you have a typo above in your expression, as there should be $X'$ in front of $epsilon epsilon'$ and behind $(X'X)^-1$.



          Start with the definition of $b$:
          $$b=(X'X)^-1X'Y.$$
          Replacing $Y$ with $Xbeta+epsilon$ in our expression above, we get
          $$b=(X'X)^-1X'(Xbeta+epsilon)=beta+(X'X)^-1X'epsilon.$$
          It follows that
          $$b-beta = (X'X)^-1X'epsilon$$



          Now turn to the defintion of $e$:
          $$e=Y-hatY=Y-Xb=Y-X(X'X)^-1X'Y.$$



          Notice $X(X'X)^-1X'$ is the projection matrix for $X$, which we will denote with $P_[X]$.
          Replacing this in our expression for $e,$ we get
          $$e=(I-P_[X])Y=M_[X]Y.$$
          Replacing $Y$ in the expression above with $Xbeta+epsilon$, we get
          $$e=M_[X](Xbeta+epsilon)=M_[X]epsilon,$$
          since $M_[X]X$ is a matrix of zeros.



          Post-multiplying $b-beta$ with $e'$, we get
          $$(b-beta)e'=(X'X)^-1X'epsilon epsilon' M_[X],$$
          since $e'=epsilon'M_[X].$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Ah. The key thing I was missing was what you wrote in the last line.
            $endgroup$
            – David
            3 hours ago















          3












          $begingroup$

          I am assuming $b$ is the OLS estimate of $beta$ and $e$ is the corresponding estimate of $epsilon$. Also I believe you have a typo above in your expression, as there should be $X'$ in front of $epsilon epsilon'$ and behind $(X'X)^-1$.



          Start with the definition of $b$:
          $$b=(X'X)^-1X'Y.$$
          Replacing $Y$ with $Xbeta+epsilon$ in our expression above, we get
          $$b=(X'X)^-1X'(Xbeta+epsilon)=beta+(X'X)^-1X'epsilon.$$
          It follows that
          $$b-beta = (X'X)^-1X'epsilon$$



          Now turn to the defintion of $e$:
          $$e=Y-hatY=Y-Xb=Y-X(X'X)^-1X'Y.$$



          Notice $X(X'X)^-1X'$ is the projection matrix for $X$, which we will denote with $P_[X]$.
          Replacing this in our expression for $e,$ we get
          $$e=(I-P_[X])Y=M_[X]Y.$$
          Replacing $Y$ in the expression above with $Xbeta+epsilon$, we get
          $$e=M_[X](Xbeta+epsilon)=M_[X]epsilon,$$
          since $M_[X]X$ is a matrix of zeros.



          Post-multiplying $b-beta$ with $e'$, we get
          $$(b-beta)e'=(X'X)^-1X'epsilon epsilon' M_[X],$$
          since $e'=epsilon'M_[X].$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Ah. The key thing I was missing was what you wrote in the last line.
            $endgroup$
            – David
            3 hours ago













          3












          3








          3





          $begingroup$

          I am assuming $b$ is the OLS estimate of $beta$ and $e$ is the corresponding estimate of $epsilon$. Also I believe you have a typo above in your expression, as there should be $X'$ in front of $epsilon epsilon'$ and behind $(X'X)^-1$.



          Start with the definition of $b$:
          $$b=(X'X)^-1X'Y.$$
          Replacing $Y$ with $Xbeta+epsilon$ in our expression above, we get
          $$b=(X'X)^-1X'(Xbeta+epsilon)=beta+(X'X)^-1X'epsilon.$$
          It follows that
          $$b-beta = (X'X)^-1X'epsilon$$



          Now turn to the defintion of $e$:
          $$e=Y-hatY=Y-Xb=Y-X(X'X)^-1X'Y.$$



          Notice $X(X'X)^-1X'$ is the projection matrix for $X$, which we will denote with $P_[X]$.
          Replacing this in our expression for $e,$ we get
          $$e=(I-P_[X])Y=M_[X]Y.$$
          Replacing $Y$ in the expression above with $Xbeta+epsilon$, we get
          $$e=M_[X](Xbeta+epsilon)=M_[X]epsilon,$$
          since $M_[X]X$ is a matrix of zeros.



          Post-multiplying $b-beta$ with $e'$, we get
          $$(b-beta)e'=(X'X)^-1X'epsilon epsilon' M_[X],$$
          since $e'=epsilon'M_[X].$






          share|cite|improve this answer









          $endgroup$



          I am assuming $b$ is the OLS estimate of $beta$ and $e$ is the corresponding estimate of $epsilon$. Also I believe you have a typo above in your expression, as there should be $X'$ in front of $epsilon epsilon'$ and behind $(X'X)^-1$.



          Start with the definition of $b$:
          $$b=(X'X)^-1X'Y.$$
          Replacing $Y$ with $Xbeta+epsilon$ in our expression above, we get
          $$b=(X'X)^-1X'(Xbeta+epsilon)=beta+(X'X)^-1X'epsilon.$$
          It follows that
          $$b-beta = (X'X)^-1X'epsilon$$



          Now turn to the defintion of $e$:
          $$e=Y-hatY=Y-Xb=Y-X(X'X)^-1X'Y.$$



          Notice $X(X'X)^-1X'$ is the projection matrix for $X$, which we will denote with $P_[X]$.
          Replacing this in our expression for $e,$ we get
          $$e=(I-P_[X])Y=M_[X]Y.$$
          Replacing $Y$ in the expression above with $Xbeta+epsilon$, we get
          $$e=M_[X](Xbeta+epsilon)=M_[X]epsilon,$$
          since $M_[X]X$ is a matrix of zeros.



          Post-multiplying $b-beta$ with $e'$, we get
          $$(b-beta)e'=(X'X)^-1X'epsilon epsilon' M_[X],$$
          since $e'=epsilon'M_[X].$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          dlnBdlnB

          81011




          81011











          • $begingroup$
            Ah. The key thing I was missing was what you wrote in the last line.
            $endgroup$
            – David
            3 hours ago
















          • $begingroup$
            Ah. The key thing I was missing was what you wrote in the last line.
            $endgroup$
            – David
            3 hours ago















          $begingroup$
          Ah. The key thing I was missing was what you wrote in the last line.
          $endgroup$
          – David
          3 hours ago




          $begingroup$
          Ah. The key thing I was missing was what you wrote in the last line.
          $endgroup$
          – David
          3 hours ago













          3












          $begingroup$

          Assuming that the coefficient estimator $b$ is calculated by OLS estimation, you have:



          $$beginequation beginaligned
          b-beta
          &= (X'X)^-1 X'y - beta \[6pt]
          &= (X'X)^-1 X'(X beta + epsilon)- beta \[6pt]
          &= (X'X)^-1 (X'X) beta + (X'X)^-1 X' epsilon - beta \[6pt]
          &= beta + (X'X)^-1 X' epsilon - beta \[6pt]
          &= (X'X)^-1 X' epsilon. \[6pt]
          endaligned endequation$$



          Presumably $e$ is the residual vector (different to the error vector $epsilon$) so we have $e = M_[X] Y = M_[X] epsilon$. Substituting this vector gives:



          $$beginequation beginaligned
          (b-beta) e'
          &= (X'X)^-1 X' epsilon (M_[X] epsilon)' \[6pt]
          &= (X'X)^-1 X' epsilon epsilon' M_[X]' \[6pt]
          &= (X'X)^-1 X' epsilon epsilon' M_[X]. \[6pt]
          endaligned endequation$$



          (The last step follows from the fact that $M_[X]$ is a symmetric matrix.) So the expression given by your professor is missing the $X'$ term.






          share|cite|improve this answer









          $endgroup$








          • 2




            $begingroup$
            Nice, I think we both must have been typing our answers at the same time. I'm glad you also found the mistake.
            $endgroup$
            – dlnB
            3 hours ago






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            3 hours ago















          3












          $begingroup$

          Assuming that the coefficient estimator $b$ is calculated by OLS estimation, you have:



          $$beginequation beginaligned
          b-beta
          &= (X'X)^-1 X'y - beta \[6pt]
          &= (X'X)^-1 X'(X beta + epsilon)- beta \[6pt]
          &= (X'X)^-1 (X'X) beta + (X'X)^-1 X' epsilon - beta \[6pt]
          &= beta + (X'X)^-1 X' epsilon - beta \[6pt]
          &= (X'X)^-1 X' epsilon. \[6pt]
          endaligned endequation$$



          Presumably $e$ is the residual vector (different to the error vector $epsilon$) so we have $e = M_[X] Y = M_[X] epsilon$. Substituting this vector gives:



          $$beginequation beginaligned
          (b-beta) e'
          &= (X'X)^-1 X' epsilon (M_[X] epsilon)' \[6pt]
          &= (X'X)^-1 X' epsilon epsilon' M_[X]' \[6pt]
          &= (X'X)^-1 X' epsilon epsilon' M_[X]. \[6pt]
          endaligned endequation$$



          (The last step follows from the fact that $M_[X]$ is a symmetric matrix.) So the expression given by your professor is missing the $X'$ term.






          share|cite|improve this answer









          $endgroup$








          • 2




            $begingroup$
            Nice, I think we both must have been typing our answers at the same time. I'm glad you also found the mistake.
            $endgroup$
            – dlnB
            3 hours ago






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            3 hours ago













          3












          3








          3





          $begingroup$

          Assuming that the coefficient estimator $b$ is calculated by OLS estimation, you have:



          $$beginequation beginaligned
          b-beta
          &= (X'X)^-1 X'y - beta \[6pt]
          &= (X'X)^-1 X'(X beta + epsilon)- beta \[6pt]
          &= (X'X)^-1 (X'X) beta + (X'X)^-1 X' epsilon - beta \[6pt]
          &= beta + (X'X)^-1 X' epsilon - beta \[6pt]
          &= (X'X)^-1 X' epsilon. \[6pt]
          endaligned endequation$$



          Presumably $e$ is the residual vector (different to the error vector $epsilon$) so we have $e = M_[X] Y = M_[X] epsilon$. Substituting this vector gives:



          $$beginequation beginaligned
          (b-beta) e'
          &= (X'X)^-1 X' epsilon (M_[X] epsilon)' \[6pt]
          &= (X'X)^-1 X' epsilon epsilon' M_[X]' \[6pt]
          &= (X'X)^-1 X' epsilon epsilon' M_[X]. \[6pt]
          endaligned endequation$$



          (The last step follows from the fact that $M_[X]$ is a symmetric matrix.) So the expression given by your professor is missing the $X'$ term.






          share|cite|improve this answer









          $endgroup$



          Assuming that the coefficient estimator $b$ is calculated by OLS estimation, you have:



          $$beginequation beginaligned
          b-beta
          &= (X'X)^-1 X'y - beta \[6pt]
          &= (X'X)^-1 X'(X beta + epsilon)- beta \[6pt]
          &= (X'X)^-1 (X'X) beta + (X'X)^-1 X' epsilon - beta \[6pt]
          &= beta + (X'X)^-1 X' epsilon - beta \[6pt]
          &= (X'X)^-1 X' epsilon. \[6pt]
          endaligned endequation$$



          Presumably $e$ is the residual vector (different to the error vector $epsilon$) so we have $e = M_[X] Y = M_[X] epsilon$. Substituting this vector gives:



          $$beginequation beginaligned
          (b-beta) e'
          &= (X'X)^-1 X' epsilon (M_[X] epsilon)' \[6pt]
          &= (X'X)^-1 X' epsilon epsilon' M_[X]' \[6pt]
          &= (X'X)^-1 X' epsilon epsilon' M_[X]. \[6pt]
          endaligned endequation$$



          (The last step follows from the fact that $M_[X]$ is a symmetric matrix.) So the expression given by your professor is missing the $X'$ term.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          BenBen

          26.8k230124




          26.8k230124







          • 2




            $begingroup$
            Nice, I think we both must have been typing our answers at the same time. I'm glad you also found the mistake.
            $endgroup$
            – dlnB
            3 hours ago






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            3 hours ago












          • 2




            $begingroup$
            Nice, I think we both must have been typing our answers at the same time. I'm glad you also found the mistake.
            $endgroup$
            – dlnB
            3 hours ago






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            3 hours ago







          2




          2




          $begingroup$
          Nice, I think we both must have been typing our answers at the same time. I'm glad you also found the mistake.
          $endgroup$
          – dlnB
          3 hours ago




          $begingroup$
          Nice, I think we both must have been typing our answers at the same time. I'm glad you also found the mistake.
          $endgroup$
          – dlnB
          3 hours ago




          1




          1




          $begingroup$
          @dlnb: Jinx! Buy me a coke!
          $endgroup$
          – Ben
          3 hours ago




          $begingroup$
          @dlnb: Jinx! Buy me a coke!
          $endgroup$
          – Ben
          3 hours ago

















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