Why does this expression simplify as such?












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:



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



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











$endgroup$

















    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:



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



    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











    $endgroup$















      3












      3








      3


      1



      $begingroup$


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



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



      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











      $endgroup$




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



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



      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




      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 Mar 21 at 22:21









      Benjamin Christoffersen

      1,291520




      1,291520










      asked Mar 21 at 21:18









      DavidDavid

      25213




      25213






















          2 Answers
          2






          active

          oldest

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



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



          Notice $X(X'X)^{-1}X'$ 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
          $$(b-beta)e'=(X'X)^{-1}X'epsilon epsilon' M_{[X]},$$
          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
            Mar 21 at 23:36



















          3












          $begingroup$

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



          $$begin{equation} begin{aligned}
          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]
          end{aligned} end{equation}$$



          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:



          $$begin{equation} begin{aligned}
          (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]
          end{aligned} end{equation}$$



          (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
            Mar 21 at 23:11






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            Mar 21 at 23:13












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          2 Answers
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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 $b$:
          $$b=(X'X)^{-1}X'Y.$$
          Replacing $Y$ with $Xbeta+epsilon$ in our expression above, we get
          $$b=(X'X)^{-1}X'(Xbeta+epsilon)=beta+(X'X)^{-1}X'epsilon.$$
          It follows that
          $$b-beta = (X'X)^{-1}X'epsilon$$



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



          Notice $X(X'X)^{-1}X'$ 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
          $$(b-beta)e'=(X'X)^{-1}X'epsilon epsilon' M_{[X]},$$
          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
            Mar 21 at 23:36
















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



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



          Notice $X(X'X)^{-1}X'$ 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
          $$(b-beta)e'=(X'X)^{-1}X'epsilon epsilon' M_{[X]},$$
          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
            Mar 21 at 23:36














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



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



          Notice $X(X'X)^{-1}X'$ 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
          $$(b-beta)e'=(X'X)^{-1}X'epsilon epsilon' M_{[X]},$$
          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 $b$:
          $$b=(X'X)^{-1}X'Y.$$
          Replacing $Y$ with $Xbeta+epsilon$ in our expression above, we get
          $$b=(X'X)^{-1}X'(Xbeta+epsilon)=beta+(X'X)^{-1}X'epsilon.$$
          It follows that
          $$b-beta = (X'X)^{-1}X'epsilon$$



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



          Notice $X(X'X)^{-1}X'$ 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
          $$(b-beta)e'=(X'X)^{-1}X'epsilon epsilon' M_{[X]},$$
          since $e'=epsilon'M_{[X]}.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 21 at 22:57









          dlnBdlnB

          1,01712




          1,01712












          • $begingroup$
            Ah. The key thing I was missing was what you wrote in the last line.
            $endgroup$
            – David
            Mar 21 at 23:36


















          • $begingroup$
            Ah. The key thing I was missing was what you wrote in the last line.
            $endgroup$
            – David
            Mar 21 at 23:36
















          $begingroup$
          Ah. The key thing I was missing was what you wrote in the last line.
          $endgroup$
          – David
          Mar 21 at 23:36




          $begingroup$
          Ah. The key thing I was missing was what you wrote in the last line.
          $endgroup$
          – David
          Mar 21 at 23:36













          3












          $begingroup$

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



          $$begin{equation} begin{aligned}
          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]
          end{aligned} end{equation}$$



          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:



          $$begin{equation} begin{aligned}
          (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]
          end{aligned} end{equation}$$



          (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
            Mar 21 at 23:11






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            Mar 21 at 23:13
















          3












          $begingroup$

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



          $$begin{equation} begin{aligned}
          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]
          end{aligned} end{equation}$$



          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:



          $$begin{equation} begin{aligned}
          (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]
          end{aligned} end{equation}$$



          (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
            Mar 21 at 23:11






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            Mar 21 at 23:13














          3












          3








          3





          $begingroup$

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



          $$begin{equation} begin{aligned}
          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]
          end{aligned} end{equation}$$



          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:



          $$begin{equation} begin{aligned}
          (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]
          end{aligned} end{equation}$$



          (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:



          $$begin{equation} begin{aligned}
          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]
          end{aligned} end{equation}$$



          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:



          $$begin{equation} begin{aligned}
          (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]
          end{aligned} end{equation}$$



          (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 Mar 21 at 23:07









          BenBen

          27.1k233126




          27.1k233126








          • 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
            Mar 21 at 23:11






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            Mar 21 at 23:13














          • 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
            Mar 21 at 23:11






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            Mar 21 at 23:13








          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
          Mar 21 at 23:11




          $begingroup$
          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
          Mar 21 at 23:11




          1




          1




          $begingroup$
          @dlnb: Jinx! Buy me a coke!
          $endgroup$
          – Ben
          Mar 21 at 23:13




          $begingroup$
          @dlnb: Jinx! Buy me a coke!
          $endgroup$
          – Ben
          Mar 21 at 23:13


















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