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





      $begingroup$


      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 4 hours ago









      Benjamin Christoffersen

      1,264519




      1,264519










      asked 5 hours ago









      DavidDavid

      24311




      24311






















          2 Answers
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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 $b$:
          $$b=(X'X)^{-1}X'Y.$$
          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
            3 hours ago



















          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
            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
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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
            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 $b$:
          $$b=(X'X)^{-1}X'Y.$$
          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
            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 $b$:
          $$b=(X'X)^{-1}X'Y.$$
          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 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:



          $$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
            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:



          $$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
            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:



          $$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 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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          Ida-Boy-Ed-Garten

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          Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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