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







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$endgroup$

















    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




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      edited yesterday









      Benjamin Christoffersen

      1,282520




      1,282520










      asked 2 days ago









      DavidDavid

      24311




      24311






















          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
            yesterday



















          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
            yesterday






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            yesterday











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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
            yesterday
















          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
            yesterday














          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 yesterday









          dlnBdlnB

          87712




          87712












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


















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
















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




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













          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
            yesterday






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            yesterday
















          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
            yesterday






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            yesterday














          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 yesterday









          BenBen

          26.9k230124




          26.9k230124








          • 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
            yesterday






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            yesterday














          • 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
            yesterday






          • 1




            $begingroup$
            @dlnb: Jinx! Buy me a coke!
            $endgroup$
            – Ben
            yesterday








          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
          yesterday




          $begingroup$
          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
          yesterday




          1




          1




          $begingroup$
          @dlnb: Jinx! Buy me a coke!
          $endgroup$
          – Ben
          yesterday




          $begingroup$
          @dlnb: Jinx! Buy me a coke!
          $endgroup$
          – Ben
          yesterday


















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          數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音...
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          When can things happen in Etherscan, such as the picture below?

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          2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  ...
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          格利澤436b

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          格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
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