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Why does this expression simplify as such?

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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

edited 3 hours ago

Benjamin Christoffersen

1,264519

asked 4 hours ago

DavidDavid

24311

$endgroup$

add a comment | 

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

edited 3 hours ago

Benjamin Christoffersen

1,264519

asked 4 hours ago

DavidDavid

24311

$endgroup$

add a comment | 

$begingroup$

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

edited 3 hours ago

Benjamin Christoffersen

1,264519

asked 4 hours ago

DavidDavid

24311

$endgroup$

share|cite|improve this question

edited 3 hours ago

Benjamin Christoffersen

1,264519

asked 4 hours ago

DavidDavid

24311

share|cite|improve this question

edited 3 hours ago

Benjamin Christoffersen

1,264519

asked 4 hours ago

DavidDavid

24311

edited 3 hours ago

Benjamin Christoffersen

1,264519

edited 3 hours ago

Benjamin Christoffersen

1,264519

asked 4 hours ago

DavidDavid

24311

asked 4 hours ago

DavidDavid

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

answered 2 hours ago

dlnBdlnB

81011

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

answered 2 hours ago

BenBen

26.8k230124

$endgroup$

• 
  
  
  

  
  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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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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

answered 2 hours ago

dlnBdlnB

81011

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

answered 2 hours ago

dlnBdlnB

81011

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

answered 2 hours ago

dlnBdlnB

81011

$endgroup$

share|cite|improve this answer

answered 2 hours ago

dlnBdlnB

81011

answered 2 hours ago

dlnBdlnB

81011

answered 2 hours ago

dlnBdlnB

81011

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

answered 2 hours ago

BenBen

26.8k230124

$endgroup$

• 
  
  
  

  
  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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

answered 2 hours ago

BenBen

26.8k230124

$endgroup$

• 
  
  
  

  
  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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

answered 2 hours ago

BenBen

26.8k230124

$endgroup$

share|cite|improve this answer

answered 2 hours ago

BenBen

26.8k230124

answered 2 hours ago

BenBen

26.8k230124

answered 2 hours ago

BenBen

26.8k230124