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?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
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.
regression multiple-regression linear-model residuals
edited Mar 21 at 22:21
Benjamin Christoffersen
1,326520
asked Mar 21 at 21:18
DavidDavid
25214
$endgroup$
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:
$$E[(b-beta)e']=E[(X'X)^-1epsilonepsilon'M_[X]]. $$
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
edited Mar 21 at 22:21
Benjamin Christoffersen
1,326520
asked Mar 21 at 21:18
DavidDavid
25214
$endgroup$
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:
$$E[(b-beta)e']=E[(X'X)^-1epsilonepsilon'M_[X]]. $$
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
edited Mar 21 at 22:21
Benjamin Christoffersen
1,326520
asked Mar 21 at 21:18
DavidDavid
25214
$endgroup$
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
regression multiple-regression linear-model residuals
edited Mar 21 at 22:21
Benjamin Christoffersen
1,326520
asked Mar 21 at 21:18
DavidDavid
25214
edited Mar 21 at 22:21
Benjamin Christoffersen
1,326520
asked Mar 21 at 21:18
DavidDavid
25214
edited Mar 21 at 22:21
Benjamin Christoffersen
1,326520
edited Mar 21 at 22:21
Benjamin Christoffersen
1,326520
edited Mar 21 at 22:21
Benjamin Christoffersen
1,326520
1,326520
asked Mar 21 at 21:18
DavidDavid
25214
asked Mar 21 at 21:18
DavidDavid
25214
asked Mar 21 at 21:18
DavidDavid
25214
25214
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $e'=epsilon'M_[X].$
answered Mar 21 at 22:57
dlnBdlnB
97713
$endgroup$
$begingroup$
Ah. The key thing I was missing was what you wrote in the last line.
$endgroup$
– David
Mar 21 at 23:36
add a comment |
$begingroup$
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$$
(The last step follows from the fact that $M_[X]$ is a symmetric matrix.) So the expression given by your professor is missing the $X'$ term.
answered Mar 21 at 23:07
BenBen
27.5k233126
$endgroup$
2
$begingroup$
Nice, I think we both must have been typing our answers at the same time. I'm glad you also found the mistake.
$endgroup$
– dlnB
Mar 21 at 23:11
1
$begingroup$
@dlnb: Jinx! Buy me a coke!
$endgroup$
– Ben
Mar 21 at 23:13
add a comment |
Your Answer
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2 Answers
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2 Answers
2
active
oldest
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oldest
votes
$begingroup$
I am assuming $b$ is the OLS estimate of $beta$ and $e$ is the corresponding estimate of $epsilon$. Also I believe you have a typo above in your expression, as there should be $X'$ in front of $epsilon epsilon'$ and behind $(X'X)^-1$.
Start with the definition of $b$:
$$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],$$
since $e'=epsilon'M_[X].$
answered Mar 21 at 22:57
dlnBdlnB
97713
$endgroup$
$begingroup$
Ah. The key thing I was missing was what you wrote in the last line.
$endgroup$
– David
Mar 21 at 23:36
add a comment |
$begingroup$
I am assuming $b$ is the OLS estimate of $beta$ and $e$ is the corresponding estimate of $epsilon$. Also I believe you have a typo above in your expression, as there should be $X'$ in front of $epsilon epsilon'$ and behind $(X'X)^-1$.
Start with the definition of $b$:
$$b=(X'X)^-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],$$
since $e'=epsilon'M_[X].$
answered Mar 21 at 22:57
dlnBdlnB
97713
$endgroup$
$begingroup$
Ah. The key thing I was missing was what you wrote in the last line.
$endgroup$
– David
Mar 21 at 23:36
add a comment |
$begingroup$
I am assuming $b$ is the OLS estimate of $beta$ and $e$ is the corresponding estimate of $epsilon$. Also I believe you have a typo above in your expression, as there should be $X'$ in front of $epsilon epsilon'$ and behind $(X'X)^-1$.
Start with the definition of $b$:
$$b=(X'X)^-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],$$
since $e'=epsilon'M_[X].$
answered Mar 21 at 22:57
dlnBdlnB
97713
$endgroup$
I am assuming $b$ is the OLS estimate of $beta$ and $e$ is the corresponding estimate of $epsilon$. Also I believe you have a typo above in your expression, as there should be $X'$ in front of $epsilon epsilon'$ and behind $(X'X)^-1$.
Start with the definition of $b$:
$$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],$$
since $e'=epsilon'M_[X].$
answered Mar 21 at 22:57
dlnBdlnB
97713
answered Mar 21 at 22:57
dlnBdlnB
97713
answered Mar 21 at 22:57
dlnBdlnB
97713
answered Mar 21 at 22:57
dlnBdlnB
97713
97713
$begingroup$
Ah. The key thing I was missing was what you wrote in the last line.
$endgroup$
– David
Mar 21 at 23:36
add a comment |
$begingroup$
Ah. The key thing I was missing was what you wrote in the last line.
$endgroup$
– David
Mar 21 at 23:36
$begingroup$
Ah. The key thing I was missing was what you wrote in the last line.
$endgroup$
– David
Mar 21 at 23:36
$begingroup$
Ah. The key thing I was missing was what you wrote in the last line.
$endgroup$
– David
Mar 21 at 23:36
add a comment |
$begingroup$
Assuming that the coefficient estimator $b$ is calculated by OLS estimation, you have:
$$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$$
(The last step follows from the fact that $M_[X]$ is a symmetric matrix.) So the expression given by your professor is missing the $X'$ term.
answered Mar 21 at 23:07
BenBen
27.5k233126
$endgroup$
2
$begingroup$
Nice, I think we both must have been typing our answers at the same time. I'm glad you also found the mistake.
$endgroup$
– dlnB
Mar 21 at 23:11
1
$begingroup$
@dlnb: Jinx! Buy me a coke!
$endgroup$
– Ben
Mar 21 at 23:13
add a comment |
$begingroup$
Assuming that the coefficient estimator $b$ is calculated by OLS estimation, you have:
$$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$$
(The last step follows from the fact that $M_[X]$ is a symmetric matrix.) So the expression given by your professor is missing the $X'$ term.
answered Mar 21 at 23:07
BenBen
27.5k233126
$endgroup$
2
$begingroup$
Nice, I think we both must have been typing our answers at the same time. I'm glad you also found the mistake.
$endgroup$
– dlnB
Mar 21 at 23:11
1
$begingroup$
@dlnb: Jinx! Buy me a coke!
$endgroup$
– Ben
Mar 21 at 23:13
add a comment |
$begingroup$
Assuming that the coefficient estimator $b$ is calculated by OLS estimation, you have:
$$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$$
(The last step follows from the fact that $M_[X]$ is a symmetric matrix.) So the expression given by your professor is missing the $X'$ term.
answered Mar 21 at 23:07
BenBen
27.5k233126
$endgroup$
Assuming that the coefficient estimator $b$ is calculated by OLS estimation, you have:
$$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.
answered Mar 21 at 23:07
BenBen
27.5k233126
answered Mar 21 at 23:07
BenBen
27.5k233126
answered Mar 21 at 23:07
BenBen
27.5k233126
answered Mar 21 at 23:07
BenBen
27.5k233126
27.5k233126
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
add a comment |
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
add a comment |
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