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

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

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.


2 Answers 2


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 $X\beta+\epsilon$ in our expression above, we get $$b=(X'X)^{-1}X'(X\beta+\epsilon)=\beta+(X'X)^{-1}X'\epsilon.$$ 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 $X\beta+\epsilon$, we get $$e=M_{[X]}(X\beta+\epsilon)=M_{[X]}\epsilon,$$ 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 $e'=\epsilon'M_{[X]}.$

  • $\begingroup$ Ah. The key thing I was missing was what you wrote in the last line. $\endgroup$
    – David
    Mar 21, 2019 at 23:36

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.

  • 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, 2019 at 23:11
  • 1
    $\begingroup$ @dlnb: Jinx! Buy me a coke! $\endgroup$
    – Ben
    Mar 21, 2019 at 23:13

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