Why does this expression simplify as such? 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. 
 A: 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 $e'=\epsilon'M_{[X]}.$
A: 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.
