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    $\begingroup$ The coefficient vector $\hat{\beta}$ is the solution to $X'Y=(X'X)^{-1}\beta$. Some algebraic manipulation reveals that this is in fact the same as the formula you give in the 2-coefficient case. Laid out nicely here: stat.purdue.edu/~jennings/stat514/stat512notes/topic3.pdf. Not sure if that helps at all. But I'd venture to guess that this is impossible in general based on that formula. $\endgroup$ Commented Jul 22, 2014 at 13:13
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    $\begingroup$ @David Did you figure out how to extend this to an arbitrary number of explanatory variables (beyond 2)? I need the expression. $\endgroup$ Commented May 16, 2016 at 7:47
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    $\begingroup$ @JaneWayne I'm not sure I understand your question: whuber gave the solution below in matrix form, $C^{-1}(\text{Cov}(X_i, y))^\prime$ $\endgroup$ Commented Jun 15, 2016 at 14:19
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    $\begingroup$ yup I studied it and he's right. $\endgroup$ Commented Jun 15, 2016 at 14:52