rearranging the normal linear regression function with one variable yields $$-b_0/b_1+1/b_1*Y_i-1/b_1*u_i=X_i$$. The question for me is why the regression coefficient in front of Y, $1/b_1$, is not equal to the regression coefficient if we run a regression of X on Y. I have already found that transforming this would not yield the correct coefficient: $$1/b_1=1/({cov(X,Y)/var(x)})=var(x)/cov(X,Y)\neq cov(X,Y)/var(y)$$ What I am interested in is the intuition behind this.

Many thanks

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    $\begingroup$ Think about the sum of squares that you minimize in each way. They are not the same. So, ... $\endgroup$ – Claude Leibovici May 10 at 17:16

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