Let $x$ and $y$ vectors $n\times1$ and $\hat{\beta}_{xy}$, $\hat{\beta}_{yx}$ the coefficients obtained by OLS of the regressions $x$ over $y$ and $y$ over $x$ respectively. Show that $\hat{\beta}_{xy}\hat{\beta}_{yx}=R^2$
I have the following:
From the regression $x$ over $y$ $x=\alpha+\beta_{xy}y+u_i$, then $$\hat{\beta}_{xy}=\frac{\sum (y_i-\bar{y})(x_i-\bar{x})}{\sum (y_i-\bar{y})^2}$$ Then $$\hat{\beta}_{xy}\hat{\beta}_{yx}=\frac{\sum (y_i-\bar{y})(x_i-\bar{x})}{\sum (y_i-\bar{y})^2}\hat{\beta}_{yx}=\frac{\sum (y_i-\bar{y})(\hat{y_i}-\bar{y})}{\sum (y_i-\bar{y})^2}$$ Then... I don't know how to manipulate the expression in order to obtain $$R^2=\frac{\sum (\hat{y_i}-\bar{y})^2}{{\sum (y_i-\bar{y})^2}}=1-\frac{\sum \hat{u_i}^2}{{\sum (y_i-\bar{y})^2}}$$
One possible way is to show that $r^2 = R^2$ for a "simple" regression (intercept and one $X$), and then to use the fact (that follows straightforward from the written $\beta_{xy}$) that $\hat{\beta}_{yx} = r_{yx}\frac{S_Y}{S_X}$. Then the wanted result will follow immediately.
$r_{xy}$ is the Pearson's correlation coefficient between $X$ and $Y$. If you got stuck with the first stage, you can see here for one way to prove $r^2 = R^2$.