Let $\mathbf{y_1, y_2}$ be standardized $N \times 1$ column vectors, such that $\mathbb{E}\mathbf{[y_1]}=\mathbb{E}\mathbf{[y_2]}=0$, $\text{Var}\mathbf{[y_1]}=\text{Var}\mathbf{[y_2]}=1$.
Furthermore, suppose that $\mathbf{y_1, y_2}$ are correlated such that $\text{Cor}(\mathbf{y_1 ,y_2}) \frac{\mathbf{y_1^Ty_2}}{\sqrt{\mathbf{y_1^Ty_1}}\sqrt{\mathbf{y_2^Ty_2}}}=\frac{\mathbf{y_1^Ty_2}}{\sqrt{N}\sqrt{N}}=r$.
Then if we perform regression on another standardized $N \times 1$ column vector $\mathbf{x}$
$$\hat{\beta_1} = \mathbf{(x^Tx)^{-1}x^Ty_1}$$ $$\hat{\beta_2} = \mathbf{(x^Tx)^{-1}x^Ty_2}$$
It seems like it should be a relationship between $\hat{\beta_1}$ and $\hat{\beta_2}$, but I'm missing something in the proof. Ideally, I'd like to do something like:
$$\hat{\beta_1} = \mathbf{(x^Tx)^{-1}x^T(\frac{y_2y_2^T}{N})y_1}$$ $$\hat{\beta_1} = \mathbf{(x^Tx)^{-1}x^Ty_2(\frac{y_2^Ty_1}{N})}$$ $$\hat{\beta_1} = \hat{\beta_2}r$$
But that doesn't work, because $\mathbf{y_2y_2^T}$ is an $N \times N$ rank-1 symmetric matrix and therefore can't be the identity.
