Sign of correlation between $y$ and $\hat{y}$ with and without intercept

Question

I am curious about the sign of $corr(y, \hat{y})$ with and without intercept.

I recall reading that when there is an intercept, the correlation is $\geq 0$, but in the case without intercept, the correlation can also be negative. From a geometric perspective, I don't see how this can be.

Regardless of whether there is an intercept, $\hat{y}$ is the orthogonal projection of $y$ onto the subspace spanned by the predictors. So shouldn't that orthogonal projection point in the same direction as $y$, in which case the correlation should be $\geq 0$ regardless of whether there is intercept or not?

Mathematically, I tried to prove this, and it seems to match my geometric intuition.

$$ \hat{y} = \hat{\beta}_{1} x + \hat{\beta}_0 \\ corr(y, \hat{y}) = corr(y, \hat{\beta}_{1} x + \hat{\beta}_0) \\ = corr(y, \hat{\beta}_{1} x) \\ = sgn(\hat{\beta}_{1}) corr(y, x) \\ $$ I believe $sgn(\hat{\beta}_{1})$ is the same as $sgn(corr(y, x))$, so the above quantity is always $\geq 0$. So we proved that $corr(y, \hat{y})$ is non-negative in the case with intercept; however, I believe this derivation also holds for the case without intercept, since $\hat{\beta}_0$ isn't really used in this derivation?

Edit: What I also want to add is, I know that when you remove the intercept, it's possible that the sign ($sgn(\hat{\beta}_{1})$) of the slope can flip; however, if the sign of the slop is flipped, doesn't the sign of $corr(y, x)$ also flip?

Answer 1

For simple linear regression (with an intercept) you have $\hat \beta_1=\frac{\sum (x_i-\bar x)(y_i-\bar y)}{\sum (x_i-\bar x)^2}=\frac{s_y}{s_x}r_{x,y}$ and, since $s_x >0$ and $s_y > 0$, this forces the slope $\hat \beta_1$ and correlation $r_{x,y}$ to have the same sign.

Without an intercept, you have $\hat \beta_1=\frac{\sum x_iy_i}{\sum x_i^2}$, which forces $\hat \beta_1$ to have the same sign as $\sum x_iy_i$. This need not have the same sign as $r_{x,y}$.

As @Glen_b commented, consider data in the first quadrant: this will inevitably have $\sum x_iy_i >0 $ and so without an intercept $\hat \beta_1 >0$ whether $r_{x,y}$ is positive or negative. As an illustration, consider the points $(0,5)$, $(2,5)$, $(4,1)$ and $(5,1)$ illustrated in black below. A fitted line with an intercept (orange) will slope in the same general direction and the fitted values have a positive correlation with the original data, while a fitted line forced through the origin (blue) need not.