Mdtryhht

The sample correlation $r$ and the sample standard deviation of $X$ (call it $s_X$) seem to be positively correlated if I simulate bivariate normal $X$, $Y$ with a positive true correlation (and seem to be negatively correlated if the true correlation between $X$ and $Y$ is negative). I found this somewhat counterintuitive. Very heuristically, I suppose it reflects the fact that $r$ represents the expected increase in Y (in units of SD(Y)) for a one-SD increase in X, and if we estimate a larger $s_X$, then $r$ reflects the change in Y associated with a larger change in X.

However, I would like to know if $Cov(r, s_x) >0$ for $r>0$ holds in general (at least for the case in which X and Y are bivariate normal and with large n). Letting $sigma$ denote a true SD, we have:

$$Cov(r, s_X) = E [ r s_X] - rho sigma_x$$

$$ approx E Bigg[ fracwidehatCov(X,Y)s_Y Bigg] - fracCov(X,Y)sigma_Y $$

I tried using a Taylor expansion on the first term, but it depends on $Cov(widehatCov(X,Y), s_Y)$, so that’s a dead end. Any ideas?

EDIT

Maybe a better direction would be to try to show that $Cov(widehatbeta, s_X)=0$, where $widehatbeta$ is the OLS coefficient of Y on X. Then we could argue that since $widehatbeta = r fracs_Ys_X$, this implies the desired result. Since $widehatbeta$ is almost like a difference of sample means, maybe we could get the former result using something like the known independence of the sample mean and variance for a normal RV?

It will depend on the joint distribution. For the example you mention, the bivariate (zero-mean) Normal distribution is characterized by the $rho, sigma_x, sigma_y$. It follows that one can have all possible combinations of values of these three parameters, implying that no relation between $rho$ and the standard deviations can be established.

For other bivariate distributions, the correlation coefficient may be fundamentally a function of the standard deviations (essentially both will be functions of more primitive parameters), in which case one can examine whether a monotonic relation exists.

Yes, it does hold asymptotically regardless of the distribution of X and Y. I was on the right track with the Taylor expansion; I just needed to make a symmetry argument: