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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:

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

$$ \approx E \Bigg[ \frac{\widehat{Cov}(X,Y)}{s_Y} \Bigg] - \frac{Cov(X,Y)}{\sigma_Y} $$

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

EDIT

Maybe a better direction would be to try to show that $Cov(\widehat{\beta}, s_X)=0$, where $\widehat{\beta}$ is the OLS coefficient of Y on X. Then we could argue that since $\widehat{\beta} = r \frac{s_Y}{s_X}$, this implies the desired result. Since $\widehat{\beta}$ 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?

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:

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

$$ \approx E \Bigg[ \frac{\widehat{Cov}(X,Y)}{s_Y} \Bigg] - \frac{Cov(X,Y)}{\sigma_Y} $$

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

EDIT

Maybe a better direction would be to try to show that $Cov(\widehat{\beta}, s_X)=0$, where $\widehat{\beta}$ is the OLS coefficient of Y on X. Then we could argue that since $\widehat{\beta} = r \frac{s_Y}{s_X}$, this implies the desired result. Since $\widehat{\beta}$ 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?

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:

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

$$ \approx E \Bigg[ \frac{\widehat{Cov}(X,Y)}{s_Y} \Bigg] - \frac{Cov(X,Y)}{\sigma_Y} $$

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

EDIT

Maybe a better direction would be to try to show that $Cov(\widehat{\beta}, s_X)=0$, where $\widehat{\beta}$ is the OLS coefficient of Y on X. Then we could argue that since $\widehat{\beta} = r \frac{s_Y}{s_X}$, this implies the desired result. Since $\widehat{\beta}$ 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?

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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:

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

$$ = E \Bigg[ \frac{\widehat{Cov}(X,Y)}{s_Y} \Bigg] - \frac{Cov(X,Y)}{\sigma_Y} $$$$ \approx E \Bigg[ \frac{\widehat{Cov}(X,Y)}{s_Y} \Bigg] - \frac{Cov(X,Y)}{\sigma_Y} $$

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

EDIT

Maybe a better direction would be to try to show that $Cov(\widehat{\beta}, s_X)=0$, where $\widehat{\beta}$ is the OLS coefficient of Y on X. Then we could argue that since $\widehat{\beta} = r \frac{s_Y}{s_X}$, this implies the desired result. Since $\widehat{\beta}$ 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?

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). Letting $\sigma$ denote a true SD, we have:

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

$$ = E \Bigg[ \frac{\widehat{Cov}(X,Y)}{s_Y} \Bigg] - \frac{Cov(X,Y)}{\sigma_Y} $$

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

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:

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

$$ \approx E \Bigg[ \frac{\widehat{Cov}(X,Y)}{s_Y} \Bigg] - \frac{Cov(X,Y)}{\sigma_Y} $$

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

EDIT

Maybe a better direction would be to try to show that $Cov(\widehat{\beta}, s_X)=0$, where $\widehat{\beta}$ is the OLS coefficient of Y on X. Then we could argue that since $\widehat{\beta} = r \frac{s_Y}{s_X}$, this implies the desired result. Since $\widehat{\beta}$ 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?

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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). Letting $\sigma$ denote a true SD, we have:

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

$$ = E \Bigg[ \frac{\widehat{Cov}(X,Y)}{s_Y} \Bigg] - \frac{Cov(X,Y)}{\sigma_Y} $$

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

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. 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). Letting $\sigma$ denote a true SD, we have:

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

$$ = E \Bigg[ \frac{\widehat{Cov}(X,Y)}{s_Y} \Bigg] - \frac{Cov(X,Y)}{\sigma_Y} $$

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

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). Letting $\sigma$ denote a true SD, we have:

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

$$ = E \Bigg[ \frac{\widehat{Cov}(X,Y)}{s_Y} \Bigg] - \frac{Cov(X,Y)}{\sigma_Y} $$

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

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