Became Hot Network Question
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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?

3 added 226 characters in body

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?

2 added 20 characters in body

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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