Is the sample correlation always positively correlated with the sample variance? The Next CEO...
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Is the sample correlation always positively correlated with the sample variance?
The Next CEO of Stack OverflowGiven known bivariate normal means and variances, update correlation estimate, $P(rho)$, with new data?Where does the correlation come from in the regression coefficient equation for simple regressionCDF of the ratio of two correlated $chi^2$ random variablesIs there a version of the correlation coefficient that is less-sensitive to outliers?Correlation in Distances of Points Within a Circle from Centre and One Other PointHow do I reproduce this distribution (with observed means, sd, kurtosis, skewness and correlation)?Is the formula of covariance right?Is my Correlation reasoning correct?Variance of $Y|x$ from regression lineIn a bivariate normal sample, why is the squared sample correlation Beta distributed?
$begingroup$
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?
correlation covariance independence
asked 5 hours ago
half-passhalf-pass
1,43441931
$endgroup$
|
show 2 more comments
$begingroup$
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?
correlation covariance independence
asked 5 hours ago
half-passhalf-pass
1,43441931
$endgroup$
$begingroup$
It would be unchanged. Hmm. I'm afraid I don't yet see the relevance, though.
$endgroup$
– half-pass
4 hours ago
$begingroup$
I should probably also note that while I wish this were a homework question, it's not... :)
$endgroup$
– half-pass
4 hours ago
1
$begingroup$
Ah, I didn't read the question carefully enough. My apologies.
$endgroup$
– jbowman
4 hours ago
$begingroup$
The first equality in your calculation is not correct. $s_x = sqrt{s^2_x}$ is consistent for the standard deviation, but is not unbiased: en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation
$endgroup$
– Andrew M
4 hours ago
$begingroup$
It's extremely close to unbiased for large n, though -- the rule-of-thumb correction factor for a normal RV is (n - 1.5) vs. (n-1).
$endgroup$
– half-pass
4 hours ago
|
show 2 more comments
$begingroup$
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?
correlation covariance independence
asked 5 hours ago
half-passhalf-pass
1,43441931
$endgroup$
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?
correlation covariance independence
correlation covariance independence
asked 5 hours ago
half-passhalf-pass
1,43441931
asked 5 hours ago
half-passhalf-pass
1,43441931
edited 3 hours ago
half-pass
asked 5 hours ago
half-passhalf-pass
1,43441931
asked 5 hours ago
half-passhalf-pass
1,43441931
asked 5 hours ago
half-passhalf-pass
1,43441931
1,43441931
$begingroup$
It would be unchanged. Hmm. I'm afraid I don't yet see the relevance, though.
$endgroup$
– half-pass
4 hours ago
$begingroup$
I should probably also note that while I wish this were a homework question, it's not... :)
$endgroup$
– half-pass
4 hours ago
1
$begingroup$
Ah, I didn't read the question carefully enough. My apologies.
$endgroup$
– jbowman
4 hours ago
$begingroup$
The first equality in your calculation is not correct. $s_x = sqrt{s^2_x}$ is consistent for the standard deviation, but is not unbiased: en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation
$endgroup$
– Andrew M
4 hours ago
$begingroup$
It's extremely close to unbiased for large n, though -- the rule-of-thumb correction factor for a normal RV is (n - 1.5) vs. (n-1).
$endgroup$
– half-pass
4 hours ago
|
show 2 more comments
$begingroup$
It would be unchanged. Hmm. I'm afraid I don't yet see the relevance, though.
$endgroup$
– half-pass
4 hours ago
$begingroup$
I should probably also note that while I wish this were a homework question, it's not... :)
$endgroup$
– half-pass
4 hours ago
1
$begingroup$
Ah, I didn't read the question carefully enough. My apologies.
$endgroup$
– jbowman
4 hours ago
$begingroup$
The first equality in your calculation is not correct. $s_x = sqrt{s^2_x}$ is consistent for the standard deviation, but is not unbiased: en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation
$endgroup$
– Andrew M
4 hours ago
$begingroup$
It's extremely close to unbiased for large n, though -- the rule-of-thumb correction factor for a normal RV is (n - 1.5) vs. (n-1).
$endgroup$
– half-pass
4 hours ago
$begingroup$
It would be unchanged. Hmm. I'm afraid I don't yet see the relevance, though.
$endgroup$
– half-pass
4 hours ago
$begingroup$
It would be unchanged. Hmm. I'm afraid I don't yet see the relevance, though.
$endgroup$
– half-pass
4 hours ago
$begingroup$
I should probably also note that while I wish this were a homework question, it's not... :)
$endgroup$
– half-pass
4 hours ago
$begingroup$
I should probably also note that while I wish this were a homework question, it's not... :)
$endgroup$
– half-pass
4 hours ago
1
1
$begingroup$
Ah, I didn't read the question carefully enough. My apologies.
$endgroup$
– jbowman
4 hours ago
$begingroup$
Ah, I didn't read the question carefully enough. My apologies.
$endgroup$
– jbowman
4 hours ago
$begingroup$
The first equality in your calculation is not correct. $s_x = sqrt{s^2_x}$ is consistent for the standard deviation, but is not unbiased: en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation
$endgroup$
– Andrew M
4 hours ago
$begingroup$
The first equality in your calculation is not correct. $s_x = sqrt{s^2_x}$ is consistent for the standard deviation, but is not unbiased: en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation
$endgroup$
– Andrew M
4 hours ago
$begingroup$
It's extremely close to unbiased for large n, though -- the rule-of-thumb correction factor for a normal RV is (n - 1.5) vs. (n-1).
$endgroup$
– half-pass
4 hours ago
$begingroup$
It's extremely close to unbiased for large n, though -- the rule-of-thumb correction factor for a normal RV is (n - 1.5) vs. (n-1).
$endgroup$
– half-pass
4 hours ago
|
show 2 more comments
2 Answers
2
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oldest
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$begingroup$
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.
answered 3 hours ago
Alecos PapadopoulosAlecos Papadopoulos
42.8k297197
$endgroup$
$begingroup$
I understand that the three parameters can have arbitrary relationships for the BVN distribution, but I don't think it follows that the sample estimates of these are asymptotically independent.
$endgroup$
– half-pass
3 hours ago
add a comment |
$begingroup$
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:
answered 1 hour ago
half-passhalf-pass
1,43441931
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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.
answered 3 hours ago
Alecos PapadopoulosAlecos Papadopoulos
42.8k297197
$endgroup$
$begingroup$
I understand that the three parameters can have arbitrary relationships for the BVN distribution, but I don't think it follows that the sample estimates of these are asymptotically independent.
$endgroup$
– half-pass
3 hours ago
add a comment |
$begingroup$
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.
answered 3 hours ago
Alecos PapadopoulosAlecos Papadopoulos
42.8k297197
$endgroup$
$begingroup$
I understand that the three parameters can have arbitrary relationships for the BVN distribution, but I don't think it follows that the sample estimates of these are asymptotically independent.
$endgroup$
– half-pass
3 hours ago
add a comment |
$begingroup$
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.
answered 3 hours ago
Alecos PapadopoulosAlecos Papadopoulos
42.8k297197
$endgroup$
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.
answered 3 hours ago
Alecos PapadopoulosAlecos Papadopoulos
42.8k297197
answered 3 hours ago
Alecos PapadopoulosAlecos Papadopoulos
42.8k297197
answered 3 hours ago
Alecos PapadopoulosAlecos Papadopoulos
42.8k297197
answered 3 hours ago
Alecos PapadopoulosAlecos Papadopoulos
42.8k297197
42.8k297197
$begingroup$
I understand that the three parameters can have arbitrary relationships for the BVN distribution, but I don't think it follows that the sample estimates of these are asymptotically independent.
$endgroup$
– half-pass
3 hours ago
add a comment |
$begingroup$
I understand that the three parameters can have arbitrary relationships for the BVN distribution, but I don't think it follows that the sample estimates of these are asymptotically independent.
$endgroup$
– half-pass
3 hours ago
$begingroup$
I understand that the three parameters can have arbitrary relationships for the BVN distribution, but I don't think it follows that the sample estimates of these are asymptotically independent.
$endgroup$
– half-pass
3 hours ago
$begingroup$
I understand that the three parameters can have arbitrary relationships for the BVN distribution, but I don't think it follows that the sample estimates of these are asymptotically independent.
$endgroup$
– half-pass
3 hours ago
add a comment |
$begingroup$
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:
answered 1 hour ago
half-passhalf-pass
1,43441931
$endgroup$
add a comment |
$begingroup$
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:
answered 1 hour ago
half-passhalf-pass
1,43441931
$endgroup$
add a comment |
$begingroup$
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:
answered 1 hour ago
half-passhalf-pass
1,43441931
$endgroup$
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:
answered 1 hour ago
half-passhalf-pass
1,43441931
answered 1 hour ago
half-passhalf-pass
1,43441931
answered 1 hour ago
half-passhalf-pass
1,43441931
answered 1 hour ago
half-passhalf-pass
1,43441931
1,43441931
add a comment |
add a comment |
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$begingroup$
It would be unchanged. Hmm. I'm afraid I don't yet see the relevance, though.
$endgroup$
– half-pass
4 hours ago
$begingroup$
I should probably also note that while I wish this were a homework question, it's not... :)
$endgroup$
– half-pass
4 hours ago
1
$begingroup$
Ah, I didn't read the question carefully enough. My apologies.
$endgroup$
– jbowman
4 hours ago
$begingroup$
The first equality in your calculation is not correct. $s_x = sqrt{s^2_x}$ is consistent for the standard deviation, but is not unbiased: en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation
$endgroup$
– Andrew M
4 hours ago
$begingroup$
It's extremely close to unbiased for large n, though -- the rule-of-thumb correction factor for a normal RV is (n - 1.5) vs. (n-1).
$endgroup$
– half-pass
4 hours ago