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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?










3












$begingroup$


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:



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



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



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?










share|cite|improve this question











$endgroup$











  • $begingroup$
    It would be unchanged. Hmm. I'm afraid I don't yet see the relevance, though.
    $endgroup$
    – half-pass
    5 hours ago










  • $begingroup$
    I should probably also note that while I wish this were a homework question, it's not... :)
    $endgroup$
    – half-pass
    5 hours ago






  • 1




    $begingroup$
    Ah, I didn't read the question carefully enough. My apologies.
    $endgroup$
    – jbowman
    5 hours ago










  • $begingroup$
    The first equality in your calculation is not correct. $s_x = sqrts^2_x$ 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















3












$begingroup$


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:



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



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



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?










share|cite|improve this question











$endgroup$











  • $begingroup$
    It would be unchanged. Hmm. I'm afraid I don't yet see the relevance, though.
    $endgroup$
    – half-pass
    5 hours ago










  • $begingroup$
    I should probably also note that while I wish this were a homework question, it's not... :)
    $endgroup$
    – half-pass
    5 hours ago






  • 1




    $begingroup$
    Ah, I didn't read the question carefully enough. My apologies.
    $endgroup$
    – jbowman
    5 hours ago










  • $begingroup$
    The first equality in your calculation is not correct. $s_x = sqrts^2_x$ 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













3












3








3





$begingroup$


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:



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



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



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?










share|cite|improve this question











$endgroup$




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:



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



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



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?







correlation covariance independence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago







half-pass

















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
    5 hours ago










  • $begingroup$
    I should probably also note that while I wish this were a homework question, it's not... :)
    $endgroup$
    – half-pass
    5 hours ago






  • 1




    $begingroup$
    Ah, I didn't read the question carefully enough. My apologies.
    $endgroup$
    – jbowman
    5 hours ago










  • $begingroup$
    The first equality in your calculation is not correct. $s_x = sqrts^2_x$ 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
    5 hours ago










  • $begingroup$
    I should probably also note that while I wish this were a homework question, it's not... :)
    $endgroup$
    – half-pass
    5 hours ago






  • 1




    $begingroup$
    Ah, I didn't read the question carefully enough. My apologies.
    $endgroup$
    – jbowman
    5 hours ago










  • $begingroup$
    The first equality in your calculation is not correct. $s_x = sqrts^2_x$ 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
5 hours ago




$begingroup$
It would be unchanged. Hmm. I'm afraid I don't yet see the relevance, though.
$endgroup$
– half-pass
5 hours ago












$begingroup$
I should probably also note that while I wish this were a homework question, it's not... :)
$endgroup$
– half-pass
5 hours ago




$begingroup$
I should probably also note that while I wish this were a homework question, it's not... :)
$endgroup$
– half-pass
5 hours ago




1




1




$begingroup$
Ah, I didn't read the question carefully enough. My apologies.
$endgroup$
– jbowman
5 hours ago




$begingroup$
Ah, I didn't read the question carefully enough. My apologies.
$endgroup$
– jbowman
5 hours ago












$begingroup$
The first equality in your calculation is not correct. $s_x = sqrts^2_x$ 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. $s_x = sqrts^2_x$ 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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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


















1












$begingroup$

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:



enter image description here






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      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















    1












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      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













    1












    1








    1





    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$



    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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
















    • $begingroup$
      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













    1












    $begingroup$

    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:



    enter image description here






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      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:



      enter image description here






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        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:



        enter image description here






        share|cite|improve this answer









        $endgroup$



        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:



        enter image description here







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        half-passhalf-pass

        1,43441931




        1,43441931



























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