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?












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:



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



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



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







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
    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
















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:



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



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



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







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
    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














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:



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



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



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







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:



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



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



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






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
    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










  • $begingroup$
    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










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









$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    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 1 hour ago









        half-passhalf-pass

        1,43441931




        1,43441931






























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            Frisyrar frisyrehårhårstrikkhårklemmerøyrethesthovudverkeuropeisk motejobbtreningfletterparykkarungdomsopprørmotkulturmusikararmotorsykkelentusiastarpolitietpolitiskforretningverksemdfiskeflette Hestehale sett bakfrå. Hestehale på ein mannleg indianarleiar. Hestehale er ein frisyre der ein samlar det meste eller alt av langt hår [1] i bakhovudet med ein hårstrikk eller liknande. Avhengig av mote kan det òg vere vanleg å sette opp hestehalen med hårklemmer eller å ha hestehalen hengande på den eine sida av hovudet ved å samle håret ved øyret. Hestehale er ein vanleg frisyre for jenter og for gutar med langt hår. Han har fått namnet sitt etter likskapen med halen på ein hest. For stramme hestehaler kan føre til hårtap, [2] i nokre høve også hovudverk. [3] Innhaldsliste 1 Hestehale på kvinner 2 Hestehale på menn 3 Galleri 4 Sjå òg 5 Bakgrunnsstoff 6 Kjelder Hestehale på kvinner | Ungjente med hestehale som held håret unna ansiktet...
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