Find the covariance of two jointly distributed RV's when you don't know the joint pmf












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I have two correlated random variables, X and Y, and I am trying to find their covariance. I know: their individual expected values $langle X rangle$ and $langle Y rangle$ and their individual variances $sigma^2_X$ and $sigma^2_Y$. If the covariance is given as
$cov(X,Y) = langle XY rangle -langle X rangle langle Y rangle$,
then I already know the second term but I have no idea how to compute the first term. I don't know the joint pmf. Do I need the joint pmf to calculate $langle XY rangle$?






probability probability-theory covariance







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    -1












    $begingroup$


    I have two correlated random variables, X and Y, and I am trying to find their covariance. I know: their individual expected values $langle X rangle$ and $langle Y rangle$ and their individual variances $sigma^2_X$ and $sigma^2_Y$. If the covariance is given as
    $cov(X,Y) = langle XY rangle -langle X rangle langle Y rangle$,
    then I already know the second term but I have no idea how to compute the first term. I don't know the joint pmf. Do I need the joint pmf to calculate $langle XY rangle$?






    probability probability-theory covariance







    share|cite|improve this question









    $endgroup$















      -1












      -1








      -1





      $begingroup$


      I have two correlated random variables, X and Y, and I am trying to find their covariance. I know: their individual expected values $langle X rangle$ and $langle Y rangle$ and their individual variances $sigma^2_X$ and $sigma^2_Y$. If the covariance is given as
      $cov(X,Y) = langle XY rangle -langle X rangle langle Y rangle$,
      then I already know the second term but I have no idea how to compute the first term. I don't know the joint pmf. Do I need the joint pmf to calculate $langle XY rangle$?






      probability probability-theory covariance







      share|cite|improve this question









      $endgroup$




      I have two correlated random variables, X and Y, and I am trying to find their covariance. I know: their individual expected values $langle X rangle$ and $langle Y rangle$ and their individual variances $sigma^2_X$ and $sigma^2_Y$. If the covariance is given as
      $cov(X,Y) = langle XY rangle -langle X rangle langle Y rangle$,
      then I already know the second term but I have no idea how to compute the first term. I don't know the joint pmf. Do I need the joint pmf to calculate $langle XY rangle$?






      probability probability-theory covariance




      probability probability-theory covariance






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      asked Dec 18 '18 at 2:12









      SabrinaChoiceSabrinaChoice

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          $begingroup$

          Consider two normal random variables $X$ and $Y$ each having mean zero and unit variance.
          If they are independent, then their covariance is $$mathbb{E}[XY]-mathbb{E}[X]mathbb{E}[Y]=mathbb{E}[X]mathbb{E}[Y]-mathbb{E}[X]mathbb{E}[Y]=0.$$
          However, if $X=Y$ (i.e., the two variables are perfectly correlated), then $$mathbb{E}[XY]-mathbb{E}[X]mathbb{E}[Y]=mathbb{E}[X^{2}]-mathbb{E}[X]mathbb{E}[Y]=1.$$



          What does this suggest?






          share|cite|improve this answer









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          • $begingroup$
            This suggests that $mathbb{E}[X^2]$=1, but I can't see much beyond that, sorry!
            $endgroup$
            – SabrinaChoice
            Dec 18 '18 at 2:21








          • 1




            $begingroup$
            But what does it suggest about the covariance? Is it uniquely determined by mean and variance?
            $endgroup$
            – parsiad
            Dec 18 '18 at 2:30










          • $begingroup$
            I'm not sure because the bottom expression you wrote is the expression for the variance. So it states that the variance is equal to the variance. I'm not seeing how it can add any other constraint.
            $endgroup$
            – SabrinaChoice
            Dec 18 '18 at 2:34






          • 4




            $begingroup$
            I was trying to get you to come to the conclusion that you have insufficient information to determine the covariance.
            $endgroup$
            – parsiad
            Dec 18 '18 at 2:37








          • 1




            $begingroup$
            Put differently: the marginals do not characterize the joint distribution. Because $p_Xtimes p_Y$ has, by definition, the same marginals as $p_{XY}$.
            $endgroup$
            – Clement C.
            Dec 18 '18 at 3:08













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          $begingroup$

          Consider two normal random variables $X$ and $Y$ each having mean zero and unit variance.
          If they are independent, then their covariance is $$mathbb{E}[XY]-mathbb{E}[X]mathbb{E}[Y]=mathbb{E}[X]mathbb{E}[Y]-mathbb{E}[X]mathbb{E}[Y]=0.$$
          However, if $X=Y$ (i.e., the two variables are perfectly correlated), then $$mathbb{E}[XY]-mathbb{E}[X]mathbb{E}[Y]=mathbb{E}[X^{2}]-mathbb{E}[X]mathbb{E}[Y]=1.$$



          What does this suggest?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This suggests that $mathbb{E}[X^2]$=1, but I can't see much beyond that, sorry!
            $endgroup$
            – SabrinaChoice
            Dec 18 '18 at 2:21








          • 1




            $begingroup$
            But what does it suggest about the covariance? Is it uniquely determined by mean and variance?
            $endgroup$
            – parsiad
            Dec 18 '18 at 2:30










          • $begingroup$
            I'm not sure because the bottom expression you wrote is the expression for the variance. So it states that the variance is equal to the variance. I'm not seeing how it can add any other constraint.
            $endgroup$
            – SabrinaChoice
            Dec 18 '18 at 2:34






          • 4




            $begingroup$
            I was trying to get you to come to the conclusion that you have insufficient information to determine the covariance.
            $endgroup$
            – parsiad
            Dec 18 '18 at 2:37








          • 1




            $begingroup$
            Put differently: the marginals do not characterize the joint distribution. Because $p_Xtimes p_Y$ has, by definition, the same marginals as $p_{XY}$.
            $endgroup$
            – Clement C.
            Dec 18 '18 at 3:08


















          1












          $begingroup$

          Consider two normal random variables $X$ and $Y$ each having mean zero and unit variance.
          If they are independent, then their covariance is $$mathbb{E}[XY]-mathbb{E}[X]mathbb{E}[Y]=mathbb{E}[X]mathbb{E}[Y]-mathbb{E}[X]mathbb{E}[Y]=0.$$
          However, if $X=Y$ (i.e., the two variables are perfectly correlated), then $$mathbb{E}[XY]-mathbb{E}[X]mathbb{E}[Y]=mathbb{E}[X^{2}]-mathbb{E}[X]mathbb{E}[Y]=1.$$



          What does this suggest?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This suggests that $mathbb{E}[X^2]$=1, but I can't see much beyond that, sorry!
            $endgroup$
            – SabrinaChoice
            Dec 18 '18 at 2:21








          • 1




            $begingroup$
            But what does it suggest about the covariance? Is it uniquely determined by mean and variance?
            $endgroup$
            – parsiad
            Dec 18 '18 at 2:30










          • $begingroup$
            I'm not sure because the bottom expression you wrote is the expression for the variance. So it states that the variance is equal to the variance. I'm not seeing how it can add any other constraint.
            $endgroup$
            – SabrinaChoice
            Dec 18 '18 at 2:34






          • 4




            $begingroup$
            I was trying to get you to come to the conclusion that you have insufficient information to determine the covariance.
            $endgroup$
            – parsiad
            Dec 18 '18 at 2:37








          • 1




            $begingroup$
            Put differently: the marginals do not characterize the joint distribution. Because $p_Xtimes p_Y$ has, by definition, the same marginals as $p_{XY}$.
            $endgroup$
            – Clement C.
            Dec 18 '18 at 3:08
















          1












          1








          1





          $begingroup$

          Consider two normal random variables $X$ and $Y$ each having mean zero and unit variance.
          If they are independent, then their covariance is $$mathbb{E}[XY]-mathbb{E}[X]mathbb{E}[Y]=mathbb{E}[X]mathbb{E}[Y]-mathbb{E}[X]mathbb{E}[Y]=0.$$
          However, if $X=Y$ (i.e., the two variables are perfectly correlated), then $$mathbb{E}[XY]-mathbb{E}[X]mathbb{E}[Y]=mathbb{E}[X^{2}]-mathbb{E}[X]mathbb{E}[Y]=1.$$



          What does this suggest?






          share|cite|improve this answer









          $endgroup$



          Consider two normal random variables $X$ and $Y$ each having mean zero and unit variance.
          If they are independent, then their covariance is $$mathbb{E}[XY]-mathbb{E}[X]mathbb{E}[Y]=mathbb{E}[X]mathbb{E}[Y]-mathbb{E}[X]mathbb{E}[Y]=0.$$
          However, if $X=Y$ (i.e., the two variables are perfectly correlated), then $$mathbb{E}[XY]-mathbb{E}[X]mathbb{E}[Y]=mathbb{E}[X^{2}]-mathbb{E}[X]mathbb{E}[Y]=1.$$



          What does this suggest?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '18 at 2:17









          parsiadparsiad

          18.3k32453




          18.3k32453












          • $begingroup$
            This suggests that $mathbb{E}[X^2]$=1, but I can't see much beyond that, sorry!
            $endgroup$
            – SabrinaChoice
            Dec 18 '18 at 2:21








          • 1




            $begingroup$
            But what does it suggest about the covariance? Is it uniquely determined by mean and variance?
            $endgroup$
            – parsiad
            Dec 18 '18 at 2:30










          • $begingroup$
            I'm not sure because the bottom expression you wrote is the expression for the variance. So it states that the variance is equal to the variance. I'm not seeing how it can add any other constraint.
            $endgroup$
            – SabrinaChoice
            Dec 18 '18 at 2:34






          • 4




            $begingroup$
            I was trying to get you to come to the conclusion that you have insufficient information to determine the covariance.
            $endgroup$
            – parsiad
            Dec 18 '18 at 2:37








          • 1




            $begingroup$
            Put differently: the marginals do not characterize the joint distribution. Because $p_Xtimes p_Y$ has, by definition, the same marginals as $p_{XY}$.
            $endgroup$
            – Clement C.
            Dec 18 '18 at 3:08




















          • $begingroup$
            This suggests that $mathbb{E}[X^2]$=1, but I can't see much beyond that, sorry!
            $endgroup$
            – SabrinaChoice
            Dec 18 '18 at 2:21








          • 1




            $begingroup$
            But what does it suggest about the covariance? Is it uniquely determined by mean and variance?
            $endgroup$
            – parsiad
            Dec 18 '18 at 2:30










          • $begingroup$
            I'm not sure because the bottom expression you wrote is the expression for the variance. So it states that the variance is equal to the variance. I'm not seeing how it can add any other constraint.
            $endgroup$
            – SabrinaChoice
            Dec 18 '18 at 2:34






          • 4




            $begingroup$
            I was trying to get you to come to the conclusion that you have insufficient information to determine the covariance.
            $endgroup$
            – parsiad
            Dec 18 '18 at 2:37








          • 1




            $begingroup$
            Put differently: the marginals do not characterize the joint distribution. Because $p_Xtimes p_Y$ has, by definition, the same marginals as $p_{XY}$.
            $endgroup$
            – Clement C.
            Dec 18 '18 at 3:08


















          $begingroup$
          This suggests that $mathbb{E}[X^2]$=1, but I can't see much beyond that, sorry!
          $endgroup$
          – SabrinaChoice
          Dec 18 '18 at 2:21






          $begingroup$
          This suggests that $mathbb{E}[X^2]$=1, but I can't see much beyond that, sorry!
          $endgroup$
          – SabrinaChoice
          Dec 18 '18 at 2:21






          1




          1




          $begingroup$
          But what does it suggest about the covariance? Is it uniquely determined by mean and variance?
          $endgroup$
          – parsiad
          Dec 18 '18 at 2:30




          $begingroup$
          But what does it suggest about the covariance? Is it uniquely determined by mean and variance?
          $endgroup$
          – parsiad
          Dec 18 '18 at 2:30












          $begingroup$
          I'm not sure because the bottom expression you wrote is the expression for the variance. So it states that the variance is equal to the variance. I'm not seeing how it can add any other constraint.
          $endgroup$
          – SabrinaChoice
          Dec 18 '18 at 2:34




          $begingroup$
          I'm not sure because the bottom expression you wrote is the expression for the variance. So it states that the variance is equal to the variance. I'm not seeing how it can add any other constraint.
          $endgroup$
          – SabrinaChoice
          Dec 18 '18 at 2:34




          4




          4




          $begingroup$
          I was trying to get you to come to the conclusion that you have insufficient information to determine the covariance.
          $endgroup$
          – parsiad
          Dec 18 '18 at 2:37






          $begingroup$
          I was trying to get you to come to the conclusion that you have insufficient information to determine the covariance.
          $endgroup$
          – parsiad
          Dec 18 '18 at 2:37






          1




          1




          $begingroup$
          Put differently: the marginals do not characterize the joint distribution. Because $p_Xtimes p_Y$ has, by definition, the same marginals as $p_{XY}$.
          $endgroup$
          – Clement C.
          Dec 18 '18 at 3:08






          $begingroup$
          Put differently: the marginals do not characterize the joint distribution. Because $p_Xtimes p_Y$ has, by definition, the same marginals as $p_{XY}$.
          $endgroup$
          – Clement C.
          Dec 18 '18 at 3:08




















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