Does a zero conditional expectation imply pairwise covariance is 0?












0












$begingroup$


Suppose in econometrics,
$$ y = beta_{0} + beta_{1}x_{1} + beta_{2}x_{2} + ... + beta_{k}x_{k} + u$$
In Gujarati's book, it says that the following equation (1)
$$ E[u | x_{1}, x_{2},..., x_{k}] = E[u] = 0 tag{1}$$
given that
$$ x_{1}, x_{2},..., x_{k}$$are independent from each other
implies (with bilinearity property of covariance)
$$ Cov(u, x_{1} + x_{2} + ... +x_{k}) = cov(u, x_{1}) = cov(u, x_{2}) = ... = cov(u,x_{k}) = 0$$
How can we derive this result?



Book's content:
enter image description here



enter image description here



Thank you very much.






statistics statistical-inference conditional-expectation conditional-probability covariance







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    0












    $begingroup$


    Suppose in econometrics,
    $$ y = beta_{0} + beta_{1}x_{1} + beta_{2}x_{2} + ... + beta_{k}x_{k} + u$$
    In Gujarati's book, it says that the following equation (1)
    $$ E[u | x_{1}, x_{2},..., x_{k}] = E[u] = 0 tag{1}$$
    given that
    $$ x_{1}, x_{2},..., x_{k}$$are independent from each other
    implies (with bilinearity property of covariance)
    $$ Cov(u, x_{1} + x_{2} + ... +x_{k}) = cov(u, x_{1}) = cov(u, x_{2}) = ... = cov(u,x_{k}) = 0$$
    How can we derive this result?



    Book's content:
    enter image description here



    enter image description here



    Thank you very much.






    statistics statistical-inference conditional-expectation conditional-probability covariance







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      Suppose in econometrics,
      $$ y = beta_{0} + beta_{1}x_{1} + beta_{2}x_{2} + ... + beta_{k}x_{k} + u$$
      In Gujarati's book, it says that the following equation (1)
      $$ E[u | x_{1}, x_{2},..., x_{k}] = E[u] = 0 tag{1}$$
      given that
      $$ x_{1}, x_{2},..., x_{k}$$are independent from each other
      implies (with bilinearity property of covariance)
      $$ Cov(u, x_{1} + x_{2} + ... +x_{k}) = cov(u, x_{1}) = cov(u, x_{2}) = ... = cov(u,x_{k}) = 0$$
      How can we derive this result?



      Book's content:
      enter image description here



      enter image description here



      Thank you very much.






      statistics statistical-inference conditional-expectation conditional-probability covariance







      share|cite|improve this question









      $endgroup$




      Suppose in econometrics,
      $$ y = beta_{0} + beta_{1}x_{1} + beta_{2}x_{2} + ... + beta_{k}x_{k} + u$$
      In Gujarati's book, it says that the following equation (1)
      $$ E[u | x_{1}, x_{2},..., x_{k}] = E[u] = 0 tag{1}$$
      given that
      $$ x_{1}, x_{2},..., x_{k}$$are independent from each other
      implies (with bilinearity property of covariance)
      $$ Cov(u, x_{1} + x_{2} + ... +x_{k}) = cov(u, x_{1}) = cov(u, x_{2}) = ... = cov(u,x_{k}) = 0$$
      How can we derive this result?



      Book's content:
      enter image description here



      enter image description here



      Thank you very much.






      statistics statistical-inference conditional-expectation conditional-probability covariance




      statistics statistical-inference conditional-expectation conditional-probability covariance






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Jan 8 at 22:31









      commentallez-vouscommentallez-vous

      2129




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          1 Answer
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          2












          $begingroup$

          Argue as follows: For the $i$th variable $x_i$ we have:
          $$E(ux_i)stackrel{(a)}=E[E(ux_imid x_1,x_2,ldots,x_k)]stackrel{(b)}=E[x_i E(umid x_1,x_2,ldots,x_k)]stackrel{(1)}=E[x_icdot 0]=0 stackrel{(1)}= E(u)E(x_i)$$
          In step (a) we use the tower property of conditional expectation; in (b) we use the fact that $x_i$ is measurable with respect to $sigma(x_1,x_2,ldots,x_k)$, so can be pulled out of the conditional expectation (i.e., we can take out what is known).



          Conclude that $E(ux_i)=E(u)E(x_i)$, which implies zero covariance between $u$ and $x_i$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you grand_chat, but I didn't get step (b) that x can be measured by $$sigma(x_{1},x_{2},...,x_{k}$$ so that we can pull it out? Could you please elaborate this a liitle bit? I appreciate it!
            $endgroup$
            – commentallez-vous
            Jan 8 at 22:51










          • $begingroup$
            @commentallez-vous Saying that "$x_i$ is measurable with respect to $sigma(x_1,ldots,x_k)$" means, roughly speaking, that $x_i$ is a function of $x_1,ldots, x_k$, and therefore we can 'take out what is known'. See my edit.
            $endgroup$
            – grand_chat
            Jan 8 at 22:54












          • $begingroup$
            Oh...so it is basically like for a given observation, this particular x is known as a constant, therefore the expected value of a constant is the constant itself, is this understanding right?
            $endgroup$
            – commentallez-vous
            Jan 8 at 22:57










          • $begingroup$
            @commentallez-vous Exactly. If you condition on $x_1,ldots, x_k$ then the value of $x_i$ is known, so $x_i$ can be treated as a constant and "factored out" of $E(ux_imid x_1,ldots,x_k)$.
            $endgroup$
            – grand_chat
            Jan 8 at 22:59










          • $begingroup$
            Thank you so much! This is a really nice proof and you saved my life! Much obliged!
            $endgroup$
            – commentallez-vous
            Jan 8 at 23:13











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          2












          $begingroup$

          Argue as follows: For the $i$th variable $x_i$ we have:
          $$E(ux_i)stackrel{(a)}=E[E(ux_imid x_1,x_2,ldots,x_k)]stackrel{(b)}=E[x_i E(umid x_1,x_2,ldots,x_k)]stackrel{(1)}=E[x_icdot 0]=0 stackrel{(1)}= E(u)E(x_i)$$
          In step (a) we use the tower property of conditional expectation; in (b) we use the fact that $x_i$ is measurable with respect to $sigma(x_1,x_2,ldots,x_k)$, so can be pulled out of the conditional expectation (i.e., we can take out what is known).



          Conclude that $E(ux_i)=E(u)E(x_i)$, which implies zero covariance between $u$ and $x_i$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you grand_chat, but I didn't get step (b) that x can be measured by $$sigma(x_{1},x_{2},...,x_{k}$$ so that we can pull it out? Could you please elaborate this a liitle bit? I appreciate it!
            $endgroup$
            – commentallez-vous
            Jan 8 at 22:51










          • $begingroup$
            @commentallez-vous Saying that "$x_i$ is measurable with respect to $sigma(x_1,ldots,x_k)$" means, roughly speaking, that $x_i$ is a function of $x_1,ldots, x_k$, and therefore we can 'take out what is known'. See my edit.
            $endgroup$
            – grand_chat
            Jan 8 at 22:54












          • $begingroup$
            Oh...so it is basically like for a given observation, this particular x is known as a constant, therefore the expected value of a constant is the constant itself, is this understanding right?
            $endgroup$
            – commentallez-vous
            Jan 8 at 22:57










          • $begingroup$
            @commentallez-vous Exactly. If you condition on $x_1,ldots, x_k$ then the value of $x_i$ is known, so $x_i$ can be treated as a constant and "factored out" of $E(ux_imid x_1,ldots,x_k)$.
            $endgroup$
            – grand_chat
            Jan 8 at 22:59










          • $begingroup$
            Thank you so much! This is a really nice proof and you saved my life! Much obliged!
            $endgroup$
            – commentallez-vous
            Jan 8 at 23:13
















          2












          $begingroup$

          Argue as follows: For the $i$th variable $x_i$ we have:
          $$E(ux_i)stackrel{(a)}=E[E(ux_imid x_1,x_2,ldots,x_k)]stackrel{(b)}=E[x_i E(umid x_1,x_2,ldots,x_k)]stackrel{(1)}=E[x_icdot 0]=0 stackrel{(1)}= E(u)E(x_i)$$
          In step (a) we use the tower property of conditional expectation; in (b) we use the fact that $x_i$ is measurable with respect to $sigma(x_1,x_2,ldots,x_k)$, so can be pulled out of the conditional expectation (i.e., we can take out what is known).



          Conclude that $E(ux_i)=E(u)E(x_i)$, which implies zero covariance between $u$ and $x_i$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you grand_chat, but I didn't get step (b) that x can be measured by $$sigma(x_{1},x_{2},...,x_{k}$$ so that we can pull it out? Could you please elaborate this a liitle bit? I appreciate it!
            $endgroup$
            – commentallez-vous
            Jan 8 at 22:51










          • $begingroup$
            @commentallez-vous Saying that "$x_i$ is measurable with respect to $sigma(x_1,ldots,x_k)$" means, roughly speaking, that $x_i$ is a function of $x_1,ldots, x_k$, and therefore we can 'take out what is known'. See my edit.
            $endgroup$
            – grand_chat
            Jan 8 at 22:54












          • $begingroup$
            Oh...so it is basically like for a given observation, this particular x is known as a constant, therefore the expected value of a constant is the constant itself, is this understanding right?
            $endgroup$
            – commentallez-vous
            Jan 8 at 22:57










          • $begingroup$
            @commentallez-vous Exactly. If you condition on $x_1,ldots, x_k$ then the value of $x_i$ is known, so $x_i$ can be treated as a constant and "factored out" of $E(ux_imid x_1,ldots,x_k)$.
            $endgroup$
            – grand_chat
            Jan 8 at 22:59










          • $begingroup$
            Thank you so much! This is a really nice proof and you saved my life! Much obliged!
            $endgroup$
            – commentallez-vous
            Jan 8 at 23:13














          2












          2








          2





          $begingroup$

          Argue as follows: For the $i$th variable $x_i$ we have:
          $$E(ux_i)stackrel{(a)}=E[E(ux_imid x_1,x_2,ldots,x_k)]stackrel{(b)}=E[x_i E(umid x_1,x_2,ldots,x_k)]stackrel{(1)}=E[x_icdot 0]=0 stackrel{(1)}= E(u)E(x_i)$$
          In step (a) we use the tower property of conditional expectation; in (b) we use the fact that $x_i$ is measurable with respect to $sigma(x_1,x_2,ldots,x_k)$, so can be pulled out of the conditional expectation (i.e., we can take out what is known).



          Conclude that $E(ux_i)=E(u)E(x_i)$, which implies zero covariance between $u$ and $x_i$.






          share|cite|improve this answer











          $endgroup$



          Argue as follows: For the $i$th variable $x_i$ we have:
          $$E(ux_i)stackrel{(a)}=E[E(ux_imid x_1,x_2,ldots,x_k)]stackrel{(b)}=E[x_i E(umid x_1,x_2,ldots,x_k)]stackrel{(1)}=E[x_icdot 0]=0 stackrel{(1)}= E(u)E(x_i)$$
          In step (a) we use the tower property of conditional expectation; in (b) we use the fact that $x_i$ is measurable with respect to $sigma(x_1,x_2,ldots,x_k)$, so can be pulled out of the conditional expectation (i.e., we can take out what is known).



          Conclude that $E(ux_i)=E(u)E(x_i)$, which implies zero covariance between $u$ and $x_i$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 22:55

























          answered Jan 8 at 22:41









          grand_chatgrand_chat

          20.3k11326




          20.3k11326












          • $begingroup$
            Thank you grand_chat, but I didn't get step (b) that x can be measured by $$sigma(x_{1},x_{2},...,x_{k}$$ so that we can pull it out? Could you please elaborate this a liitle bit? I appreciate it!
            $endgroup$
            – commentallez-vous
            Jan 8 at 22:51










          • $begingroup$
            @commentallez-vous Saying that "$x_i$ is measurable with respect to $sigma(x_1,ldots,x_k)$" means, roughly speaking, that $x_i$ is a function of $x_1,ldots, x_k$, and therefore we can 'take out what is known'. See my edit.
            $endgroup$
            – grand_chat
            Jan 8 at 22:54












          • $begingroup$
            Oh...so it is basically like for a given observation, this particular x is known as a constant, therefore the expected value of a constant is the constant itself, is this understanding right?
            $endgroup$
            – commentallez-vous
            Jan 8 at 22:57










          • $begingroup$
            @commentallez-vous Exactly. If you condition on $x_1,ldots, x_k$ then the value of $x_i$ is known, so $x_i$ can be treated as a constant and "factored out" of $E(ux_imid x_1,ldots,x_k)$.
            $endgroup$
            – grand_chat
            Jan 8 at 22:59










          • $begingroup$
            Thank you so much! This is a really nice proof and you saved my life! Much obliged!
            $endgroup$
            – commentallez-vous
            Jan 8 at 23:13


















          • $begingroup$
            Thank you grand_chat, but I didn't get step (b) that x can be measured by $$sigma(x_{1},x_{2},...,x_{k}$$ so that we can pull it out? Could you please elaborate this a liitle bit? I appreciate it!
            $endgroup$
            – commentallez-vous
            Jan 8 at 22:51










          • $begingroup$
            @commentallez-vous Saying that "$x_i$ is measurable with respect to $sigma(x_1,ldots,x_k)$" means, roughly speaking, that $x_i$ is a function of $x_1,ldots, x_k$, and therefore we can 'take out what is known'. See my edit.
            $endgroup$
            – grand_chat
            Jan 8 at 22:54












          • $begingroup$
            Oh...so it is basically like for a given observation, this particular x is known as a constant, therefore the expected value of a constant is the constant itself, is this understanding right?
            $endgroup$
            – commentallez-vous
            Jan 8 at 22:57










          • $begingroup$
            @commentallez-vous Exactly. If you condition on $x_1,ldots, x_k$ then the value of $x_i$ is known, so $x_i$ can be treated as a constant and "factored out" of $E(ux_imid x_1,ldots,x_k)$.
            $endgroup$
            – grand_chat
            Jan 8 at 22:59










          • $begingroup$
            Thank you so much! This is a really nice proof and you saved my life! Much obliged!
            $endgroup$
            – commentallez-vous
            Jan 8 at 23:13
















          $begingroup$
          Thank you grand_chat, but I didn't get step (b) that x can be measured by $$sigma(x_{1},x_{2},...,x_{k}$$ so that we can pull it out? Could you please elaborate this a liitle bit? I appreciate it!
          $endgroup$
          – commentallez-vous
          Jan 8 at 22:51




          $begingroup$
          Thank you grand_chat, but I didn't get step (b) that x can be measured by $$sigma(x_{1},x_{2},...,x_{k}$$ so that we can pull it out? Could you please elaborate this a liitle bit? I appreciate it!
          $endgroup$
          – commentallez-vous
          Jan 8 at 22:51












          $begingroup$
          @commentallez-vous Saying that "$x_i$ is measurable with respect to $sigma(x_1,ldots,x_k)$" means, roughly speaking, that $x_i$ is a function of $x_1,ldots, x_k$, and therefore we can 'take out what is known'. See my edit.
          $endgroup$
          – grand_chat
          Jan 8 at 22:54






          $begingroup$
          @commentallez-vous Saying that "$x_i$ is measurable with respect to $sigma(x_1,ldots,x_k)$" means, roughly speaking, that $x_i$ is a function of $x_1,ldots, x_k$, and therefore we can 'take out what is known'. See my edit.
          $endgroup$
          – grand_chat
          Jan 8 at 22:54














          $begingroup$
          Oh...so it is basically like for a given observation, this particular x is known as a constant, therefore the expected value of a constant is the constant itself, is this understanding right?
          $endgroup$
          – commentallez-vous
          Jan 8 at 22:57




          $begingroup$
          Oh...so it is basically like for a given observation, this particular x is known as a constant, therefore the expected value of a constant is the constant itself, is this understanding right?
          $endgroup$
          – commentallez-vous
          Jan 8 at 22:57












          $begingroup$
          @commentallez-vous Exactly. If you condition on $x_1,ldots, x_k$ then the value of $x_i$ is known, so $x_i$ can be treated as a constant and "factored out" of $E(ux_imid x_1,ldots,x_k)$.
          $endgroup$
          – grand_chat
          Jan 8 at 22:59




          $begingroup$
          @commentallez-vous Exactly. If you condition on $x_1,ldots, x_k$ then the value of $x_i$ is known, so $x_i$ can be treated as a constant and "factored out" of $E(ux_imid x_1,ldots,x_k)$.
          $endgroup$
          – grand_chat
          Jan 8 at 22:59












          $begingroup$
          Thank you so much! This is a really nice proof and you saved my life! Much obliged!
          $endgroup$
          – commentallez-vous
          Jan 8 at 23:13




          $begingroup$
          Thank you so much! This is a really nice proof and you saved my life! Much obliged!
          $endgroup$
          – commentallez-vous
          Jan 8 at 23:13


















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