Conditional expectation $E(X_1 mid overline{X}_n)$ if $X_1,dots,X_n$ are i.i.d. Am I correct?












2












$begingroup$


Contitional expectation $E(X_1 mid overline{X}_n)$ if $X_1,dots,X_n$ are i.i.d.



Since $X_1,dots,X_n$ are i.i.d, then $E(X_1 mid overline{X}_n) = E(X_1)=overline{X}_n)$



Am I correct in thinking this?
Thanks!



(Just to overclearify $overline{X}_n$ is the sample mean)
Question is from Van Der Vaart: Asymptotic Statistics.






asymptotics statistical-inference







share|cite|improve this question











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  • 1




    $begingroup$
    The answer from @SiongThyeGoh is right. Nevertheless, is not true that $bar X_n=E(X_1)$ (and then is also not true that $E(X_1|bar X_n)=E(X_1)$ ). Check that $E(X_1|bar X_n)$ and $bar X_n$ are random variables (the same r.v., indeed), but $E(X_1)$ is a constant.
    $endgroup$
    – Alejandro Nasif Salum
    Dec 8 '18 at 14:59


















2












$begingroup$


Contitional expectation $E(X_1 mid overline{X}_n)$ if $X_1,dots,X_n$ are i.i.d.



Since $X_1,dots,X_n$ are i.i.d, then $E(X_1 mid overline{X}_n) = E(X_1)=overline{X}_n)$



Am I correct in thinking this?
Thanks!



(Just to overclearify $overline{X}_n$ is the sample mean)
Question is from Van Der Vaart: Asymptotic Statistics.






asymptotics statistical-inference







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The answer from @SiongThyeGoh is right. Nevertheless, is not true that $bar X_n=E(X_1)$ (and then is also not true that $E(X_1|bar X_n)=E(X_1)$ ). Check that $E(X_1|bar X_n)$ and $bar X_n$ are random variables (the same r.v., indeed), but $E(X_1)$ is a constant.
    $endgroup$
    – Alejandro Nasif Salum
    Dec 8 '18 at 14:59
















2












2








2


1



$begingroup$


Contitional expectation $E(X_1 mid overline{X}_n)$ if $X_1,dots,X_n$ are i.i.d.



Since $X_1,dots,X_n$ are i.i.d, then $E(X_1 mid overline{X}_n) = E(X_1)=overline{X}_n)$



Am I correct in thinking this?
Thanks!



(Just to overclearify $overline{X}_n$ is the sample mean)
Question is from Van Der Vaart: Asymptotic Statistics.






asymptotics statistical-inference







share|cite|improve this question











$endgroup$




Contitional expectation $E(X_1 mid overline{X}_n)$ if $X_1,dots,X_n$ are i.i.d.



Since $X_1,dots,X_n$ are i.i.d, then $E(X_1 mid overline{X}_n) = E(X_1)=overline{X}_n)$



Am I correct in thinking this?
Thanks!



(Just to overclearify $overline{X}_n$ is the sample mean)
Question is from Van Der Vaart: Asymptotic Statistics.






asymptotics statistical-inference




asymptotics statistical-inference






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 14:31









Bernard

119k740113




119k740113










asked Dec 8 '18 at 14:14









L200123L200123

855




855








  • 1




    $begingroup$
    The answer from @SiongThyeGoh is right. Nevertheless, is not true that $bar X_n=E(X_1)$ (and then is also not true that $E(X_1|bar X_n)=E(X_1)$ ). Check that $E(X_1|bar X_n)$ and $bar X_n$ are random variables (the same r.v., indeed), but $E(X_1)$ is a constant.
    $endgroup$
    – Alejandro Nasif Salum
    Dec 8 '18 at 14:59
















  • 1




    $begingroup$
    The answer from @SiongThyeGoh is right. Nevertheless, is not true that $bar X_n=E(X_1)$ (and then is also not true that $E(X_1|bar X_n)=E(X_1)$ ). Check that $E(X_1|bar X_n)$ and $bar X_n$ are random variables (the same r.v., indeed), but $E(X_1)$ is a constant.
    $endgroup$
    – Alejandro Nasif Salum
    Dec 8 '18 at 14:59










1




1




$begingroup$
The answer from @SiongThyeGoh is right. Nevertheless, is not true that $bar X_n=E(X_1)$ (and then is also not true that $E(X_1|bar X_n)=E(X_1)$ ). Check that $E(X_1|bar X_n)$ and $bar X_n$ are random variables (the same r.v., indeed), but $E(X_1)$ is a constant.
$endgroup$
– Alejandro Nasif Salum
Dec 8 '18 at 14:59






$begingroup$
The answer from @SiongThyeGoh is right. Nevertheless, is not true that $bar X_n=E(X_1)$ (and then is also not true that $E(X_1|bar X_n)=E(X_1)$ ). Check that $E(X_1|bar X_n)$ and $bar X_n$ are random variables (the same r.v., indeed), but $E(X_1)$ is a constant.
$endgroup$
– Alejandro Nasif Salum
Dec 8 '18 at 14:59












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$begingroup$

We have $$E(X_i |bar{X}_n)= E(X_j |bar{X}_n)$$



Summing them up $$sum_{i=1}^n E(X_i|bar{X}_n)=E(sum_{i=1}^n X_i|bar{X}_n)=E(nbar{X}_n|bar{X}_n)=nbar{X}_n$$



$$nE(X_1|bar{X}_n)=nbar{X}_n$$
Hence,
$$E(X_1|bar{X}_n)=bar{X}_n$$






share|cite|improve this answer









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    $begingroup$

    We have $$E(X_i |bar{X}_n)= E(X_j |bar{X}_n)$$



    Summing them up $$sum_{i=1}^n E(X_i|bar{X}_n)=E(sum_{i=1}^n X_i|bar{X}_n)=E(nbar{X}_n|bar{X}_n)=nbar{X}_n$$



    $$nE(X_1|bar{X}_n)=nbar{X}_n$$
    Hence,
    $$E(X_1|bar{X}_n)=bar{X}_n$$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      We have $$E(X_i |bar{X}_n)= E(X_j |bar{X}_n)$$



      Summing them up $$sum_{i=1}^n E(X_i|bar{X}_n)=E(sum_{i=1}^n X_i|bar{X}_n)=E(nbar{X}_n|bar{X}_n)=nbar{X}_n$$



      $$nE(X_1|bar{X}_n)=nbar{X}_n$$
      Hence,
      $$E(X_1|bar{X}_n)=bar{X}_n$$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        We have $$E(X_i |bar{X}_n)= E(X_j |bar{X}_n)$$



        Summing them up $$sum_{i=1}^n E(X_i|bar{X}_n)=E(sum_{i=1}^n X_i|bar{X}_n)=E(nbar{X}_n|bar{X}_n)=nbar{X}_n$$



        $$nE(X_1|bar{X}_n)=nbar{X}_n$$
        Hence,
        $$E(X_1|bar{X}_n)=bar{X}_n$$






        share|cite|improve this answer









        $endgroup$



        We have $$E(X_i |bar{X}_n)= E(X_j |bar{X}_n)$$



        Summing them up $$sum_{i=1}^n E(X_i|bar{X}_n)=E(sum_{i=1}^n X_i|bar{X}_n)=E(nbar{X}_n|bar{X}_n)=nbar{X}_n$$



        $$nE(X_1|bar{X}_n)=nbar{X}_n$$
        Hence,
        $$E(X_1|bar{X}_n)=bar{X}_n$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 14:18









        Siong Thye GohSiong Thye Goh

        101k1466117




        101k1466117






























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