Proving $mathbb{E}(S_{tau}|tau)=tau mathbb{E}(xi_1)$ and $mathbb{E}S_{tau}=mathbb{E}tau mathbb{E}xi_1$












2












$begingroup$


Let $xi_1, x_2,...,xi_n$ and $tau$ be independent random variables in $(Omega, mathcal{F}, mathbb{P})$. $xi_1, x_2,...,xi_n$ are uniformly distributed and $tau={0,1,2,...,n}$. Let $S_{tau}=xi_1+xi_2+...+xi_{tau}$ be a random sum of random variables.



I need to show that



1) $mathbb{E}(S_{tau}|tau)=tau mathbb{E}(xi_1)$ and $mathbb{E}S_{tau}=mathbb{E}tau mathbb{E}xi_1$.



2) $mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1$ and $mathbb{D}S_{tau}=mathbb{E}tau mathbb{D}xi_1+mathbb{D}tau (mathbb{E}xi_1)^2$.



So for the first part I think I can use this
$mathbb{E}S_{tau}=mathbb{E}mathbb{E}(S_{tau}|tau)=sum_{ngeq1}mathbb{E}(xi_1+xi_2+...+xi_{tau}|tau)mathbb{P}(tau=n)=sum_{ngeq 1}nmathbb{E}xi_imathbb{P}(tau=n)=mathbb{E}xi_i sum_{ngeq1}mathbb{P}(tau=n)=mathbb{E}xi_1mathbb{E}tau .$



So am I right?



For the second part I am really not sure what to do.






probability probability-theory random-variables conditional-expectation







share|cite|improve this question









$endgroup$












  • $begingroup$
    What do you mean by $mathbb{D}(X)$?
    $endgroup$
    – Lundborg
    Jan 3 at 10:55
















2












$begingroup$


Let $xi_1, x_2,...,xi_n$ and $tau$ be independent random variables in $(Omega, mathcal{F}, mathbb{P})$. $xi_1, x_2,...,xi_n$ are uniformly distributed and $tau={0,1,2,...,n}$. Let $S_{tau}=xi_1+xi_2+...+xi_{tau}$ be a random sum of random variables.



I need to show that



1) $mathbb{E}(S_{tau}|tau)=tau mathbb{E}(xi_1)$ and $mathbb{E}S_{tau}=mathbb{E}tau mathbb{E}xi_1$.



2) $mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1$ and $mathbb{D}S_{tau}=mathbb{E}tau mathbb{D}xi_1+mathbb{D}tau (mathbb{E}xi_1)^2$.



So for the first part I think I can use this
$mathbb{E}S_{tau}=mathbb{E}mathbb{E}(S_{tau}|tau)=sum_{ngeq1}mathbb{E}(xi_1+xi_2+...+xi_{tau}|tau)mathbb{P}(tau=n)=sum_{ngeq 1}nmathbb{E}xi_imathbb{P}(tau=n)=mathbb{E}xi_i sum_{ngeq1}mathbb{P}(tau=n)=mathbb{E}xi_1mathbb{E}tau .$



So am I right?



For the second part I am really not sure what to do.






probability probability-theory random-variables conditional-expectation







share|cite|improve this question









$endgroup$












  • $begingroup$
    What do you mean by $mathbb{D}(X)$?
    $endgroup$
    – Lundborg
    Jan 3 at 10:55














2












2








2





$begingroup$


Let $xi_1, x_2,...,xi_n$ and $tau$ be independent random variables in $(Omega, mathcal{F}, mathbb{P})$. $xi_1, x_2,...,xi_n$ are uniformly distributed and $tau={0,1,2,...,n}$. Let $S_{tau}=xi_1+xi_2+...+xi_{tau}$ be a random sum of random variables.



I need to show that



1) $mathbb{E}(S_{tau}|tau)=tau mathbb{E}(xi_1)$ and $mathbb{E}S_{tau}=mathbb{E}tau mathbb{E}xi_1$.



2) $mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1$ and $mathbb{D}S_{tau}=mathbb{E}tau mathbb{D}xi_1+mathbb{D}tau (mathbb{E}xi_1)^2$.



So for the first part I think I can use this
$mathbb{E}S_{tau}=mathbb{E}mathbb{E}(S_{tau}|tau)=sum_{ngeq1}mathbb{E}(xi_1+xi_2+...+xi_{tau}|tau)mathbb{P}(tau=n)=sum_{ngeq 1}nmathbb{E}xi_imathbb{P}(tau=n)=mathbb{E}xi_i sum_{ngeq1}mathbb{P}(tau=n)=mathbb{E}xi_1mathbb{E}tau .$



So am I right?



For the second part I am really not sure what to do.






probability probability-theory random-variables conditional-expectation







share|cite|improve this question









$endgroup$




Let $xi_1, x_2,...,xi_n$ and $tau$ be independent random variables in $(Omega, mathcal{F}, mathbb{P})$. $xi_1, x_2,...,xi_n$ are uniformly distributed and $tau={0,1,2,...,n}$. Let $S_{tau}=xi_1+xi_2+...+xi_{tau}$ be a random sum of random variables.



I need to show that



1) $mathbb{E}(S_{tau}|tau)=tau mathbb{E}(xi_1)$ and $mathbb{E}S_{tau}=mathbb{E}tau mathbb{E}xi_1$.



2) $mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1$ and $mathbb{D}S_{tau}=mathbb{E}tau mathbb{D}xi_1+mathbb{D}tau (mathbb{E}xi_1)^2$.



So for the first part I think I can use this
$mathbb{E}S_{tau}=mathbb{E}mathbb{E}(S_{tau}|tau)=sum_{ngeq1}mathbb{E}(xi_1+xi_2+...+xi_{tau}|tau)mathbb{P}(tau=n)=sum_{ngeq 1}nmathbb{E}xi_imathbb{P}(tau=n)=mathbb{E}xi_i sum_{ngeq1}mathbb{P}(tau=n)=mathbb{E}xi_1mathbb{E}tau .$



So am I right?



For the second part I am really not sure what to do.






probability probability-theory random-variables conditional-expectation




probability probability-theory random-variables conditional-expectation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 3 at 10:40









AtstovasAtstovas

1159




1159












  • $begingroup$
    What do you mean by $mathbb{D}(X)$?
    $endgroup$
    – Lundborg
    Jan 3 at 10:55


















  • $begingroup$
    What do you mean by $mathbb{D}(X)$?
    $endgroup$
    – Lundborg
    Jan 3 at 10:55
















$begingroup$
What do you mean by $mathbb{D}(X)$?
$endgroup$
– Lundborg
Jan 3 at 10:55




$begingroup$
What do you mean by $mathbb{D}(X)$?
$endgroup$
– Lundborg
Jan 3 at 10:55










1 Answer
1






active

oldest

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2












$begingroup$

For the first question, note that



$$
S_tau = sum_{i=1}^n 1_{(tau leq i)} xi_i
$$



therefore



$$
mathbb{E}(S_tau | tau) = mathbb{E}left(sum_{i=1}^n 1_{(tau leq i)} xi_i Big| tau right) = sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right)
$$



Note now that $1_{(tau leq i)}$ is measurable when given $tau$ and that $xi_i$ are all independent from $tau$ so



$$
mathbb{E}left( 1_{(tau leq i)} xi_i | tau right) = 1_{(tau leq i)} mathbb{E}(xi_i)
$$



In total we get (since $xi_i$ are identically distributed)



$$
mathbb{E}(S_tau | tau) =sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right) = sum_{i=1}^n 1_{(tau leq i)} mathbb{E}(xi_i) = mathbb{E}(xi_1)sum_{i=1}^n 1_{(tau leq i)} = mathbb{E}(xi_1)tau
$$



Taking expectations on both sides yields the second part of the first question.



Assuming by $mathbb{D}$ you mean the variance, you can perform the exact same calculations as above to get



$$
mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1
$$



Then using the law of total variance, we get



$$
mathbb{D}(S_tau) = mathbb{D}(mathbb{E}(S_tau | tau)) + mathbb{E}(mathbb{D}(S_tau | tau)) = mathbb{D}(tau) (mathbb{E}(xi_1))^2 + mathbb{E}(tau) mathbb{D}(xi_1)
$$



as desired.






share|cite|improve this answer









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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    For the first question, note that



    $$
    S_tau = sum_{i=1}^n 1_{(tau leq i)} xi_i
    $$



    therefore



    $$
    mathbb{E}(S_tau | tau) = mathbb{E}left(sum_{i=1}^n 1_{(tau leq i)} xi_i Big| tau right) = sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right)
    $$



    Note now that $1_{(tau leq i)}$ is measurable when given $tau$ and that $xi_i$ are all independent from $tau$ so



    $$
    mathbb{E}left( 1_{(tau leq i)} xi_i | tau right) = 1_{(tau leq i)} mathbb{E}(xi_i)
    $$



    In total we get (since $xi_i$ are identically distributed)



    $$
    mathbb{E}(S_tau | tau) =sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right) = sum_{i=1}^n 1_{(tau leq i)} mathbb{E}(xi_i) = mathbb{E}(xi_1)sum_{i=1}^n 1_{(tau leq i)} = mathbb{E}(xi_1)tau
    $$



    Taking expectations on both sides yields the second part of the first question.



    Assuming by $mathbb{D}$ you mean the variance, you can perform the exact same calculations as above to get



    $$
    mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1
    $$



    Then using the law of total variance, we get



    $$
    mathbb{D}(S_tau) = mathbb{D}(mathbb{E}(S_tau | tau)) + mathbb{E}(mathbb{D}(S_tau | tau)) = mathbb{D}(tau) (mathbb{E}(xi_1))^2 + mathbb{E}(tau) mathbb{D}(xi_1)
    $$



    as desired.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      For the first question, note that



      $$
      S_tau = sum_{i=1}^n 1_{(tau leq i)} xi_i
      $$



      therefore



      $$
      mathbb{E}(S_tau | tau) = mathbb{E}left(sum_{i=1}^n 1_{(tau leq i)} xi_i Big| tau right) = sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right)
      $$



      Note now that $1_{(tau leq i)}$ is measurable when given $tau$ and that $xi_i$ are all independent from $tau$ so



      $$
      mathbb{E}left( 1_{(tau leq i)} xi_i | tau right) = 1_{(tau leq i)} mathbb{E}(xi_i)
      $$



      In total we get (since $xi_i$ are identically distributed)



      $$
      mathbb{E}(S_tau | tau) =sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right) = sum_{i=1}^n 1_{(tau leq i)} mathbb{E}(xi_i) = mathbb{E}(xi_1)sum_{i=1}^n 1_{(tau leq i)} = mathbb{E}(xi_1)tau
      $$



      Taking expectations on both sides yields the second part of the first question.



      Assuming by $mathbb{D}$ you mean the variance, you can perform the exact same calculations as above to get



      $$
      mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1
      $$



      Then using the law of total variance, we get



      $$
      mathbb{D}(S_tau) = mathbb{D}(mathbb{E}(S_tau | tau)) + mathbb{E}(mathbb{D}(S_tau | tau)) = mathbb{D}(tau) (mathbb{E}(xi_1))^2 + mathbb{E}(tau) mathbb{D}(xi_1)
      $$



      as desired.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        For the first question, note that



        $$
        S_tau = sum_{i=1}^n 1_{(tau leq i)} xi_i
        $$



        therefore



        $$
        mathbb{E}(S_tau | tau) = mathbb{E}left(sum_{i=1}^n 1_{(tau leq i)} xi_i Big| tau right) = sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right)
        $$



        Note now that $1_{(tau leq i)}$ is measurable when given $tau$ and that $xi_i$ are all independent from $tau$ so



        $$
        mathbb{E}left( 1_{(tau leq i)} xi_i | tau right) = 1_{(tau leq i)} mathbb{E}(xi_i)
        $$



        In total we get (since $xi_i$ are identically distributed)



        $$
        mathbb{E}(S_tau | tau) =sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right) = sum_{i=1}^n 1_{(tau leq i)} mathbb{E}(xi_i) = mathbb{E}(xi_1)sum_{i=1}^n 1_{(tau leq i)} = mathbb{E}(xi_1)tau
        $$



        Taking expectations on both sides yields the second part of the first question.



        Assuming by $mathbb{D}$ you mean the variance, you can perform the exact same calculations as above to get



        $$
        mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1
        $$



        Then using the law of total variance, we get



        $$
        mathbb{D}(S_tau) = mathbb{D}(mathbb{E}(S_tau | tau)) + mathbb{E}(mathbb{D}(S_tau | tau)) = mathbb{D}(tau) (mathbb{E}(xi_1))^2 + mathbb{E}(tau) mathbb{D}(xi_1)
        $$



        as desired.






        share|cite|improve this answer









        $endgroup$



        For the first question, note that



        $$
        S_tau = sum_{i=1}^n 1_{(tau leq i)} xi_i
        $$



        therefore



        $$
        mathbb{E}(S_tau | tau) = mathbb{E}left(sum_{i=1}^n 1_{(tau leq i)} xi_i Big| tau right) = sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right)
        $$



        Note now that $1_{(tau leq i)}$ is measurable when given $tau$ and that $xi_i$ are all independent from $tau$ so



        $$
        mathbb{E}left( 1_{(tau leq i)} xi_i | tau right) = 1_{(tau leq i)} mathbb{E}(xi_i)
        $$



        In total we get (since $xi_i$ are identically distributed)



        $$
        mathbb{E}(S_tau | tau) =sum_{i=1}^n mathbb{E}left( 1_{(tau leq i)} xi_i big| tau right) = sum_{i=1}^n 1_{(tau leq i)} mathbb{E}(xi_i) = mathbb{E}(xi_1)sum_{i=1}^n 1_{(tau leq i)} = mathbb{E}(xi_1)tau
        $$



        Taking expectations on both sides yields the second part of the first question.



        Assuming by $mathbb{D}$ you mean the variance, you can perform the exact same calculations as above to get



        $$
        mathbb{D}(S_{tau}|tau)= tau mathbb{D}xi_1
        $$



        Then using the law of total variance, we get



        $$
        mathbb{D}(S_tau) = mathbb{D}(mathbb{E}(S_tau | tau)) + mathbb{E}(mathbb{D}(S_tau | tau)) = mathbb{D}(tau) (mathbb{E}(xi_1))^2 + mathbb{E}(tau) mathbb{D}(xi_1)
        $$



        as desired.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 11:07









        LundborgLundborg

        857617




        857617






























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