how to prove $mathbb{E}[UVmid mathcal{B} ] = Umathbb{E}[Vmid mathcal{B} ]$ if $U$ is $mathcal{B}$-measurable?












0












$begingroup$


How can we prove
$$mathbb{E}[UVmid mathcal{B} ] = Umathbb{E}[Vmid mathcal{B} ]$$



given that both r.v's are positive and $U$ being $mathcal{B}$-measurable?



I use this property pretty often but I don't know how to actually prove it. Either we have to show that $forall B in mathcal{B}$



$$int_{B}UVdmathbb{P} =int_{B}Umathbb{E}[V mid mathcal{B} ] dmathbb{P} $$



or $forall Z text{ that is } mathcal{B}$-measurable and positive



$$mathbb{E}[UVZ] = mathbb{E}[Umathbb{E}[Vmid mathcal{B} ]Z ]$$



but how?






probability-theory conditional-expectation







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  • 1




    $begingroup$
    Please replace every $backslash$ by mid .
    $endgroup$
    – Did
    Jan 8 at 16:52


















0












$begingroup$


How can we prove
$$mathbb{E}[UVmid mathcal{B} ] = Umathbb{E}[Vmid mathcal{B} ]$$



given that both r.v's are positive and $U$ being $mathcal{B}$-measurable?



I use this property pretty often but I don't know how to actually prove it. Either we have to show that $forall B in mathcal{B}$



$$int_{B}UVdmathbb{P} =int_{B}Umathbb{E}[V mid mathcal{B} ] dmathbb{P} $$



or $forall Z text{ that is } mathcal{B}$-measurable and positive



$$mathbb{E}[UVZ] = mathbb{E}[Umathbb{E}[Vmid mathcal{B} ]Z ]$$



but how?






probability-theory conditional-expectation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Please replace every $backslash$ by mid .
    $endgroup$
    – Did
    Jan 8 at 16:52
















0












0








0





$begingroup$


How can we prove
$$mathbb{E}[UVmid mathcal{B} ] = Umathbb{E}[Vmid mathcal{B} ]$$



given that both r.v's are positive and $U$ being $mathcal{B}$-measurable?



I use this property pretty often but I don't know how to actually prove it. Either we have to show that $forall B in mathcal{B}$



$$int_{B}UVdmathbb{P} =int_{B}Umathbb{E}[V mid mathcal{B} ] dmathbb{P} $$



or $forall Z text{ that is } mathcal{B}$-measurable and positive



$$mathbb{E}[UVZ] = mathbb{E}[Umathbb{E}[Vmid mathcal{B} ]Z ]$$



but how?






probability-theory conditional-expectation







share|cite|improve this question











$endgroup$




How can we prove
$$mathbb{E}[UVmid mathcal{B} ] = Umathbb{E}[Vmid mathcal{B} ]$$



given that both r.v's are positive and $U$ being $mathcal{B}$-measurable?



I use this property pretty often but I don't know how to actually prove it. Either we have to show that $forall B in mathcal{B}$



$$int_{B}UVdmathbb{P} =int_{B}Umathbb{E}[V mid mathcal{B} ] dmathbb{P} $$



or $forall Z text{ that is } mathcal{B}$-measurable and positive



$$mathbb{E}[UVZ] = mathbb{E}[Umathbb{E}[Vmid mathcal{B} ]Z ]$$



but how?






probability-theory conditional-expectation




probability-theory conditional-expectation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 21:17









Shashi

7,1731528




7,1731528










asked Jan 8 at 15:54









rapidracimrapidracim

1,5401319




1,5401319








  • 1




    $begingroup$
    Please replace every $backslash$ by mid .
    $endgroup$
    – Did
    Jan 8 at 16:52
















  • 1




    $begingroup$
    Please replace every $backslash$ by mid .
    $endgroup$
    – Did
    Jan 8 at 16:52










1




1




$begingroup$
Please replace every $backslash$ by mid .
$endgroup$
– Did
Jan 8 at 16:52






$begingroup$
Please replace every $backslash$ by mid .
$endgroup$
– Did
Jan 8 at 16:52












1 Answer
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$begingroup$

Notice that there exists a monotone sequence of simple random variables $U_n$ which are $mathcal B$ - measurable converging point wise to $U$. The random variables $U_n$ can be written as
$$U_n=sum_{k=1}^{N_n}mathbf{1}_{A_{k,n}}$$
where $A_{k, n}inmathcal B $.



Now we have using MCT
begin{align}
int_B UV, dmathbb P&=int_{ B}lim_{ntoinfty} sum_{k=1}^{N_n}mathbf{1}_{A_{k, n}} V, dmathbb P \
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} V, dmathbb P \
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} mathbb E(Vmid mathcal B), dmathbb P\
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{ B} mathbf{1}_{A_{k,n}} mathbb E(Vmid mathcal B), dmathbb P
end{align}
And this implies indeed using MCT again that...



You need to figure out how to use MCT yourself though.






share|cite|improve this answer











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    $begingroup$

    Notice that there exists a monotone sequence of simple random variables $U_n$ which are $mathcal B$ - measurable converging point wise to $U$. The random variables $U_n$ can be written as
    $$U_n=sum_{k=1}^{N_n}mathbf{1}_{A_{k,n}}$$
    where $A_{k, n}inmathcal B $.



    Now we have using MCT
    begin{align}
    int_B UV, dmathbb P&=int_{ B}lim_{ntoinfty} sum_{k=1}^{N_n}mathbf{1}_{A_{k, n}} V, dmathbb P \
    &=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} V, dmathbb P \
    &=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} mathbb E(Vmid mathcal B), dmathbb P\
    &=lim_{ntoinfty} sum_{k=1}^{N_n}int_{ B} mathbf{1}_{A_{k,n}} mathbb E(Vmid mathcal B), dmathbb P
    end{align}
    And this implies indeed using MCT again that...



    You need to figure out how to use MCT yourself though.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Notice that there exists a monotone sequence of simple random variables $U_n$ which are $mathcal B$ - measurable converging point wise to $U$. The random variables $U_n$ can be written as
      $$U_n=sum_{k=1}^{N_n}mathbf{1}_{A_{k,n}}$$
      where $A_{k, n}inmathcal B $.



      Now we have using MCT
      begin{align}
      int_B UV, dmathbb P&=int_{ B}lim_{ntoinfty} sum_{k=1}^{N_n}mathbf{1}_{A_{k, n}} V, dmathbb P \
      &=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} V, dmathbb P \
      &=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} mathbb E(Vmid mathcal B), dmathbb P\
      &=lim_{ntoinfty} sum_{k=1}^{N_n}int_{ B} mathbf{1}_{A_{k,n}} mathbb E(Vmid mathcal B), dmathbb P
      end{align}
      And this implies indeed using MCT again that...



      You need to figure out how to use MCT yourself though.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Notice that there exists a monotone sequence of simple random variables $U_n$ which are $mathcal B$ - measurable converging point wise to $U$. The random variables $U_n$ can be written as
        $$U_n=sum_{k=1}^{N_n}mathbf{1}_{A_{k,n}}$$
        where $A_{k, n}inmathcal B $.



        Now we have using MCT
        begin{align}
        int_B UV, dmathbb P&=int_{ B}lim_{ntoinfty} sum_{k=1}^{N_n}mathbf{1}_{A_{k, n}} V, dmathbb P \
        &=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} V, dmathbb P \
        &=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} mathbb E(Vmid mathcal B), dmathbb P\
        &=lim_{ntoinfty} sum_{k=1}^{N_n}int_{ B} mathbf{1}_{A_{k,n}} mathbb E(Vmid mathcal B), dmathbb P
        end{align}
        And this implies indeed using MCT again that...



        You need to figure out how to use MCT yourself though.






        share|cite|improve this answer











        $endgroup$



        Notice that there exists a monotone sequence of simple random variables $U_n$ which are $mathcal B$ - measurable converging point wise to $U$. The random variables $U_n$ can be written as
        $$U_n=sum_{k=1}^{N_n}mathbf{1}_{A_{k,n}}$$
        where $A_{k, n}inmathcal B $.



        Now we have using MCT
        begin{align}
        int_B UV, dmathbb P&=int_{ B}lim_{ntoinfty} sum_{k=1}^{N_n}mathbf{1}_{A_{k, n}} V, dmathbb P \
        &=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} V, dmathbb P \
        &=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} mathbb E(Vmid mathcal B), dmathbb P\
        &=lim_{ntoinfty} sum_{k=1}^{N_n}int_{ B} mathbf{1}_{A_{k,n}} mathbb E(Vmid mathcal B), dmathbb P
        end{align}
        And this implies indeed using MCT again that...



        You need to figure out how to use MCT yourself though.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 8 at 21:15

























        answered Jan 8 at 16:05









        ShashiShashi

        7,1731528




        7,1731528






























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