Conditional Expectation, product of random variables












0












$begingroup$


I struggle with showing the following.



Let $tau$ be an exponentially distributed random variable, that means $mathbb P(tau > x)=mathrm{exp}(-x)$ and $xi$ a bounded random variable.



Show that $$mathbb Eleft[xifrac{mathbb 1_{(x<tauleq y)}}{tau}right]=0quad text{for all $x,y$ with $0<x<y$}$$
implies
$$mathbb Eleft[ximidtauright]=0.$$



Maybe you can help me with that.



Thank you very much!






random-variables conditional-expectation







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I struggle with showing the following.



    Let $tau$ be an exponentially distributed random variable, that means $mathbb P(tau > x)=mathrm{exp}(-x)$ and $xi$ a bounded random variable.



    Show that $$mathbb Eleft[xifrac{mathbb 1_{(x<tauleq y)}}{tau}right]=0quad text{for all $x,y$ with $0<x<y$}$$
    implies
    $$mathbb Eleft[ximidtauright]=0.$$



    Maybe you can help me with that.



    Thank you very much!






    random-variables conditional-expectation







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      I struggle with showing the following.



      Let $tau$ be an exponentially distributed random variable, that means $mathbb P(tau > x)=mathrm{exp}(-x)$ and $xi$ a bounded random variable.



      Show that $$mathbb Eleft[xifrac{mathbb 1_{(x<tauleq y)}}{tau}right]=0quad text{for all $x,y$ with $0<x<y$}$$
      implies
      $$mathbb Eleft[ximidtauright]=0.$$



      Maybe you can help me with that.



      Thank you very much!






      random-variables conditional-expectation







      share|cite|improve this question











      $endgroup$




      I struggle with showing the following.



      Let $tau$ be an exponentially distributed random variable, that means $mathbb P(tau > x)=mathrm{exp}(-x)$ and $xi$ a bounded random variable.



      Show that $$mathbb Eleft[xifrac{mathbb 1_{(x<tauleq y)}}{tau}right]=0quad text{for all $x,y$ with $0<x<y$}$$
      implies
      $$mathbb Eleft[ximidtauright]=0.$$



      Maybe you can help me with that.



      Thank you very much!






      random-variables conditional-expectation




      random-variables conditional-expectation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 27 '18 at 10:07







      Agnetha Timara

















      asked Dec 27 '18 at 10:00









      Agnetha TimaraAgnetha Timara

      1086




      1086






















          1 Answer
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          $begingroup$

          The result is true for any positive random variable $tau$. Exponential distribution is note required. $E(xi |tau)=0$ iff $E(frac {xi} {tau} |tau)=0$ iff $E(frac {xi} {tau} I_A)=0$ for any set $A$ of the type ${tau in E}$ where $E$ is a Borel set in $mathbb R$. It is enough
          to prove this when $E$ is an open interval and in this case what we need is exactly the hypothesis.



          PS: there is no guarantee that $frac {xi} {tau}$ is integrable. To avoid this problem we can replace $xi$ by $xi I_{{tau >epsilon}}$ in the proof and let $epsilonto 0$ at the end.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much for your quick reply. Isn't it necessary for $E(ximidtau)=0iff Eleft(fracxitaumidtauright)=0$ to hold that we have $Eleft|fracxitauright|<infty$ which does not have to be the case here.
            $endgroup$
            – Agnetha Timara
            Dec 27 '18 at 10:29










          • $begingroup$
            @AgnethaTimara You are right but that can easily be circumvented. It is enough to to show that $E( xi I_{tau >epsilon}|tau)=0$ for each $epsilon >0$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 27 '18 at 10:37












          • $begingroup$
            Thanks again for clarifying that. Can you also tell me why it suffices to show that for a generating set of the $sigma$ algebra $sigma(tau)$. Is there a theorem I can refer to or can I use dominated convergence?
            $endgroup$
            – Agnetha Timara
            Dec 27 '18 at 11:44










          • $begingroup$
            @AgnethaTimara The $pi -lambda$ Theorem or Dynkin's Theorem is ideally suited for this kind of situation. It tells you immediately why it is enough to consider intervals.
            $endgroup$
            – Kavi Rama Murthy
            Dec 27 '18 at 11:47










          • $begingroup$
            Sorry for the many questions. I really appreciate your help. These theorems tell me that the intervals generate the Borel $sigma$ algebra. I don't understand yet why $int_{{x<tauleq y}} Z=0$ implies $int_{{tauin E}} Z=0$ for $Ein mathcal B$ and $Z$ a random variable. Obviously $E$ can be represented as a union / intersection of intervals but for example $int_A Z=0$ for $Ain{A_1,ldots A_n}$ does not imply $int_{bigcup_i A_i}Z=0$
            $endgroup$
            – Agnetha Timara
            Dec 27 '18 at 12:16











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          $begingroup$

          The result is true for any positive random variable $tau$. Exponential distribution is note required. $E(xi |tau)=0$ iff $E(frac {xi} {tau} |tau)=0$ iff $E(frac {xi} {tau} I_A)=0$ for any set $A$ of the type ${tau in E}$ where $E$ is a Borel set in $mathbb R$. It is enough
          to prove this when $E$ is an open interval and in this case what we need is exactly the hypothesis.



          PS: there is no guarantee that $frac {xi} {tau}$ is integrable. To avoid this problem we can replace $xi$ by $xi I_{{tau >epsilon}}$ in the proof and let $epsilonto 0$ at the end.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much for your quick reply. Isn't it necessary for $E(ximidtau)=0iff Eleft(fracxitaumidtauright)=0$ to hold that we have $Eleft|fracxitauright|<infty$ which does not have to be the case here.
            $endgroup$
            – Agnetha Timara
            Dec 27 '18 at 10:29










          • $begingroup$
            @AgnethaTimara You are right but that can easily be circumvented. It is enough to to show that $E( xi I_{tau >epsilon}|tau)=0$ for each $epsilon >0$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 27 '18 at 10:37












          • $begingroup$
            Thanks again for clarifying that. Can you also tell me why it suffices to show that for a generating set of the $sigma$ algebra $sigma(tau)$. Is there a theorem I can refer to or can I use dominated convergence?
            $endgroup$
            – Agnetha Timara
            Dec 27 '18 at 11:44










          • $begingroup$
            @AgnethaTimara The $pi -lambda$ Theorem or Dynkin's Theorem is ideally suited for this kind of situation. It tells you immediately why it is enough to consider intervals.
            $endgroup$
            – Kavi Rama Murthy
            Dec 27 '18 at 11:47










          • $begingroup$
            Sorry for the many questions. I really appreciate your help. These theorems tell me that the intervals generate the Borel $sigma$ algebra. I don't understand yet why $int_{{x<tauleq y}} Z=0$ implies $int_{{tauin E}} Z=0$ for $Ein mathcal B$ and $Z$ a random variable. Obviously $E$ can be represented as a union / intersection of intervals but for example $int_A Z=0$ for $Ain{A_1,ldots A_n}$ does not imply $int_{bigcup_i A_i}Z=0$
            $endgroup$
            – Agnetha Timara
            Dec 27 '18 at 12:16
















          1












          $begingroup$

          The result is true for any positive random variable $tau$. Exponential distribution is note required. $E(xi |tau)=0$ iff $E(frac {xi} {tau} |tau)=0$ iff $E(frac {xi} {tau} I_A)=0$ for any set $A$ of the type ${tau in E}$ where $E$ is a Borel set in $mathbb R$. It is enough
          to prove this when $E$ is an open interval and in this case what we need is exactly the hypothesis.



          PS: there is no guarantee that $frac {xi} {tau}$ is integrable. To avoid this problem we can replace $xi$ by $xi I_{{tau >epsilon}}$ in the proof and let $epsilonto 0$ at the end.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much for your quick reply. Isn't it necessary for $E(ximidtau)=0iff Eleft(fracxitaumidtauright)=0$ to hold that we have $Eleft|fracxitauright|<infty$ which does not have to be the case here.
            $endgroup$
            – Agnetha Timara
            Dec 27 '18 at 10:29










          • $begingroup$
            @AgnethaTimara You are right but that can easily be circumvented. It is enough to to show that $E( xi I_{tau >epsilon}|tau)=0$ for each $epsilon >0$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 27 '18 at 10:37












          • $begingroup$
            Thanks again for clarifying that. Can you also tell me why it suffices to show that for a generating set of the $sigma$ algebra $sigma(tau)$. Is there a theorem I can refer to or can I use dominated convergence?
            $endgroup$
            – Agnetha Timara
            Dec 27 '18 at 11:44










          • $begingroup$
            @AgnethaTimara The $pi -lambda$ Theorem or Dynkin's Theorem is ideally suited for this kind of situation. It tells you immediately why it is enough to consider intervals.
            $endgroup$
            – Kavi Rama Murthy
            Dec 27 '18 at 11:47










          • $begingroup$
            Sorry for the many questions. I really appreciate your help. These theorems tell me that the intervals generate the Borel $sigma$ algebra. I don't understand yet why $int_{{x<tauleq y}} Z=0$ implies $int_{{tauin E}} Z=0$ for $Ein mathcal B$ and $Z$ a random variable. Obviously $E$ can be represented as a union / intersection of intervals but for example $int_A Z=0$ for $Ain{A_1,ldots A_n}$ does not imply $int_{bigcup_i A_i}Z=0$
            $endgroup$
            – Agnetha Timara
            Dec 27 '18 at 12:16














          1












          1








          1





          $begingroup$

          The result is true for any positive random variable $tau$. Exponential distribution is note required. $E(xi |tau)=0$ iff $E(frac {xi} {tau} |tau)=0$ iff $E(frac {xi} {tau} I_A)=0$ for any set $A$ of the type ${tau in E}$ where $E$ is a Borel set in $mathbb R$. It is enough
          to prove this when $E$ is an open interval and in this case what we need is exactly the hypothesis.



          PS: there is no guarantee that $frac {xi} {tau}$ is integrable. To avoid this problem we can replace $xi$ by $xi I_{{tau >epsilon}}$ in the proof and let $epsilonto 0$ at the end.






          share|cite|improve this answer











          $endgroup$



          The result is true for any positive random variable $tau$. Exponential distribution is note required. $E(xi |tau)=0$ iff $E(frac {xi} {tau} |tau)=0$ iff $E(frac {xi} {tau} I_A)=0$ for any set $A$ of the type ${tau in E}$ where $E$ is a Borel set in $mathbb R$. It is enough
          to prove this when $E$ is an open interval and in this case what we need is exactly the hypothesis.



          PS: there is no guarantee that $frac {xi} {tau}$ is integrable. To avoid this problem we can replace $xi$ by $xi I_{{tau >epsilon}}$ in the proof and let $epsilonto 0$ at the end.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 27 '18 at 11:52

























          answered Dec 27 '18 at 10:11









          Kavi Rama MurthyKavi Rama Murthy

          60.5k42161




          60.5k42161












          • $begingroup$
            Thank you very much for your quick reply. Isn't it necessary for $E(ximidtau)=0iff Eleft(fracxitaumidtauright)=0$ to hold that we have $Eleft|fracxitauright|<infty$ which does not have to be the case here.
            $endgroup$
            – Agnetha Timara
            Dec 27 '18 at 10:29










          • $begingroup$
            @AgnethaTimara You are right but that can easily be circumvented. It is enough to to show that $E( xi I_{tau >epsilon}|tau)=0$ for each $epsilon >0$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 27 '18 at 10:37












          • $begingroup$
            Thanks again for clarifying that. Can you also tell me why it suffices to show that for a generating set of the $sigma$ algebra $sigma(tau)$. Is there a theorem I can refer to or can I use dominated convergence?
            $endgroup$
            – Agnetha Timara
            Dec 27 '18 at 11:44










          • $begingroup$
            @AgnethaTimara The $pi -lambda$ Theorem or Dynkin's Theorem is ideally suited for this kind of situation. It tells you immediately why it is enough to consider intervals.
            $endgroup$
            – Kavi Rama Murthy
            Dec 27 '18 at 11:47










          • $begingroup$
            Sorry for the many questions. I really appreciate your help. These theorems tell me that the intervals generate the Borel $sigma$ algebra. I don't understand yet why $int_{{x<tauleq y}} Z=0$ implies $int_{{tauin E}} Z=0$ for $Ein mathcal B$ and $Z$ a random variable. Obviously $E$ can be represented as a union / intersection of intervals but for example $int_A Z=0$ for $Ain{A_1,ldots A_n}$ does not imply $int_{bigcup_i A_i}Z=0$
            $endgroup$
            – Agnetha Timara
            Dec 27 '18 at 12:16


















          • $begingroup$
            Thank you very much for your quick reply. Isn't it necessary for $E(ximidtau)=0iff Eleft(fracxitaumidtauright)=0$ to hold that we have $Eleft|fracxitauright|<infty$ which does not have to be the case here.
            $endgroup$
            – Agnetha Timara
            Dec 27 '18 at 10:29










          • $begingroup$
            @AgnethaTimara You are right but that can easily be circumvented. It is enough to to show that $E( xi I_{tau >epsilon}|tau)=0$ for each $epsilon >0$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 27 '18 at 10:37












          • $begingroup$
            Thanks again for clarifying that. Can you also tell me why it suffices to show that for a generating set of the $sigma$ algebra $sigma(tau)$. Is there a theorem I can refer to or can I use dominated convergence?
            $endgroup$
            – Agnetha Timara
            Dec 27 '18 at 11:44










          • $begingroup$
            @AgnethaTimara The $pi -lambda$ Theorem or Dynkin's Theorem is ideally suited for this kind of situation. It tells you immediately why it is enough to consider intervals.
            $endgroup$
            – Kavi Rama Murthy
            Dec 27 '18 at 11:47










          • $begingroup$
            Sorry for the many questions. I really appreciate your help. These theorems tell me that the intervals generate the Borel $sigma$ algebra. I don't understand yet why $int_{{x<tauleq y}} Z=0$ implies $int_{{tauin E}} Z=0$ for $Ein mathcal B$ and $Z$ a random variable. Obviously $E$ can be represented as a union / intersection of intervals but for example $int_A Z=0$ for $Ain{A_1,ldots A_n}$ does not imply $int_{bigcup_i A_i}Z=0$
            $endgroup$
            – Agnetha Timara
            Dec 27 '18 at 12:16
















          $begingroup$
          Thank you very much for your quick reply. Isn't it necessary for $E(ximidtau)=0iff Eleft(fracxitaumidtauright)=0$ to hold that we have $Eleft|fracxitauright|<infty$ which does not have to be the case here.
          $endgroup$
          – Agnetha Timara
          Dec 27 '18 at 10:29




          $begingroup$
          Thank you very much for your quick reply. Isn't it necessary for $E(ximidtau)=0iff Eleft(fracxitaumidtauright)=0$ to hold that we have $Eleft|fracxitauright|<infty$ which does not have to be the case here.
          $endgroup$
          – Agnetha Timara
          Dec 27 '18 at 10:29












          $begingroup$
          @AgnethaTimara You are right but that can easily be circumvented. It is enough to to show that $E( xi I_{tau >epsilon}|tau)=0$ for each $epsilon >0$.
          $endgroup$
          – Kavi Rama Murthy
          Dec 27 '18 at 10:37






          $begingroup$
          @AgnethaTimara You are right but that can easily be circumvented. It is enough to to show that $E( xi I_{tau >epsilon}|tau)=0$ for each $epsilon >0$.
          $endgroup$
          – Kavi Rama Murthy
          Dec 27 '18 at 10:37














          $begingroup$
          Thanks again for clarifying that. Can you also tell me why it suffices to show that for a generating set of the $sigma$ algebra $sigma(tau)$. Is there a theorem I can refer to or can I use dominated convergence?
          $endgroup$
          – Agnetha Timara
          Dec 27 '18 at 11:44




          $begingroup$
          Thanks again for clarifying that. Can you also tell me why it suffices to show that for a generating set of the $sigma$ algebra $sigma(tau)$. Is there a theorem I can refer to or can I use dominated convergence?
          $endgroup$
          – Agnetha Timara
          Dec 27 '18 at 11:44












          $begingroup$
          @AgnethaTimara The $pi -lambda$ Theorem or Dynkin's Theorem is ideally suited for this kind of situation. It tells you immediately why it is enough to consider intervals.
          $endgroup$
          – Kavi Rama Murthy
          Dec 27 '18 at 11:47




          $begingroup$
          @AgnethaTimara The $pi -lambda$ Theorem or Dynkin's Theorem is ideally suited for this kind of situation. It tells you immediately why it is enough to consider intervals.
          $endgroup$
          – Kavi Rama Murthy
          Dec 27 '18 at 11:47












          $begingroup$
          Sorry for the many questions. I really appreciate your help. These theorems tell me that the intervals generate the Borel $sigma$ algebra. I don't understand yet why $int_{{x<tauleq y}} Z=0$ implies $int_{{tauin E}} Z=0$ for $Ein mathcal B$ and $Z$ a random variable. Obviously $E$ can be represented as a union / intersection of intervals but for example $int_A Z=0$ for $Ain{A_1,ldots A_n}$ does not imply $int_{bigcup_i A_i}Z=0$
          $endgroup$
          – Agnetha Timara
          Dec 27 '18 at 12:16




          $begingroup$
          Sorry for the many questions. I really appreciate your help. These theorems tell me that the intervals generate the Borel $sigma$ algebra. I don't understand yet why $int_{{x<tauleq y}} Z=0$ implies $int_{{tauin E}} Z=0$ for $Ein mathcal B$ and $Z$ a random variable. Obviously $E$ can be represented as a union / intersection of intervals but for example $int_A Z=0$ for $Ain{A_1,ldots A_n}$ does not imply $int_{bigcup_i A_i}Z=0$
          $endgroup$
          – Agnetha Timara
          Dec 27 '18 at 12:16


















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