Conditional Expectation, product of random variables
$begingroup$
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
asked Dec 27 '18 at 10:00
Agnetha TimaraAgnetha Timara
1086
$endgroup$
add a comment |
$begingroup$
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
asked Dec 27 '18 at 10:00
Agnetha TimaraAgnetha Timara
1086
$endgroup$
add a comment |
$begingroup$
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
asked Dec 27 '18 at 10:00
Agnetha TimaraAgnetha Timara
1086
$endgroup$
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
asked Dec 27 '18 at 10:00
Agnetha TimaraAgnetha Timara
1086
asked Dec 27 '18 at 10:00
Agnetha TimaraAgnetha Timara
1086
edited Dec 27 '18 at 10:07
Agnetha Timara
asked Dec 27 '18 at 10:00
Agnetha TimaraAgnetha Timara
1086
asked Dec 27 '18 at 10:00
Agnetha TimaraAgnetha Timara
1086
asked Dec 27 '18 at 10:00
Agnetha TimaraAgnetha Timara
1086
1086
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
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.
answered Dec 27 '18 at 10:11
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
$endgroup$
$begingroup$
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
|
show 1 more comment
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1 Answer
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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.
answered Dec 27 '18 at 10:11
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
answered Dec 27 '18 at 10:11
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
answered Dec 27 '18 at 10:11
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
$endgroup$
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.
answered Dec 27 '18 at 10:11
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
edited Dec 27 '18 at 11:52
answered Dec 27 '18 at 10:11
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
answered Dec 27 '18 at 10:11
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
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