Definition of the conditional expectation operator $E^Q_{t,z}$?












0












$begingroup$


I'm studying the book "Arbitrage Theory in Continuous Time" by Bjork, and the authore uses a lot the notation $E^Q_{t,z}$, where $Q$ is a probability measure and $z=Z_t$ a stochastic process, but he did not give the definition.



Form the exercise below (taken from the book) I guess that $E^Q_{t,z}(Z_u)=E^Q(Z_u|Z_t)$. So for $u = t$ we should have $E^Q_{t,z}(Z_u)=Z_t$, is this correct?



enter image description here






stochastic-calculus expected-value







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $E^Q_{t,z}$ must be $E^Q(cdot|Z_t=z)$. But that doesn't mean $E^Q_{t,z}(Z_u)=Z_t$, which would be true if $Z$ were also a martingale. In any case, are you sure there's no definition in the book? I'd have another look at the section where the author first discusses Markov processes.
    $endgroup$
    – AddSup
    Dec 26 '18 at 14:30












  • $begingroup$
    Thank you very much for the help. So in the exercise $$int_t^T dlog Z_t^2=log Z_T^2-log Z_t^2,$$ but then $$E^Q_{t,z}(log Z_t^2)=E^Q(log Z_t^2|Z_t=z)=log z^2?$$
    $endgroup$
    – sound wave
    Dec 26 '18 at 14:41






  • 1




    $begingroup$
    Sure. And yes, that's correct.
    $endgroup$
    – AddSup
    Dec 26 '18 at 14:44










  • $begingroup$
    Ok thanks. To be rigorous, in the integral should I change the time letter of $Z$, i.e. writing $int_t^T dlog Z_s^2$ (with $s$ instead of $t$) ?
    $endgroup$
    – sound wave
    Dec 26 '18 at 14:57








  • 1




    $begingroup$
    Yes. More like to be correct.
    $endgroup$
    – AddSup
    Dec 26 '18 at 18:21
















0












$begingroup$


I'm studying the book "Arbitrage Theory in Continuous Time" by Bjork, and the authore uses a lot the notation $E^Q_{t,z}$, where $Q$ is a probability measure and $z=Z_t$ a stochastic process, but he did not give the definition.



Form the exercise below (taken from the book) I guess that $E^Q_{t,z}(Z_u)=E^Q(Z_u|Z_t)$. So for $u = t$ we should have $E^Q_{t,z}(Z_u)=Z_t$, is this correct?



enter image description here






stochastic-calculus expected-value







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $E^Q_{t,z}$ must be $E^Q(cdot|Z_t=z)$. But that doesn't mean $E^Q_{t,z}(Z_u)=Z_t$, which would be true if $Z$ were also a martingale. In any case, are you sure there's no definition in the book? I'd have another look at the section where the author first discusses Markov processes.
    $endgroup$
    – AddSup
    Dec 26 '18 at 14:30












  • $begingroup$
    Thank you very much for the help. So in the exercise $$int_t^T dlog Z_t^2=log Z_T^2-log Z_t^2,$$ but then $$E^Q_{t,z}(log Z_t^2)=E^Q(log Z_t^2|Z_t=z)=log z^2?$$
    $endgroup$
    – sound wave
    Dec 26 '18 at 14:41






  • 1




    $begingroup$
    Sure. And yes, that's correct.
    $endgroup$
    – AddSup
    Dec 26 '18 at 14:44










  • $begingroup$
    Ok thanks. To be rigorous, in the integral should I change the time letter of $Z$, i.e. writing $int_t^T dlog Z_s^2$ (with $s$ instead of $t$) ?
    $endgroup$
    – sound wave
    Dec 26 '18 at 14:57








  • 1




    $begingroup$
    Yes. More like to be correct.
    $endgroup$
    – AddSup
    Dec 26 '18 at 18:21














0












0








0





$begingroup$


I'm studying the book "Arbitrage Theory in Continuous Time" by Bjork, and the authore uses a lot the notation $E^Q_{t,z}$, where $Q$ is a probability measure and $z=Z_t$ a stochastic process, but he did not give the definition.



Form the exercise below (taken from the book) I guess that $E^Q_{t,z}(Z_u)=E^Q(Z_u|Z_t)$. So for $u = t$ we should have $E^Q_{t,z}(Z_u)=Z_t$, is this correct?



enter image description here






stochastic-calculus expected-value







share|cite|improve this question









$endgroup$




I'm studying the book "Arbitrage Theory in Continuous Time" by Bjork, and the authore uses a lot the notation $E^Q_{t,z}$, where $Q$ is a probability measure and $z=Z_t$ a stochastic process, but he did not give the definition.



Form the exercise below (taken from the book) I guess that $E^Q_{t,z}(Z_u)=E^Q(Z_u|Z_t)$. So for $u = t$ we should have $E^Q_{t,z}(Z_u)=Z_t$, is this correct?



enter image description here






stochastic-calculus expected-value




stochastic-calculus expected-value






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 26 '18 at 13:22









sound wavesound wave

28619




28619








  • 1




    $begingroup$
    $E^Q_{t,z}$ must be $E^Q(cdot|Z_t=z)$. But that doesn't mean $E^Q_{t,z}(Z_u)=Z_t$, which would be true if $Z$ were also a martingale. In any case, are you sure there's no definition in the book? I'd have another look at the section where the author first discusses Markov processes.
    $endgroup$
    – AddSup
    Dec 26 '18 at 14:30












  • $begingroup$
    Thank you very much for the help. So in the exercise $$int_t^T dlog Z_t^2=log Z_T^2-log Z_t^2,$$ but then $$E^Q_{t,z}(log Z_t^2)=E^Q(log Z_t^2|Z_t=z)=log z^2?$$
    $endgroup$
    – sound wave
    Dec 26 '18 at 14:41






  • 1




    $begingroup$
    Sure. And yes, that's correct.
    $endgroup$
    – AddSup
    Dec 26 '18 at 14:44










  • $begingroup$
    Ok thanks. To be rigorous, in the integral should I change the time letter of $Z$, i.e. writing $int_t^T dlog Z_s^2$ (with $s$ instead of $t$) ?
    $endgroup$
    – sound wave
    Dec 26 '18 at 14:57








  • 1




    $begingroup$
    Yes. More like to be correct.
    $endgroup$
    – AddSup
    Dec 26 '18 at 18:21














  • 1




    $begingroup$
    $E^Q_{t,z}$ must be $E^Q(cdot|Z_t=z)$. But that doesn't mean $E^Q_{t,z}(Z_u)=Z_t$, which would be true if $Z$ were also a martingale. In any case, are you sure there's no definition in the book? I'd have another look at the section where the author first discusses Markov processes.
    $endgroup$
    – AddSup
    Dec 26 '18 at 14:30












  • $begingroup$
    Thank you very much for the help. So in the exercise $$int_t^T dlog Z_t^2=log Z_T^2-log Z_t^2,$$ but then $$E^Q_{t,z}(log Z_t^2)=E^Q(log Z_t^2|Z_t=z)=log z^2?$$
    $endgroup$
    – sound wave
    Dec 26 '18 at 14:41






  • 1




    $begingroup$
    Sure. And yes, that's correct.
    $endgroup$
    – AddSup
    Dec 26 '18 at 14:44










  • $begingroup$
    Ok thanks. To be rigorous, in the integral should I change the time letter of $Z$, i.e. writing $int_t^T dlog Z_s^2$ (with $s$ instead of $t$) ?
    $endgroup$
    – sound wave
    Dec 26 '18 at 14:57








  • 1




    $begingroup$
    Yes. More like to be correct.
    $endgroup$
    – AddSup
    Dec 26 '18 at 18:21








1




1




$begingroup$
$E^Q_{t,z}$ must be $E^Q(cdot|Z_t=z)$. But that doesn't mean $E^Q_{t,z}(Z_u)=Z_t$, which would be true if $Z$ were also a martingale. In any case, are you sure there's no definition in the book? I'd have another look at the section where the author first discusses Markov processes.
$endgroup$
– AddSup
Dec 26 '18 at 14:30






$begingroup$
$E^Q_{t,z}$ must be $E^Q(cdot|Z_t=z)$. But that doesn't mean $E^Q_{t,z}(Z_u)=Z_t$, which would be true if $Z$ were also a martingale. In any case, are you sure there's no definition in the book? I'd have another look at the section where the author first discusses Markov processes.
$endgroup$
– AddSup
Dec 26 '18 at 14:30














$begingroup$
Thank you very much for the help. So in the exercise $$int_t^T dlog Z_t^2=log Z_T^2-log Z_t^2,$$ but then $$E^Q_{t,z}(log Z_t^2)=E^Q(log Z_t^2|Z_t=z)=log z^2?$$
$endgroup$
– sound wave
Dec 26 '18 at 14:41




$begingroup$
Thank you very much for the help. So in the exercise $$int_t^T dlog Z_t^2=log Z_T^2-log Z_t^2,$$ but then $$E^Q_{t,z}(log Z_t^2)=E^Q(log Z_t^2|Z_t=z)=log z^2?$$
$endgroup$
– sound wave
Dec 26 '18 at 14:41




1




1




$begingroup$
Sure. And yes, that's correct.
$endgroup$
– AddSup
Dec 26 '18 at 14:44




$begingroup$
Sure. And yes, that's correct.
$endgroup$
– AddSup
Dec 26 '18 at 14:44












$begingroup$
Ok thanks. To be rigorous, in the integral should I change the time letter of $Z$, i.e. writing $int_t^T dlog Z_s^2$ (with $s$ instead of $t$) ?
$endgroup$
– sound wave
Dec 26 '18 at 14:57






$begingroup$
Ok thanks. To be rigorous, in the integral should I change the time letter of $Z$, i.e. writing $int_t^T dlog Z_s^2$ (with $s$ instead of $t$) ?
$endgroup$
– sound wave
Dec 26 '18 at 14:57






1




1




$begingroup$
Yes. More like to be correct.
$endgroup$
– AddSup
Dec 26 '18 at 18:21




$begingroup$
Yes. More like to be correct.
$endgroup$
– AddSup
Dec 26 '18 at 18:21










1 Answer
1






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0












$begingroup$

The notation $mathbb E^Q_{s,a}big[B_tbig]$, where $B$ a stochastic process, $Q$ is the probability measure, $sinmathbb R^+$ a given time, $ainmathbb R$ a given value for the process $B$, is a shortcut for the longer notation for the conditional expectation of the stochastic process $mathbb E^Qbig[B_t|B_s=abig]$.



If the two times coincide, i.e. $tequiv s$, then the result is:
$$mathbb E^Q_{s,a}big[B_sbig]= mathbb E^Qbig[B_s|B_s=abig]=a.$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    You can find more about the notation here: stat.auckland.ac.nz/~fewster/325/notes/ch3.pdf
    $endgroup$
    – user636387
    Feb 2 at 12:48











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









0












$begingroup$

The notation $mathbb E^Q_{s,a}big[B_tbig]$, where $B$ a stochastic process, $Q$ is the probability measure, $sinmathbb R^+$ a given time, $ainmathbb R$ a given value for the process $B$, is a shortcut for the longer notation for the conditional expectation of the stochastic process $mathbb E^Qbig[B_t|B_s=abig]$.



If the two times coincide, i.e. $tequiv s$, then the result is:
$$mathbb E^Q_{s,a}big[B_sbig]= mathbb E^Qbig[B_s|B_s=abig]=a.$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    You can find more about the notation here: stat.auckland.ac.nz/~fewster/325/notes/ch3.pdf
    $endgroup$
    – user636387
    Feb 2 at 12:48
















0












$begingroup$

The notation $mathbb E^Q_{s,a}big[B_tbig]$, where $B$ a stochastic process, $Q$ is the probability measure, $sinmathbb R^+$ a given time, $ainmathbb R$ a given value for the process $B$, is a shortcut for the longer notation for the conditional expectation of the stochastic process $mathbb E^Qbig[B_t|B_s=abig]$.



If the two times coincide, i.e. $tequiv s$, then the result is:
$$mathbb E^Q_{s,a}big[B_sbig]= mathbb E^Qbig[B_s|B_s=abig]=a.$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    You can find more about the notation here: stat.auckland.ac.nz/~fewster/325/notes/ch3.pdf
    $endgroup$
    – user636387
    Feb 2 at 12:48














0












0








0





$begingroup$

The notation $mathbb E^Q_{s,a}big[B_tbig]$, where $B$ a stochastic process, $Q$ is the probability measure, $sinmathbb R^+$ a given time, $ainmathbb R$ a given value for the process $B$, is a shortcut for the longer notation for the conditional expectation of the stochastic process $mathbb E^Qbig[B_t|B_s=abig]$.



If the two times coincide, i.e. $tequiv s$, then the result is:
$$mathbb E^Q_{s,a}big[B_sbig]= mathbb E^Qbig[B_s|B_s=abig]=a.$$






share|cite|improve this answer









$endgroup$



The notation $mathbb E^Q_{s,a}big[B_tbig]$, where $B$ a stochastic process, $Q$ is the probability measure, $sinmathbb R^+$ a given time, $ainmathbb R$ a given value for the process $B$, is a shortcut for the longer notation for the conditional expectation of the stochastic process $mathbb E^Qbig[B_t|B_s=abig]$.



If the two times coincide, i.e. $tequiv s$, then the result is:
$$mathbb E^Q_{s,a}big[B_sbig]= mathbb E^Qbig[B_s|B_s=abig]=a.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Feb 2 at 11:52







user636387















  • 1




    $begingroup$
    You can find more about the notation here: stat.auckland.ac.nz/~fewster/325/notes/ch3.pdf
    $endgroup$
    – user636387
    Feb 2 at 12:48














  • 1




    $begingroup$
    You can find more about the notation here: stat.auckland.ac.nz/~fewster/325/notes/ch3.pdf
    $endgroup$
    – user636387
    Feb 2 at 12:48








1




1




$begingroup$
You can find more about the notation here: stat.auckland.ac.nz/~fewster/325/notes/ch3.pdf
$endgroup$
– user636387
Feb 2 at 12:48




$begingroup$
You can find more about the notation here: stat.auckland.ac.nz/~fewster/325/notes/ch3.pdf
$endgroup$
– user636387
Feb 2 at 12:48


















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