Definition of the conditional expectation operator $E^Q_{t,z}$?
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?
stochastic-calculus expected-value
asked Dec 26 at 13:22
sound wave
1598
add a comment |
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?
stochastic-calculus expected-value
asked Dec 26 at 13:22
sound wave
1598
1
$E^Q_{t,z}$ 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.
– AddSup
Dec 26 at 14:30
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 $$E^Q_{t,z}(log Z_t^2)=E^Q(log Z_t^2|Z_t=z)=log z^2?$$
– sound wave
Dec 26 at 14:41
1
Sure. And yes, that's correct.
– AddSup
Dec 26 at 14:44
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 $t$) ?
– sound wave
Dec 26 at 14:57
1
Yes. More like to be correct.
– AddSup
Dec 26 at 18:21
add a comment |
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?
stochastic-calculus expected-value
asked Dec 26 at 13:22
sound wave
1598
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?
stochastic-calculus expected-value
stochastic-calculus expected-value
asked Dec 26 at 13:22
sound wave
1598
asked Dec 26 at 13:22
sound wave
1598
asked Dec 26 at 13:22
sound wave
1598
asked Dec 26 at 13:22
sound wave
1598
asked Dec 26 at 13:22
sound wave
1598
1598
1
$E^Q_{t,z}$ 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.
– AddSup
Dec 26 at 14:30
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 $$E^Q_{t,z}(log Z_t^2)=E^Q(log Z_t^2|Z_t=z)=log z^2?$$
– sound wave
Dec 26 at 14:41
1
Sure. And yes, that's correct.
– AddSup
Dec 26 at 14:44
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 $t$) ?
– sound wave
Dec 26 at 14:57
1
Yes. More like to be correct.
– AddSup
Dec 26 at 18:21
add a comment |
1
$E^Q_{t,z}$ 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.
– AddSup
Dec 26 at 14:30
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 $$E^Q_{t,z}(log Z_t^2)=E^Q(log Z_t^2|Z_t=z)=log z^2?$$
– sound wave
Dec 26 at 14:41
1
Sure. And yes, that's correct.
– AddSup
Dec 26 at 14:44
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 $t$) ?
– sound wave
Dec 26 at 14:57
1
Yes. More like to be correct.
– AddSup
Dec 26 at 18:21
1
1
$E^Q_{t,z}$ 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.
– AddSup
Dec 26 at 14:30
$E^Q_{t,z}$ 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.
– AddSup
Dec 26 at 14:30
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 $$E^Q_{t,z}(log Z_t^2)=E^Q(log Z_t^2|Z_t=z)=log z^2?$$
– sound wave
Dec 26 at 14:41
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 $$E^Q_{t,z}(log Z_t^2)=E^Q(log Z_t^2|Z_t=z)=log z^2?$$
– sound wave
Dec 26 at 14:41
1
1
Sure. And yes, that's correct.
– AddSup
Dec 26 at 14:44
Sure. And yes, that's correct.
– AddSup
Dec 26 at 14:44
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 $t$) ?
– sound wave
Dec 26 at 14:57
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 $t$) ?
– sound wave
Dec 26 at 14:57
1
1
Yes. More like to be correct.
– AddSup
Dec 26 at 18:21
Yes. More like to be correct.
– AddSup
Dec 26 at 18:21
add a comment |
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1
$E^Q_{t,z}$ 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.
– AddSup
Dec 26 at 14:30
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 $$E^Q_{t,z}(log Z_t^2)=E^Q(log Z_t^2|Z_t=z)=log z^2?$$
– sound wave
Dec 26 at 14:41
1
Sure. And yes, that's correct.
– AddSup
Dec 26 at 14:44
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 $t$) ?
– sound wave
Dec 26 at 14:57
1
Yes. More like to be correct.
– AddSup
Dec 26 at 18:21