Definition of the conditional expectation operator $E^Q_{t,z}$?
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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 '18 at 13:22
sound wavesound wave
28619
$endgroup$
add a comment |
$begingroup$
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?
stochastic-calculus expected-value
asked Dec 26 '18 at 13:22
sound wavesound wave
28619
$endgroup$
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
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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
add a comment |
$begingroup$
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?
stochastic-calculus expected-value
asked Dec 26 '18 at 13:22
sound wavesound wave
28619
$endgroup$
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?
stochastic-calculus expected-value
stochastic-calculus expected-value
asked Dec 26 '18 at 13:22
sound wavesound wave
28619
asked Dec 26 '18 at 13:22
sound wavesound wave
28619
asked Dec 26 '18 at 13:22
sound wavesound wave
28619
asked Dec 26 '18 at 13:22
sound wavesound wave
28619
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
add a comment |
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
add a comment |
1 Answer
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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:
$$mathbb E^Q_{s,a}big[B_sbig]= mathbb E^Qbig[B_s|B_s=abig]=a.$$
$endgroup$
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
add a comment |
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$begingroup$
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:
$$mathbb E^Q_{s,a}big[B_sbig]= mathbb E^Qbig[B_s|B_s=abig]=a.$$
$endgroup$
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
add a comment |
$begingroup$
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:
$$mathbb E^Q_{s,a}big[B_sbig]= mathbb E^Qbig[B_s|B_s=abig]=a.$$
$endgroup$
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
add a comment |
$begingroup$
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:
$$mathbb E^Q_{s,a}big[B_sbig]= mathbb E^Qbig[B_s|B_s=abig]=a.$$
$endgroup$
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:
$$mathbb E^Q_{s,a}big[B_sbig]= mathbb E^Qbig[B_s|B_s=abig]=a.$$
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
add a comment |
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
add a comment |
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$begingroup$
$E^Q_{t,z}$ 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