Why $P^mu[X_{S+t} in Gamma mid mathcal{F}_{S+}]=P^mu[X_{S+t} in Gamma mid X_{S}]=0 text{ ,} P^mutext{-a.s. on...
For any progressively measurable process $X$ and any optional time $S$ of ${mathcal{F}_t}$ why do we have that
$$P^mu[X_{S+t} in Gamma mid mathcal{F}_{S+}]=P^mu[X_{S+t} in Gamma mid X_{S}]=0 text{ ,} P^mutext{-a.s. on } {S=infty}$$
This is remark $2.6.4$ in Section 2.6 of Karatzas and Shreve's Brownian Motion and stochastic Calculus
The author mentions at the beginning of the remark that ${X_{S+t} in Gamma }:= {X_{S+t} in Gamma,S<infty }$.
Now since $S$ is an ${mathcal{F}_t}$ optional time, $S$ is $ mathcal{F}_{S+}$ measurable
So we can write
begin{equation}
begin{split}
P^mu[X_{S+t} in Gamma mid mathcal{F}_{S+}] 1_{{S=infty}}=E^mu[1_{{X_{S+t} in Gamma}} 1_{{S=infty}} mid mathcal{F}_{S+}]=E^mu[1_{{X_{S+t} in Gamma,S<infty}}} 1_{{S=infty}} mid mathcal{F}_{S+}]\=0
end{split}
end{equation}
since the product of the two indicators functions is zero everywhere
Is this reasoning correct? If not , could you tell me what is problem. Furthermore how could I show that $P^mu[X_{S+t} in Gamma mid X_{S}]=0$ on ${S=infty }$. Where is the progressive measurability of $X$ being used?
stochastic-processes stochastic-calculus markov-process stochastic-analysis
asked Nov 30 at 13:47
user3503589
1,2011721
add a comment |
For any progressively measurable process $X$ and any optional time $S$ of ${mathcal{F}_t}$ why do we have that
$$P^mu[X_{S+t} in Gamma mid mathcal{F}_{S+}]=P^mu[X_{S+t} in Gamma mid X_{S}]=0 text{ ,} P^mutext{-a.s. on } {S=infty}$$
This is remark $2.6.4$ in Section 2.6 of Karatzas and Shreve's Brownian Motion and stochastic Calculus
The author mentions at the beginning of the remark that ${X_{S+t} in Gamma }:= {X_{S+t} in Gamma,S<infty }$.
Now since $S$ is an ${mathcal{F}_t}$ optional time, $S$ is $ mathcal{F}_{S+}$ measurable
So we can write
begin{equation}
begin{split}
P^mu[X_{S+t} in Gamma mid mathcal{F}_{S+}] 1_{{S=infty}}=E^mu[1_{{X_{S+t} in Gamma}} 1_{{S=infty}} mid mathcal{F}_{S+}]=E^mu[1_{{X_{S+t} in Gamma,S<infty}}} 1_{{S=infty}} mid mathcal{F}_{S+}]\=0
end{split}
end{equation}
since the product of the two indicators functions is zero everywhere
Is this reasoning correct? If not , could you tell me what is problem. Furthermore how could I show that $P^mu[X_{S+t} in Gamma mid X_{S}]=0$ on ${S=infty }$. Where is the progressive measurability of $X$ being used?
stochastic-processes stochastic-calculus markov-process stochastic-analysis
asked Nov 30 at 13:47
user3503589
1,2011721
add a comment |
For any progressively measurable process $X$ and any optional time $S$ of ${mathcal{F}_t}$ why do we have that
$$P^mu[X_{S+t} in Gamma mid mathcal{F}_{S+}]=P^mu[X_{S+t} in Gamma mid X_{S}]=0 text{ ,} P^mutext{-a.s. on } {S=infty}$$
This is remark $2.6.4$ in Section 2.6 of Karatzas and Shreve's Brownian Motion and stochastic Calculus
The author mentions at the beginning of the remark that ${X_{S+t} in Gamma }:= {X_{S+t} in Gamma,S<infty }$.
Now since $S$ is an ${mathcal{F}_t}$ optional time, $S$ is $ mathcal{F}_{S+}$ measurable
So we can write
begin{equation}
begin{split}
P^mu[X_{S+t} in Gamma mid mathcal{F}_{S+}] 1_{{S=infty}}=E^mu[1_{{X_{S+t} in Gamma}} 1_{{S=infty}} mid mathcal{F}_{S+}]=E^mu[1_{{X_{S+t} in Gamma,S<infty}}} 1_{{S=infty}} mid mathcal{F}_{S+}]\=0
end{split}
end{equation}
since the product of the two indicators functions is zero everywhere
Is this reasoning correct? If not , could you tell me what is problem. Furthermore how could I show that $P^mu[X_{S+t} in Gamma mid X_{S}]=0$ on ${S=infty }$. Where is the progressive measurability of $X$ being used?
stochastic-processes stochastic-calculus markov-process stochastic-analysis
asked Nov 30 at 13:47
user3503589
1,2011721
For any progressively measurable process $X$ and any optional time $S$ of ${mathcal{F}_t}$ why do we have that
$$P^mu[X_{S+t} in Gamma mid mathcal{F}_{S+}]=P^mu[X_{S+t} in Gamma mid X_{S}]=0 text{ ,} P^mutext{-a.s. on } {S=infty}$$
This is remark $2.6.4$ in Section 2.6 of Karatzas and Shreve's Brownian Motion and stochastic Calculus
The author mentions at the beginning of the remark that ${X_{S+t} in Gamma }:= {X_{S+t} in Gamma,S<infty }$.
Now since $S$ is an ${mathcal{F}_t}$ optional time, $S$ is $ mathcal{F}_{S+}$ measurable
So we can write
begin{equation}
begin{split}
P^mu[X_{S+t} in Gamma mid mathcal{F}_{S+}] 1_{{S=infty}}=E^mu[1_{{X_{S+t} in Gamma}} 1_{{S=infty}} mid mathcal{F}_{S+}]=E^mu[1_{{X_{S+t} in Gamma,S<infty}}} 1_{{S=infty}} mid mathcal{F}_{S+}]\=0
end{split}
end{equation}
since the product of the two indicators functions is zero everywhere
Is this reasoning correct? If not , could you tell me what is problem. Furthermore how could I show that $P^mu[X_{S+t} in Gamma mid X_{S}]=0$ on ${S=infty }$. Where is the progressive measurability of $X$ being used?
stochastic-processes stochastic-calculus markov-process stochastic-analysis
stochastic-processes stochastic-calculus markov-process stochastic-analysis
asked Nov 30 at 13:47
user3503589
1,2011721
asked Nov 30 at 13:47
user3503589
1,2011721
asked Nov 30 at 13:47
user3503589
1,2011721
asked Nov 30 at 13:47
user3503589
1,2011721
asked Nov 30 at 13:47
user3503589
1,2011721
1,2011721
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
I think your reasoning of $P^mu[X_{S+t}inGamma|mathcal F_{S+}]=0$ on ${S=infty}$ is correct.
According to the definition of $sigma(X_S)$ given in Problem 1.1.17, ${S=infty}insigma(X_S)$. So I think you can apply the same reasoning as above to verify that $P^mu[X_{S+t}inGamma|X_S]=0$ on ${S=infty}$.
To me it seems, therefore, that in asserting
$$
P^mu[X_{S+t}inGamma|mathcal F_{S+}]
=P^mu[X_{S+t}inGamma|X_S]
=0text{ on }{S=infty},
$$
we don't need the progressive measurability of $X$. I guess that the authors say "for any progressively measurable process $X$" because immediately preceding the remark, they've defined Markov processes as progressively measurable processes satisfying the Markov property (Definitions 2.6.2 and 2.6.3). That is, "for any progressively measurable process $X$," it seems to me, is to say "for any progressively measurable process $X$ (whether it's a Markov process or not)."
$quadquad$Markov processes are defined among progressively measurable processes probably in view of Proposition 1.2.18, i.e., to make sure $X_Sinmathcal F_{S+}$ (the Markov property is an assertion that a conditional expectation given $mathcal F_{S+}$ be equal to that given $X_S$, so among other things $X_S$ must be measurable with respect to $mathcal F_{S+}$).
answered Dec 1 at 10:26
AddSup
371214
Thank you for very nice descriptive answer . But when you say I think you are correct do you mean maybe or you are convinced ? What bothers me is that this solution of mine to show that the conditional expectation is zero ( and yours too ) seems to be just a convenience because of the way the authors choose to define $X_{S+ t} in Gamma$ which made me wonder at the first place if my explanation is actually corrected.
– user3503589
Dec 1 at 11:11
I am not really sure if I understood your last paragraph . First because Markov property talks about conditional probabilities ( and what you said about conditional expectation is just a consequence) . Second I don’t see why would be need that$ X_S$ is $mathcal{F}_{S+} measurable ?
– user3503589
Dec 1 at 11:22
1
Hi @user3503589 1. I am convinced. If $X1_A=0$, $X$ must be zero on $A$. 2. I don't understand what you mean by "a convenience." A definition is a definition. 3. A conditional probability is a conditional expectation, as you know. And the definition of conditional expectation is such that a version must be measurable with respect to the conditioning $sigma$-algebra.
– AddSup
Dec 1 at 11:34
By a convenience I meant quite trivial. I am sorry but what I meant was, where do you get ("Markov property is an assertion that the conditional expectation given $mathcal{F}_{S+}$ , is equal to that $X_S$") when you explain the reasoning for progressive measurability.
– user3503589
Dec 1 at 11:48
1
Sorry for the ambiguity, I was simply reading a conditional probability as a conditional expectation whenever I see one. So by "the conditional expectation given $mathcal F_{S+}$" I meant "the conditional expectation of $1_{X_{S+t}inGamma}$ given $mathcal F_{S+}$.'' But it's true that the Markov property defined in terms of conditional probabilities is equivalent to the other well-known characterization involving conditional expectations of arbitrary bounded functions (Proposition 2.6.7(e')).
– AddSup
Dec 1 at 11:56
|
show 1 more comment
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020108%2fwhy-p-mux-st-in-gamma-mid-mathcalf-s-p-mux-st-in-gamma-m%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I think your reasoning of $P^mu[X_{S+t}inGamma|mathcal F_{S+}]=0$ on ${S=infty}$ is correct.
According to the definition of $sigma(X_S)$ given in Problem 1.1.17, ${S=infty}insigma(X_S)$. So I think you can apply the same reasoning as above to verify that $P^mu[X_{S+t}inGamma|X_S]=0$ on ${S=infty}$.
To me it seems, therefore, that in asserting
$$
P^mu[X_{S+t}inGamma|mathcal F_{S+}]
=P^mu[X_{S+t}inGamma|X_S]
=0text{ on }{S=infty},
$$
we don't need the progressive measurability of $X$. I guess that the authors say "for any progressively measurable process $X$" because immediately preceding the remark, they've defined Markov processes as progressively measurable processes satisfying the Markov property (Definitions 2.6.2 and 2.6.3). That is, "for any progressively measurable process $X$," it seems to me, is to say "for any progressively measurable process $X$ (whether it's a Markov process or not)."
$quadquad$Markov processes are defined among progressively measurable processes probably in view of Proposition 1.2.18, i.e., to make sure $X_Sinmathcal F_{S+}$ (the Markov property is an assertion that a conditional expectation given $mathcal F_{S+}$ be equal to that given $X_S$, so among other things $X_S$ must be measurable with respect to $mathcal F_{S+}$).
answered Dec 1 at 10:26
AddSup
371214
Thank you for very nice descriptive answer . But when you say I think you are correct do you mean maybe or you are convinced ? What bothers me is that this solution of mine to show that the conditional expectation is zero ( and yours too ) seems to be just a convenience because of the way the authors choose to define $X_{S+ t} in Gamma$ which made me wonder at the first place if my explanation is actually corrected.
– user3503589
Dec 1 at 11:11
I am not really sure if I understood your last paragraph . First because Markov property talks about conditional probabilities ( and what you said about conditional expectation is just a consequence) . Second I don’t see why would be need that$ X_S$ is $mathcal{F}_{S+} measurable ?
– user3503589
Dec 1 at 11:22
1
Hi @user3503589 1. I am convinced. If $X1_A=0$, $X$ must be zero on $A$. 2. I don't understand what you mean by "a convenience." A definition is a definition. 3. A conditional probability is a conditional expectation, as you know. And the definition of conditional expectation is such that a version must be measurable with respect to the conditioning $sigma$-algebra.
– AddSup
Dec 1 at 11:34
By a convenience I meant quite trivial. I am sorry but what I meant was, where do you get ("Markov property is an assertion that the conditional expectation given $mathcal{F}_{S+}$ , is equal to that $X_S$") when you explain the reasoning for progressive measurability.
– user3503589
Dec 1 at 11:48
1
Sorry for the ambiguity, I was simply reading a conditional probability as a conditional expectation whenever I see one. So by "the conditional expectation given $mathcal F_{S+}$" I meant "the conditional expectation of $1_{X_{S+t}inGamma}$ given $mathcal F_{S+}$.'' But it's true that the Markov property defined in terms of conditional probabilities is equivalent to the other well-known characterization involving conditional expectations of arbitrary bounded functions (Proposition 2.6.7(e')).
– AddSup
Dec 1 at 11:56
|
show 1 more comment
I think your reasoning of $P^mu[X_{S+t}inGamma|mathcal F_{S+}]=0$ on ${S=infty}$ is correct.
According to the definition of $sigma(X_S)$ given in Problem 1.1.17, ${S=infty}insigma(X_S)$. So I think you can apply the same reasoning as above to verify that $P^mu[X_{S+t}inGamma|X_S]=0$ on ${S=infty}$.
To me it seems, therefore, that in asserting
$$
P^mu[X_{S+t}inGamma|mathcal F_{S+}]
=P^mu[X_{S+t}inGamma|X_S]
=0text{ on }{S=infty},
$$
we don't need the progressive measurability of $X$. I guess that the authors say "for any progressively measurable process $X$" because immediately preceding the remark, they've defined Markov processes as progressively measurable processes satisfying the Markov property (Definitions 2.6.2 and 2.6.3). That is, "for any progressively measurable process $X$," it seems to me, is to say "for any progressively measurable process $X$ (whether it's a Markov process or not)."
$quadquad$Markov processes are defined among progressively measurable processes probably in view of Proposition 1.2.18, i.e., to make sure $X_Sinmathcal F_{S+}$ (the Markov property is an assertion that a conditional expectation given $mathcal F_{S+}$ be equal to that given $X_S$, so among other things $X_S$ must be measurable with respect to $mathcal F_{S+}$).
answered Dec 1 at 10:26
AddSup
371214
Thank you for very nice descriptive answer . But when you say I think you are correct do you mean maybe or you are convinced ? What bothers me is that this solution of mine to show that the conditional expectation is zero ( and yours too ) seems to be just a convenience because of the way the authors choose to define $X_{S+ t} in Gamma$ which made me wonder at the first place if my explanation is actually corrected.
– user3503589
Dec 1 at 11:11
I am not really sure if I understood your last paragraph . First because Markov property talks about conditional probabilities ( and what you said about conditional expectation is just a consequence) . Second I don’t see why would be need that$ X_S$ is $mathcal{F}_{S+} measurable ?
– user3503589
Dec 1 at 11:22
1
Hi @user3503589 1. I am convinced. If $X1_A=0$, $X$ must be zero on $A$. 2. I don't understand what you mean by "a convenience." A definition is a definition. 3. A conditional probability is a conditional expectation, as you know. And the definition of conditional expectation is such that a version must be measurable with respect to the conditioning $sigma$-algebra.
– AddSup
Dec 1 at 11:34
By a convenience I meant quite trivial. I am sorry but what I meant was, where do you get ("Markov property is an assertion that the conditional expectation given $mathcal{F}_{S+}$ , is equal to that $X_S$") when you explain the reasoning for progressive measurability.
– user3503589
Dec 1 at 11:48
1
Sorry for the ambiguity, I was simply reading a conditional probability as a conditional expectation whenever I see one. So by "the conditional expectation given $mathcal F_{S+}$" I meant "the conditional expectation of $1_{X_{S+t}inGamma}$ given $mathcal F_{S+}$.'' But it's true that the Markov property defined in terms of conditional probabilities is equivalent to the other well-known characterization involving conditional expectations of arbitrary bounded functions (Proposition 2.6.7(e')).
– AddSup
Dec 1 at 11:56
|
show 1 more comment
I think your reasoning of $P^mu[X_{S+t}inGamma|mathcal F_{S+}]=0$ on ${S=infty}$ is correct.
According to the definition of $sigma(X_S)$ given in Problem 1.1.17, ${S=infty}insigma(X_S)$. So I think you can apply the same reasoning as above to verify that $P^mu[X_{S+t}inGamma|X_S]=0$ on ${S=infty}$.
To me it seems, therefore, that in asserting
$$
P^mu[X_{S+t}inGamma|mathcal F_{S+}]
=P^mu[X_{S+t}inGamma|X_S]
=0text{ on }{S=infty},
$$
we don't need the progressive measurability of $X$. I guess that the authors say "for any progressively measurable process $X$" because immediately preceding the remark, they've defined Markov processes as progressively measurable processes satisfying the Markov property (Definitions 2.6.2 and 2.6.3). That is, "for any progressively measurable process $X$," it seems to me, is to say "for any progressively measurable process $X$ (whether it's a Markov process or not)."
$quadquad$Markov processes are defined among progressively measurable processes probably in view of Proposition 1.2.18, i.e., to make sure $X_Sinmathcal F_{S+}$ (the Markov property is an assertion that a conditional expectation given $mathcal F_{S+}$ be equal to that given $X_S$, so among other things $X_S$ must be measurable with respect to $mathcal F_{S+}$).
answered Dec 1 at 10:26
AddSup
371214
I think your reasoning of $P^mu[X_{S+t}inGamma|mathcal F_{S+}]=0$ on ${S=infty}$ is correct.
According to the definition of $sigma(X_S)$ given in Problem 1.1.17, ${S=infty}insigma(X_S)$. So I think you can apply the same reasoning as above to verify that $P^mu[X_{S+t}inGamma|X_S]=0$ on ${S=infty}$.
To me it seems, therefore, that in asserting
$$
P^mu[X_{S+t}inGamma|mathcal F_{S+}]
=P^mu[X_{S+t}inGamma|X_S]
=0text{ on }{S=infty},
$$
we don't need the progressive measurability of $X$. I guess that the authors say "for any progressively measurable process $X$" because immediately preceding the remark, they've defined Markov processes as progressively measurable processes satisfying the Markov property (Definitions 2.6.2 and 2.6.3). That is, "for any progressively measurable process $X$," it seems to me, is to say "for any progressively measurable process $X$ (whether it's a Markov process or not)."
$quadquad$Markov processes are defined among progressively measurable processes probably in view of Proposition 1.2.18, i.e., to make sure $X_Sinmathcal F_{S+}$ (the Markov property is an assertion that a conditional expectation given $mathcal F_{S+}$ be equal to that given $X_S$, so among other things $X_S$ must be measurable with respect to $mathcal F_{S+}$).
answered Dec 1 at 10:26
AddSup
371214
answered Dec 1 at 10:26
AddSup
371214
answered Dec 1 at 10:26
AddSup
371214
answered Dec 1 at 10:26
AddSup
371214
371214
Thank you for very nice descriptive answer . But when you say I think you are correct do you mean maybe or you are convinced ? What bothers me is that this solution of mine to show that the conditional expectation is zero ( and yours too ) seems to be just a convenience because of the way the authors choose to define $X_{S+ t} in Gamma$ which made me wonder at the first place if my explanation is actually corrected.
– user3503589
Dec 1 at 11:11
I am not really sure if I understood your last paragraph . First because Markov property talks about conditional probabilities ( and what you said about conditional expectation is just a consequence) . Second I don’t see why would be need that$ X_S$ is $mathcal{F}_{S+} measurable ?
– user3503589
Dec 1 at 11:22
1
Hi @user3503589 1. I am convinced. If $X1_A=0$, $X$ must be zero on $A$. 2. I don't understand what you mean by "a convenience." A definition is a definition. 3. A conditional probability is a conditional expectation, as you know. And the definition of conditional expectation is such that a version must be measurable with respect to the conditioning $sigma$-algebra.
– AddSup
Dec 1 at 11:34
By a convenience I meant quite trivial. I am sorry but what I meant was, where do you get ("Markov property is an assertion that the conditional expectation given $mathcal{F}_{S+}$ , is equal to that $X_S$") when you explain the reasoning for progressive measurability.
– user3503589
Dec 1 at 11:48
1
Sorry for the ambiguity, I was simply reading a conditional probability as a conditional expectation whenever I see one. So by "the conditional expectation given $mathcal F_{S+}$" I meant "the conditional expectation of $1_{X_{S+t}inGamma}$ given $mathcal F_{S+}$.'' But it's true that the Markov property defined in terms of conditional probabilities is equivalent to the other well-known characterization involving conditional expectations of arbitrary bounded functions (Proposition 2.6.7(e')).
– AddSup
Dec 1 at 11:56
|
show 1 more comment
Thank you for very nice descriptive answer . But when you say I think you are correct do you mean maybe or you are convinced ? What bothers me is that this solution of mine to show that the conditional expectation is zero ( and yours too ) seems to be just a convenience because of the way the authors choose to define $X_{S+ t} in Gamma$ which made me wonder at the first place if my explanation is actually corrected.
– user3503589
Dec 1 at 11:11
I am not really sure if I understood your last paragraph . First because Markov property talks about conditional probabilities ( and what you said about conditional expectation is just a consequence) . Second I don’t see why would be need that$ X_S$ is $mathcal{F}_{S+} measurable ?
– user3503589
Dec 1 at 11:22
1
Hi @user3503589 1. I am convinced. If $X1_A=0$, $X$ must be zero on $A$. 2. I don't understand what you mean by "a convenience." A definition is a definition. 3. A conditional probability is a conditional expectation, as you know. And the definition of conditional expectation is such that a version must be measurable with respect to the conditioning $sigma$-algebra.
– AddSup
Dec 1 at 11:34
By a convenience I meant quite trivial. I am sorry but what I meant was, where do you get ("Markov property is an assertion that the conditional expectation given $mathcal{F}_{S+}$ , is equal to that $X_S$") when you explain the reasoning for progressive measurability.
– user3503589
Dec 1 at 11:48
1
Sorry for the ambiguity, I was simply reading a conditional probability as a conditional expectation whenever I see one. So by "the conditional expectation given $mathcal F_{S+}$" I meant "the conditional expectation of $1_{X_{S+t}inGamma}$ given $mathcal F_{S+}$.'' But it's true that the Markov property defined in terms of conditional probabilities is equivalent to the other well-known characterization involving conditional expectations of arbitrary bounded functions (Proposition 2.6.7(e')).
– AddSup
Dec 1 at 11:56
Thank you for very nice descriptive answer . But when you say I think you are correct do you mean maybe or you are convinced ? What bothers me is that this solution of mine to show that the conditional expectation is zero ( and yours too ) seems to be just a convenience because of the way the authors choose to define $X_{S+ t} in Gamma$ which made me wonder at the first place if my explanation is actually corrected.
– user3503589
Dec 1 at 11:11
Thank you for very nice descriptive answer . But when you say I think you are correct do you mean maybe or you are convinced ? What bothers me is that this solution of mine to show that the conditional expectation is zero ( and yours too ) seems to be just a convenience because of the way the authors choose to define $X_{S+ t} in Gamma$ which made me wonder at the first place if my explanation is actually corrected.
– user3503589
Dec 1 at 11:11
I am not really sure if I understood your last paragraph . First because Markov property talks about conditional probabilities ( and what you said about conditional expectation is just a consequence) . Second I don’t see why would be need that$ X_S$ is $mathcal{F}_{S+} measurable ?
– user3503589
Dec 1 at 11:22
I am not really sure if I understood your last paragraph . First because Markov property talks about conditional probabilities ( and what you said about conditional expectation is just a consequence) . Second I don’t see why would be need that$ X_S$ is $mathcal{F}_{S+} measurable ?
– user3503589
Dec 1 at 11:22
1
1
Hi @user3503589 1. I am convinced. If $X1_A=0$, $X$ must be zero on $A$. 2. I don't understand what you mean by "a convenience." A definition is a definition. 3. A conditional probability is a conditional expectation, as you know. And the definition of conditional expectation is such that a version must be measurable with respect to the conditioning $sigma$-algebra.
– AddSup
Dec 1 at 11:34
Hi @user3503589 1. I am convinced. If $X1_A=0$, $X$ must be zero on $A$. 2. I don't understand what you mean by "a convenience." A definition is a definition. 3. A conditional probability is a conditional expectation, as you know. And the definition of conditional expectation is such that a version must be measurable with respect to the conditioning $sigma$-algebra.
– AddSup
Dec 1 at 11:34
By a convenience I meant quite trivial. I am sorry but what I meant was, where do you get ("Markov property is an assertion that the conditional expectation given $mathcal{F}_{S+}$ , is equal to that $X_S$") when you explain the reasoning for progressive measurability.
– user3503589
Dec 1 at 11:48
By a convenience I meant quite trivial. I am sorry but what I meant was, where do you get ("Markov property is an assertion that the conditional expectation given $mathcal{F}_{S+}$ , is equal to that $X_S$") when you explain the reasoning for progressive measurability.
– user3503589
Dec 1 at 11:48
1
1
Sorry for the ambiguity, I was simply reading a conditional probability as a conditional expectation whenever I see one. So by "the conditional expectation given $mathcal F_{S+}$" I meant "the conditional expectation of $1_{X_{S+t}inGamma}$ given $mathcal F_{S+}$.'' But it's true that the Markov property defined in terms of conditional probabilities is equivalent to the other well-known characterization involving conditional expectations of arbitrary bounded functions (Proposition 2.6.7(e')).
– AddSup
Dec 1 at 11:56
Sorry for the ambiguity, I was simply reading a conditional probability as a conditional expectation whenever I see one. So by "the conditional expectation given $mathcal F_{S+}$" I meant "the conditional expectation of $1_{X_{S+t}inGamma}$ given $mathcal F_{S+}$.'' But it's true that the Markov property defined in terms of conditional probabilities is equivalent to the other well-known characterization involving conditional expectations of arbitrary bounded functions (Proposition 2.6.7(e')).
– AddSup
Dec 1 at 11:56
|
show 1 more comment
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020108%2fwhy-p-mux-st-in-gamma-mid-mathcalf-s-p-mux-st-in-gamma-m%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
