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Is this the definition of stochastic integral up to a stopping/explosion time?
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Edit: If anyone has any tips how to improve this question to get an answer or can tell me if some part is not understandable that would be very nice.
I always had problems understanding the definition of the stochastic integral up to a stopping time. I now present how I understand it and what problems I have in this understanding.
My understanding
1. I see two reasons for the stochastic integral $int_0^T G_s dB_s$ being only defined up to a stopping time. Either the integral explodes at $T$ or $G_s$ is only defined up to $T$ (for example $G_s = sigma(X_s)$ in the solution of a SDE and $X_s$ leaves the domain of $sigma$ at time $T$).
2. If it is defined up to this stopping time $T$ let $T_k$ be an increasing sequence of stopping times with $T = sup T_k$ s.t. $G^k_s = G_s mathbb1_s le T_k in L^2([0, infty], dt)$ and thus $I^k(G)_t :=int_0^t G^k_s dB_t$ exists and is a square integrable martingale (assuming $G_s$ is adapted). Now we choose one representative $hatI^k(G)$ out of the equivalence class $I^k(G)$(the integral is only defined as $L^2$ equivalence class) for each $k$ and since for $n > m$ we know that $G^n_s wedge T_m$ and $G^m_s wedge T_m$ are indistinguishable we know that $hatI^n(G)$ and $hatI^m(G)$ coincide on almost every path $omega$ up to $T_m(omega)$. So we can define some $hatI(G)$ which is defined up to $T = sup T_k$ by just throwing the countably many measure-zero sets (one for each $k$) away and just setting $hatI(G)(omega) = 0$ on those paths.
This is now well defined up to measure $0$ and can be seen as representative of our stochastic integral up to $T$.
Questions I have with my understanding
- (i) Can this be made less complicated?
- (ii) I just assumed that these stopping times $T_k$ exist. Do they always exist? The integral $int_0^T G_s dB_s$ is defined up to stopping time $T$ - how would I construct these $T_k$ that are so crucial to my definition.
- (iii) What kind of topology do I know have on these stochastic integrals up to stopping time $T$? Somehow for normal stochastic integrals I know exactly in which "space" my integrals are ($M_c^2([0,t])$, the space of square integrable continous martingales endowed with the norm $||M_s|| = mathbbE[sup_s le t |M_s|^2]$). I somehow can't grasp what this $int_0^T G_s dB_s$ actually is. Can I see them as element of any topological or even banach space?
probability-theory stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis
asked Jul 18 at 19:20
ImprobableQuestions
62
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up vote
1
down vote
favorite
Edit: If anyone has any tips how to improve this question to get an answer or can tell me if some part is not understandable that would be very nice.
I always had problems understanding the definition of the stochastic integral up to a stopping time. I now present how I understand it and what problems I have in this understanding.
My understanding
1. I see two reasons for the stochastic integral $int_0^T G_s dB_s$ being only defined up to a stopping time. Either the integral explodes at $T$ or $G_s$ is only defined up to $T$ (for example $G_s = sigma(X_s)$ in the solution of a SDE and $X_s$ leaves the domain of $sigma$ at time $T$).
2. If it is defined up to this stopping time $T$ let $T_k$ be an increasing sequence of stopping times with $T = sup T_k$ s.t. $G^k_s = G_s mathbb1_s le T_k in L^2([0, infty], dt)$ and thus $I^k(G)_t :=int_0^t G^k_s dB_t$ exists and is a square integrable martingale (assuming $G_s$ is adapted). Now we choose one representative $hatI^k(G)$ out of the equivalence class $I^k(G)$(the integral is only defined as $L^2$ equivalence class) for each $k$ and since for $n > m$ we know that $G^n_s wedge T_m$ and $G^m_s wedge T_m$ are indistinguishable we know that $hatI^n(G)$ and $hatI^m(G)$ coincide on almost every path $omega$ up to $T_m(omega)$. So we can define some $hatI(G)$ which is defined up to $T = sup T_k$ by just throwing the countably many measure-zero sets (one for each $k$) away and just setting $hatI(G)(omega) = 0$ on those paths.
This is now well defined up to measure $0$ and can be seen as representative of our stochastic integral up to $T$.
Questions I have with my understanding
- (i) Can this be made less complicated?
- (ii) I just assumed that these stopping times $T_k$ exist. Do they always exist? The integral $int_0^T G_s dB_s$ is defined up to stopping time $T$ - how would I construct these $T_k$ that are so crucial to my definition.
- (iii) What kind of topology do I know have on these stochastic integrals up to stopping time $T$? Somehow for normal stochastic integrals I know exactly in which "space" my integrals are ($M_c^2([0,t])$, the space of square integrable continous martingales endowed with the norm $||M_s|| = mathbbE[sup_s le t |M_s|^2]$). I somehow can't grasp what this $int_0^T G_s dB_s$ actually is. Can I see them as element of any topological or even banach space?
probability-theory stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis
asked Jul 18 at 19:20
ImprobableQuestions
62
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Edit: If anyone has any tips how to improve this question to get an answer or can tell me if some part is not understandable that would be very nice.
I always had problems understanding the definition of the stochastic integral up to a stopping time. I now present how I understand it and what problems I have in this understanding.
My understanding
1. I see two reasons for the stochastic integral $int_0^T G_s dB_s$ being only defined up to a stopping time. Either the integral explodes at $T$ or $G_s$ is only defined up to $T$ (for example $G_s = sigma(X_s)$ in the solution of a SDE and $X_s$ leaves the domain of $sigma$ at time $T$).
2. If it is defined up to this stopping time $T$ let $T_k$ be an increasing sequence of stopping times with $T = sup T_k$ s.t. $G^k_s = G_s mathbb1_s le T_k in L^2([0, infty], dt)$ and thus $I^k(G)_t :=int_0^t G^k_s dB_t$ exists and is a square integrable martingale (assuming $G_s$ is adapted). Now we choose one representative $hatI^k(G)$ out of the equivalence class $I^k(G)$(the integral is only defined as $L^2$ equivalence class) for each $k$ and since for $n > m$ we know that $G^n_s wedge T_m$ and $G^m_s wedge T_m$ are indistinguishable we know that $hatI^n(G)$ and $hatI^m(G)$ coincide on almost every path $omega$ up to $T_m(omega)$. So we can define some $hatI(G)$ which is defined up to $T = sup T_k$ by just throwing the countably many measure-zero sets (one for each $k$) away and just setting $hatI(G)(omega) = 0$ on those paths.
This is now well defined up to measure $0$ and can be seen as representative of our stochastic integral up to $T$.
Questions I have with my understanding
- (i) Can this be made less complicated?
- (ii) I just assumed that these stopping times $T_k$ exist. Do they always exist? The integral $int_0^T G_s dB_s$ is defined up to stopping time $T$ - how would I construct these $T_k$ that are so crucial to my definition.
- (iii) What kind of topology do I know have on these stochastic integrals up to stopping time $T$? Somehow for normal stochastic integrals I know exactly in which "space" my integrals are ($M_c^2([0,t])$, the space of square integrable continous martingales endowed with the norm $||M_s|| = mathbbE[sup_s le t |M_s|^2]$). I somehow can't grasp what this $int_0^T G_s dB_s$ actually is. Can I see them as element of any topological or even banach space?
probability-theory stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis
asked Jul 18 at 19:20
ImprobableQuestions
62
Edit: If anyone has any tips how to improve this question to get an answer or can tell me if some part is not understandable that would be very nice.
I always had problems understanding the definition of the stochastic integral up to a stopping time. I now present how I understand it and what problems I have in this understanding.
My understanding
1. I see two reasons for the stochastic integral $int_0^T G_s dB_s$ being only defined up to a stopping time. Either the integral explodes at $T$ or $G_s$ is only defined up to $T$ (for example $G_s = sigma(X_s)$ in the solution of a SDE and $X_s$ leaves the domain of $sigma$ at time $T$).
2. If it is defined up to this stopping time $T$ let $T_k$ be an increasing sequence of stopping times with $T = sup T_k$ s.t. $G^k_s = G_s mathbb1_s le T_k in L^2([0, infty], dt)$ and thus $I^k(G)_t :=int_0^t G^k_s dB_t$ exists and is a square integrable martingale (assuming $G_s$ is adapted). Now we choose one representative $hatI^k(G)$ out of the equivalence class $I^k(G)$(the integral is only defined as $L^2$ equivalence class) for each $k$ and since for $n > m$ we know that $G^n_s wedge T_m$ and $G^m_s wedge T_m$ are indistinguishable we know that $hatI^n(G)$ and $hatI^m(G)$ coincide on almost every path $omega$ up to $T_m(omega)$. So we can define some $hatI(G)$ which is defined up to $T = sup T_k$ by just throwing the countably many measure-zero sets (one for each $k$) away and just setting $hatI(G)(omega) = 0$ on those paths.
This is now well defined up to measure $0$ and can be seen as representative of our stochastic integral up to $T$.
Questions I have with my understanding
- (i) Can this be made less complicated?
- (ii) I just assumed that these stopping times $T_k$ exist. Do they always exist? The integral $int_0^T G_s dB_s$ is defined up to stopping time $T$ - how would I construct these $T_k$ that are so crucial to my definition.
- (iii) What kind of topology do I know have on these stochastic integrals up to stopping time $T$? Somehow for normal stochastic integrals I know exactly in which "space" my integrals are ($M_c^2([0,t])$, the space of square integrable continous martingales endowed with the norm $||M_s|| = mathbbE[sup_s le t |M_s|^2]$). I somehow can't grasp what this $int_0^T G_s dB_s$ actually is. Can I see them as element of any topological or even banach space?
probability-theory stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis
asked Jul 18 at 19:20
ImprobableQuestions
62
edited Jul 20 at 7:37
asked Jul 18 at 19:20
ImprobableQuestions
62
asked Jul 18 at 19:20
ImprobableQuestions
62
asked Jul 18 at 19:20
ImprobableQuestions
62
62
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