Showing the convergence in probabilty of two Ito integrals The Next CEO of Stack OverflowConvergence in $L_1$ and Convergence of the IntegralsQuestions about expectation of stochastic integralsComputing an Ito Integral using the DefinitionFinding a probability measure such that the time-shifted Brownian motion is also a Brownian motionAre Ito Integrals adapted to the Brownian Motion FiltrationHelp with change of measure and martingalesAre there examples where these two stochastic integrals are not independent?Convergence of Stochastic IntegralIto isometry with two independent Brownian motionsAre Ito Integrals adapted to the Brownian Motion Filtration 2
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Showing the convergence in probabilty of two Ito integrals
The Next CEO of Stack OverflowConvergence in $L_1$ and Convergence of the IntegralsQuestions about expectation of stochastic integralsComputing an Ito Integral using the DefinitionFinding a probability measure such that the time-shifted Brownian motion is also a Brownian motionAre Ito Integrals adapted to the Brownian Motion FiltrationHelp with change of measure and martingalesAre there examples where these two stochastic integrals are not independent?Convergence of Stochastic IntegralIto isometry with two independent Brownian motionsAre Ito Integrals adapted to the Brownian Motion Filtration 2
$begingroup$
Consider a Brownian Motion, $B(t)$, which are all $mathcal F_t$ measurable. Let $f,f_n$ be in $L_ad^2[a,b] times Omega$, which are the set of stochastic processes that are adapted to the filtration $mathcal F_t$ and satisfy $int_a^bEleft(f(t)^2right)dt < infty$. Suppose that $int_a^b|f(t)-f_n(t)|dtto0$ almost surely. Prove that $int_a^bf_n(t)dB(t) to int_a^bf_n(t)dB(t)$ in probability.
I am trying to show this two different ways. First by using $L^2$ convergence which will then imply convergence in probability. What I have for this way is the following:
Since $f$ and $f_n$ are in $L_ad^2$ then the ito integral $int_a^bf(t)-f_n(t)dB(t)$ exists and $Eleft|int_a^bf(t)-f_n(t)dB(t)right|^2=int_a^bEleft|f(t)-f_n(t)right|^2dt$, so it will suffice to show that the right hand side of this equal sign converges to $0$ which i'm unsure how to do.
Second, I am try to show this directly using the definition of convergence in probability.
$Pleft(left|int_a^bf(t)-f_n(t)dB(t)right|>epsilonright)lefrac1epsilonEleft|int_a^bf(t)-f_n(t)dB(t)right|$, hence if I could show that $int_a^bf(t)dB(t) to int_a^bf_n(t)dB(t)$ in $L^1$ then I would be done.
Similarly,
$ Pleft(left|int_a^bf(t)-f_n(t)dB(t)right|>epsilonright)lefrac1epsilon^2Eleft|int_a^bf(t)-f_n(t)dB(t)right|^2= frac1epsilon^2int_a^bEleft|f(t)-f_n(t)right|^2dt$ so if we could show this last term converged to $0$ then we would be done.
probability-theory convergence stochastic-calculus
asked Mar 18 at 20:13
alpastoralpastor
485314
$endgroup$
add a comment |
$begingroup$
Consider a Brownian Motion, $B(t)$, which are all $mathcal F_t$ measurable. Let $f,f_n$ be in $L_ad^2[a,b] times Omega$, which are the set of stochastic processes that are adapted to the filtration $mathcal F_t$ and satisfy $int_a^bEleft(f(t)^2right)dt < infty$. Suppose that $int_a^b|f(t)-f_n(t)|dtto0$ almost surely. Prove that $int_a^bf_n(t)dB(t) to int_a^bf_n(t)dB(t)$ in probability.
I am trying to show this two different ways. First by using $L^2$ convergence which will then imply convergence in probability. What I have for this way is the following:
Since $f$ and $f_n$ are in $L_ad^2$ then the ito integral $int_a^bf(t)-f_n(t)dB(t)$ exists and $Eleft|int_a^bf(t)-f_n(t)dB(t)right|^2=int_a^bEleft|f(t)-f_n(t)right|^2dt$, so it will suffice to show that the right hand side of this equal sign converges to $0$ which i'm unsure how to do.
Second, I am try to show this directly using the definition of convergence in probability.
$Pleft(left|int_a^bf(t)-f_n(t)dB(t)right|>epsilonright)lefrac1epsilonEleft|int_a^bf(t)-f_n(t)dB(t)right|$, hence if I could show that $int_a^bf(t)dB(t) to int_a^bf_n(t)dB(t)$ in $L^1$ then I would be done.
Similarly,
$ Pleft(left|int_a^bf(t)-f_n(t)dB(t)right|>epsilonright)lefrac1epsilon^2Eleft|int_a^bf(t)-f_n(t)dB(t)right|^2= frac1epsilon^2int_a^bEleft|f(t)-f_n(t)right|^2dt$ so if we could show this last term converged to $0$ then we would be done.
probability-theory convergence stochastic-calculus
asked Mar 18 at 20:13
alpastoralpastor
485314
$endgroup$
$begingroup$
For stochastic integrals $M_t := int_a^b f(s) , dB_s$ it holds that $mathbbE(M_t)=0$, note that this does not imply $mathbbE(|M_t|)=0$.
$endgroup$
– saz
Mar 19 at 10:41
$begingroup$
@saz I noticed that after I posted the question and will delete that from above, thank you though. Do you know how I should proceed with this problem?
$endgroup$
– alpastor
Mar 19 at 15:14
$begingroup$
As formulated, the statement is false. Maybe the assumption is that $int_a^b|f(t)-f_n(t)|^2dtto0$ a.s.?
$endgroup$
– zhoraster
Mar 19 at 16:49
$begingroup$
@zhoraster then could use the Generalized Lesbegue Dominated convergence and say $Eint_a^b|f(t)-f_n(t)|^2dtto0$ because $int_a^b|f(t)-f_n(t)|^2dt le 2int_a^b|f(t)|^2+|f_n(t)|^2dt$ and $2int_a^b|f(t)|^2+|f_n(t)|^2dt$ is in $L^1$ since $f,f_n in L_ad^2$ and if we assume $int_a^b|f(t)-f_n(t)|^2dtto0$ then $int_a^b|f_n(t)|^2dtto int_a^b|f(t)|^2dt$ which means $2int_a^b|f(t)|^2+|f_n(t)|^2dt to 4int_a^b|f(t)|^2dt$ which is also $L^1$.
$endgroup$
– alpastor
Mar 19 at 17:05
$begingroup$
I don't get your point. Currently the statement is false. You might need to edit the post.
$endgroup$
– zhoraster
Mar 20 at 4:30
add a comment |
$begingroup$
Consider a Brownian Motion, $B(t)$, which are all $mathcal F_t$ measurable. Let $f,f_n$ be in $L_ad^2[a,b] times Omega$, which are the set of stochastic processes that are adapted to the filtration $mathcal F_t$ and satisfy $int_a^bEleft(f(t)^2right)dt < infty$. Suppose that $int_a^b|f(t)-f_n(t)|dtto0$ almost surely. Prove that $int_a^bf_n(t)dB(t) to int_a^bf_n(t)dB(t)$ in probability.
I am trying to show this two different ways. First by using $L^2$ convergence which will then imply convergence in probability. What I have for this way is the following:
Since $f$ and $f_n$ are in $L_ad^2$ then the ito integral $int_a^bf(t)-f_n(t)dB(t)$ exists and $Eleft|int_a^bf(t)-f_n(t)dB(t)right|^2=int_a^bEleft|f(t)-f_n(t)right|^2dt$, so it will suffice to show that the right hand side of this equal sign converges to $0$ which i'm unsure how to do.
Second, I am try to show this directly using the definition of convergence in probability.
$Pleft(left|int_a^bf(t)-f_n(t)dB(t)right|>epsilonright)lefrac1epsilonEleft|int_a^bf(t)-f_n(t)dB(t)right|$, hence if I could show that $int_a^bf(t)dB(t) to int_a^bf_n(t)dB(t)$ in $L^1$ then I would be done.
Similarly,
$ Pleft(left|int_a^bf(t)-f_n(t)dB(t)right|>epsilonright)lefrac1epsilon^2Eleft|int_a^bf(t)-f_n(t)dB(t)right|^2= frac1epsilon^2int_a^bEleft|f(t)-f_n(t)right|^2dt$ so if we could show this last term converged to $0$ then we would be done.
probability-theory convergence stochastic-calculus
asked Mar 18 at 20:13
alpastoralpastor
485314
$endgroup$
Consider a Brownian Motion, $B(t)$, which are all $mathcal F_t$ measurable. Let $f,f_n$ be in $L_ad^2[a,b] times Omega$, which are the set of stochastic processes that are adapted to the filtration $mathcal F_t$ and satisfy $int_a^bEleft(f(t)^2right)dt < infty$. Suppose that $int_a^b|f(t)-f_n(t)|dtto0$ almost surely. Prove that $int_a^bf_n(t)dB(t) to int_a^bf_n(t)dB(t)$ in probability.
I am trying to show this two different ways. First by using $L^2$ convergence which will then imply convergence in probability. What I have for this way is the following:
Since $f$ and $f_n$ are in $L_ad^2$ then the ito integral $int_a^bf(t)-f_n(t)dB(t)$ exists and $Eleft|int_a^bf(t)-f_n(t)dB(t)right|^2=int_a^bEleft|f(t)-f_n(t)right|^2dt$, so it will suffice to show that the right hand side of this equal sign converges to $0$ which i'm unsure how to do.
Second, I am try to show this directly using the definition of convergence in probability.
$Pleft(left|int_a^bf(t)-f_n(t)dB(t)right|>epsilonright)lefrac1epsilonEleft|int_a^bf(t)-f_n(t)dB(t)right|$, hence if I could show that $int_a^bf(t)dB(t) to int_a^bf_n(t)dB(t)$ in $L^1$ then I would be done.
Similarly,
$ Pleft(left|int_a^bf(t)-f_n(t)dB(t)right|>epsilonright)lefrac1epsilon^2Eleft|int_a^bf(t)-f_n(t)dB(t)right|^2= frac1epsilon^2int_a^bEleft|f(t)-f_n(t)right|^2dt$ so if we could show this last term converged to $0$ then we would be done.
probability-theory convergence stochastic-calculus
probability-theory convergence stochastic-calculus
asked Mar 18 at 20:13
alpastoralpastor
485314
asked Mar 18 at 20:13
alpastoralpastor
485314
edited Mar 19 at 15:14
alpastor
asked Mar 18 at 20:13
alpastoralpastor
485314
asked Mar 18 at 20:13
alpastoralpastor
485314
asked Mar 18 at 20:13
alpastoralpastor
485314
485314
$begingroup$
For stochastic integrals $M_t := int_a^b f(s) , dB_s$ it holds that $mathbbE(M_t)=0$, note that this does not imply $mathbbE(|M_t|)=0$.
$endgroup$
– saz
Mar 19 at 10:41
$begingroup$
@saz I noticed that after I posted the question and will delete that from above, thank you though. Do you know how I should proceed with this problem?
$endgroup$
– alpastor
Mar 19 at 15:14
$begingroup$
As formulated, the statement is false. Maybe the assumption is that $int_a^b|f(t)-f_n(t)|^2dtto0$ a.s.?
$endgroup$
– zhoraster
Mar 19 at 16:49
$begingroup$
@zhoraster then could use the Generalized Lesbegue Dominated convergence and say $Eint_a^b|f(t)-f_n(t)|^2dtto0$ because $int_a^b|f(t)-f_n(t)|^2dt le 2int_a^b|f(t)|^2+|f_n(t)|^2dt$ and $2int_a^b|f(t)|^2+|f_n(t)|^2dt$ is in $L^1$ since $f,f_n in L_ad^2$ and if we assume $int_a^b|f(t)-f_n(t)|^2dtto0$ then $int_a^b|f_n(t)|^2dtto int_a^b|f(t)|^2dt$ which means $2int_a^b|f(t)|^2+|f_n(t)|^2dt to 4int_a^b|f(t)|^2dt$ which is also $L^1$.
$endgroup$
– alpastor
Mar 19 at 17:05
$begingroup$
I don't get your point. Currently the statement is false. You might need to edit the post.
$endgroup$
– zhoraster
Mar 20 at 4:30
add a comment |
$begingroup$
For stochastic integrals $M_t := int_a^b f(s) , dB_s$ it holds that $mathbbE(M_t)=0$, note that this does not imply $mathbbE(|M_t|)=0$.
$endgroup$
– saz
Mar 19 at 10:41
$begingroup$
@saz I noticed that after I posted the question and will delete that from above, thank you though. Do you know how I should proceed with this problem?
$endgroup$
– alpastor
Mar 19 at 15:14
$begingroup$
As formulated, the statement is false. Maybe the assumption is that $int_a^b|f(t)-f_n(t)|^2dtto0$ a.s.?
$endgroup$
– zhoraster
Mar 19 at 16:49
$begingroup$
@zhoraster then could use the Generalized Lesbegue Dominated convergence and say $Eint_a^b|f(t)-f_n(t)|^2dtto0$ because $int_a^b|f(t)-f_n(t)|^2dt le 2int_a^b|f(t)|^2+|f_n(t)|^2dt$ and $2int_a^b|f(t)|^2+|f_n(t)|^2dt$ is in $L^1$ since $f,f_n in L_ad^2$ and if we assume $int_a^b|f(t)-f_n(t)|^2dtto0$ then $int_a^b|f_n(t)|^2dtto int_a^b|f(t)|^2dt$ which means $2int_a^b|f(t)|^2+|f_n(t)|^2dt to 4int_a^b|f(t)|^2dt$ which is also $L^1$.
$endgroup$
– alpastor
Mar 19 at 17:05
$begingroup$
I don't get your point. Currently the statement is false. You might need to edit the post.
$endgroup$
– zhoraster
Mar 20 at 4:30
$begingroup$
For stochastic integrals $M_t := int_a^b f(s) , dB_s$ it holds that $mathbbE(M_t)=0$, note that this does not imply $mathbbE(|M_t|)=0$.
$endgroup$
– saz
Mar 19 at 10:41
$begingroup$
For stochastic integrals $M_t := int_a^b f(s) , dB_s$ it holds that $mathbbE(M_t)=0$, note that this does not imply $mathbbE(|M_t|)=0$.
$endgroup$
– saz
Mar 19 at 10:41
$begingroup$
@saz I noticed that after I posted the question and will delete that from above, thank you though. Do you know how I should proceed with this problem?
$endgroup$
– alpastor
Mar 19 at 15:14
$begingroup$
@saz I noticed that after I posted the question and will delete that from above, thank you though. Do you know how I should proceed with this problem?
$endgroup$
– alpastor
Mar 19 at 15:14
$begingroup$
As formulated, the statement is false. Maybe the assumption is that $int_a^b|f(t)-f_n(t)|^2dtto0$ a.s.?
$endgroup$
– zhoraster
Mar 19 at 16:49
$begingroup$
As formulated, the statement is false. Maybe the assumption is that $int_a^b|f(t)-f_n(t)|^2dtto0$ a.s.?
$endgroup$
– zhoraster
Mar 19 at 16:49
$begingroup$
@zhoraster then could use the Generalized Lesbegue Dominated convergence and say $Eint_a^b|f(t)-f_n(t)|^2dtto0$ because $int_a^b|f(t)-f_n(t)|^2dt le 2int_a^b|f(t)|^2+|f_n(t)|^2dt$ and $2int_a^b|f(t)|^2+|f_n(t)|^2dt$ is in $L^1$ since $f,f_n in L_ad^2$ and if we assume $int_a^b|f(t)-f_n(t)|^2dtto0$ then $int_a^b|f_n(t)|^2dtto int_a^b|f(t)|^2dt$ which means $2int_a^b|f(t)|^2+|f_n(t)|^2dt to 4int_a^b|f(t)|^2dt$ which is also $L^1$.
$endgroup$
– alpastor
Mar 19 at 17:05
$begingroup$
@zhoraster then could use the Generalized Lesbegue Dominated convergence and say $Eint_a^b|f(t)-f_n(t)|^2dtto0$ because $int_a^b|f(t)-f_n(t)|^2dt le 2int_a^b|f(t)|^2+|f_n(t)|^2dt$ and $2int_a^b|f(t)|^2+|f_n(t)|^2dt$ is in $L^1$ since $f,f_n in L_ad^2$ and if we assume $int_a^b|f(t)-f_n(t)|^2dtto0$ then $int_a^b|f_n(t)|^2dtto int_a^b|f(t)|^2dt$ which means $2int_a^b|f(t)|^2+|f_n(t)|^2dt to 4int_a^b|f(t)|^2dt$ which is also $L^1$.
$endgroup$
– alpastor
Mar 19 at 17:05
$begingroup$
I don't get your point. Currently the statement is false. You might need to edit the post.
$endgroup$
– zhoraster
Mar 20 at 4:30
$begingroup$
I don't get your point. Currently the statement is false. You might need to edit the post.
$endgroup$
– zhoraster
Mar 20 at 4:30
add a comment |
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$begingroup$
For stochastic integrals $M_t := int_a^b f(s) , dB_s$ it holds that $mathbbE(M_t)=0$, note that this does not imply $mathbbE(|M_t|)=0$.
$endgroup$
– saz
Mar 19 at 10:41
$begingroup$
@saz I noticed that after I posted the question and will delete that from above, thank you though. Do you know how I should proceed with this problem?
$endgroup$
– alpastor
Mar 19 at 15:14
$begingroup$
As formulated, the statement is false. Maybe the assumption is that $int_a^b|f(t)-f_n(t)|^2dtto0$ a.s.?
$endgroup$
– zhoraster
Mar 19 at 16:49
$begingroup$
@zhoraster then could use the Generalized Lesbegue Dominated convergence and say $Eint_a^b|f(t)-f_n(t)|^2dtto0$ because $int_a^b|f(t)-f_n(t)|^2dt le 2int_a^b|f(t)|^2+|f_n(t)|^2dt$ and $2int_a^b|f(t)|^2+|f_n(t)|^2dt$ is in $L^1$ since $f,f_n in L_ad^2$ and if we assume $int_a^b|f(t)-f_n(t)|^2dtto0$ then $int_a^b|f_n(t)|^2dtto int_a^b|f(t)|^2dt$ which means $2int_a^b|f(t)|^2+|f_n(t)|^2dt to 4int_a^b|f(t)|^2dt$ which is also $L^1$.
$endgroup$
– alpastor
Mar 19 at 17:05
$begingroup$
I don't get your point. Currently the statement is false. You might need to edit the post.
$endgroup$
– zhoraster
Mar 20 at 4:30