Interchanging Malliavin derivative with Lebesgue integralLinear transformation of Levy processesRoyden - section 4.2, page 73 - linearityMalliavin derivative of a Lebesgue integral.Further Reading on Stochastic Calculus/AnalysisWhat does it mean to integrate a Brownian motion with respect to time?Malliavin derivative with respect to part of samples.Product of functions with finite chaos expansion is in $L^2(P)$Hilbert space-valued $L^2$ random variables.Semimartingale characteristics for stochastic integral?A question about Malliavin derivatives of Lebesgue integrals of processes

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Interchanging Malliavin derivative with Lebesgue integral


Linear transformation of Levy processesRoyden - section 4.2, page 73 - linearityMalliavin derivative of a Lebesgue integral.Further Reading on Stochastic Calculus/AnalysisWhat does it mean to integrate a Brownian motion with respect to time?Malliavin derivative with respect to part of samples.Product of functions with finite chaos expansion is in $L^2(P)$Hilbert space-valued $L^2$ random variables.Semimartingale characteristics for stochastic integral?A question about Malliavin derivatives of Lebesgue integrals of processes













4












$begingroup$


I am reading Oksendal's book "Malliavin calculus for Levy processes with application to finance". In the proof of Lemma 4.9 (page 47), the author interchanges the Malliavin derivative $D_t$ with the Lebesgue integral $ds$.
$$D_tint_0^T u^2(s),ds = 2int_0^T u(s)D_tu(s),ds$$
Could anyone shed any light?










share|cite|improve this question











$endgroup$











  • $begingroup$
    As $D_t$ is a derivative, $D_t$ is linear operator on $mathbb D_1,2$ and $D_t[u^2(s)]= 2 u(s) D_t[u(s)]$.
    $endgroup$
    – Zbigniew
    Nov 28 '15 at 13:41











  • $begingroup$
    @Zbigniew but why does the linearity guarantee that we can do the interchange? Could you please be more specific?
    $endgroup$
    – ting
    Nov 28 '15 at 16:05






  • 1




    $begingroup$
    @Zbigniew why can we differentiate it inside
    $endgroup$
    – ting
    Dec 1 '15 at 13:45















4












$begingroup$


I am reading Oksendal's book "Malliavin calculus for Levy processes with application to finance". In the proof of Lemma 4.9 (page 47), the author interchanges the Malliavin derivative $D_t$ with the Lebesgue integral $ds$.
$$D_tint_0^T u^2(s),ds = 2int_0^T u(s)D_tu(s),ds$$
Could anyone shed any light?










share|cite|improve this question











$endgroup$











  • $begingroup$
    As $D_t$ is a derivative, $D_t$ is linear operator on $mathbb D_1,2$ and $D_t[u^2(s)]= 2 u(s) D_t[u(s)]$.
    $endgroup$
    – Zbigniew
    Nov 28 '15 at 13:41











  • $begingroup$
    @Zbigniew but why does the linearity guarantee that we can do the interchange? Could you please be more specific?
    $endgroup$
    – ting
    Nov 28 '15 at 16:05






  • 1




    $begingroup$
    @Zbigniew why can we differentiate it inside
    $endgroup$
    – ting
    Dec 1 '15 at 13:45













4












4








4


1



$begingroup$


I am reading Oksendal's book "Malliavin calculus for Levy processes with application to finance". In the proof of Lemma 4.9 (page 47), the author interchanges the Malliavin derivative $D_t$ with the Lebesgue integral $ds$.
$$D_tint_0^T u^2(s),ds = 2int_0^T u(s)D_tu(s),ds$$
Could anyone shed any light?










share|cite|improve this question











$endgroup$




I am reading Oksendal's book "Malliavin calculus for Levy processes with application to finance". In the proof of Lemma 4.9 (page 47), the author interchanges the Malliavin derivative $D_t$ with the Lebesgue integral $ds$.
$$D_tint_0^T u^2(s),ds = 2int_0^T u(s)D_tu(s),ds$$
Could anyone shed any light?







probability-theory stochastic-processes lebesgue-integral stochastic-analysis malliavin-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 28 '18 at 1:22







user223391

















asked Nov 27 '15 at 22:39









tingting

212




212











  • $begingroup$
    As $D_t$ is a derivative, $D_t$ is linear operator on $mathbb D_1,2$ and $D_t[u^2(s)]= 2 u(s) D_t[u(s)]$.
    $endgroup$
    – Zbigniew
    Nov 28 '15 at 13:41











  • $begingroup$
    @Zbigniew but why does the linearity guarantee that we can do the interchange? Could you please be more specific?
    $endgroup$
    – ting
    Nov 28 '15 at 16:05






  • 1




    $begingroup$
    @Zbigniew why can we differentiate it inside
    $endgroup$
    – ting
    Dec 1 '15 at 13:45
















  • $begingroup$
    As $D_t$ is a derivative, $D_t$ is linear operator on $mathbb D_1,2$ and $D_t[u^2(s)]= 2 u(s) D_t[u(s)]$.
    $endgroup$
    – Zbigniew
    Nov 28 '15 at 13:41











  • $begingroup$
    @Zbigniew but why does the linearity guarantee that we can do the interchange? Could you please be more specific?
    $endgroup$
    – ting
    Nov 28 '15 at 16:05






  • 1




    $begingroup$
    @Zbigniew why can we differentiate it inside
    $endgroup$
    – ting
    Dec 1 '15 at 13:45















$begingroup$
As $D_t$ is a derivative, $D_t$ is linear operator on $mathbb D_1,2$ and $D_t[u^2(s)]= 2 u(s) D_t[u(s)]$.
$endgroup$
– Zbigniew
Nov 28 '15 at 13:41





$begingroup$
As $D_t$ is a derivative, $D_t$ is linear operator on $mathbb D_1,2$ and $D_t[u^2(s)]= 2 u(s) D_t[u(s)]$.
$endgroup$
– Zbigniew
Nov 28 '15 at 13:41













$begingroup$
@Zbigniew but why does the linearity guarantee that we can do the interchange? Could you please be more specific?
$endgroup$
– ting
Nov 28 '15 at 16:05




$begingroup$
@Zbigniew but why does the linearity guarantee that we can do the interchange? Could you please be more specific?
$endgroup$
– ting
Nov 28 '15 at 16:05




1




1




$begingroup$
@Zbigniew why can we differentiate it inside
$endgroup$
– ting
Dec 1 '15 at 13:45




$begingroup$
@Zbigniew why can we differentiate it inside
$endgroup$
– ting
Dec 1 '15 at 13:45










1 Answer
1






active

oldest

votes


















1












$begingroup$

We'll prove this in two steps. First, we pull the derivative operator $mathfrakD$ inside the (Lebesgue) integral, and then apply a "Chain Rule". Since the book by Øksendal et al. is referenced, I won't prove the "Chain Rule", but establish the interchange of the derivative and the integral through some density arguments.



The proof is not that hard, but the notation annoying though quite intuitive; I'll provide more clarification if needed.



Consider any process $UinmathscrL^2(mathopen[0,inftymathclose[;mathscrW^1,2)$, where $mathscrW^1,2$ is the usual Watanabe–Sobolev space of Malliavin-differentiable functions (Nualart's and Øksendal's $mathbbD^1,2$, but I hate the hollow 'D', and this is what Hairer uses). Note that this is the Hilbert space of processes $UinmathscrL^2(varOmegatimesmathopen[0,inftymathclose[;mathbbR)$ for which




  • $U(t)in mathscrW^1,2$, for (almost) all $tinmathopen[0,inftymathclose[$;

  • there exists a version of the process $(omega,tau,t)mapstomathfrakD_tau U(omega,t)$ in $mathscrL^2(varOmegatimesmathopen[0,inftymathclose[^2;mathbbR)$.

It can be shown that processes $U_nu$ of the form
$$U_nu(t):= sum_j=1^nu X_j h_j(t),$$
where $X_j$ is a smooth and cylindrical random variable and $h_jinmathscrL^2(mathopen[0,inftymathclose[;mathbbR)$, are dense in $mathscrL^2(mathopen[0,inftymathclose[;mathscrW^1,2)$. In fact, $mathscrL^2(mathopen[0,inftymathclose[;mathscrW^1,2)$ is defined as the closure of the space of such process under a suitable norm which we will not need here.



This means that we can find a sequence of such processes $(U_nu)_nuinmathbbN$ such that
$$U_nuto U~textin~mathscrL^2(varOmega;mathscrL^2(mathopen[0,inftymathclose[;mathbbR)),~textas~nutoinfty,$$
and
$$mathfrakD_cdot U_nuto mathfrakD_cdot U~textin~mathscrL^2(varOmega;mathscrL^2(mathopen[0,inftymathclose[;mathbbR)^otimes 2),~textas~nutoinfty.$$



A consequence of the above is that
$$int_0^T U_nu(t)~!mathrmdtto int_0^T U(t)~!mathrmdt~textin~mathscrL^2(varOmega;mathbbR),~textas~nutoinfty,$$
and
$$mathfrakD_cdotint_0^T U_nu(t)~!mathrmdtto mathfrakD_cdotint_0^T U(t)~!mathrmdt~textin~mathscrL^2(varOmega;mathscrL^2(mathopen[0,inftymathclose[;mathbbR)),~textas~nutoinfty.$$



And these imply,
$$mathfrakD_tauint_0^T U(t)~!mathrmdt = int_0^T mathfrakD_tau U(t)~!mathrmdt.$$



Now, set $U(t) = u(t)^2$. Then, by the "Chain Rule" for Malliavin derivatives (Theorem 3.5 in Øksendal et al.),
$$mathfrakD_tauint_0^T u(t)^2~!mathrmdt = int_0^T mathfrakD_tau big(u(t)^2big)~!mathrmdt = 2int_0^T u(t)mathfrakD_tau u(t)~!mathrmdt.$$






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    $begingroup$

    We'll prove this in two steps. First, we pull the derivative operator $mathfrakD$ inside the (Lebesgue) integral, and then apply a "Chain Rule". Since the book by Øksendal et al. is referenced, I won't prove the "Chain Rule", but establish the interchange of the derivative and the integral through some density arguments.



    The proof is not that hard, but the notation annoying though quite intuitive; I'll provide more clarification if needed.



    Consider any process $UinmathscrL^2(mathopen[0,inftymathclose[;mathscrW^1,2)$, where $mathscrW^1,2$ is the usual Watanabe–Sobolev space of Malliavin-differentiable functions (Nualart's and Øksendal's $mathbbD^1,2$, but I hate the hollow 'D', and this is what Hairer uses). Note that this is the Hilbert space of processes $UinmathscrL^2(varOmegatimesmathopen[0,inftymathclose[;mathbbR)$ for which




    • $U(t)in mathscrW^1,2$, for (almost) all $tinmathopen[0,inftymathclose[$;

    • there exists a version of the process $(omega,tau,t)mapstomathfrakD_tau U(omega,t)$ in $mathscrL^2(varOmegatimesmathopen[0,inftymathclose[^2;mathbbR)$.

    It can be shown that processes $U_nu$ of the form
    $$U_nu(t):= sum_j=1^nu X_j h_j(t),$$
    where $X_j$ is a smooth and cylindrical random variable and $h_jinmathscrL^2(mathopen[0,inftymathclose[;mathbbR)$, are dense in $mathscrL^2(mathopen[0,inftymathclose[;mathscrW^1,2)$. In fact, $mathscrL^2(mathopen[0,inftymathclose[;mathscrW^1,2)$ is defined as the closure of the space of such process under a suitable norm which we will not need here.



    This means that we can find a sequence of such processes $(U_nu)_nuinmathbbN$ such that
    $$U_nuto U~textin~mathscrL^2(varOmega;mathscrL^2(mathopen[0,inftymathclose[;mathbbR)),~textas~nutoinfty,$$
    and
    $$mathfrakD_cdot U_nuto mathfrakD_cdot U~textin~mathscrL^2(varOmega;mathscrL^2(mathopen[0,inftymathclose[;mathbbR)^otimes 2),~textas~nutoinfty.$$



    A consequence of the above is that
    $$int_0^T U_nu(t)~!mathrmdtto int_0^T U(t)~!mathrmdt~textin~mathscrL^2(varOmega;mathbbR),~textas~nutoinfty,$$
    and
    $$mathfrakD_cdotint_0^T U_nu(t)~!mathrmdtto mathfrakD_cdotint_0^T U(t)~!mathrmdt~textin~mathscrL^2(varOmega;mathscrL^2(mathopen[0,inftymathclose[;mathbbR)),~textas~nutoinfty.$$



    And these imply,
    $$mathfrakD_tauint_0^T U(t)~!mathrmdt = int_0^T mathfrakD_tau U(t)~!mathrmdt.$$



    Now, set $U(t) = u(t)^2$. Then, by the "Chain Rule" for Malliavin derivatives (Theorem 3.5 in Øksendal et al.),
    $$mathfrakD_tauint_0^T u(t)^2~!mathrmdt = int_0^T mathfrakD_tau big(u(t)^2big)~!mathrmdt = 2int_0^T u(t)mathfrakD_tau u(t)~!mathrmdt.$$






    share|cite|improve this answer











    $endgroup$

















      1












      $begingroup$

      We'll prove this in two steps. First, we pull the derivative operator $mathfrakD$ inside the (Lebesgue) integral, and then apply a "Chain Rule". Since the book by Øksendal et al. is referenced, I won't prove the "Chain Rule", but establish the interchange of the derivative and the integral through some density arguments.



      The proof is not that hard, but the notation annoying though quite intuitive; I'll provide more clarification if needed.



      Consider any process $UinmathscrL^2(mathopen[0,inftymathclose[;mathscrW^1,2)$, where $mathscrW^1,2$ is the usual Watanabe–Sobolev space of Malliavin-differentiable functions (Nualart's and Øksendal's $mathbbD^1,2$, but I hate the hollow 'D', and this is what Hairer uses). Note that this is the Hilbert space of processes $UinmathscrL^2(varOmegatimesmathopen[0,inftymathclose[;mathbbR)$ for which




      • $U(t)in mathscrW^1,2$, for (almost) all $tinmathopen[0,inftymathclose[$;

      • there exists a version of the process $(omega,tau,t)mapstomathfrakD_tau U(omega,t)$ in $mathscrL^2(varOmegatimesmathopen[0,inftymathclose[^2;mathbbR)$.

      It can be shown that processes $U_nu$ of the form
      $$U_nu(t):= sum_j=1^nu X_j h_j(t),$$
      where $X_j$ is a smooth and cylindrical random variable and $h_jinmathscrL^2(mathopen[0,inftymathclose[;mathbbR)$, are dense in $mathscrL^2(mathopen[0,inftymathclose[;mathscrW^1,2)$. In fact, $mathscrL^2(mathopen[0,inftymathclose[;mathscrW^1,2)$ is defined as the closure of the space of such process under a suitable norm which we will not need here.



      This means that we can find a sequence of such processes $(U_nu)_nuinmathbbN$ such that
      $$U_nuto U~textin~mathscrL^2(varOmega;mathscrL^2(mathopen[0,inftymathclose[;mathbbR)),~textas~nutoinfty,$$
      and
      $$mathfrakD_cdot U_nuto mathfrakD_cdot U~textin~mathscrL^2(varOmega;mathscrL^2(mathopen[0,inftymathclose[;mathbbR)^otimes 2),~textas~nutoinfty.$$



      A consequence of the above is that
      $$int_0^T U_nu(t)~!mathrmdtto int_0^T U(t)~!mathrmdt~textin~mathscrL^2(varOmega;mathbbR),~textas~nutoinfty,$$
      and
      $$mathfrakD_cdotint_0^T U_nu(t)~!mathrmdtto mathfrakD_cdotint_0^T U(t)~!mathrmdt~textin~mathscrL^2(varOmega;mathscrL^2(mathopen[0,inftymathclose[;mathbbR)),~textas~nutoinfty.$$



      And these imply,
      $$mathfrakD_tauint_0^T U(t)~!mathrmdt = int_0^T mathfrakD_tau U(t)~!mathrmdt.$$



      Now, set $U(t) = u(t)^2$. Then, by the "Chain Rule" for Malliavin derivatives (Theorem 3.5 in Øksendal et al.),
      $$mathfrakD_tauint_0^T u(t)^2~!mathrmdt = int_0^T mathfrakD_tau big(u(t)^2big)~!mathrmdt = 2int_0^T u(t)mathfrakD_tau u(t)~!mathrmdt.$$






      share|cite|improve this answer











      $endgroup$















        1












        1








        1





        $begingroup$

        We'll prove this in two steps. First, we pull the derivative operator $mathfrakD$ inside the (Lebesgue) integral, and then apply a "Chain Rule". Since the book by Øksendal et al. is referenced, I won't prove the "Chain Rule", but establish the interchange of the derivative and the integral through some density arguments.



        The proof is not that hard, but the notation annoying though quite intuitive; I'll provide more clarification if needed.



        Consider any process $UinmathscrL^2(mathopen[0,inftymathclose[;mathscrW^1,2)$, where $mathscrW^1,2$ is the usual Watanabe–Sobolev space of Malliavin-differentiable functions (Nualart's and Øksendal's $mathbbD^1,2$, but I hate the hollow 'D', and this is what Hairer uses). Note that this is the Hilbert space of processes $UinmathscrL^2(varOmegatimesmathopen[0,inftymathclose[;mathbbR)$ for which




        • $U(t)in mathscrW^1,2$, for (almost) all $tinmathopen[0,inftymathclose[$;

        • there exists a version of the process $(omega,tau,t)mapstomathfrakD_tau U(omega,t)$ in $mathscrL^2(varOmegatimesmathopen[0,inftymathclose[^2;mathbbR)$.

        It can be shown that processes $U_nu$ of the form
        $$U_nu(t):= sum_j=1^nu X_j h_j(t),$$
        where $X_j$ is a smooth and cylindrical random variable and $h_jinmathscrL^2(mathopen[0,inftymathclose[;mathbbR)$, are dense in $mathscrL^2(mathopen[0,inftymathclose[;mathscrW^1,2)$. In fact, $mathscrL^2(mathopen[0,inftymathclose[;mathscrW^1,2)$ is defined as the closure of the space of such process under a suitable norm which we will not need here.



        This means that we can find a sequence of such processes $(U_nu)_nuinmathbbN$ such that
        $$U_nuto U~textin~mathscrL^2(varOmega;mathscrL^2(mathopen[0,inftymathclose[;mathbbR)),~textas~nutoinfty,$$
        and
        $$mathfrakD_cdot U_nuto mathfrakD_cdot U~textin~mathscrL^2(varOmega;mathscrL^2(mathopen[0,inftymathclose[;mathbbR)^otimes 2),~textas~nutoinfty.$$



        A consequence of the above is that
        $$int_0^T U_nu(t)~!mathrmdtto int_0^T U(t)~!mathrmdt~textin~mathscrL^2(varOmega;mathbbR),~textas~nutoinfty,$$
        and
        $$mathfrakD_cdotint_0^T U_nu(t)~!mathrmdtto mathfrakD_cdotint_0^T U(t)~!mathrmdt~textin~mathscrL^2(varOmega;mathscrL^2(mathopen[0,inftymathclose[;mathbbR)),~textas~nutoinfty.$$



        And these imply,
        $$mathfrakD_tauint_0^T U(t)~!mathrmdt = int_0^T mathfrakD_tau U(t)~!mathrmdt.$$



        Now, set $U(t) = u(t)^2$. Then, by the "Chain Rule" for Malliavin derivatives (Theorem 3.5 in Øksendal et al.),
        $$mathfrakD_tauint_0^T u(t)^2~!mathrmdt = int_0^T mathfrakD_tau big(u(t)^2big)~!mathrmdt = 2int_0^T u(t)mathfrakD_tau u(t)~!mathrmdt.$$






        share|cite|improve this answer











        $endgroup$



        We'll prove this in two steps. First, we pull the derivative operator $mathfrakD$ inside the (Lebesgue) integral, and then apply a "Chain Rule". Since the book by Øksendal et al. is referenced, I won't prove the "Chain Rule", but establish the interchange of the derivative and the integral through some density arguments.



        The proof is not that hard, but the notation annoying though quite intuitive; I'll provide more clarification if needed.



        Consider any process $UinmathscrL^2(mathopen[0,inftymathclose[;mathscrW^1,2)$, where $mathscrW^1,2$ is the usual Watanabe–Sobolev space of Malliavin-differentiable functions (Nualart's and Øksendal's $mathbbD^1,2$, but I hate the hollow 'D', and this is what Hairer uses). Note that this is the Hilbert space of processes $UinmathscrL^2(varOmegatimesmathopen[0,inftymathclose[;mathbbR)$ for which




        • $U(t)in mathscrW^1,2$, for (almost) all $tinmathopen[0,inftymathclose[$;

        • there exists a version of the process $(omega,tau,t)mapstomathfrakD_tau U(omega,t)$ in $mathscrL^2(varOmegatimesmathopen[0,inftymathclose[^2;mathbbR)$.

        It can be shown that processes $U_nu$ of the form
        $$U_nu(t):= sum_j=1^nu X_j h_j(t),$$
        where $X_j$ is a smooth and cylindrical random variable and $h_jinmathscrL^2(mathopen[0,inftymathclose[;mathbbR)$, are dense in $mathscrL^2(mathopen[0,inftymathclose[;mathscrW^1,2)$. In fact, $mathscrL^2(mathopen[0,inftymathclose[;mathscrW^1,2)$ is defined as the closure of the space of such process under a suitable norm which we will not need here.



        This means that we can find a sequence of such processes $(U_nu)_nuinmathbbN$ such that
        $$U_nuto U~textin~mathscrL^2(varOmega;mathscrL^2(mathopen[0,inftymathclose[;mathbbR)),~textas~nutoinfty,$$
        and
        $$mathfrakD_cdot U_nuto mathfrakD_cdot U~textin~mathscrL^2(varOmega;mathscrL^2(mathopen[0,inftymathclose[;mathbbR)^otimes 2),~textas~nutoinfty.$$



        A consequence of the above is that
        $$int_0^T U_nu(t)~!mathrmdtto int_0^T U(t)~!mathrmdt~textin~mathscrL^2(varOmega;mathbbR),~textas~nutoinfty,$$
        and
        $$mathfrakD_cdotint_0^T U_nu(t)~!mathrmdtto mathfrakD_cdotint_0^T U(t)~!mathrmdt~textin~mathscrL^2(varOmega;mathscrL^2(mathopen[0,inftymathclose[;mathbbR)),~textas~nutoinfty.$$



        And these imply,
        $$mathfrakD_tauint_0^T U(t)~!mathrmdt = int_0^T mathfrakD_tau U(t)~!mathrmdt.$$



        Now, set $U(t) = u(t)^2$. Then, by the "Chain Rule" for Malliavin derivatives (Theorem 3.5 in Øksendal et al.),
        $$mathfrakD_tauint_0^T u(t)^2~!mathrmdt = int_0^T mathfrakD_tau big(u(t)^2big)~!mathrmdt = 2int_0^T u(t)mathfrakD_tau u(t)~!mathrmdt.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Mar 20 at 23:05

























        answered Mar 20 at 19:33









        sami.spricht.sprachesami.spricht.sprache

        15312




        15312



























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