• $(\Omega,\mathcal A,\operatorname P)$ be a complete probability space
  • $T>0$
  • $I:=(0,T]$
  • $(\mathcal F_t)_{t\in\overline I}$ be a complete and right-continuous filtration on $(\Omega,\mathcal A)$
  • $U$ be a separable $\mathbb R$-Hilbert space
  • $M$ be a $U$-valued continuous square-integrable $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$
  • $\mathfrak L_1(U)$ denote the space of nuclear operators on $U$ and $\mathfrak L_1^+(U)$ denote the subspace of nuclear, nonnegative and self-adjoint operators on $U$
  • $[\![M]\!]$ denote the unique (up to indistinguishability) $\mathfrak L_1(H)$-valued continuous $\mathcal F$-adapted process of locally bounded variation such that $M\otimes M-[\![M]\!]$ is an $\mathcal F$-martingale
  • $\operatorname{Cov}(M)$ denote the unique (up to indistinguishability) $\mathfrak L_1^+(H)$-valued $\mathcal F$-predictable process on $(\Omega,\mathcal A,\operatorname P)$ with $$[\![M]\!]_t=\int_{(0,\:t]}\operatorname{Cov}(M)\:{\rm d}[M]\;\;\;\text{for all }t\in\overline I\text{ almost surely}\tag1\;,$$ where $[M]$ denotes the quadratic variation of $M$

There is a real-valued $\mathcal F$-predictable process $\lambda^n$ and a $U$-valued $\mathcal F$-predictable process $e^n$ on $(\Omega,\mathcal A,\operatorname P)$ for $n\in\mathbb N$ such that

  • $(e^n_t(\omega))_{n\in\mathbb N}$ is an orthonormal basis of $\overline{Q_t(\omega)U}$ with $$Q_t(\omega)e^n_t(\omega)=\lambda^n_t(\omega)e_t^n(\omega)\;\;\;\text{for all }n\in\mathbb N\cap[0,\operatorname{rank}Q_t(\omega)]\tag2$$ for all $(\omega,t)\in\Omega\times\overline I$
  • $(\lambda^n_t(\omega))_{n\in\mathbb N}\subseteq[0,\infty)$ is nondecreasing with $$\lambda^n_t(\omega)>0\;\;\;\text{for all }n\in\mathbb N\cap[0,\operatorname{rank}Q_t(\omega)]\tag3$$ and $$\lim_{n\to\infty}\lambda_t^n(\omega)=0\tag4$$ for all $(\omega,t)\in\Omega\times\overline I$

Now, let $H$ be a separable $\mathbb R$-Hilbert space and $\mathcal E_M(H)$ denote the space of elementary $\mathcal F$-predictable processes (defined as usual) with values in $\mathfrak L(U,H)$ equipped with the seminorm $$\left\|\Phi\right\|_{\mathcal E_M(H)}^2:=\operatorname E\left[\int\left\|\Phi_sQ^{1/2}_s\right\|_{\operatorname{HS}(U,\:H)}^2\:{\rm d}[M]_s\right]\;\;\;\text{for }\Phi\in\mathcal E_M(H)\;.$$ Let $\mathcal I^2_M(H)$ denote the completion of $\mathcal E_M(H)$ with respect to that seminorm. Then we can characterize $\mathcal I^2_M(H)$ as follows: If $\mathcal H$ is any separable $\mathbb R$-Hilbert space, $(\mathcal h_n)_{n\in\mathbb N}$ is an orthonormal basis of $\mathcal H$ and $$A_t(\omega):=\sum_{n\in\mathbb N}\sqrt{\lambda^n_t(\omega)}e_t^n(\omega)\otimes\mathcal h_n\in\mathfrak L(Q_t^{1/2}(\omega)U,\mathcal H)\;\;\;\text{for }(\omega,t)\in\Omega\times\overline I\;,$$ then $$\mathcal I^2_M(H)=\left\{\Psi A:\Psi\in\mathcal L^2(\operatorname P\otimes{\rm d}[M];\operatorname{HS}(\mathcal H,H))\text{ is }\mathcal F\text{-predictable}\right\}\tag5$$ and $$\left\|\Psi A\right\|_{\mathcal I^2_M}=\left\|\Psi\right\|_{\mathcal L^2(\operatorname P\otimes{\rm d}[M];\:\operatorname{HS}(\mathcal H,\:H))}\;\;\;\text{for all }\Psi\in\mathcal L^2(\operatorname P\otimes{\rm d}[M];\operatorname{HS}(\mathcal H,H))\;.\tag6$$

Now, assume $\Psi\in\mathcal I^2_M(H)$ and $\Phi\in\mathcal I^2_{\Psi\:\cdot\:M}(\tilde H)$, where $\Psi\cdot M$ denotes the integral process and $\tilde H$ is another separable $\mathbb R$-Hilbert space. In order to prove the associativity known from the real-valued case, i.e. $\Phi\Psi\in\mathcal I^2_M(\tilde H)$ and $$\Phi\Psi\cdot M=\Phi\cdot(\Psi\cdot M)\;\;\;\text{almost surely}\tag7\;,$$ we first need to show that the composition $\Phi\Psi$ is even well-defined. We know that $\Psi=\tilde \Psi A$ for some $\tilde\Psi\in\mathcal L^2(\operatorname P\otimes{\rm d}[M];\operatorname{HS}(\mathcal H,H))$. So, in order to show the well-definedness, we need at least to show that $$\tilde\Psi_t(\omega)\mathcal h\in\operatorname{Cov}(\Psi\cdot M)^{1/2}H\;\;\;\text{for all }\mathcal h\in\mathcal H\text{ and }(\omega,t)\in\Omega\times\overline I\tag8\;.$$

How can we do that?


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.