How can we prove that $\langle\int_0^t\Phi_s{\rm d}W_s,x\rangle_H=\sum_{n\in\mathbb N}\int_0^t\langle\sqrt{λ_n}\Phi_se_n,x\rangle_H{\rm d}B_s^{(n)}$?

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Let$^1$

  • $U$ and $H$ be separable $\mathbb R$-Hilbert spaces
  • $Q\in\mathfrak L(U,H)$ be nonnegative and symmetric operator on $U$ with finite trace
  • $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $U$ with $$Qe_n=\lambda_ne_n\;\;\;\text{for all }n\in\mathbb N$$ for some $(\lambda_n)_{n\in\mathbb N}\subseteq[0,\infty)$
  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space and $(\mathcal F_t)_{t\ge 0}$ be a filtration of $\mathcal A$
  • $W$ be a $Q$-Wiener process on $(\Omega,\mathcal A,\operatorname P)$ with respect to $\mathcal F$, $$B^{(n)}:=\begin{cases}\frac{\langle W,e_n\rangle_U}{\sqrt{\lambda_n}}&\text{, if }n\in\mathbb N\text{ with }\lambda_n>0\\0&\text{, else}\end{cases}$$ and $$W^{(n)}:=\sum_{i=1}^n\sqrt{\lambda_i}B^{(i)}e_i\;\;\;\text{for }n\in\mathbb N$$

Note that the $B^{(n)}$ are independent $\mathcal F$-Brownian motions on $(\Omega,\mathcal A,\operatorname P)$, $$\left\|W_t^{(n)}-W_t\right\|_{L^2(\operatorname P,\;U)}\stackrel{n\to\infty}\to 0\;\;\;\text{for all }t\ge 0\tag 1$$ and $$\operatorname P\left[\left\|W^{(n)}-W\right\|_{C^0([0,\;t],\;U)}\stackrel{n\to\infty}\to 0\right]=1\;\;\;\text{for all }t\ge 0\tag 2$$

Now, let $(\Phi_t)_{t\ge 0}$ be a $\mathfrak L(U,H)$-valued $\mathcal F$-adapted and locally bounded$^2$ stochastic process on $(\Omega,\mathcal A,\operatorname P)$ with $$\Phi_t=\sum_{i=1}^n\xi_{i-1}1_{(t_{i-1},t_i]}\;\;\;\text{for all }t\ge 0\tag 3$$ for some $\mathfrak L(U,H)$-valued random variables $\xi_0,\ldots,\xi_{n-1}$ on $(\Omega,\mathcal A,\operatorname P)$ and some $0=t_0<\cdots<t_n$. Moreover, let $$\int_0^t\Phi_s{\rm d}W_s:=\sum_{i=1}^n\xi_{i-1}\left(W_{t_i\wedge t}-W_{t_{i-1}\wedge t}\right)\;\;\;\text{for }t\ge 0\;.$$

Let $x\in H$. How can we prove that $$\langle\int_0^t\Phi_s{\rm d}W_s,x\rangle_H=\sum_{n\in\mathbb N}\int_0^t\langle\sqrt{\lambda_n}\Phi_se_n,x\rangle_H{\rm d}B_s^{(n)}\tag 4$$ in $L^2(\operatorname P)$, for all $t\ge 0$?

Let $$X^{(n)}_t:=\langle\sqrt{\lambda_n}\Phi_te_n,x\rangle_H\;\;\;\text{for }n\in\mathbb N\text{ and }t\ge 0\;.$$ By definition of the Itō integral with respect to a Brownian motion, $$\int_0^tX^{(n)}_s{\rm d}B_s=\sqrt{\lambda_n}\sum_{i=1}^n\langle\xi_{i-1}e_n,x\rangle_H\left(B^{(n)}_{t_i\wedge t}-B^{(n)}_{t_{i-1}\wedge t}\right)\;\;\;\text{for all }n\in\mathbb N\text{ and }t\ge 0\;.\tag 5$$

$\color{red}{\text{Now comes the crucial part!}}$ In the following equation chain, why do $(6)$ and $(7)$ hold?

\begin{equation} \begin{split} \langle\int_0^t\Phi_s{\rm d}W_s,x\rangle_H&=\sum_{i=1}^n\langle\xi_{i-1}\left(W_{t_i\wedge t}-W_{t_{i-1}\wedge t}\right),x\rangle_H\\ &\stackrel{(6)}{\color{red}=}\sum_{i=1}^n\langle\xi_{i-1}\sum_{k\in\mathbb N}\sqrt{\lambda_k}\left(B^{(k)}_{t_i\wedge t}-B^{(k)}_{t_{i-1}\wedge t}\right)e_k,x\rangle_H\\ &=\sum_{i=1}^n\sum_{k\in\mathbb N}\sqrt{\lambda_k}\langle\xi_{i-1}e_k,x\rangle_H\left(B^{(k)}_{t_i\wedge t}-B^{(k)}_{t_{i-1}\wedge t}\right)\\ &\stackrel{(7)}{\color{red}=}\sum_{k\in\mathbb N}\sum_{i=1}^n\sqrt{\lambda_k}\langle\xi_{i-1}e_k,x\rangle_H\left(B^{(k)}_{t_i\wedge t}-B^{(k)}_{t_{i-1}\wedge t}\right)\\ &=\sum_{k\in\mathbb N}\int_0^tX^{(k)}_s{\rm d}B^{(k)}_s \end{split}\tag 8 \end{equation}

for all $t\ge 0$. In particular, I don't really understand in which sense the equation chain $(8)$ holds. In the sense of $L^2(\operatorname P)$-convergence or in the sense of $\operatorname P$-almost sure convergence?


$^1$ Let $\mathfrak L(U,H)$ denote the space of bounded, linear operators from $U$ to $H$.

$^2$ i.e. $$\sup_\Omega\left\|\Phi_t\right\|_{\mathfrak L(U,\;H)}<\infty\;\;\;\text{for all }t\ge 0\;.$$