If $(Φ^x)_{x∈ℝ}$ is a family of real-valued stochastic processes and $B$ is a Brownian motion, then $\int_0^tΦ^x_s\:dB_s=(\int_0^t\Phi_s\:dB_s)(x)$

Let

• $T>0$
• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(\mathcal F_t)_{t\in[0,\:T]}$ be a complete filtration on $(\Omega,\mathcal A)$
• $B$ be a (standard, real-valued) $\mathcal F$-Brownian motion on $(\Omega,\mathcal A,\operatorname P)$

Moreover, let $$\mathcal E^2_{\text{loc}}(H):=\left\{\Phi:\Omega\times[0,T]\to H:\Phi\text{ is }\mathcal F\text{-progressively measurable with }\operatorname P\left[\int_0^t\left\|\Phi_s\right\|_H^2{\rm d}s<\infty\right]=1\right\}$$ for any separable $\mathbb R$-Hilbert space $H$.

Let

• $d\in\mathbb N$
• $\Lambda\subseteq\mathbb R^d$ be Borel measurable
• $\left\{\Phi^x:x\in\Lambda\right\}\subseteq\mathcal E^2_{\text{loc}}(\mathbb R)$ with $$\left(\Lambda\ni x\mapsto\Phi_t^x(\omega)\right)\in L^2(\Lambda)\;\;\;\text{for all }(\omega,t)\in\Omega\times[0,T]\tag 1$$

Moreover, let $$\Psi:\Omega\times[0,T]\to L^2(\Lambda)\;,\;\;\;(\omega,t)\mapsto\left(\Lambda\ni x\mapsto\Phi_t^x(\omega)\right)\;.$$ I want to show that $$\int_0^t\Phi_s^x\:{\rm d}B_s=\left(\int_0^t\Psi_s\:{\rm d}B_s\right)(x)\;\;\;\text{for all }x\in\Lambda\text{ and }t\in[0,T]\tag 2\;.$$

Is that possible, and if so, how? Of course, the usual approach would be to start with considering the case where each $\Phi^x$ is an elementary process (in the usual sense), but it's not trivial to show that then $\Psi$ is an elementary process (in the corresponding sense) too. So, maybe we need additional assumptions. Please take note of my related comment below.

• $(2)$ is motivated by the following problem: Let $U$ be a separable $ℝ$-Hilbert space, $Q∈\mathfrak L(U)$ be nonnegative and self-adjoint with $\text{tr}Q<∞$, $W$ be a $Q$-Wiener process on $U$, $x_0∈Λ$, $F:[0,T]×ℝ^d→ℝ^d$, $G:[0,T]×ℝ^d→\mathfrak L(U,ℝ^d)$ and $X^{x_0}$ be a strong solution of $$X_t=x_0+\int_0^tF(s,X_s)\:{\rm d}s+\int_0^tG(s,X_s)∘{\rm d}W_s\;\;\;\text{for all }t∈[0,T]\;.\tag 3$$ Then we obtain $${\rm d}F(t,X_t^{x_0})=\left[\frac{∂F}{∂t}(t,X_t^{x_0})+(F(t,X_t^{x_0})⋅∇)F(t,X_t^{x_0})\right]{\rm d}t+(G(t,X_t^{x_0})⋅∇)F(t,X_t^{x_0})∘{\rm d}W_t\tag 4$$ for all $t∈[0,T]$. – 0xbadf00d Dec 5 '16 at 21:05
• Now, I want to eliminate the dependence on $x_0$ in $(4)$ and transform this SPDE into a SDE on $L^2(Λ,ℝ^d)$. I've asked a special case of this question in a separate thread. The problem in this question seems to be a starting point for a solution to the other question. – 0xbadf00d Dec 5 '16 at 21:05
• Do you have any assumptions on the joint measurability of $\Phi^x_t(\omega)$ with respect to $(x,t,\omega)$? As it stands I only see separate measurability in $x$ and in $(t,\omega)$, and I suspect that's not enough to make sense of the integral of $\Psi$. – Nate Eldredge Dec 5 '16 at 21:58
• @NateEldredge I didn't want to bloat the question with too much details. You're free to assume any technical assumption you need (just write these assumption down). – 0xbadf00d Dec 6 '16 at 11:03