Independence of Solution of SDE $S^{(x_0, \sigma, \mu)}_t$ of Initial Information $\mathcal{G}_0$

Question

Question

Consider the following stochastic differential equation, given as an equivalent stochastic integral equation, where the multidimensional integrals are to be read componentwise:

\begin{equation} S_t = x_0 + \int_{0}^{t} \mu(S_t) ds + \int_{0}^{t} \sigma (S_t) dB_s. \end{equation}

Under our assumptions (see below), it is the case that an (up to indistinguishability) unique solution process

$$ S^{(x_0, \sigma, \mu)} :[0,T] \times \Omega \rightarrow \mathbb{R}^d, \quad (t, \omega) \mapsto S_t(\omega),$$

for this equation exists (to see this, consider for example Theorem 8.3. in Brownian Motion, Martingales and Stochastic Calculus from Le Gall).

Now, I suspect, that, for all $t \in [0,T]$, the random variable $S^{(x_0, \sigma, \mu)}_t$ is independent of $\mathcal{G}_0$, which is the initial $\sigma$-algebra of the underlying filtration .

Is this true? How could I show this rigorously? It is trivial for the case $t = 0$, but what if $t >0$?

I suspect, the result follows somehow from the fact, that $(B_t)_{t \in [0,T]}$ has independent increments.

I am stuck and I would be extremely grateful for any advice!

Preliminaries and Standard Technical Framework

• Let $T \in (0, \infty)$ be fixed.

• Let $d \in \mathbb{N}_{\geq 1}$ be fixed.

• Let $$(\Omega, \mathcal{G}, (\mathcal{G}_t)_{t \in [0,T]}, \mathbb{P})$$ be a complete probability space with a complete, right-continuous filtration $(\mathcal{G}_t)_{t \in [0,T]}$.

• Let $$B : [0,T] \times \Omega \rightarrow \mathbb{R}^d, \quad (t,\omega) \mapsto B_t(\omega)$$ be a standard $d$-dimensional $(\mathcal{G}_t)_{t \in [0,T]}$-adapted Brownian motion on $\mathbb{R}^d$ such that, for every pair $(t,s) \in \mathbb{R}^2$ with $0 \leq t < s$, the random variable $B_s-B_t$ is independent of $\mathcal{G}_t$.

• Let \begin{align} &\sigma: \mathbb{R}^d \rightarrow \mathbb{R}^{d \times d}, \\ &\mu: \mathbb{R}^d \rightarrow \mathbb{R}^{d}, \end{align} be globally Lipschitz.

• Let $x_0 \in \mathbb{R}^d$ be fixed.

Answer 1

The solution $S_t$ is adapted to the completed canonical filtration $(\bar{\mathcal{F}}_t)_{t \geq 0}$ of the driving Brownian motion $(B_t)_{t \geq 0}$. Consequently, it suffices to show that $\bar{\mathcal{F}}_{t}$ is independent from $\mathcal{G}_0$ for all $t \geq 0$.

The admissibility of the filtration $(\mathcal{G}_t)_{t \geq 0}$ implies that $$(B_{t_n}-B_{t_{n-1}},\ldots,B_{t_1}-B_0)$$ is independent from $\mathcal{G}_0$ for any $0<t_1<\ldots<t_n$, $n \in \mathbb{N}$. Since sets of the form $$\bigcap_{j=1}^n \{B_{t_j}-B_{t_{j-1}} \in A_j\}$$ with $n \in \mathbb{N}$, $A_j$ measurable and $0=:t_0<t_1<\ldots < t_n \leq t$, generate the canonical filtration $\mathcal{F}_t$, this implies that $\mathcal{F}_t$ is independent of $\mathcal{G}_0$. Finally, we remark that adding (subsets of) nullsets doesn't destroy independence, and therefore $\bar{\mathcal{F}}_t$ is independent of $\mathcal{G}_0$.