# Is it true that $F(X_0, \cdot) = X_0 + \int_0^T \sigma(s, X_0) \, \mathrm d B_s$ a.s.

measure-theorystochastic-calculusstochastic-processes

$$\newcommand{\RR}{\mathbb{R}} \newcommand{\TT}{\mathbb{T}} \newcommand{\NN}{\mathbb{N}} \newcommand{\PP}{\mathbb{P}} \newcommand{\EE}{\mathbb{E}} \newcommand{\FF}{\mathbb{F}} \newcommand{\PPP}{\mathscr{P}} \newcommand{\KKK}{\mathscr{K}} \newcommand{\eps}{\varepsilon} \newcommand{\diff}{\mathop{}\!\mathrm{d}}$$

We fix $$T \in (0, \infty)$$ and let $$\TT$$ be the interval $$[0, T]$$. Let $$\sigma : \TT \times \RR^d \to \RR^d \otimes \RR^m$$ be bounded and measurable. Let $$(B_t, t \ge 0)$$ be a $$m$$-dimensional Brownian motion and $$\FF := (\mathcal F_t, t \ge 0)$$ an admissible filtration on a probability space $$(\Omega, \mathcal A, \PP)$$. We assume $$(\Omega, \mathcal A, \FF, \PP)$$ satisfies the usual conditions. We define a map $$F: \RR^d \times \Omega \to \RR^d$$ by
$$F(x, \cdot) := x + \int_0^T \sigma(s, x) \diff B_s.$$

Let $$X_0 : \Omega \to \RR^d$$ be a $$\mathcal F_0$$-measurable random variable. Inspired by this answer, I would like to ask:

Is it true that $$F(X_0, \cdot) = X_0 + \int_0^T \sigma(s, X_0) \diff B_s$$ a.s.?

Thank you so much for your elaboration!

My attempt: For $$n \in \NN^*$$ and $$k\in \{0, 1, \cdots, n\}$$, let $$t^n_k := \frac{kT}{n}$$. We define random variables
\begin{align} A^n (x) &:= \sum_{k=0}^{n-1} \sigma (t^n_k,x) (B_{t^n_{k+1}} – B_{t^n_{k}}), \\ B^n &:= \sum_{k=0}^{n-1} \sigma (t^n_k, X_0) (B_{t^n_{k+1}} – B_{t^n_{k}}). \end{align}

By properties of ItÃ´ stochastic integral, for each $$x \in \RR^d$$ there is a subsequence $$n \mapsto\varphi_n^x$$ such that $$A^{\varphi_n^x}(x) \xrightarrow{n \to \infty} F(x, \cdot)$$ a.s. Similarly, there is a subsequence $$n \mapsto\psi_n$$ such that $$B^{\psi_n} \xrightarrow{n \to \infty} X_0 + \int_0^T \sigma(s, X_0) \diff B_s$$ a.s.

Yes. We fix some notation. Write $$Y := F(X_0, \cdot)$$, $$Z := X_0 + \int_0^T \sigma(s, X_0) \, dB_s$$, and consider the process $$W_t := B_t - B_0$$ together with its completed natural filtration $$\mathcal W_t$$. By construction $$\mathcal W_t$$ and $$\mathcal F_0$$ are independent, and further we have $$\mathcal F_t = \mathcal W_t \vee \mathcal F_0$$.

We will show that for all events $$E \in \mathcal F_T$$, we have

$$\mathbb E[\mathbf 1_E Y] = \mathbb E[\mathbf 1_E Z],$$ which will imply the desired conclusion that $$Y = Z$$ almost surely, since otherwise at least one of $$Y > Z$$ or $$Z < Y$$ happens with positive probability, and setting $$E$$ above to be either of these events we obtain a contradiction.

Since $$\mathcal F_t = \mathcal W_t \vee \mathcal F_0$$, by a standard but tedious monotone class argument it is enough to check the desired equality on events $$E$$ of the form $$A \cap B$$ for $$A \in \mathcal W_T$$ and $$B \in \mathcal F_0$$.

We write

$$\mathbb E[\mathbf 1_E Z] = \mathbb E[\mathbf 1_A \mathbf 1_B Z] = \mathbb E[\mathbf 1_B \mathbb E[\mathbf 1_A Z| \mathcal F_0]].$$

By considering the regular conditional probability with respect to $$\mathcal F_0$$ and applying the freezing lemma with the independence of $$\mathcal W_T$$ and $$\mathcal F_0$$ (here we may have to assume some regularity on $$\sigma$$ to ensure the freezing lemma applies), we have

$$\mathbb E[\mathbf 1_A Z| \mathcal F_0] = \mathbb E[\mathbf 1_A F(x, \cdot)]_{|x = X_0} = \mathbb E[1_A Y| \mathcal F_0]$$

and so

$$\mathbb E[\mathbf 1_E Z] = \mathbb E [\mathbf 1_B \mathbb E[\mathbf 1_A Y| \mathcal F_0 ]] = \mathbb E[\mathbf 1_E Y],$$
as desired.