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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.

## Best Answer

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.