$ \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 Itô 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.


1 Answer 1


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.

  • $\begingroup$ @Akira It should be said that the manipulation in the second to last equation line is the hardest to justify. If possible you could try to work it out yourself, possibly assuming some regularity on $\sigma$ to apply the freezing lemma. $\endgroup$
    – Nate River
    Commented May 15 at 21:18
  • $\begingroup$ It seems the section on conditional probabilities in Chapter 4 of Stochastic Calculus by Baldi contains a more or less complete justification for said manipulation. $\endgroup$
    – Nate River
    Commented May 15 at 21:29
  • $\begingroup$ Is it clear that $F$ is (jointly) measurable (probably yes, and I'm just being paranoid) or do we have to be careful how we actually define it? If we had more regularity (e.g. Lipschitzness, ...) we could even have a version where $x \mapsto x + \int_0^T \, \sigma(s, x) \, \mathrm{d} B_s$ is a random continuous map. In the same vein, I think the original question is solved by OP's attempt if $A^n$ and $F$ are continuous in $x$ (if $A^n, F$ are random continuous maps, then $A^n(x) \to F(x)$ in prob. for all $x$ implies $A^n(X_0) \to F(X_0)$ in prob. for independent $X_0$$). $\endgroup$
    – unwissen
    Commented May 16 at 7:15
  • $\begingroup$ Do you mean by $\mathcal W_t \vee \mathcal F_0$ the $\sigma$-algebra generated by the union of $W_t$ and $\mathcal F_0$? $\endgroup$
    – Akira
    Commented May 16 at 7:31
  • $\begingroup$ @Akira Indeed, the smallest sigma algebra containing both $\mathcal W_t$ and $\mathcal F_0$. $\endgroup$
    – Nate River
    Commented May 16 at 7:33

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