$ \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 measurable. Let $a := \sigma \sigma^\top$. For simplicity, we denote $\sigma_t(x) := \sigma (t, x)$ and $a_t(x) := a (t, x)$. We assume there exist constants $C>0, 0 < \beta < 1$ such that for $t \in \TT$ and $x, y \in \RR^d$: \begin{align} \frac{1}{C} |y|^2 \le \langle a_t (x) y, y \rangle & \le C |y|^2, \label{main_assmpt1:ineq1} \\ |\sigma_t (x) - \sigma_t (y)| &\le C |x-y|^\beta. \label{main_assmpt1:ineq2} \end{align}

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 $(\Omega, \mathcal A, \PP)$. Let $X_0 : \Omega \to \RR^d$ be a $\mathcal F_0$-measurable random variable. Let $$ X_t := X_0 + \int_0^t \sigma(s, X_0) \diff B_s, \quad t \in \TT. $$

Let $t \in (0, T]$. Is the distribution of $X_t$ absolutely continuous w.r.t. the Lebesgue measure on $\RR^d$?

Thank you so much for your elaboration!

  • $\begingroup$ Yes, this is a very classical result (Friedman, if I remember correctly). $\endgroup$
    – zhoraster
    Commented May 14 at 11:04

1 Answer 1


I believe the answer is yes. By independence of $X_0$ from the Brownian increments, by the claim here we can write $X_t = F(X_0, \omega)$, where $F(x, \omega) := x + \int_0^t\sigma(s, x) \, dB_s$.

By standard results, for each $x$, the random variable $F(x, \cdot)$ is Gaussian with covariance $\int a(s, x) \, ds$. By the uniform ellipticity condition, the covariance matrix is nondegenerate, and in particular it admits a density. Further, the uniform ellipticity implies their densities are uniformly bounded say by $M > 0$, and thus we have the crude "uniform absolute continuity" bound

$$\mathbb P(F(x, \cdot) \in A) \leq M \mu(A),$$

for all $A \subset \mathbb R^d$, uniformly in $x$. Thus

$$\mathbb P(X_t \in A) = \int_{\mathbb R} \mathbb P(F(x, \cdot) \in A) \, d\mu_{X_0} \leq M \mu(A),$$

and we conclude absolute continuity of $X_t$ as desired.


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