2
$\begingroup$

$ \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!

$\endgroup$
1
  • $\begingroup$ Yes, this is a very classical result (Friedman, if I remember correctly). $\endgroup$
    – zhoraster
    Commented May 14 at 11:04

1 Answer 1

3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.