# Does $X_t$ with $t>0$ admit a density?

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

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!

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

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.