Suppose we have an sde of the form:

\begin{eqnarray} dX_t=b(X_t)dX_t + \sigma (X_t)dB_t \end{eqnarray} where $b$ and $\sigma$ are Lipschitz. Then we have existence and uniqueness of the solution $X_t$, which is an Ito diffusion.

When can we say that the distribution of $X_t$ for a particular t is absolutely continuous wrt Lebesgue measure (besides perhaps the initial distibution at $t=0$)?

  • $\begingroup$ One can show that $X_t$ has a density on the set $\{\sigma>0\}$. So, in particular, if $\sigma>0$, then $X_t$ has a density (wrt Lebesgue measure). $\endgroup$
    – saz
    Commented Dec 17, 2014 at 16:00
  • $\begingroup$ @Saz any proof or reference?Is it because the diffusion term creates constant movement so we can't have a fixed point? $\endgroup$
    – TKM
    Commented Dec 17, 2014 at 18:46
  • $\begingroup$ @TKM I have added a reference. $\endgroup$
    – saz
    Commented Dec 25, 2014 at 7:40

1 Answer 1


There is the following statement:

Theorem: Let $(X_t)_{t \geq 0}$ be a solution of the SDE $$dX_t = b(X_t) \, dt + \sigma(X_t) \, dB_t$$ where $\sigma$ is Lipschitz continuous and $b$ of linear growth (i.e. $|b(x)| \leq C (1+|x|)$). Then $X_t$ has a density (with respect to Lebesgue measure) on the set $\{x \in \mathbb{R}; \sigma(x) \neq 0\}$ for each $t>0$.

For a proof see Nicolas Fournier & Jacques Printems: Absolute continuity for some one-dimensional processes. (The relevant part of this paper is highly readable; you need some basics on Fourier transform as well as SDEs.)

  • $\begingroup$ Very good. Thank you for this great answer. $\endgroup$
    – Todd
    Commented Dec 11, 2019 at 16:06
  • $\begingroup$ Dear @saz, what if $b$ and $\sigma$ are time dependent as well (and Lipschitz with linear growth)? $\endgroup$
    – Nobody
    Commented Jun 13, 2022 at 21:16

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