**Short version.** If $X$ is a diffusion with generator $L$ and the Lebesgue measure is invariant for $X$, then $L^\*$ has no term of order zero and it corresponds to another diffusion $X^\*$. Denoting by $p$ and $p^\*$ their kernels,



> 
> do we have $p^\*\_t(x,y)=p\_t(y,x)$?
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> 
> 


I am interested in full answers, but I would assume there is a reference for this.


**In details.** Let $X$ be a diffusion, solution to
$$\mathrm dX\_t = b(X\_t)\mathrm dt + \sum\_i\sigma\_i(X\_t)\mathrm dW^i\_t.$$
I am in a somewhat regular case: $b$ and all the $\sigma\_i$ are smooth, the diffusion is Feller (the semigroup acts on continuous functions vanishing at infinity) and hypoelliptic, i.e. the vector fields satisfy the [parabolic Hörmander condition](https://en.wikipedia.org/wiki/H%C3%B6rmander%27s_condition) when written in Stratonovich form (so the kernel is well-defined and smooth). I am also interested in cases with stronger hypotheses.


This diffusion has generator $L:f\mapsto \sum\_kb^k\partial\_kf+\frac12\sum\_i\sum\_{k\ell}\sigma\_i^k\sigma\_i^\ell\partial\_{k\ell}f$, with (formal) adjoint
$$ L^\*:f\mapsto
- \sum\_k\partial\_k\big(b^kf\big)+\frac12\sum\_i\sum\_{k\ell}\partial\_{k\ell}\big(\sigma\_i^k\sigma\_i^\ell f\big)
= \frac12\sum\_i\sum\_{k\ell}\sigma\_i^k\sigma\_i^\ell\partial\_{k\ell}f + (\text{lower order}).$$
If $L^\*1=0$, i.e. there are no terms of order zero (the Lebesgue measure is preserved), this $L^\*$ is the generator of another diffusion
$$\mathrm dX^\*\_t = b^\*(X\_t)\mathrm dt + \sum\_i\sigma\_i(X\_t)\mathrm dW^i\_t$$
with the same noise components. (I am abusing notation a bit, in the sense that $b^\*$ is not the adjoint of $b$, since the former depends on the $\sigma\_i$.)


Since $L$ is hypoelliptic, $X$ started at $x$ admits a smooth kernel $(t,y)\mapsto p\_t(x,y)$, i.e.
$$\mathbb P\_x(X\_t\in A)=\int\_Ap\_t(x,y)\mathrm dy.$$
We can check (easier in Stratonovich form) that hypoellipticity of $L$ equivalent to that of $L^\*$, so $X^\*$ also has a smooth kernel $p^\*$.



> 
> **Question.**
> Is it true that for any $f,g$ smooth with compact support (say), we have
> $$ \int f(x)p^\*\_t(x,y)g(y)\mathrm dx\mathrm dy = \int f(x)p\_t(y,x)g(y)\mathrm dx\mathrm dy? $$
> 
> 
> 


I call this a question about time reversals because it should be true that $t\in[0,T]\mapsto X\_{T-t}$ has the same distribution as $t\in[0,T]\mapsto X^\*\_t$ when the distribution of the initial point is the Lebesgue measure (whatever that means), so $p\_t(y,x)$ is the time reversal of the kernel $p\_t(x,y)$ of $X$, and $p^\*$ is the kernel of the time reversal $X^\*$ of $X$.


[Haussmann and Pardoux](https://projecteuclid.org/journals/annals-of-probability/volume-14/issue-4/Time-Reversal-of-Diffusions/10.1214/aop/1176992362.full) (open access) prove results about time reversals that can be used to answer my question in the positive if we have instead a finite equilibrium measure $\mu$ (i.e. $L^\*\mu=0$ when we consider the adjoint with respect to $\mu$).


**Formal proof.** At the formal level, I see the following reasoning as evidence that the result should be true.


Let $f,g$ be nice functions, and consider the derivative of the integral
$$I:s\in(0,t)\mapsto \int p^\*\_s(x,y)p\_{t-s}(z,y)g(z)\mathrm dy.$$
The Fokker-Planck equation asserts that $\partial\_tp = L^\*\_yp$ and $\partial\_tp^\* = L^{\*\*}\_yp^\* = L\_yp^\*$ ($y$ is the second variable), so
$$I'(s) = \int\Big((L\_yp^\*)\_s(x,y)p\_{t-s}(z,y) - p^\*\_s(x,y)(L^\*\_yp)\_{t-s}(z,y)\Big)\mathrm dy = 0 $$
since $L^\*$ is the adjoint of $L$ (this is more or less the same as saying that $s\mapsto p\_{t-s}(z,X\_s)$ is a local martingale, for any initial condition $X^\*\_0=x$ and point $z$, although this latter fact is actually true and easy to prove).


Hence we must have $I(0)=I(t)$ with
$$ I(0) = \int \delta\_x(y)p\_t(z,y)\mathrm dy
 = p\_t(z,x) $$
and
$$ I(t) = \int p^\*\_t(x,y)\delta\_z(y)\mathrm dy
 = p^\*\_t(x,z). $$
Maybe there are standard arguments that can prove the derivative and limits taken above are valid, but I have not been able to find them either myself or in the literature.


**Edit: The *dual process* approach.**
If I am not mistaken, one approach suggested by Mateusz Kwaśnicki in his answer is to go in the direction $X\mapsto\hat X\mapsto \hat L$ rather than $X\mapsto L^\*\mapsto X^\*$. It would be sufficient to carry through the following steps.


1. Prove that there exists a dual process $\hat X$ (more or less the resolvent of the processes are in duality), in the sense of section 13.2 of [CW].
2. Prove that the semigroup of $\hat X$ is given by $P\_tf(x)=\int p\_t(y,x)f(y)\mathrm dy$.
3. Prove that $\hat X$ satisfies the dual SDE, so in fact $\hat X=X^\*$.


These all come with some level of difficulty, but there are a few results in [CW] that may help. For instance, for 1. we can use Theorem 15.5, and all that remains to show is that the Lebesgue measure is invariant (in the strong sense $\int\mathbb P\_x(X\_t\in A)\mathrm dx=\mathrm{leb}(A)$) and that there exists a “cofine topology” in the sense of Definition 15.4. Point 2. is very close to the duality of the semigroups, as in Proposition 13.6, and point 3., once we have point 2., reduces to having enough functions in the domain of $\hat L$.


Although the route seems sound, altogether the results in [CW] look very measure-theoretic to me (unsurprisingly) and I am having trouble relating them to the smooth quantities I am considering.


[CW] Chung, Walsh, [Markov Processes, Brownian Motion, and Time Symmetry](https://doi.org/10.1007/0-387-28696-9).



Is $p_t(y,x)$ the kernel of the time reversal of the diffusion $X$, for $p_t(x,y)$ the kernel of $X$?
This is an immediate corollary of Theorem 3.50 in



> 
> D. W. Stroock, [An introduction to the analysis of paths on a Riemannian manifold](https://bookstore.ams.org/surv-74-s) (2000).
> 
> 
> Mathematical Surveys and Monographs. 74. Providence, RI: American Mathematical Society (AMS). xvii, 269 p.
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> 
> 


See also Theorem 4.17 for a manifold version of it.


