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The following question is bothering me. I think it is probably known but I cannot find any reference...

Let $(X_t)$, $(Y_t)$, $(Z_t)$ denote 3 Feller processes with respective infinitesimal generators $A_X, A_Y, A_Z$.

Assume that $A_X = A_Y + A_Z$ (i.e. $X$ is a 'mix' of $Y$ and $Z$).

Moreover, at any time $t$, the law of $Y_t$ is stationary for the process $Z$.

How can it be proved that, if $X_0$ has the same law as $Y_0$, then $X_t$ and $Y_t$ have the same law at all time $t>0$ ?

Any argument/reference is welcome.

Thanks!

EDIT

@michael I don't think it is a relevant example. If I am not mistaken, Ornstein-Uhlenbeck processes have only one stationary measure.

Here, the basic framework is that the set $\mathcal{S}_Z$ of stationary distributions for $Z$ is not trivial and the law of $Y_t$ moves inside this set. Note that if the law of $Y_t$ does not change (i.e. $Y$ is stationary), then the result is immediate.

For example, you could take $Z$ to be a (totally asymmetric) exclusion process so that any Bernoulli product measure is stationary and take $Y$ a process that increase the probability of an occupied site as time goes by.

@Ilya The statement means that, at any time $t$, the law of $Y_t$ is a stationary distribution for the process $Z$.

@MJ73550 That would be too easy :-) Indeed, if $A_Y$ and $A_Z$ commute, then we can first run the dynamic of $Y$ up to time $t$ after that run $Z$ for another period of length $t$.

Here, we cannot assume such a strong property. At all time, both dynamics must be run concurrently yet I think that the result should still hold...

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  • $\begingroup$ have you worked through this with Y and Z being independent but not identically distributed Ornstein-uhlenbeck processes ? $\endgroup$
    – user83457
    Commented Apr 21, 2016 at 6:53
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    $\begingroup$ What exactly does your "moreover" statement mean? $\endgroup$
    – SBF
    Commented Apr 21, 2016 at 15:55
  • $\begingroup$ do $A_Y$ and $A_Z$ commute ? $\endgroup$
    – MJ73550
    Commented Apr 22, 2016 at 7:22

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