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Let $M$ be an almost surely continuous martingale that is not almost surely constant in time - that is, it is not the case that almost surely, $M_t = M_0$ for all $t$.

Assume further that $M$ is a time homogeneous Markov process, and that it is transitive, in the sense that for any measurable subset $U$ of $\mathbb R$ with nonzero Lebesgue measure, we have

$$\mathbb E\big[\int_0^\infty \mathbf 1_U (M_t) \, dt \big] > 0.$$

Suppose $f: \mathbb R \to \mathbb R$ is a continuous function such that $f(M_t)$ is a martingale.

Question: Does if follow that $f$ is necessarily an affine linear function? That is, $f(x) = a + bx$ for some $a, b \in \mathbb R$.

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    $\begingroup$ $M$ could be bounded. $\endgroup$
    – Martin Hairer
    Commented Oct 20, 2021 at 11:03
  • $\begingroup$ Ah yes.. I need to impose some kind of transitivity condition.. I had mistakenly thought that a time homogeneous Markov process would already be transitive but this is not the case. $\endgroup$
    – Nate River
    Commented Oct 20, 2021 at 12:40
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    $\begingroup$ When $f \in C^2(\mathbb{R})$, the Ito formular (see Theorem 4.57 in 'Limit Theorems for Stochastic Processes') gives, that $\int_0^t f''(M_s) \, d\langle M,M \rangle_s$ is a continuous martingale of locally bounded variation, which means is is already constant. We thus have $\int_0^t f''(M_s) \, d\langle M,M \rangle_s = 0$ almost surely for all $t \geq 0$. If we now also knew the variation $(\langle M,M\rangle_t)$ to be strictly growing almost surely, we would have the wanted result. (We already know it to be growing surely.) $\endgroup$
    – Ben Deitmar
    Commented Oct 20, 2021 at 15:07

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If you allow for an arbitrary starting point, then just use the optional stopping theorem for $f(M_t)$: $$\begin{aligned} f(x) & = \mathbb E^x f(M(\tau_{(a,b)})) \\ & = \mathbb P^x(M(\tau_{(a,b)}) = b) f(b) + P^x(M(\tau_{(a,b)}) = a) f(a) \\ & = \frac{x-a}{b-a} f(b) + \frac{b-x}{b-a} f(a) , \end{aligned}$$ as desired. Here $\tau_{(a,b)}$ is the first hitting time of $a$ or $b$; it is finite almost surely because $M$ is transitive.

If the starting point (or distribution) is fixed, follow the same argument after the first hitting time of any given $x$ (the latter is almost surely finite because the process is transitive) and use the strong Markov property.

Note: one-dimensional diffusions have been studied by E.B. Dynkin and others, and an essentially complete description is available in terms of the speed measure and the scale function. I think the above argument belongs to this theory (although it is very simple, of course).

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  • $\begingroup$ @IosifPinelis: $M$ is a martingale, that is enough (just apply the optional stopping theorem). Not sure what $b(y)$ stands for in your comment, but if it is the drift coefficient (the linear term), then it necessarily vanishes if $M$ is a martingale. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Oct 20, 2021 at 19:19
  • $\begingroup$ Oh thats a very nice argument. Thank you for your answer! $\endgroup$
    – Nate River
    Commented Oct 20, 2021 at 23:52
  • $\begingroup$ I may be missing something really obvious, but how do we get that $\mathbb P^x (M(\tau_{(a, b)} = b) = \frac{x-a}{b-a}?$ And similarly for $\mathbb P^x (M(\tau_{(a, b)} = a)$. I know such a formula holds for Brownian motion, but I wasn’t aware it held for general continuous Markov martingales. $\endgroup$
    – Nate River
    Commented Oct 20, 2021 at 23:57
  • $\begingroup$ @NateRiver: That's just the optional stopping theorem: $$x = \mathbb E^x M(\tau_{(a,b)}) = a \mathbb P^x(\tau_{(a,b)} = a) + b \mathbb P^x(\tau_{(a,b)} = b).$$ Combine this with $\mathbb P^x(\tau_{(a,b)} = a) + \mathbb P^x(\tau_{(a,b)} = b) = 1$ to get the above expression. No need for the Markov property here, really. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Oct 21, 2021 at 0:22
  • $\begingroup$ Also, every continuous martingale is a time-changed Brownian motion, and the hitting distribution is not affected by a random time-change. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Oct 21, 2021 at 0:24

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