**Motivation:**


If a continuous function on the real line is periodic with periods $p\_1, p\_2 > 0$ such that $\frac{p\_1}{p\_2}$ is irrational, then the function is constant. Is there a probabilistic analogue of this statement?


**Problem set up:**


Let $X$ be a continuous stochastic process with $X\_t$ in $L^1$ for all $t \in [0, \infty)$. Denote by $\mathcal F\_t$ its natural filtration.


Suppose there exist $p\_1, p\_2 > 0$, with $\frac{p\_1}{p\_2}$ irrational such that for all $t \in [0, \infty)$, $E[X\_{t + p\_i}| \mathcal F\_t] = X\_t$, for each $i = 1, 2$.



> 
> **Question:** Is $X$ an $\mathcal F\_t$-martingale?
> 
> 
> 



If a process is periodic on average with mutually incommensurable periods, is the process a martingale?
**Yes**, it is. For each $t$, let $S\_t$ be the set of those $s \geqslant t$ for which $X\_t = \mathbb E(X\_s | \mathcal F\_t)$. By assumption (and induction), for all $k, n \geqslant 0$ such that $k p\_1 - n p\_2 \geqslant 0$ we have
\[ X\_t = \mathbb E(X\_{t + k p\_1} | \mathcal F\_t) \qquad \text{and} \qquad X\_{t + k p\_1 - n p\_2} = \mathbb E(X\_{t + k p\_1} | \mathcal F\_{t + k p\_1 - n p\_2}) , \]
and therefore
\[ X\_t = \mathbb E(X\_{t + k p\_1} | \mathcal F\_t) = X\_t = \mathbb E(\mathbb E(X\_{t + k p\_1} | \mathcal F\_{t + k p\_1 - n p\_2}) | \mathcal F\_t) = \mathbb E(X\_{t + k p\_1 - n p\_2} | \mathcal F\_t) . \]
In other words, $S\_t$ contains $t + k p\_1 - n p\_2$ for every $k, n \geqslant 0$ such that $k p\_1 - n p\_2 \geqslant 0$. Plus, of course, $S\_t$ is closed. We conclude that $S\_t = [t, \infty)$, as desired.




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*Edit:* As pointed out in the comment, pathwise continuity of $X\_t$ need not imply $L^1$-continuity, so it is not clear that $S\_t$ is closed.


So let us take a slightly different path. Define
\[T = {k p\_1 - n p\_2 : k, n \geqslant 0} \cap [0, \infty) .\]
We already know (by the first paragraph of this answer) that $(X\_t : t \in T)$ is a martingale with respect to $\mathcal F\_t$. By Doob's regularisation theorem (or strictly speaking, by the argument used in the proof of this result), there is a right-continuous martingale $(\tilde X\_t : t \in [0, \infty))$ such that (almost surely)
\[ \tilde X\_t = X\_t \qquad \text{for all } t \in T . \]
But $(X\_t : t \in [0, \infty))$ is right-continuous, too, and hence necessarily (almost surely)
\[ \tilde X\_t = X\_t \qquad \text{for all } t \in [0, \infty) . \]
Thus, $X\_t$ is indeed a martingale.


