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show/hide this revision's text 3 Added to my answer an argument I sketched in the comments.

By repeating this argument, it follows that if there is any point $s$ with $\mathbb{E}(X_s) \neq 0$ then there are points $s$ for which $|\mathbb{E}(X_s)|$ is arbitrarily large. In fact, since by cyclic exchangeability, $\mathbb{E}(X_{s/n})=\mathbb{E}(X_s)/n$, it then follows that there are points arbitrarily close to zero for which $|\mathbb{E}(X_s)$ |\mathbb{E}(X_s)|$ is arbitrarily large.

But

Edit: (This is an expansion of the argument I sketched in the comments.) Note that for any stochastic proces $(X_t)=(X_t)_{0 \leq t \leq 1}$ if $(Z_s)_{0 U$ is a uniform $[0,1]$ random variable, independent of $(X_t)$, then the process $(X_t')$ obtained from $(X_t)$ by cyclically shifting $(X_t)$ by $U$, has cyclically exchangeable increments. Furthermore, if $(X_t)$ itself has cyclically exchangeable increments, then $(X_t)$ and $(X_t')$ have the same distribution.

Now let $(Z_s)=(Z_s)_{0 \leq s \leq t}$ is 1}$ be a Lévy processthen . Let $(Z_s - (s/t)Z_t)_{0 U$ be uniform on $[0,1]$ and independent of $(Z_s)$, and let $(Y_s)=(Y_s)_{0 \leq s \leq t}$ 1}$ be the process you get by cyclically shifting $(Z_s)$ by U. Then $Y_1=Z_1$, and $(Y_s)$ has the same distribution as $(Z_s)$.

Conditional upon $Z_1$ (which equals $Y_1$), we don't automatically know the distribution of $(Z_s)$. However, we know the following facts.

  • Conditional on $Z_1$, $(Y_s)$ is distributed as a Lévy bridge) uniformly random cyclic shift of the conditioned process $(Z_s)$ (conditioned on $Z_1$), so still has has cyclically exchangeable increments. (To prove that

  • Since $Z_1=Y_1$, $(Y_s)$ conditioned on $Z_1$ is the same as $(Y_s)$ conditioned on $Y_1$. But $(Y_s)$ and $(Z_s)$ have the same distribution so $(Y_s)$ conditioned on $Y_1$ is distributed as $(Z_s)$ conditioned on $Z_1$.

    • Putting these facts together, just note we see that conditional on $Z_1$, $(Z_s)$ still has cyclically exchangeable increments, and thus (still conditional on $Z_1$) $(Z_s - sZ_1)$ does as well. But then $(Z_s - sZ_1)$ is a process with c.e. increments and equal to zero at $s=0$, $s=1$. By the distribution first three paragraphs of my answer, it follows that if $(Z_s)_{0 \sup_{0 \leq s \leq t}$ 1} |\mathbb{E}(Z_s -sZ_1 | Z_1)|$ is itself invariant under cyclic shifts.almost surely finite, then almost surely $\mathbb{E}(Z_s|Z_1)=sZ_1$.

    But $\mathbb{E}(Z_s -sZ_1 | Z_1)

    Edit: So this proof works for the Lévy process under = \mathbb{E}(Z_s|Z_1)+sZ_1$ so the assumption that requirement boils down to $\mathbb{E}(sup_{s \sup_{0 \in (0,1)} leq s \leq 1} |Z_s|) < \mathbb{E}(Z_s|Z_1)|$ almost surely finite. Using the tower law, this holds as long as $\mathbb{E}(\sup_{0 \infty$ (which leq s \leq 1} |Z_s|)$ is almost surely finite. I think should be this is equivalent to assuming requiring that $\mathbb{E}|Z_1| < \infty$ but I still haven't checked)checked.
    show/hide this revision's text 2 added 1066 characters in body

    I think the right way to phrase this discussion is as follows. Let $(X_s)_{0 \leq s \leq t}$ be a real stochastic process with cyclically exchangeable increments: for all $u \in [0,t]$, the process $(X'_s)_{0\leq s \leq t}$ obtained by a cyclic shift by $u$, has the same distribution as the original process.

    Suppose that $X_0=X_t=0$ with probability one. Then for all $s$, $\mathbb{E}(X_s)=0$. (As in Ori's argument, for this step a continuity argument is needed, which you may not like. On the other hand, this kind of continuity argument is bog-standard -- it is a basic procedure when you study infinitely divisible distributions via their characteristic functions.)

    Edit: Here is an argument to replace the continuity argument but which requires an additional assumption. Suppose for simplicity that $t=1$. The additional assumption is that $\sup_{s \in (0,1)} |\mathbb{E}(X_s)| < \infty$. Suppose there is $s$ s.t. $\mathbb{E}(X_s) = z > 0$. Then by cyclic exchangeability, $\mathbb{E}(X_{1-s}) = -z$. Again by cyclic exchangeability, $|\mathbb{E}(X_{|2s-1|})| = 2z$, the sign depending on the sign of $2s-1$.

    By repeating this argument, it follows that if there is any point $s$ with $\mathbb{E}(X_s) \neq 0$ then there are points $s$ for which $|\mathbb{E}(X_s)|$ is arbitrarily large. In fact, since by cyclic exchangeability, $\mathbb{E}(X_{s/n})=\mathbb{E}(X_s)/n$, it then follows that there are points arbitrarily close to zero for which $|\mathbb{E}(X_s)$ is arbitrarily large.

    But if $(Z_s)_{0 \leq s \leq t}$ is a Lévy process then $(Z_s - (s/t)Z_t)_{0 \leq s \leq t}$ (a Lévy bridge) has cyclically exchangeable increments. (To prove that, just note that the distribution of $(Z_s)_{0 \leq s \leq t}$ is itself invariant under cyclic shifts.)

    Edit: So this proof works for the Lévy process under the assumption that $\mathbb{E}(sup_{s \in (0,1)} |Z_s|) < \infty$ (which I think should be equivalent to assuming $\mathbb{E}|Z_1| < \infty$ but I haven't checked).

    show/hide this revision's text 1

    I think the right way to phrase this discussion is as follows. Let $(X_s)_{0 \leq s \leq t}$ be a real stochastic process with cyclically exchangeable increments: for all $u \in [0,t]$, the process $(X'_s)_{0\leq s \leq t}$ obtained by a cyclic shift by $u$, has the same distribution as the original process.

    Suppose that $X_0=X_t=0$ with probability one. Then for all $s$, $\mathbb{E}(X_s)=0$. (As in Ori's argument, for this step a continuity argument is needed, which you may not like. On the other hand, this kind of continuity argument is bog-standard -- it is a basic procedure when you study infinitely divisible distributions via their characteristic functions.)

    But if $(Z_s)_{0 \leq s \leq t}$ is a Lévy process then $(Z_s - (s/t)Z_t)_{0 \leq s \leq t}$ (a Lévy bridge) has cyclically exchangeable increments. (To prove that, just note that the distribution of $(Z_s)_{0 \leq s \leq t}$ is itself invariant under cyclic shifts.)