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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$but I still haven't checked)checked. 2 added 1066 characters in body I think the right way to phrase this discussion is as follows. Let$(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). 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.)