TITLE: Two martingales whose distributions agree for each time have the same overall distribution
QUESTION [8 upvotes]: Let $\{X_n\}$ and $\{Y_n\}$ be two martingales. Suppose that for each fixed $n \in \mathbb Z_+$, $X_n$ and $Y_n$ have the same distribution. Must it hold that the random sequences $\{X_n\}$ and $\{Y_n\}$ have the same distribution? At first glance this appears to be false, and I tried constructing a counterexample using various combinations of random walks and expectations of a fixed random variable with respect to an increasing collection of $\sigma$-algebras. However, I was not successful. Does anyone have a a counterexample (or a proof that the answer to my question is affirmative)?

REPLY [3 votes]: For $p\in [0,\frac 13]$, let $(Z_n^p)_{n\in\{0,1\}}$ be a process with law determined by 
$$Z^p_0 = \begin{cases}-1 & \text{ w.p. } \frac 13\\0 & \text{ w.p. } \frac 13\\1 & \text{ w.p. } \frac 13\\
\end{cases}
$$
and $\mathbb{P}[Z^p_1=j|Z^p_0=i] = q_i^j$, where
$$\begin{align}
q_0^0 &= p\\
q_0^2 = q_0^{-2} &= \frac 12 -\frac p2\\
q_1^2 = q_{-1}^{-2} &= \frac 23 + \frac p4\\
q_1^0 = q_{-1}^0 &= \frac 16 - \frac p2\\
q_1^{-2} = q_{-1}^2 &= \frac 16 + \frac p4
\end{align}
$$
We may check directly that $Z^p$ is a martingale, and that the law of $Z^p_1$ is independent of $p$:
$$
Z^p_1 = \begin{cases}-2 & \text{ w.p. } \frac 49\\0 & \text{ w.p. } \frac 19\\2 & \text{ w.p. } \frac 49\\
\end{cases}
$$
Defining $(X_n)_{n\in\mathbb{N}} = Z^0_{\min(n,1)}$ and $(Y_n)_{n\in\mathbb{N}} = Z^\frac{1}{3}_{\min(n,1)}$, we are done.