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)?


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.


A construction due to Hamza and Klebaner suggests the following example. Consider some independent i.i.d. sequences $(R_n)_{n\geqslant1}$ and $(W_n)_{n\geqslant0}$ such that the distribution of $W_n$ is standard normal and $|R_n|\leqslant1$ almost surely. Assume that $\varrho=\mathbb E[R_n]$ is not zero. Let $(\mathcal F_n)_{n\geqslant0}$ the filtration defined by $\mathcal F_0=\sigma(W_0)$ and $\mathcal F_{n+1}=\mathcal F_n\vee\sigma(W_{n+1},R_{n+1})$ for every $n\geqslant0$.

Define the sequence $(Z_n)_{n\geqslant0}$ recursively by $Z_0=W_0$ and, for every $n\geqslant0$, $$ Z_{n+1}=R_{n+1}\cdot Z_n+\sqrt{1-R_{n+1}^2}\cdot W_{n+1}. $$ Then each $Z_n$ is standard normal, the process $(Z_n)_{n\geqslant0}$ is adapted to the filtration $(\mathcal F_n)_{n\geqslant0}$ and $\mathbb E[Z_{n+1}\mid\mathcal F_n]=\varrho Z_n$.

Hence the process $(X_n)_{n\geqslant0}$ defined by $X_n=\varrho^{-n}Z_n$ for every $n\geqslant0$ is a martingale and, for each $n\geqslant0$, the distribution of $X_n$ is centered normal with variance $\varrho^{-2n}$. However, the distribution of $X_{n+1}$ conditionally on $X_n$ depends on the full distribution of $R_1$, and not only on the parameter $\varrho$.

For example, if $R_1=\frac12$ almost surely, then $\varrho=\frac12$ and $X_{n+1}$ conditionally on $X_n$ is gaussian with mean $X_n$ and variance $3\cdot4^{n}$. On the other hand, if $R=0$ or $R=1$ with equal probabilities then $\varrho=\frac12$ but $X_{n+1}=2X_n$ or $X_{n+1}=2^{n+1}W_{n+1}$ with equal probabilities hence the distribution of $X_{n+1}$ conditionally on $X_n$ is the barycenter of a Dirac measure at $2X_n$ and of the centered gaussian measure with variance $4^{n+1}$.

To sum up, one can consider $X_0=Y_0=W_0$, $X_{n+1}=X_n+2^n\sqrt{3}\cdot W_{n+1}$ and $Y_{n+1}=2Y_n$ or $Y_{n+1}=2^{n+1}W_{n+1}$ with equal probabilities, with $(W_n)_{n\geqslant0}$ i.i.d. standard normal. Then $(X_n)_{n\geqslant0}$ and $(Y_n)_{n\geqslant0}$ are two martingales such that, for every $n\geqslant0$, the distributions of $X_n$ and $Y_n$ are both centered normal with variance $4^n$ but the distributions of $(X_n,X_{n+1})$ and $(Y_n,Y_{n+1})$ do not coincide.


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