8
$\begingroup$

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)?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.