Describe all martingales $(X_n)_{n\in\mathbb{N}}$, such that $X_n\in\{-1,0,1\}$ for all $n\in\mathbb{N}$ with an arbitrary sample space $\Omega$.

Question

Describe all martingales $(X_n)_{n\in\mathbb{N}}$, such that $X_n\in\{-1,0,1\}$ for all $n\in\mathbb{N}$ with an arbitrary sample space $\Omega$.

This Question evolved out of this Question where $\Omega$ is not arbitrary. I racked my head about how to approach this Problem. A hint was to look at $\mathbb{P}(X_{n+k}=X_k|X_n=1)=\dots$. But this does not get me any further.

Any assistance or thoughts would be much appreciated.

Answer 1

Let $(X_n)$ be a Martingal that meets the above conditions. Then the following properties apply:

Claim 1: $\mathbb{P}(X_{n+k}=X_n\forall k\in\mathbb{N}|X_n=1)=1$

Assume $\mathbb{P}(X_{n+1}\neq X_n|X_n=1)>0$. Then for all $\omega\in\{X_n=1\}$ the following inequality holds: $$E[X_{n+1}|\mathcal{F}_n]=\mathbb{P}(X_{n+1}=X_n|X_n=1)-\mathbb{P}(X_{n+1}=-1|X_n=1)<1=X_n$$ With induction this can be shown for all $k\in\mathbb{N}$. Which makes the sequence $(X_n)$ not a martingale.

Claim 2: $\mathbb{P}(X_{n+k}=X_n\forall k\in\mathbb{N}|X_n=-1)=1$

can be proven analogously to claim 1.

Claim 3: Let $T:=\min\{n\in\mathbb{N}_0\cup\{\infty\}|X_n\in\{-1,1\}\}$. Then: $$\forall x<T:X_n=0 \text{ (a.s) }\wedge \forall n\geq T:X_n=X_T\text{ (a.s) }$$

The first property follows from the construction of $T$ itself. So let $\mathbb{P}(X_n\neq X_T)>0$ for a particular $n>T$. Then either Claim 1 or Claim 2 does not apply. Which proves the second property.

So $(T,X_T)$ describes the searched Martingale. The only restriction on the distribution of this random vector in $(\mathbb{N}_0\cup\{\infty\})\times\{-1,1\}$ is the following:

Claim 4:$\forall n\in\mathbb{N}: \mathbb{P}(X_T=1,T=n)=\mathbb{P}(X_T=-1,T=n)=\frac{1}{2}$

First of all we note that $E[X_{n+1}|\mathcal{F_n}]=0$ holds in $\{X_n=0\}$.Furthermore, it also applies that: $$E[X_{n+1}|\mathcal{F_n}]=\mathbb{P}(T=n+1,X_{n+1}=1)-\mathbb{P}(T=n+1,X_{n+1}=-1)$$ With which the claim follows.

The last point that still has to be proven is that such a sequence always forms a martingale.

Let $(T,X_T)$ be an arbitrary random vector which suffices claim 4. Then the sequence $(X_n)$ is integrable (because it is bounded) and is adapted (because $T$ is a stopping time). Further $E[X_{n+1}|\mathcal{F}_n]=X_n$ applies because $X_{n+1}=X_n=X_T$ (a.s) holds in $\{T\leq n\}$ by Definition, such that $E[X_{n+1}|\mathcal{F}_n]=X_T=X_n$. In $\{T=n+1\}$ claim 4 holds, such that $E[X_{n+1}|\mathcal{F}_n]=0=X_n$. Moreover $X_{n+1}=X_n=0$ (a.s.) holds in $\{T>n+1\}$, such that $E[X_{n+1}|\mathcal{F}_n]=0=X_n$.