Restriction martingale to a measurable set

Question

If we have that $M_n$ is a martingale adapted to $\mathscr{F}_n\subset\mathscr{F}$ on a probability space $(\Omega,\mathscr{F},\mu)$ then if $A$ is an $\mathscr{F}_0$ measurable set we have $$M_n 1_A=E[M_{n+1} \mid \mathscr{F_n}]1_A=E[M_{n+1}1_A \mid \mathscr{F}_n].$$ So we have that $\tilde{M_n}:=M_n1_A$ is a martingale.

If $A$ is instead an $\mathscr{F}$-measurable set is it true that $\tilde{M_n}:=M_nE[1_A \mid \mathscr{F}_n]$ is a martingale? I am having trouble proving it. If it's not true, is there a concise counterexample?

Is there a "better" version of a restriction of a martingale to an $\mathscr{F}$-measurable set in some sense?

Answer 1

Remark 1. If $A\in \mathcal{F}_{m-1}$ for some (finite) $m$, then, the process is a martingale after the occurrance of $A$, i.e., $\left(\overline{M}_{n+m}\right)_{n\geq 0}$ is a martingale. Thus, if you wait some finite time, the process becomes a martingale.

Remark 2. Despite the slight distinction between $\overline{M}_n=M_n 1_A$ and $\overline{M}_n=M_n E\left(1_A\left|\mathcal{F}_n\right.\right)$, as in general $1_A \neq E\left(1_A\left|\mathcal{F}_n\right.\right)$, saz example works for both constructions.

I will just complement saz' answer and exhibit a counter-example with $A$ intersecting the tail of $\mathcal{F}$, i.e., $A\notin \mathcal{F}_m$ for all $m$, to show that $\overline{M}_n=M_n E\left(1_A\left|\mathcal{F}_n\right.\right)$ is not necessarily a martingale (nor any finite time-shift counter-part).

Counter-example. Let $M_n$ be the random walk as in saz' reply, except that it stops when it hits either $k\in\mathbb{N}$ or $-k$. Let $T=\min\left\{m\in\mathbb{N}\,:\, M_m\in\left\{k,-k\right\}\right\}$ be the stopping time. Note that we are considering the martingale $M^{T}_n\overset{\Delta}=M_{n\wedge T}$, but I will refer to it simply as $M_n$. Let $A=\left\{\omega\in\Omega\,:\, M_n(\omega) = k,\,\mbox{eventually}\right\}$. Define $E^N=\left\{\omega\in\Omega\,:\, (T(\omega)=N) \wedge (M_N(\omega)=k)\right\}$ and note that for each $N\geq k$, with the same parity as $k$, $\mathbb{P}(E^N)>0$. We have

(i) $E\left(\overline{M}_N\left|\mathcal{F}_{N-1}\right.\right)(\omega)= E\left(M_N E\left(1_A\left|\mathcal{F}_N\right.\right)\left|\mathcal{F}_{N-1}\right.\right)(\omega)=E\left(M_N 1_A\left|\mathcal{F}_{N-1}\right.\right)(\omega)$ $=k\times \mathbb{P}\left(M_N=k\left|M_{N-1}=k-1\right.\right)+(k-2)\times \mathbb{P}\left(M_N=k-2,A\left|M_{N-1}=k-1\right.\right)$
$= k\times \frac{1}{2}+(k-2)\times p_1$ for almost all $\omega\in E^N$, where $p_1<\frac{1}{2}$ is the probability that the walker moves to the left from $k-1$, but it eventually hits $k$.

(ii) $\overline{M}_{N-1}(\omega)=M_{N-1}(\omega) E\left(1_A\left|\mathcal{F}_{N-1}\right.\right)(\omega)=(k-1)\times \left(\frac{1}{2}+p_1\right)$.

Now, one can check that equality between (i) and (ii) holds if and only if $p_1=\frac{1}{2}$, which is a contradiction. In other words, $E\left(\overline{M}_N\left|\mathcal{F}_{N-1}\right.\right)(\omega)\neq \overline{M}_{N-1}(\omega)$, for almost all $\omega\in E^N$. Since, $\mathbb{P}(E^N)>0$, then, $\overline{M}_n$ is not a martingale.

Answer 2

No, the restriction is, in general, not a martingale. Consider for instance a simple random walk $(M_n)_{n \in \mathbb{N}}$, i.e. $$M_n = \sum_{j=1}^n X_j$$

where $X_j$, $j \geq 1$, are independent identically distributed random variables taking the values $\pm 1$ with probability $1/2$. Clearly, $(M_n)_{n \geq 1}$ is a martingale with respect to the canonical filtration. If we define

$$A := \{X_1=1\}$$

then $\tilde{M}_n := M_n 1_A$ fails to be a martingale since

$$\mathbb{E}(\tilde{M}_0)=0 \neq \frac{1}{2} = \mathbb{E}(\tilde{M}_1).$$

Remark: In a similar fashion you can construct counterexamples for other martingales $(M_n)_{n \geq 1}$ by considering

$$A := \{M_n \geq (1+\epsilon) \mathbb{E}(M_0)\}$$

for suitable $\epsilon>0$.