# Can optional stopping hold for $\mathcal{L}^1$ bounded martingales?

As part of a larger problem I am trying to figure out if it possible for a $$\mathcal{L}^1$$-bounded martingale $$(M_n)_{n\geqslant 1}, M_0=0$$ to satisfy the following:

For some almost surely finite stopping time $$\tau$$, we have $$\mathbb{E}[M_\tau]>0$$.

I believe it is not possible since bounded in $$\mathcal{L}^1$$ implies convergence to some integrable $$M_\infty$$, and so for an a.s. finite stopping time $$\tau$$ we can write

$$\mathbb{E}[M_\tau]=\mathbb{E}[M_\tau \mathbb{1}_{\{\tau < \infty\}}]+ \mathbb{E}[M_\infty]$$

We know the second term on the RHS is $$0$$ since it equals the expectation of $$M_0$$, and the first term on the RHS will be the expectation of some random variable in the sequence which is again that of $$M_0$$.

Is this correct?

I am concerned that it is not true since I cannot seem to find a variant of optional stopping that would be satisfied for almost surely finite stopping times and boundedness in $$\mathcal{L}^1$$. Perhaps this is a case to show that the key deduction of optional stopping is not an if statement?

EDIT: As mentioned in the comment, I was mistaken in thinking that the martingale property holds the limit, which it doesn't in this case.

• I agree that this probably isn't true, but I don't think your argument works. In particular, I don't see why $\mathbb{E}[M_\infty] = \mathbb{E}[M_0]$. This is true of uniformly integrable martingales, but not necessarily $L^1$ bounded martingales. Apr 25, 2023 at 17:58
• There are some things to consider. For example, we should think of a stopping time without finite mean. Also $M_{n}$ cannot be uniformly integrable as then $M_{T\wedge n}$ is uniformly integrable. The most basic example of an $L^{1}$ bounded but non uniformly integrable family is $n\mathbf{1}_{(0,\frac{1}{n})}$ . Maybe you can create a non uniformly integrable martingale out of these. Apr 25, 2023 at 18:47

This is indeed possible. For an example, let $$(X_n)$$ be a sequence of i.i.d. random variables with $$\mathbb{P}(X_n = 2) = \mathbb{P}(X_n = 0) = \frac 12$$. Observe that $$\mathbb{E}[X_n] = 1$$, so the process $$\widehat M_n := \prod_{k=1}^n X_k$$ is a martingale, with $$\widehat M_0 = 1$$. Note that $$\mathbb{E}[|\widehat M_n|] = 1$$, so $$\widehat M_n$$ is $$L^1$$ bounded.
Now, let $$M_n := 1-\widehat M_n$$ (so that $$M_0 = 0$$), which is also an $$L^1$$ bounded martingale.
Define the stopping time $$\tau := \inf\{n : X_n = 0\}$$, which is finite a.s. because $$\mathbb{P}(\tau = \infty) \le \mathbb{P}(\tau > n) = 2^{-n} \rightarrow 0$$. However, $$M_\tau = 1$$, so $$\mathbb{E}[M_\tau] = 1 > 0$$.