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.

  • 1
    $\begingroup$ 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. $\endgroup$ Apr 25, 2023 at 17:58
  • $\begingroup$ 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. $\endgroup$ Apr 25, 2023 at 18:47

1 Answer 1


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 $\mathbb{E}[M_\tau] = 1 > 0$.


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