This tiny post is devoted to a proof of the almost sure convergence of martingales bounded in $\mathrm{L}^1$. This proof that we give below relies on the almost sure convergence of martingales bounded in $\mathrm{L}^2$, after a truncation step. In order to keep the martingale property after truncation, we truncate with a stopping time. The boundedness in $\mathrm{L}^1$ is used to show via the maximal inequality that the martingale is almost surely bounded. Note that this proof differs from the classical and historical proof from scratch which is based on up-crossing or oscillations.

The martingales are either in discrete time or in continuous time with continuous paths.

The theorem. Let $M={(M_t)}_{t\geq0}$ be a continuous martingale bounded in $\mathrm{L}^1$. Then there exists $M_\infty\in\mathrm{L}^1$ such that $\lim_{t\to\infty}M_t=M_\infty$ almost surely. Moreover the convergence holds in $\mathrm{L}^1$ if and only if $M$ is uniformly integrable.

A proof. The fact that $M_\infty\in\mathrm{L}^1$ follows without effort from the almost sure convergence, the boundedness in $\mathrm{L}^1$, and the Fatou lemma, namely
$\mathbb{E}(|M_\infty|) =\mathbb{E}(\varliminf_{t\to\infty}|M_t|) \leq\varliminf_{t\to\infty}\mathbb{E}(|M_t|) \leq C<\infty.$ Moreover, it is a general fact that a sequence of random variables that converges almost surely to a limit belonging to $\mathrm{L}^1$ does converge in $\mathrm{L}^1$ if and only if it is uniformly integrable.

It remains to prove a.s. convergence. We can assume that $M_0=0$, otherwise consider the martingale $M-M_0={(M_t-M_0)}_{t\geq0}$ which is also bounded in $\mathrm{L}^1$, making $M$ converge a.s. to $M_0+(M-M_0)_\infty$. We proceed by truncation and reduction to the square integrable case.

By the Doob maximal inequality with $p=1$, and $r>0$,

$$\mathbb{P}\Bigr(\sup_{s\in[0,t]}|M_s|\geq r\Bigr) \leq\frac{\mathbb{E}(|M_t|)}{r}.$$
By monotone convergence, with $C:=\sup_{t\geq0}\mathbb{E}(|M_t|)<\infty$, for all $r>0$,
$\mathbb{P}\Bigr(\sup_{t\geq0}|M_t|\geq r\Bigr) \leq\frac{C}{r}.$
It follows that $\mathbb{P}\Bigr(\sup_{t\geq0}|M_t|=\infty\Bigr)\leq\lim_{r\to\infty}\mathbb{P}\Bigr(\sup_{t\geq0}|M_t|\geq r\Bigr)=0.$
In other words almost surely ${(M_t)}_{t\geq0}$ is bounded.
As a consequence, on an almost sure event, say $\Omega’$, for large enough $n$,
$T_n:=\inf\{t\geq0:|M_t|\geq n\}=\infty.$

On the other hand, by the Doob stopping theorem, for all $n\geq0$, ${(M_{t\wedge T_n})}_{t\geq0}$ is a martingale. Moreover, since $M_0=0$, we have $\sup_{t\geq0}|M_{t\wedge T_n}|\leq n$. Since ${(M_{t\wedge T_n})}_{t\geq0}$ is bounded in $\mathrm{L}^2$, there exists $M^{(n)}_\infty\in\mathrm{L}^2$ such that $\lim_{t\to\infty}M_{t\wedge T_n}=M^{(n)}_\infty$ almost surely (and in $\mathrm{L}^2$ but this is useless here). Let us denote by $\Omega_n$ the almost sure event on which this convergence holds. Then, on the almost sure event $\Omega’\cap(\cap_n\Omega_n)$, we have, for all $m,n$, $M^{(n)}_\infty=M^{(m)}_\infty=:M_\infty$, and

$\lim_{t\to\infty}M_t=M_\infty.$

Truncation. Truncation is very natural to increase integrability. It is for instance used in the proof of the strong law of large numbers for independent random variables in $\mathrm{L}^1$ in order to reduce the problem to variables in $\mathrm{L}^p$ with $p>1$, the case $p=4$ being particularly simple.

Prerequisites. The ingredients should be established before and without using this theorem namely maximal inequalities for martingales, almost sure convergence of martingales bounded in $\mathrm{L}^2$, and stopping theorem for martingales and arbitrary stopping times.

Note. Since $M$ is continuous, on $\{M_0=0\}$, we have $T_n=\inf\{t\geq0:|M_t|=n\}$. How to adapt this proof to discrete time martingales ? In this case, $\sup_{t\in\mathbb{N}}|M_{t\wedge T_n}|$ is not necessarily $\leq n$ but it can be bounded by the random variable $n+M_{T_n}\mathbf{1}_{T_n<\infty}$ which is integrable, and this provides uniform integrability. Note that it is not the discreteness of time that poses a problem but rather the discontinuity of the trajectories. Continuous martingales are more convenient to handle.

Acknowledgements. Thanks to my friend and colleague Nathaël Gozlan who pointed out a bug in an earlier version which was related to a missing handling of $M_0$ (see also discrete time case).

Last Updated on 2020-09-25

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