## Lecture 8. The Doob’s stopping theorem

The next four lectures will be devoted to the foundational theorems of the theory of continuous time martingales. All of these theorems are due to Joseph Doob.

The following first theorem shows that martingales behave in a very nice way with respect to stopping times.

Theorem (Doob’s stopping theorem) Let $(\mathcal{F}_t)_{t \ge 0}$ be a filtration defined on a probability space $(\Omega, \mathcal{F},\mathbb{P})$ and let $(M_t)_{t \ge 0}$ be a stochastic process that is adapted to the filtration $(\mathcal{F}_t)_{t \ge 0}$, whose paths are right continuous and locally bounded. The following properties are equivalent:

• $(M_t)_{t \ge 0}$ is a martingale with respect to the filtration $(\mathcal{F}_t)_{t \ge 0}$;
• For any, almost surely bounded stopping time $T$ of the filtration $(\mathcal{F}_t)_{t \ge 0}$ such that $\mathbb{E}(\mid M_T \mid)<+\infty$, we have $\mathbb{E} (M_T)=\mathbb{E} (M_0)$.

Proof:

Let us assume that $(M_t)_{t \ge 0}$ is a martingale with respect to the filtration $(\mathcal{F}_t)_{t \ge 0}$, whose paths are right continuous and locally bounded. Let now $T$ be a stopping time of the filtration $(\mathcal{F}_t)_{t \ge 0}$ that is almost surely bounded by $K>0$. Let us first assume that $T$ takes its values in a finite set: $0 \le t_1 <.... Thanks to the martingale property, we have
$\mathbb{E} (M_T)$
$= \mathbb{E} (\sum_{i=1}^n M_T 1_{T=t_i})$
$=\sum_{i=1}^n \mathbb{E} (M_{t_i} 1_{T=t_i})$
$=\sum_{i=1}^n \mathbb{E} (M_{t_n} 1_{T=t_i})$
$=\mathbb{E} (M_{t_n})$
$=\mathbb{E} (M_{0})$.

The theorem is therefore proved if $T$ takes its values in a finite set. If $T$ takes an infinite number of values, we approximate $T$ by the following sequence of stopping times:

$\tau_n =\sum_{k=1}^{2^n} \frac{kK}{2^n} 1_{ \left\{\frac{(k-1)K}{2^n} \le T < \frac{kK}{2^n} \right\} }.$

The stopping time $\tau_n$ takes its values in a finite set and when $n \rightarrow +\infty$, $\tau_n \rightarrow T$. To conclude the proof of the first part of the proposition, it is therefore enough to show that

$\lim_{n \rightarrow +\infty} \mathbb{E} ( M_{\tau_n} )=\mathbb{E} (M_{T} ).$
For this, we are going to prove that the family $(M_{\tau_n})_{n \in\mathbb{N}}$ is uniformly integrable. Let $A \ge 0$.

Since $\tau_n$ takes its values in a finite set, by using the martingale property and Jensen’s inequality, it is easily checked that

$\mathbb{E} (| M_{K} |1_{M_{\tau_n} \ge A }) \ge \mathbb{E} (| M_{\tau_n}| 1_{M_{\tau_n} \ge A }).$

Therefore, we have

$\mathbb{E} ( M_{\tau_n} 1_{M_{\tau_n} \ge A }) \le \mathbb{E} ( M_{K} 1_{\sup_{0 \le s \le K} M_s \ge A }) \rightarrow_{A \rightarrow + \infty} 0.$

By uniform integrability and convergence in probability, we deduce that

$\lim_{n \rightarrow +\infty} \mathbb{E} ( M_{\tau_n} )=\mathbb{E} (M_{T} ),$

from which it is concluded that

$\mathbb{E} (M_T)=\mathbb{E} (M_0).$

Conversely, let us now assume that for any, almost surely bounded stopping time $T$ of the filtration $(\mathcal{F}_t)_{t \ge 0}$ such that $\mathbb{E}(\mid M_T \mid)<+\infty$, we have $\mathbb{E}(M_T)=\mathbb{E} (M_0)$.

Let $0 \le s \le t$ and $A \in \mathcal{F}_s$. By using the stopping time

$T=s 1_A +t 1_{^c A},$

we are led to

$\mathbb{E} \left( (M_t - M_s) 1_A \right)=0,$

which implies the martingale property for $(M_t)_{t \ge 0}$. $\square$

The hypothesis that the paths of $(M_t)_{t \ge 0}$ be right continuous and locally bounded is actually not strictly necessary, however the hypothesis that the stopping time $T$ be almost surely bounded is essential, as it is proved in the following exercise.

Exercise : Let $(\mathcal{F}_t)_{t \ge 0}$ be a filtration defined on a probability space $(\Omega, \mathcal{F},\mathbb{P})$ and let $(M_t)_{t \ge 0}$ be a continuous martingale (that is a martingale with continuous paths) with respect to the filtration $(\mathcal{F}_t)_{t \ge 0}$ such that $M_0=0$ almost surely. For $a>0$, we denote $T_a=\inf \{ t >0, M_t=a \}$. Show that $T_a$ is a stopping time of the filtration $(\mathcal{F}_t)_{t \ge 0}$. Prove that $T_a$ is not almost surely bounded.

Exercise : Let $(\mathcal{F}_t)_{t \ge 0}$ be a filtration defined on a probability space $(\Omega, \mathcal{F},\mathbb{P})$ and let $(M_t)_{t \ge 0}$ be a continuous martingale with respect to the filtration $(\mathcal{F}_t)_{t \ge 0}$. By mimicking the proof of Doob’s stopping theorem, show that if $T_1$ and $T_2$ are two almost surely bounded stopping times of the filtration $(\mathcal{F}_t)_{t \ge 0}$ such that $T_1 \le T_2$ and $\mathbb{E}(\mid M_{T_1} \mid)<+\infty$, $\mathbb{E}(\mid M_{T_2} \mid<+\infty$, then, $\mathbb{E} (M_{T_2} \mid \mathcal{F}_{T_1})=M_{T_1}.$

Deduce that the stochastic process $(M_{t\wedge T_2})_{t \ge 0}$ is a martingale with respect to the filtration $(\mathcal{F}_{t \wedge T_2})_{t \ge 0}$.

Exercise : Let $(\mathcal{F}_t)_{t \ge 0}$ be a filtration defined on a probability space $(\Omega, \mathcal{F},\mathbb{P})$ and let $(M_t)_{t \ge 0}$ be a submartingale with respect to the filtration $(\mathcal{F}_t)_{t \ge 0}$ whose paths are continuous. By mimicking the proof of Doob’s stopping theorem, show that if $T_1$ and $T_2$ are two almost surely bounded stopping times of the filtration $(\mathcal{F}_t)_{t \ge 0}$ such that $T_1 \le T_2$ and $\mathbb{E}(\mid M_{T_1} \mid)<+\infty$, $\mathbb{E}(\mid M_{T_2} \mid)<+\infty$, then, $\mathbb{E} (M_{T_1} ) \le \mathbb{E} (M_{T_2} )$.D

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