## Lecture 9. The Doob’s convergence theorem

Let us first remind some basic facts about the notion of uniform integrability which is a crucial tool in the study of continuous time martingales.

Definition. Let $(X_i)_{i \in \mathcal{I}}$ be a family of random variables. We say that the family $(X_i)_{i \in \mathcal{I}}$ is uniformly integrable if for every $\varepsilon >0$, there exists $K \ge 0$ such that

$\forall i \in \mathcal{I},\quad \mathbb{E}(\mid X_i \mid 1_{\mid X_i \mid >K})<\varepsilon.$

We have the following properties:

• A finite family of integrable random variables is uniformly integrable ;
• If the family $(X_i)_{i \in \mathcal{I}}$ is uniformly integrable then it is bounded in $L^1$, that is $\sup_\mathcal{I} \mathbb{E}(\mid X_i \mid ) <+\infty$;
• If the family $(X_i)_{i \in \mathcal{I}}$ is bounded in $L^p$ with $p>1$, that is $\sup_\mathcal{I} \mathbb{E}(\mid X_i \mid^p ) <+\infty$, then it is uniformly integrable.

The notion of uniform integrability is often used to prove a convergence in $L^1$ thanks to the following result:

Proposition. Let $(X_n)_{n\in \mathbb{N}}$ be a sequence of integrable random variables. Let $X$ be an integrable random variable. The sequence $(X_n)_{n\in \mathbb{N}}$ converges toward $X$ in $L^1$, that is $\lim_{n \to +\infty} \mathbb{E}(\mid X_n-X \mid )=0$, if and only if:

• In probability, $X_n \rightarrow_{n \rightarrow +\infty} X$, that is for every $\varepsilon >0$, $\lim_{n \to +\infty} \mathbb{P}( \mid X_n -X \mid \ge \varepsilon) =0$;
• The family $(X_n)_{n\in \mathbb{N}}$ is uniformly integrable.

It is clear that if $X$ is an integrable random variable defined on a filtered probability space $(\Omega, (\mathcal{F}_t)_{t \ge 0}, \mathcal{F},\mathbb{P})$ then the process $\left(\mathbb{E}(X\mid \mathcal{F}_t) \right)_{t \ge 0}$ is a martingale with respect to the filtration $(\mathcal{F}_t)_{t \ge 0}$. The following theorem characterizes the martingales that are of this form.

Theorem (Doob’s convergence 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 martingale with respect to the filtration $(\mathcal{F}_t)_{t \ge 0}$ whose paths are left limited and right continuous. The following properties are equivalent:

• When $t \to +\infty$, $(M_t)_{t \ge 0}$ converges in $L^1$;
• When $t \to +\infty$, $(M_t)_{t \ge 0}$ converges almost surely toward an integrable and $\mathcal{F}$-measurable random variable $X$ that satisfies

$M_t =\mathbb{E}(X\mid \mathcal{F}_t), t \ge 0;$

• The family $(M_t)_{t \ge 0}$ is uniformly integrable.

Proof:

As a first step, we show that if the martingale $(M_t)_{t \ge 0}$ is bounded in $L^1$, that is

$\sup_{t \ge 0} \mathbb{E} ( \mid M_t \mid)<+\infty$

then $(M_t)_{t \ge 0}$ almost surely converges toward an integrable and $\mathcal{F}$-measurable random variable $X$.

Let us first observe that

$\left\{ \omega \in \Omega, M_t (\omega) \text{ converges} \right\}=\left\{ \omega \in \Omega, \lim \sup_{t \rightarrow +\infty} M_t(\omega)= \lim \inf_{t \rightarrow +\infty} M_t(\omega)\right\}$

Therefore, in order to show that $(M_t)_{t \ge 0}$ almost surely converges when $t \to +\infty$, we may prove that

$\mathbb{P} \left( \left\{ \omega \in \Omega, \lim \sup_{t \rightarrow +\infty} M_t(\omega) >\lim \inf_{t \rightarrow +\infty} M_t (\omega)\right\} \right)=0.$

Let us assume that

$\mathbb{P} \left( \left\{ \omega \in \Omega, \lim \sup_{t \rightarrow +\infty} M_t(\omega) >\lim \inf_{t \rightarrow +\infty} M_t (\omega)\right\} \right)>0.$

In that case we may find $a such that:

$\mathbb{P} \left( \left\{ \omega \in \Omega, \lim \sup_{t \rightarrow +\infty} M_t(\omega) >a>b>\lim \inf_{t \rightarrow +\infty} M_t (\omega)\right\} \right)>0.$

The idea now is to study the oscillations of $(M_t)_{t \ge 0}$ between $a$ and $b$. For $N \in \mathbb{N}$, $N>0$ and $n\in \mathbb{N}$, we denote

$\mathcal{D}_{n,N}=\left\{ \frac{kN}{2^n}, 0 \le k \le 2^n \right\},$

and

$\mathcal{D}=\cup_{n,N} \mathcal{D}_{n,N}.$

Let $\mathcal{N} (a,b,n,N)$ be the greatest integer $k$ for which we may find elements of $\mathcal{D}_{n,N}$,

$0 \le q_1

that satisfy

$M_{q_i} b.$

Let now

$Y_{n,N}=\sum_{k=1}^{2^n} C_{ \frac{kN}{2^n}} (M_{\frac{kN}{2^n}}-M_{\frac{(k-1)N}{2^n}}),$

where $C_k \in \{0,1\}$ is recursively defined by:

$C_{ 1}=1_{ M_0 < a},$

$C_{k}=1_{C_{k-1}=1} 1_{ M_{\frac{(k-1)N}{2^n}} \le b}+1_{C_{k-1}=0} 1_{ M_{\frac{(k-1)N}{2^n}} < a}.$

Since $(M_t)_{t \ge 0}$ is martingale, it is easily checked that

$\mathbb{E} (Y_{n,N})=0.$

Furthermore, thanks to the very definition of $\mathcal{N} (a,b,n,N)$, we have

$Y_{n,N} \ge (b-a)\mathcal{N} (a,b,n,N) -\max( a-M_N,0).$

Therefore,

$(b-a) \mathbb{E} \left( \mathcal{N} (a,b,n,N) \right) \le \mathbb{E}\left(\max( a-M_N,0)\right)\le\mid a \mid+ \mathbb{E} ( \mid M_N\mid ) \le \mid a \mid+ \sup_{t>0} \mathbb{E} (\mid M_t \mid ),$

and thus

$(b-a) \mathbb{E} \left( \sup_{n,N} \mathcal{N} (a,b,n,N) \right) \le \mid a \mid+ \sup_{t>0} \mathbb{E} ( \mid M_t \mid ).$

This implies that almost surely $\sup_{n,N} \mathcal{N}(a,b,n,N)<+\infty$, from which we deduce

$\mathbb{P} \left( \left\{ \omega \in \Omega, \lim \sup_{t \rightarrow +\infty,t\in \mathcal{D}} M_t(\omega) >a>b>\lim \inf_{t \rightarrow +\infty,t\in \mathcal{D}} M_t (\omega)\right\}\right)=0.$

Since the paths of $(M_t)_{t \ge 0}$ are right continuous, we have

$\mathbb{P} \left( \left\{ \omega \in \Omega, \lim \sup_{t\rightarrow +\infty,t\in \mathcal{D}} M_t(\omega) >a>b>\lim \inf_{t\rightarrow +\infty,t\in \mathcal{D}} M_t (\omega)\right\}\right)$
$=\mathbb{P} \left( \left\{ \omega \in \Omega, \lim \sup_{t\rightarrow +\infty} M_t(\omega) >a>b>\lim \inf_{t \rightarrow +\infty} M_t (\omega)\right\} \right) .$

This is absurd.

Thus, if $(M_t)_{t \ge 0}$ is bounded in $L^1$, it almost surely converges toward an $\mathcal{F}$-measurable random variable $X$. Fatou’s lemma provides the integrability of $X$.

With this preliminary result in hands, we can now turn to the proof of the theorem.

Let us assume that $(M_t)_{t \ge 0}$ converges in $L^1$. In that case, it is of course bounded in $L^1$, and thus almost surely converges toward an $\mathcal{F}$-measurable and integrable random variable $X$. Let $t \ge 0$ and $A \in \mathcal{F}_t$, we have for $s \ge t$,

$\mathbb{E} ( M_s 1_A)=\mathbb{E} ( M_t 1_A)$

By letting $s \rightarrow +\infty$, the dominated convergence theorem yields

$\mathbb{E} (X 1_A)=\mathbb{E} ( M_t 1_A).$

Therefore, as expected, we obtain

$\mathbb{E} (X \mid \mathcal{F}_t)= M_t.$

Let us now assume that $(M_t)_{t \ge 0}$ almost surely converges toward an $\mathcal{F}$-measurable and integrable random variable $X$ that satisfies

$M_t =\mathbb{E}(X\mid \mathcal{F}_t), t \ge 0.$

We almost surely have $\sup_{t \ge 0} \mid M_t \mid <+\infty$ and thus for $A \ge 0$,

$\mathbb{E} ( \mid M_t \mid 1_{\mid M_t \mid \ge A} )=\mathbb{E} ( \mid \mathbb{E}(X\mid \mathcal{F}_t) \mid 1_{\mid M_t \mid \ge A})$
$\le \mathbb{E} ( \mid X \mid 1_{\mid M_t \mid \ge A} )$
$\le \mathbb{E} ( \mid X \mid 1_{\sup_{t \ge 0} \mid M_t \mid \ge A}).$

This implies the uniform integrability for the family $(M_t)_{t \ge 0}$ .

Finally, if the family $(M_t)_{t \ge 0}$ is uniformly integrable, then it is bounded in $L^1$ and therefore almost surely converges. The almost sure convergence, together with the uniform integrability, provides the convergence in $L^1$.

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