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<b 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 <r_1 <q_2 <r_2 <...<q_k<r_k\le N

that satisfy

M_{q_i} <a, M_{r_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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