## Lecture 10. The Doob’s regularization theorem

When dealing with stochastic processes, it is often important to work with versions of these processes whose paths are as regular as possible. In that direction, the Kolmogorov’s continuity theorem (see Lecture 6) provided a sufficient condition allowing to work with continuous versions of stochastic processes. For martingales, the possibility of working with regular versions, is related to the regularity properties of the filtration with respect to which the martingale property is satisfied.

Definition: Let $(\mathcal{F}_t)_{t \ge 0}$ be a filtration on a probability space $(\Omega, \mathcal{F},\mathbb{P})$. If the following assumptions are fulfilled:

• If $A \in \mathcal{F}$ satisfies $\mathbb{P} (A) =0$, then every subset of $A$ is in $\mathcal{F}_0$;
• The filtration $(\mathcal{F}_t)_{t \ge 0}$ is right continuous, that is for every $t \ge 0$

$\mathcal{F}_t=\cap_{\varepsilon >0} \mathcal{F}_{t+\varepsilon},$

the filtered probability space

$(\Omega, (\mathcal{F}_t)_{t \ge 0},\mathcal{F},\mathbb{P})$

is said to satisfy the usual conditions.

The above set of assumptions are called the usual conditions because, as we will see it in the next Lectures, these are the conditions under which it is convenient to work in order to properly define the stochastic integral.

Let $(\mathcal{F}_t)_{t \ge 0}$ be a filtration on a probability space $(\Omega, \mathcal{F},\mathbb{P})$ and let $(M_t)_{t \ge 0}$ be a (sub, super) martingale with respect to the filtration $(\mathcal{F}_t)_{t \ge 0}$ whose paths are right continuous and left limited. The filtered probability space

$(\Omega, (\mathcal{F}_t)_{t \ge 0},\mathcal{F},\mathbb{P})$

may canonically be enlarged into a filtered probability space

$(\Omega, (\mathcal{G}_t)_{t \ge 0},\mathcal{G},\mathbb{P})$

that satisfies the usual conditions. Indeed, $\mathcal{G}$ can be taken to be the $\mathbb{P}$-completion of $\mathcal{F}$ and

$\mathcal{G}_t=\cap_{u >t} \sigma ( \mathcal{F}_u, \mathcal{N} )$

where $\mathcal{N}$ is the set of events whose probability is zero. Moreover $(M_t)_{t \ge 0}$ is a (sub, super) martingale with respect to the filtration $(\mathcal{G}_t)_{t \ge 0}$ (this is not straightforward and let to the reader as an exercise ). The filtered probability space

$(\Omega, (\mathcal{G}_t)_{t \ge 0},\mathcal{G},\mathbb{P})$

is called the usual completion of

$(\Omega, (\mathcal{F}_t)_{t \ge 0},\mathcal{F},\mathbb{P}).$

Exercise: Let $(\Omega, (\mathcal{F}_t)_{t \ge 0},\mathcal{F},\mathbb{P})$ be a filtered probability space that satisfies the usual conditions and let $(X_t)_{t \ge 0}$ be stochastic process adapted to the filtration $(\mathcal{F}_t)_{t \ge 0}$ whose paths are left limited and right continuous. Let $K$ be a compact subset of $\mathbb{R}$. Show that the random time

$T=\inf \{ t \ge 0, X_t \in K \}$

is a stopping time of the filtration $(\mathcal{F}_t)_{t \ge 0}$.

Theorem (Doob’s regularization theorem): Let $(\Omega, (\mathcal{F}_t)_{t \ge 0},\mathcal{F},\mathbb{P})$ be a filtered probability space that satisfies the usual conditions and let $(M_t)_{t \ge 0}$ be a supermartingale with respect to the filtration $(\mathcal{F}_t)_{t \ge 0}$. Let us assume that the function $t \rightarrow \mathbb{E} (M_t)$ is right continuous. There exists a modification $(\tilde{M}_t)_{t \ge 0}$ of $(M_t)_{t \ge 0}$ such that:

• $(\tilde{M}_t)_{t \ge 0}$ is adapted to the filtration $(\mathcal{F}_t)_{t \ge 0}$;
• The paths of $(\tilde{M}_t)_{t \ge 0}$ are locally bounded, right continuous and left limited;
• $(\tilde{M}_t)_{t \ge 0}$ is a supermartingale with respect to the filtration $(\mathcal{F}_t)_{t \ge 0}$.

Proof:

As for the proof of Doob’s convergence theorem (see Lecture 9), the idea is to study the oscillations of $(M_t)_{t\ge 0}$. In what follows, we will use the notations introduced in the proof of this theorem that we remind below.

For $N \in \mathbb{N}$, $N>0$ and $n\in \mathbb{N}$, we denote

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

$\mathcal{D}_{N}=\cup_n \mathcal{D}_{n,N}$

and

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

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

$0 \le q_1

such that

$M_{q_i} < a, M_{r_i} > b.$

Let now $\Omega^*$ be the set of $\omega \in \Omega$ such that $\forall t\ge 0, \lim_{s \rightarrow t, s>t, s \in \mathcal{D}} M_s (\omega)$ exists and is finite.

It is easily seen that:

$\Omega^* =\cap_{a,b \in \mathbb{Q}} \cap_{N \in \mathbb{N}^*} \left\{ \omega \in \Omega, \sup_{t \in \mathcal{D}_N} \mid M_t (\omega) \mid <+\infty \text{ et } \sup_{n\in \mathbb{N}} \mathcal{N}(a,b,n,N) <+\infty \right\}.$

Therefore we have $\Omega^* \in \mathcal{F}$. We may prove, as we proved the Doob's convergence theorem that $\mathbb{P} (\Omega^*)=1$.

For $t \ge 0$, we define $(\tilde{M}_t)_{ t \ge 0}$ in the following way:

• If $\omega \in \Omega^*$, $\tilde{M}_t (\omega)=\lim_{s \rightarrow t, s>t, s \in \mathcal{D}} M_s(\omega)$
• If $\omega \notin \Omega^*$, $\tilde{M}_t (\omega)=0.$

It is clear that the paths of $(\tilde{M}_t)_{t \ge 0}$ are locally bounded, right continuous and left limited. Let us now show that this process is the expected modification of $(M_t)_{t \ge 0}$.

We first observe that for $t \ge 0$, the random variable $\lim_{s \rightarrow t, s>t, s \in \mathcal{D}} M_s$ is measurable with respect to $\cap_{s >t} \mathcal{F}_s=\mathcal{F}_t$. Furthermore, $\Omega \backslash \Omega^*$ has a zero probability and is therefore in $\mathcal{F}_0$, according to the usual conditions. This shows that the process $(\tilde{M}_t)_{t \ge 0}$ is adapted to the filtration $(\mathcal{F}_t)_{t \ge 0}$.

We now show that $(\tilde{M}_t)_{t \ge 0}$ is a modification of $(M_t)_{t \ge 0}$.

Let $t \ge 0$. We have almost surely

$\lim_{s \rightarrow t, s>t, s \in \mathcal{D}} M_s=\tilde{M}_t.$

Let us prove that this convergence also holds in $L^1$. To prove this, it is enough to check that for every decreasing family $(s_n)_{n \in \mathbb{N}}$ such that $s_n \in \mathcal{D}$ and that converges toward $t$, the family $(M_{s_n})_{n \in \mathbb{N}}$ is uniformly integrable.

Let $\varepsilon >0$. Since $u \rightarrow \mathbb{E} (M_u)$ is assumed to be right continuous, we can find $s \in \mathbb{R}$ such that $t < s$ and such that for every $s > u > t$,,

$0 \le \mathbb{E} (M_u) - \mathbb{E} (M_s) \le \frac{\varepsilon}{2}.$

For $s>u>t$ and $\lambda >0$, we have:
$\mathbb{E} ( \mid M_u \mid 1_{ \mid M_u \mid > \lambda})=-\mathbb{E} ( M_u 1_{ M_u < -\lambda})+\mathbb{E}(M_u)-\mathbb{E} ( M_u 1_{ M_u \le \lambda} )$
$\le \mathbb{E} (\mid M_s \mid 1_{ \mid M_u \mid > \lambda})+\frac{\varepsilon}{2}.$

Now, since $M_s \in L^1$, we can find $\delta >0$ such that for every $F \in \mathcal{F}$ that satisfies $\mathbb{P} (F) < \delta$, we have $\mathbb{E} ( \mid M_s \mid 1_F)<\frac{\varepsilon}{2}.$ But for $t < u < s$,

$\mathbb{P}(\mid M_u \mid > \lambda) \le \frac{\mathbb{E} ( \mid M_u \mid )}{\lambda} = \frac{\mathbb{E} ( M_u )+2\mathbb{E}(\max (-M_u,0))}{\lambda}.$

From Jensen inequality, it is then seen that the process $(\max (-M_u,0))_{t is a submartingale, and therefore

$\mathbb{E}(\max (-M_u,0))\le\mathbb{E}(\max (-M_{s},0)).$

We deduce that for $t < u < s$,

$\mathbb{P}(\mid M_u \mid > \lambda) \le \frac{ \mathbb{E} ( M_t )+2\mathbb{E}(\max (-M_{s},0))}{\lambda}.$
It is thus possible to find $A > 0$ such that for every $t < u < s$,

$\mathbb{P}(\mid M_u \mid > A) < \delta,$

For $t < u < s$, we have then

$\mathbb{E} ( \mid M_u \mid 1_{ \mid M_u \mid > \lambda}) < \varepsilon.$

This implies that for every decreasing family $(s_n)_{n \in \mathbb{N}}$ such that $s_n \in \mathcal{D}$ and that converges toward $t$, the family $(M_{s_n})_{n \in \mathbb{N}}$ is uniformly integrable. The convergence

$\lim_{s \rightarrow t, s>t, s \in \mathcal{D}} M_s=\tilde{M}_t.$

thus also holds in $L^1$. Now, since $(M_t)_{t \ge 0}$ is a supermartingale, for $s \ge t$ we have

$\mathbb{E} \left( M_s \mid \mathcal{F}_t \right) \le M_t.$

This implies,

$\lim_{s \rightarrow t, s>t, s \in \mathcal{D}} \mathbb{E} \left( M_s \mid \mathcal{F}_t \right) \le M_t,$

and

$\mathbb{E} \left( \tilde{M}_t \mid \mathcal{F}_t \right) \le M_t.$

Hence, since $\tilde{M}_t$ is adapted to the filtration $\mathcal{F}_t$

$\tilde{M}_t \le M_t,$

Due to the fact that the function $u \rightarrow \mathbb{E} (M_u)$ is right continuous, we have

$\lim_{s \rightarrow t, s>t, s \in \mathcal{D}} \mathbb{E} (M_s)=\mathbb{E} (M_t).$

But from the $L^1$ convergence, we also have

$\lim_{s \rightarrow t, s>t, s \in \mathcal{D}} \mathbb{E}(M_s)=\mathbb{E} \left(\lim_{s \rightarrow t, s>t, s \in \mathcal{D}} M_s\right)=\mathbb{E} (\tilde{M}_t),$

This gives

$\mathbb{E} (\tilde{M}_t)=\mathbb{E} (M_t).$

The random variable $M_t -\tilde{M}_t$ is therefore non-negative and has a zero expectation. This implies that almost surely $M_t=\tilde{M}_t$. The stochastic process $(\tilde{M}_t)_{t \ge 0}$ is therefore a modification of $(M_t)_{t \ge 0}$. Finally, since a modification of a supermartingale is still a supermartingale, this concludes the proof of the theorem.

The following exercise shows that martingales naturally appear when studying equivalent measures on a filtered probability space. We assume here the reader familiar with the Radon-Nikodym theorem.

Exercise. Let $(\Omega, (\mathcal{F}_t)_{t \ge 0},\mathcal{F},\mathbb{P})$ be a filtered probability space that satisfies the usual conditions. We denote

$\mathcal{F}_{\infty}=\sigma \left( \mathcal{F}_t , t \ge 0 \right)$

and for $t\ge 0$, $\mathbb{P}_{/\mathcal{F}_t}$ is the restriction of $\mathbb{P}$ to $\mathcal{F}_t$. Let $\mathbb{Q}$ be a probability measure on $\mathcal{F}_{\infty}$ such that for every $t \ge 0$,

$\mathbb{Q}_{/\mathcal{F}_t} \ll \mathbb{P}_{/\mathcal{F}_t}.$

• Show that there exists a right continuous and left limited martingale $(D_t)_{t \ge 0}$ such that for every $t \ge 0$,

$D_t=\frac{d\mathbb{Q}_{/\mathcal{F}_t}}{d\mathbb{P}_{/\mathcal{F}_t}},\text{ }\mathbb{P}-a.s.$

• Show that the following properties are equivalent:
1) $\mathbb{Q}_{/\mathcal{F}_\infty} \ll \mathbb{P}_{/\mathcal{F}_\infty}$;
2) The martingale $(D_t)_{t \ge 0}$ is uniformly integrable;
3) $(D_t)_{t \ge 0}$ converges in $L^1$;
4) $(D_t)_{t \ge 0}$ almost surely converges to an integrable and $\mathcal{F}_\infty$ measurable random variable $D$ such that $D_t =\mathbb{E}(D\mid \mathcal{F}_t), \quad t \ge 0.$

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