7 Filtrations, processes and martingales

7.1 Basic definitions

First, recall the definitions of a filtration, an adapted process, a (sub)martingale, a stopping time and a stopped process, which are already in Mathlib.

Definition 7.1 Filtration

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A filtration on a measurable space $(\Omega , \mathcal{A})$ with measure $P$ indexed by a preordered set $T$ is a family of sigma-algebras $\mathcal{F} = (\mathcal{F}_t)_{t \in T}$ such that for all $i \le j$, $\mathcal{F}_i \subseteq \mathcal{F}_j$ and for all $t \in T$, $\mathcal{F}_t \subseteq \mathcal{A}$.

Definition 7.2

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A process $X : T \to \Omega \to E$ is said to be adapted with respect to a filtration $\mathcal{F}$ if for all $t \in T$, $X_t$ is $\mathcal{F}_t$-measurable.

Definition 7.3 Progressively measurable

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A stochastic process $X$ is said to be progressively measurable with respect to a filtration $\mathcal{F}$ if at each point in time $i$, $X$ restricted to $(-\infty , i] \times \Omega $ is measurable with respect to the product $\sigma $-algebra where the $\sigma $-algebra used for $\Omega $ is $\mathcal{F}_i$.

Lemma 7.4

If a stochastic process $(X_t)_{t \in T}$ is right continuous and adapted, then it is progressively measurable.

Proof ▶

Fixing $t \in T$, we need to show that $X$ restricted to $[0, t] \times \Omega $ is measurable with respect to $\mathcal{B}([0, t]) \otimes \mathcal{F}_t$. To this end, we define a left continuous discrete approximation of $X$ on $[0, t]$ by

\[ X^n_s = X_{(k + 1)t 2^{-n}} \text{ for } s \in (k t 2^{-n}, (k + 1) t 2^{-n}] \]

and $X^n_0 = X_0$. As $X$ is right continuous, it is easy to see that $X^n \to X$ pointwise as $n \to \infty $. Thus, as each $X^n$ is progressively measurable, it follows that $X$ is also progressively measurable (by e.g. using MeasureTheory.progMeasurable_of_tendsto).

Definition 7.5 Martingale

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Let $\mathcal{F}$ be a filtration on a measurable space $\Omega $ with measure $P$ indexed by $T$. A family of functions $M : T \to \Omega \to E$ is a martingale with respect to a filtration $\mathcal{F}$ if $M$ is adapted with respect to $\mathcal{F}$ and for all $i \le j$, $P[M_j \mid \mathcal{F}_i] = M_i$ almost surely.

Definition 7.6 Submartingale

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Let $\mathcal{F}$ be a filtration on a measurable space $\Omega $ with measure $P$ indexed by $T$. A family of functions $M : T \to \Omega \to E$ is a submartingale with respect to a filtration $\mathcal{F}$ if $M$ is adapted with respect to $\mathcal{F}$ and for all $i \le j$, $P[M_j \mid \mathcal{F}_i] \ge M_i$ almost surely.

Lemma 7.7

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Let $X$ be a real-valued submartingale with respect to a filtration $\mathcal{F}$. Then for all $i \le j$, we have $0 \le P[M_j - M_i \mid \mathcal{F}_i]$ almost surely.

Proof ▶

Lemma 7.8

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Let $X$ be a submartingale. Then for all bounded stopping times $\tau $, the stopped value $X_\tau $ is integrable.

Proof ▶

Lemma 7.9

If $X$ is a martingale and $Y$ is an adapted modification of $X$, then $Y$ is a martingale.

Proof ▶

Let $i \le j$ in $T$. We want to show that $P[Y_j \mid \mathcal{F}_i] = Y_i$ almost surely. It suffices to show that $\int _A Y_j \: dP = \int _A Y_i \: dP$ for all $A \in \mathcal{F}_i$. Let then $A \in \mathcal{F}_i$.

\begin{align*} \int _A Y_j \: dP & = \int _A X_j \: dP = \int _A X_i \: dP = \int _A Y_i \: dP \: . \end{align*}

Lemma 7.10

If $X$ is a submartingale and $Y$ is an adapted modification of $X$, then $Y$ is a submartingale.

Proof ▶

Let $i \le j$ in $T$. We want to show that $P[Y_j \mid \mathcal{F}_i] \ge Y_i$ almost surely. It suffices to show that $\int _A Y_j \: dP \ge \int _A Y_i \: dP$ for all $A \in \mathcal{F}_i$. Let then $A \in \mathcal{F}_i$.

\begin{align*} \int _A Y_j \: dP & = \int _A X_j \: dP \ge \int _A X_i \: dP = \int _A Y_i \: dP \: . \end{align*}

Lemma 7.11 Jensen’s inequality for the conditional expectation

Let $X : \Omega \to E$ be an integrable random variable with values in a normed space $E$ and let $\phi : E \to \mathbb {R}$ be a convex function such that $\phi \circ X$ is integrable. Then, for any sub-$\sigma $-algebra $\mathcal{G}$, we have

\begin{align*} \phi \left( \mathbb {E}[X \mid \mathcal{G}] \right) & \le \mathbb {E}[\phi (X) \mid \mathcal{G}] \quad \text{a.s.} \end{align*}

Proof ▶

Done in a Mathlib PR for finite measures: #27953.

Corollary 7.12

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Let $X : \Omega \to E$ be an integrable random variable with values in a normed space $E$. Then, for any sub-$\sigma $-algebra $\mathcal{G}$, we have

\begin{align*} \Vert \mathbb {E}[X \mid \mathcal{G}] \Vert & \le \mathbb {E}[\Vert X \Vert \mid \mathcal{G}] \quad \text{a.s.} \end{align*}

Proof ▶

Lemma 7.13

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Let $X : T \rightarrow \Omega \rightarrow E$ a martingale with values in a normed space $E$. Let $\phi : E \rightarrow \mathbb {R}$ convex and continuous such that $\phi (X_t)\in L^1(\Omega )$ for every $t\in T$. Then $\phi (X)$ is a sub-martingale.

Proof ▶

By the conditional Jensen inequality (Lemma 7.11), $\phi (X_t) = \phi \left( \mathbb {E}[X_T\ |\ \mathcal{F}_t] \right)\leq \mathbb {E}[\phi (X_T)\ |\ \mathcal{F}_t]$.

Corollary 7.14

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Let $X : T \rightarrow \Omega \rightarrow E$ a martingale with values in a normed space $E$. Then $\Vert X \Vert $ is a sub-martingale.

Proof ▶

Same proof as Lemma 7.13, specialized to $\phi = \Vert \cdot \Vert $, for which we can use Corollary 7.12.

Lemma 7.15

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Let $X : T \rightarrow \Omega \rightarrow E$ a sub-martingale. Let $\phi :E \rightarrow \mathbb {R}$ convex, continuous, and increasing such that $\phi (X_t)\in L^1(\Omega )$ for every $t\in T$. Then $\phi (X)$ is a sub-martingale.

Proof ▶

By Jensen and the fact that $\phi $ is increasing $\phi (X_t) \leq \phi \left( \mathbb {E}[X_T\ |\ \mathcal{F}_t] \right)\leq \mathbb {E}[\phi (X_T)\ |\ \mathcal{F}_t]$.

Definition 7.16 Stopping time

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A stopping time with respect to some filtration $\mathcal{F}$ indexed by $T$ is a function $\tau : \Omega \to T \cup \{ \infty \} $ such that for all $i$, the preimage of $\{ j \mid j \le i\} $ along $\tau $ is measurable with respect to $\mathcal{F}_i$.

Definition 7.17 $\sigma $-algebra generated by a stopping time

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Given a stopping time $\tau $ on a time index $T$, define $\mathcal{F}_\tau = \bigcup _{t \in T} \{ A \in \mathcal{F} \mid A \cap \{ \tau \le t\} \in \mathcal{F}_t\} .$

Lemma 7.18

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Let $\tau , \sigma $ be stopping times such that $\tau \le \sigma $. Then, $\mathcal{F}_\tau \subseteq \mathcal{F}_\sigma $.

Proof ▶

Definition 7.19 Stopped process

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Let $X : T \to \Omega \to E$ be a stochastic process and let $\tau : \Omega \to T$. The stopped process with respect to $\tau $ is defined by

\begin{align*} (X^{\tau })_t = \begin{cases} X_t & \text{if } t \le \tau \\ X_{\tau } & \text{otherwise} \end{cases}\end{align*}

Lemma 7.20

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Let $X : \mathbb {N} \to \Omega \to \mathbb {R}$ be a sub-martingale and $\tau $ a stopping time with respect to the filtration $\mathcal{F}$. Then, the stopped process $X^{\tau }$ is a sub-martingale with respect to the filtration $\mathcal{F}$.

Proof ▶

Definition 7.21 Hitting time

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For $X : T \to \Omega \to E$ a stochastic process, $B$ a subset of $E$ and $t_0 \in T$, the hitting time of $X$ in $B$ after $t_0$ is the random variable $\Omega \to T\cup \{ \infty \} $ defined by

\begin{align*} \tau _{B, t_0}(\omega ) = \inf \{ t \in T \mid t \ge t_0, \: X_t(\omega ) \in B\} \: , \end{align*}

in which the infimum is infinite if the set is empty.

7.2 Right-continuous filtrations

We now give the definition of a filtered probability space satisfying the usual conditions.

Definition 7.22 Left continuation

Assume that $T$ is a partial order. For $\mathcal{F}$ a filtration indexed by $T$ and $t \in T$, we define the left continuation as

\[ \mathcal{F}_{t-} = \begin{cases} \mathcal{F}_t, & \text{if $t$ is isolated on the left in the order topology}; \\ \bigsqcup _{s {\lt} t} \mathcal{F}_s, & \text{otherwise}. \end{cases} \]

Note that $\bigsqcup $ denotes the supremum in the lattice of sigma-algebras on $\Omega $.

Definition 7.23 Right continuation

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Assume that $T$ is a partial order. For $\mathcal{F}$ a filtration indexed by $T$ and $t \in T$, we define the right continuation as

\[ \mathcal{F}_{t+} = \begin{cases} \mathcal{F}_t, & \text{if $t$ is isolated on the right in the order topology}; \\ \bigsqcap _{s {\gt} t} \mathcal{F}_s, & \text{otherwise}. \end{cases} \]

Note that $\bigsqcap $ denotes the infimum in the lattice of sigma-algebras on $\Omega $.

Lemma 7.24

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The right continuation of $\mathcal{F}$ is a filtration.

Proof ▶

We endow $T$ with the order topology. Let us first prove that the right continuation is nondecreasing. Let $s, t \in T$ such that $s \le t$.

Suppose $s$ is isolated on the right. Then $\mathcal{F}_{s+} = \mathcal{F}_s$.
- If $t$ is isolated on the right, then $\mathcal{F}_{t+} = \mathcal{F}_t$. Because $\mathcal{F}$ is a filtration, we have $\mathcal{F}_s \subseteq \mathcal{F}_t$, and thus $\mathcal{F}_{s+} \subseteq \mathcal{F}_{t+}$.
- If $t$ is not isolated on the right, then $\mathcal{F}_{t+} = \bigsqcap _{u {\gt} t} \mathcal{F}_{u}$. Let $u {\gt} t$. Then $s \le u$, thus $\mathcal{F}_s \subseteq \mathcal{F}_u$. This proves that $\mathcal{F}_s \subseteq \bigsqcap _{u {\gt} t}, \mathcal{F}_u$, and thus $\mathcal{F}_{s+} \subseteq \mathcal{F}_{t+}$.
Suppose now that $s$ is not isolated on the right, so that $\mathcal{F}_{s+} = \bigsqcup _{u {\gt} s} \mathcal{F}_u$.
- If $t$ is isolated on the right, then $\mathcal{F}_{t+} = \mathcal{F}_t$. As $s$ is not isolated on the right, we deduce that $s \ne t$, and thus $s {\lt} t$ because $T$ is a partial order. Therefore $\bigsqcap _{u {\gt} s} \mathcal{F}_u \subseteq \mathcal{F}_t$, and thus $\mathcal{F}_{s+} \subseteq \mathcal{F}_{t+}$.
- If $t$ is not isolated on the right, then $\mathcal{F}_{t+} = \bigsqcap _{u {\gt} t} \mathcal{F}_u$. For any $u {\gt} t$, we have $u {\gt} s$ and thus $\bigsqcap _{v {\gt} s} \mathcal{F}_v \subseteq \mathcal{F}_u$, proving that $\bigsqcap _{u {\gt} s} \mathcal{F}_u \subseteq \bigsqcap _{u {\gt} t} \mathcal{F}_u$, and thus $\mathcal{F}_{s+} \subseteq \mathcal{F}_{t+}$.

Turn now to the proof that for all $t \in T$, $\mathcal{F}_{t+} \subseteq \mathcal{A}$. Let $t \in T$. If $t$ is isolated on the right, then $\mathcal{F}_{t+} = \mathcal{F}_t \subseteq \mathcal{A}$ by definition of a filtration. Otherwise there exists $u {\gt} t$, and $\mathcal{F}_{t+} \subseteq \mathcal{F}_u \subseteq \mathcal{A}$, so we are done.

Lemma 7.25

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Suppose $T$ is a topological space with the topology being the order topology. For $t \in T$, the right continuation of $\mathcal{F}$ at $t$ is given by

\[ \mathcal{F}_{t+} = \begin{cases} \mathcal{F}_t, & \text{if $t$ is isolated on the right}; \\ \bigsqcap _{s {\gt} t} \mathcal{F}_s, & \text{otherwise}. \end{cases} \]

Proof ▶

This follows from Definition 7.23 because the topology on $T$ agrees with the one used to define the right continuation.

Lemma 7.26

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Suppose $T$ is a topological space with the topology being the order topology. Assume that $t \in T$ is isolated on the right, meaning that there exists a neighbourhood $\mathcal{V}$ of $t$ such that $\mathcal{V} \cap \{ u \mid u {\gt} t\} = \emptyset $. Then $\mathcal{F}_{t+} = \mathcal{F}_t$.

Proof ▶

This is a direct consequence of Lemma 7.25.

Lemma 7.27

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Assume that $T$ is a linear order with successor. This means that for any $t$, there is an element $succ(t) \ge t$ such that for any $u {\gt} t$, $u \ge succ(t)$, and if $succ(t) \le t$, then $t$ is maximal. Then the right continuation of $\mathcal{F}$ is equal to $\mathcal{F}$.

Proof ▶

Endow $T$ with the order topology. In a linear order with successor equipped with the order topology, every point is isolated on the right, so we can conclude by Lemma 7.26.

Lemma 7.28

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If $t \in T$ is maximal, $\mathcal{F}_{t+} = \mathcal{F}_t$.

Proof ▶

Endow $T$ with the order topology. As $t$ is maximal, it is isolated on the right for this topology, so we can conclude by Lemma 7.26.

Lemma 7.29

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If $T$ is a linear order and there exists $u {\gt} t$ such that $(t, u) = \emptyset $, then $\mathcal{F}_{t+} = \mathcal{F}_t$.

Proof ▶

Endow $T$ with the order topology. The hypothesis implies that $t$ is isolated on the right in this topology, so we can conclude by Lemma 7.26.

Lemma 7.30

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Suppose $T$ is a topological space with the topology being the order topology. Assume that $t \in T$ is not isolated on the right, meaning that for all neighbourhood $\mathcal{V}$ of $t$, $\mathcal{V} \cap \{ u | u {\gt} t\} \ne \emptyset $. Then $\mathcal{F}_{t+} = \bigsqcap _{u {\gt} t} \mathcal{F}_u$.

Proof ▶

This is a direct consequence of Lemma 7.25.

Lemma 7.31

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Assume that $T$ is a densely ordered linear order, meaning that for all $s {\lt} t$, there exists $u$ such that $s {\lt} u {\lt} t$. If $t$ is not maximal, then $\mathcal{F}_{t+} = \bigsqcap _{u {\gt} t} \mathcal{F}_u$.

Proof ▶

Endow $T$ with the order topology. In a densely ordered linear order, a point which is not maximal is not isolated on the right, so we can conclude by Lemma 7.30.

Lemma 7.32

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If $T$ is a densely ordered linear order with no maximal element, then forall $t \in T$ we have $\mathcal{F}_{t+} = \bigsqcap _{u {\gt} t} \mathcal{F}_u$.

Proof ▶

For all $t$, $t$ is not maximal, so we can conclude by Lemma 7.31.

Lemma 7.33

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The filtration $\mathcal{F}$ is contained in its right continuation.

Proof ▶

Endow $T$ with the order topology, and consider $t \in T$. Using Lemma 7.25, we split into two cases. If $t$ is isolated on the right, then $\mathcal{F}_{t+} = \mathcal{F}_t \supseteq \mathcal{F}_t$ and we are done. Otherwise, for all $u {\gt} t$, $\mathcal{F}_t \subseteq \mathcal{F}_u$, therefore $\mathcal{F}_t \subseteq \bigsqcap _{u {\gt} t} \mathcal{F}_u$, and we are done.

Lemma 7.34

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The right continuation of the right continuation of $\mathcal{F}$ is equal to the right continuation of $\mathcal{F}$.

Proof ▶

Let $t \in T$. From Lemma 7.33, we already now that $\mathcal{F}_{t+} \subseteq \mathcal{F}_{t++}$. Endow $T$ with the order topology and split according to Lemma 7.25. If $t$ is isolated on the right, then $\mathcal{F}_{t++} = \mathcal{F}_{t+}$ and we are done. Otherwise consider $u {\gt} t$. Then there exists $v \in T$ such that $t {\lt} v {\lt} u$. If $v$ is not isolated on the right then

\[ \mathcal{F}_{t++} = \bigsqcap _{s {\gt} t} \mathcal{F}_{s+} \subseteq \mathcal{F}_{v+} = \bigsqcap _{s {\gt} v} \mathcal{F}_s \subseteq \mathcal{F}_u, \]

and otherwise

\[ \mathcal{F}_{t++} = \bigsqcap _{s {\gt} t} \mathcal{F}_{s+} \subseteq \mathcal{F}_{v+} = \mathcal{F}_v \subseteq \mathcal{F}_u, \]

thus $\mathcal{F}_{t++} \subseteq \bigsqcap _{s {\gt} t} \mathcal{F}_s = \mathcal{F}_{t+}$, which concludes the proof.

Lemma 7.35 Basic properties of the right continuation

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Fake lemma for the dependency graph. Import this to depend on Definition 7.23.

Proof ▶

Definition 7.36 Right-continuous filtration

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We say that the filtration is right-continuous if for all $t \in T$, $\mathcal{F}_{t+} \subseteq \mathcal{F}_t$.

Lemma 7.37

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If $\mathcal{F}$ is right-continuous, then for all $t \in T$, $\mathcal{F}_t = \mathcal{F}_{t+}$.

Proof ▶

This is a direct consequence of Definition 7.36 and Lemma 7.33.

Lemma 7.38

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The right continuation of $\mathcal{F}$ is right-continuous.

Proof ▶

This follows immediately from Lemma 7.34.

Lemma 7.39

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If $\mathcal{F}$ is right-continuous, then for all $t \in T$, any set $A \subseteq \Omega $ which is $\mathcal{F}_t$-measurable is also $\mathcal{F}_{t+}$-measurable.

Proof ▶

This is a direct consequence of Definition 7.36.

Lemma 7.40 Basic properties of right continuous filtrations

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Fake lemma for the dependency graph. Import this to depend on Definition 7.36.

Proof ▶

Definition 7.41 Usual conditions

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We say that a filtered probability space $(\Omega , \mathcal{F}, P)$ satisfies the usual conditions if the filtration is right-continuous and if $\mathcal{F}_0$ contains all the $P$-null sets.

7.3 Predictable processes

Definition 7.42 Predictable $\sigma $-algebra

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Let $\mathcal{F}$ be a filtration on a measurable space indexed $\Omega $ by a linearly ordered set $T$. Let $S = \{ \{ \bot \} \times A \mid A \in \mathcal{F}_\bot \} $ if $T$ has a bottom element and $S = \emptyset $ otherwise. The predictable sigma-algebra on $T \times \Omega $ is the sigma-algebra generated by the set of sets $\{ (t, \infty ] \times A \mid t \in T, \: A \in \mathcal{F}_t\} \cup S$.

Definition 7.43 Predictable process

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A process $X : T \to \Omega \to E$ is said to be predictable with respect to a filtration $\mathcal{F}$ if it is measurable with respect to the predictable sigma-algebra on $T \times \Omega $.

Lemma 7.44

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A predictable process is progressively measurable.

Proof ▶

Let $X : T \times \Omega \to E$ be a predictable process, we will show that it is progressively measurable. Namely, fixing $t \in T$, denoting

\[ \iota _t : [0, t] \to T : s \mapsto s \]

we need to show that $\iota _t \circ X : [0, t] \times \Omega \to E$ is measurable with respect to $\mathcal{B}([0, t]) \otimes \mathcal{F}_t$.

Denoting $\Sigma _{\mathcal{F}}$ for the predictable $\sigma $-algebra generated by $\mathcal{F}$, as $u$ is predictable, we have that $X^{-1}(\mathcal{B}(E)) \le \Sigma _{\mathcal{F}}$. Thus, to show that $\iota _t \circ X$ is $\mathcal{B}([0, t]) \otimes \mathcal{F}_t$-measurable, it suffices to show that $\iota _t^{-1}(\Sigma _{\mathcal{F}}) \le \mathcal{B}([0, t]) \otimes \mathcal{F}_t$. In particular, as

\[ \Sigma _{\mathcal{F}} = \sigma (\{ (s, \infty ) \times A \mid A \in \mathcal{F}_s\} \cup \{ \{ \perp \} \times A \mid A \in \mathcal{F}_\perp \} ) \]

is suffices to show that sets of the form $\iota _t^{-1}((s, \infty ) \times A)$ for some $s \in T, A \in \mathcal{F}_s$ and $\iota _t^{-1}(\{ \bot \} \times A)$ for some $A \in \mathcal{F}_\bot $ are $\mathcal{B}([0, t]) \otimes \mathcal{F}_t$-measurable.

Indeed, if $A \in \mathcal{F}_\bot $

\[ \iota _t^{-1}(\{ \bot \} \times A) = \{ \bot \} \times A \]

while for any $s \in T$ and $A \in \mathcal{F}_s$,

\[ \iota _t^{-1}((s, \infty ) \times A) = \begin{cases} \varnothing , & t {\lt} s\\ (s, t] \times A, & s \le t. \end{cases} \]

By the monotonicity of the filtration $\mathcal{F}$, all of these cases are $\mathcal{B}([0, t]) \otimes \mathcal{F}_t$-measurable allowing us to conclude.

Lemma 7.45

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Sets of the form $(s, t] \times A$ for any $A \in \mathcal{F}_s$ is measurable with respect to the predictable $\sigma $-algebra.

Proof ▶

For $t \le s$, the set in question is empty and thusly, trivially measurable. On the other hand, for $s {\lt} t$, measurability follows as $(s, t] \times A = (s, \infty ) \times A \setminus (t, \infty ) \times A$.

Lemma 7.46

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Let $X : \mathbb {N} \to \Omega \to E$ be a stochastic process and let $\mathcal{F}$ be a filtration indexed by $\mathbb {N}$. Then $X$ is predictable if and only if $X_0$ is $\mathcal{F}_0$-measurable and for all $n \in \mathbb {N}$, $X_{n+1}$ is $\mathcal{F}_n$-measurable.

Proof ▶

Suppose first that $X$ is predictable. Straightaway, $X_0$ is $\mathcal{F}_0$-measurable as predictable implies progressively measurable which in turn implies adapted.

Fixing $n$, we observe that for any $S \in \mathcal{B}(E)$,

\[ X_{n + 1}^{-1}(S) = \{ \omega \mid (n + 1, \omega ) \in X^{-1}(S)\} = \pi ^{-1}(\iota ^{-1}(X^{-1}(S))) \]

where

\[ \pi : \Omega \to \{ n + 1\} \times \Omega : \omega \mapsto (n + 1, \omega ) \]

and

\[ \iota : \{ n + 1\} \times \Omega \to T \times \Omega : (n + 1, \omega ) \mapsto (n + 1, \omega ). \]

As $X^{-1}(S) \in \Sigma _{\mathcal{F}}$ – the predictable $\sigma $-algebra, it suffices to show that $\pi ^{-1}(\iota ^{-1}(\Sigma _{\mathcal{F}})) \in \mathcal{F}_n$. To this end, we again only need to show these for the generating sets of $\Sigma _{\mathcal{F}}$:

For $A \in \mathcal{F}_0$, measurability is clear as $\iota ^{-1}(\{ 0\} \times A) = \varnothing $.
Similarly, for $m {\gt} n$ and $A \in \mathcal{F}_m$, $\iota ^{-1}((m, \infty ) \times A) = \varnothing $.
For $m \le n$ and $A \in \mathcal{F}_m \le \mathcal{F}_n$ we have that $\pi ^{-1}(\iota ^{-1}((m, \infty ) \times A)) = A$ which is $\mathcal{F}_n$ measurable by the monotonicity of the filtration.

Now, supposing $X_0$ is $\mathcal{F}_0$-measurable and $X_{n + 1}$ is $\mathcal{F}_n$-measurable, we will show that $X$ is predictable. Indeed, fixing $S \in \mathcal{B}(E)$, we have

\[ X^{-1}(S) = \bigcup _{n \in \mathbb {N}} \{ n\} \times X_n^{-1}(S) = {0} \times X_0^{-1}(S) \cup \bigcup _{n \in \mathbb {N}} \{ n + 1\} \times X_{n + 1}^{-1}(S). \]

Thus, as $\{ 0\} \times X_0^{-1}(S) \in \Sigma _{\mathcal{F}}$ by construction and $\{ n + 1\} \times X_{n + 1}^{-1}(S) = (n, n + 1] \times X_{n + 1}^{-1}(S) \in \Sigma _{\mathcal{F}}$ by Lemma 7.45 and the fact that $X_{n + 1}^{-1}(S) \in \mathcal{F}_n$, we have that $X^{-1}(S) \in \Sigma _{\mathcal{F}}$ as required.

7.4 Uniformly integrable

Lemma 7.47

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If $(X_i)_{i \in \iota }$ is a family of (probabilistically) uniformly integrable functions and $(\mathcal{F}_j)_{j \in \kappa }$ is a family of $\sigma $-algebras, then the family $(P[X_i \mid \mathcal{F}_j])_{i \in \iota , j \in \kappa }$ is uniformly integrable.

Proof ▶

Since $(X_i)_{i \in \iota }$ is uniformly integrable, it is uniformly bounded in $L^1$, thus so is $(P[X_i \mid \mathcal{F}_j])_{i \in \iota , j \in \kappa }$. Moreover, for any $\epsilon {\gt} 0$, there exists some $\delta {\gt} 0$ such that for any measurable set $A$ with $P(A) {\lt} \delta $, we have that $\sup _{i \in \iota } P[|X_i| \mathbb {I}_A] {\lt} \epsilon $.

On the other hand, by Markov’s inequality, for any $\lambda {\gt} 0$, $i \in \iota $ and $j \in \kappa $ we have that

\[ P(|P[X_i \mid \mathcal{F}_j]| \ge \lambda ) \le \lambda ^{-1}P[|P[X_i \mid \mathcal{F}_j]|] \le \lambda ^{-1}P[|X_i|]. \]

Now set $\lambda := \delta ^{-1} \sup _{i \in \iota } P[|X_i|] + 1$. Then for any $i \in \iota $ and $j \in \kappa $ we have that

\[ P(|P[X_i \mid \mathcal{F}_j]| \ge \lambda ) \le \frac{P[|X_i|]}{\delta ^{-1} \sup _{k \in \iota } P[|X_k|] + 1} {\lt} \delta , \]

and so,

\begin{align*} P[|P[X_i \mid \mathcal{F}_j]| \mathbb {I}_{|P[X_i \mid \mathcal{F}_i]| \ge \lambda }] & = P[|P[X_i \mid \mathcal{F}_j] \mathbb {I}_{|P[X_i \mid \mathcal{F}_i]| \ge \lambda }|] \\ & = P[|P[X_i \mathbb {I}_{|P[X_i \mid \mathcal{F}_i]| \ge \lambda } \mid \mathcal{F}_j]|] \\ & \le P[P[|X_i|\mathbb {I}_{|P[X_i \mid \mathcal{F}_j]| \ge \lambda } \mid \mathcal{F}_j]] \\ & = P[|X_i|\mathbb {I}_{|P[X_i \mid \mathcal{F}_j]| \ge \lambda }] {\lt} \epsilon , \end{align*}

showing that $(P[X_i \mid \mathcal{F}_j])_{i \in \iota , j \in \kappa }$ is uniformly integrable.

Lemma 7.48

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Let $X$ be a martingale on a discrete index set and let $(\tau _k)_{k \in \mathbb {N}}$ be a sequence of stopping times that are uniformly bounded by $n$. Then, the family of stopped values $\{ X_{\tau _k}\} _{k \in \mathbb {N}}$ is uniformly integrable.

Proof ▶

By optional sampling (Lemma 7.63), we have that for each $k$, $X_{\tau _k} = P[X_n \mid \mathcal{F}_{\tau _k}]$. Thus, the result follows by Lemma 7.47 as $\{ X_n\} $ is uniformly integrable.

Lemma 7.49

Let $X$ be a martingale and let $(\tau _k)_{k \in \mathbb {N}}$ be a sequence of stopping times that are uniformly bounded by $n$. Then, the family of stopped values $\{ X_{\tau _k}\} _{k \in \mathbb {N}}$ is uniformly integrable if for each $k$, $\tau _k$ takes value in a countable set.

Proof ▶

Same proof as in Lemma 7.48.

Lemma 7.50

✓

Let $(X_t)_{t \in T}$ and $(Y_t)_{t \in T}$ be two families of uniformly integrable random variables. Then the family $(X_t + Y_t)_{t \in T}$ is uniformly integrable.

Proof ▶

The families $X$ and $Y$ are uniformly integrable in the measure-theoretic sense and almost-everywhere strongly measurable, so $X + Y$ is too (see MeasureTheory.UnifIntegrable.add). Moreover, $X$ and $Y$ are bounded in $L^p$, so $X + Y$ is too. So $X + Y$ is uniformly integrable.

Lemma 7.51

Let $(X_s)_{s \in S}$ be a family of random variables and $(Y_t)_{t \in T}$ be a family of uniformly integrable random variables.. If for all $s$, there exists $t$ such that $\| X_t\| \le \| Y_s\| $ almost surely, then $X$ is uniformly integrable.

Proof ▶

Let $\epsilon {\gt} 0$. The family $Y$ is uniformly integrable, thus there exists $C \ge 0$ such that for $t \in T$, $P[\| Y_t\| ^p \mathbb {I}_{\| Y_t\| \ge C}]^{1/p} \le \epsilon $. For all $s$, there exists $t$ such that $\| X_s\| ^p \le \| Y_t\| ^p$, so $P[\| X_s\| ^p \mathbb {I}_{\| X_s\| \ge C}]^{1/p} \le \epsilon $. Thus $X$ is uniformly integrable.

Lemma 7.52

Let $(X_t)_{t \in T}$ be a family of random variables and $Y$ be a real random variable in $L^p$. If for all $t$, $\| X_t\| \le Y$ almost surely, then $X$ is uniformly integrable.

Proof ▶

Because $Y$ is in $L^p$, we deduce that $\{ Y\} $ is uniformly integrable. The conclusion then follows from Lemma 7.51.

Lemma 7.53

✓

If $(X_t)_{t \in T}$ is a family of uniformly integrable random variables, then so is $(\| X_t\| )_{t \in T}$.

Proof ▶

Apply Lemma 7.51 with $Y := X$.

Lemma 7.54

✓

Let $(X_t)_{t \in T}$ be a family of uniformly integrable random variables. It is uniformly integrable if and only if $(\| X_t\| )_{t \in T}$ is.

Proof ▶

The forward direction is Lemma 7.53. The converse direction follows from Lemma 7.51 with $Y := (\| X_t\| )_{t \in T}$.

Lemma 7.55

✓

If $(X_t)_{t \in T}$ is uniformly integrable and $\phi : S \to T$, then $(X_{\phi (s)})_{s \in S}$ is uniformly integrable.

Proof ▶

This is immediate from the definition.

Lemma 7.56

Let $X$ be a submartingale on a discrete index set and let $(\tau _k)_{k \in \mathbb {N}}$ be a sequence of stopping times that are uniformly bounded by $p$. Then, the family of stopped values $\{ X_{\tau _k}\} _{k \in \mathbb {N}}$ is uniformly integrable.

Proof ▶

Use Doob decomposition to write $X_n = M_n + A_n$, where $M$ (Definition 12.5) is a martingale (Lemma 12.7) and $A$ (Definition 12.1) is a predictable process (Lemma 12.6). We know from Lemma 7.48 that $(M_{\tau _k})_{k \in \mathbb {N}}$ is uniformly integrable. Combining Lemma 7.50 and Lemma 7.52, it suffices to show that $(A_{\tau _k})_{k \in \mathbb {N}}$ is dominated. It is dominated by $A_p$ thanks to Lemma 12.2 and Lemma 12.8.

Lemma 7.57

✓

Let $(X_n)_{n \in \mathbb {N}}$ be a sequence of $p$-uniformly integrable stochastic processes and suppose $X_n \to X$ in probability as $n \to \infty $. Then, $X$ is $L^p$.

Proof ▶

Since $X_n \to X$ in probability, it has a subsequence $(X_{n_k}) \subseteq (X_n)$ which converges to $X$ almost surely. Thus, we have by Fatou’s lemma that

\[ P[|X|^p] = P[\liminf _{k \to \infty } |X_{n_k}|^p] \le \liminf _{k \to \infty } P[|X_{n_k}|^p] {\lt} \infty \]

where the last inequality follows as uniform integrability implies that $(X_n)$ is uniform bounded in $L^p$.

Lemma 7.58

✓

Let $(X_t)_{t \in T}$ be a family of $p$-uniformly integrable stochastic processes. Then the family of limits in probability of sequences of $X$ is uniformly integrable.

Proof ▶

Let $\epsilon {\gt} 0$. There exists $\delta {\gt} 0$ such that for all $t\in T$ and all measurable set $S$ such that $P(S){\lt}\delta $,

\[ P[\| X_t\| ^p\mathbb {I}_S]^{1/p}\le \varepsilon . \]

Let $(t_n)_{n \in \mathbb {N}}$ be a sequence in $T$ such that $X_{t_n}$ converges in probability to $Y$. Then it has a subsequence $(X_{t_{n_k}})$ which converges to $Y$ almost surely. Thus, we have by Fatou’s lemma that

\[ P[\| Y\| ^p \mathbb {I}_{S}]^{1/p} = P[\liminf _{k \to \infty } \| X_{t_{n_k}}\| ^p \mathbb {I}_{S}]^{1/p} \le \liminf _{k \to \infty } P[\| X_{t_{n_k}}\| ^p \mathbb {I}_{S}]^{1/p} \le \epsilon . \]

This proves that the family of limits in probability of sequences of $X$ is uniformly integrable in the measure theory sense. One can prove uniform boundedness of this family by using Fatou’s lemma and the existence of an almost everywhere convergent subsequence in a similar way.

Lemma 7.59 Vitali convergence theorem

✓

A sequence of functions converges in $L^1$ if and only if it converges in probability and is uniformly integrable.

Proof ▶

7.5 Optional sampling

Definition 7.60 Ordered Monoid

✓

Let $(M, +)$ be a commutative monoid that is also a partial order. It is said to be an ordered monoid if for all $a, b, c \in M$, we have the following implication:

\[ a \le b \implies a + c \le b + c. \]

Definition 7.61 Ordered Module

✓

Let $\alpha , \beta $ be preorders with $0$ elements and such that there is a scalar multiplication $(\_ \cdot \_ ) : \alpha \times \beta \to \beta $. Then $\beta $ is said to be an ordered $\alpha $-module (or ordered module if $\alpha $ is clear from the context) if the following hold:

$\forall a \in \alpha , \forall b_1, b_2 \in \beta , 0 \le a \implies b_1 \le b_2 \implies a \cdot b_1 \le a \cdot b_2$;
$\forall a_1, a_2 \in \alpha , \forall b \in \beta , 0 \le b \implies a_1 \le a_2 \implies a_1 \cdot b \le a_2 \cdot b$.

Definition 7.62 Order-closed topology

✓

Let $X$ be a topological space that is also a preorder. The space $X$ is set to be order-closed, or to have order-closed topology, if the set $\{ (x, y) \in X \times X \mid x \le y\} $ is closed.

Lemma 7.63 Optional sampling (discrete time)

✓

Let $X$ be a discrete time martingale with respect to the filtration $\mathcal{F}$ and let $\tau , \sigma $ be stopping times. Then, if $\tau $ is bounded, we have that almost surely, $X_{\tau \wedge \sigma } = P[X_{\tau } \mid \mathcal{F}_{\sigma }]$.

Proof ▶

Lemma 7.64

Let $X$ be a discrete time submartingale with respect to the filtration $\mathcal{F}$ taking values in a real Banach space $E$. Assume $E$ is an order-closed partial order, an ordered monoid and an ordered module. Let $\tau , \sigma $ be stopping times. Then, if $\tau $ is bounded, we have that almost surely, $X_{\tau \wedge \sigma } \le P[X_{\tau } \mid \mathcal{F}_{\sigma }]$.

Proof ▶

Use Doob decomposition to write $X_n = M_n + A_n$, where $M$ (Definition 12.5) is a martingale (Lemma 12.7) and $A$ (Definition 12.1) is a predictable process (Lemma 12.6). By Lemma 7.63, we have that almost surely, $M_{\tau \wedge \sigma } = P[M_{\tau } \mid \mathcal{F}_{\sigma }]$. Because $A$ is predictable and $\tau \wedge \sigma \le \sigma $, we deduce that almost surely, $A_{\tau \wedge \sigma } = P[A_{\tau \wedge \sigma } \mid \mathcal{F}_\sigma ]$. Moreover, by Lemma 12.8, we know that almost surely, $A$ is nondecreasing. Therefore, using the fact $\tau \wedge \sigma \le \tau $, we get that $P[A_{\tau \wedge \sigma } \mid \mathcal{F}_\sigma ] \le P[A_\tau \mid \mathcal{F}_\sigma ]$. We deduce that almost surely,

\[ X_{\tau \wedge \sigma } = M_{\tau \wedge \sigma } + A_{\tau \wedge \sigma } \le P[M_\tau \mid \mathcal{F}_\sigma ] + P[A_\tau \mid \mathcal{F}_\sigma ] = P[X_\tau \mid \mathcal{F}_\sigma ], \]

concluding the proof.

Lemma 7.65

Let $X$ be a discrete time supermartingale with respect to the filtration $\mathcal{F}$ taking values in a real Banach space $E$. Assume $E$ is an order-closed partial order, an ordered monoid and an ordered module. Let $\tau , \sigma $ be stopping times. Then, if $\tau $ is bounded, we have that almost surely, $X_{\tau \wedge \sigma } \ge P[X_{\tau } \mid \mathcal{F}_{\sigma }]$.

Proof ▶

We know that $-X$ is a submartingale, so from Lemma 7.64 we obtain that almost surely, $-X_{\tau \wedge \sigma } \le P[-X_{\tau } \mid \mathcal{F}_{\sigma }]$. Multiplying by $-1$ yields the desired result.

Definition 7.66 Discrete approximation sequence

✓

Given a stopping time $\tau : \Omega \to T \cup \{ \infty \} $, a sequence of stopping times $(\tau _n)_{n \in \mathbb {N}}$ is called an discrete approximation of $\tau $ if $\tau _n(\Omega )$ is countable for each $n$ and $\tau _n \downarrow \tau $ a.s. as $n \to \infty $.

Definition 7.67 Approximable time index

✓

A time index set $T$ is said to be approximable if for any stopping time $\tau : \Omega \to T \cup \{ \infty \} $, there exists a discrete approximation sequence $(\tau _n)$ of $\tau $.

Lemma 7.68

Given a right continuous process $X$ and a discrete approximation sequence $(\tau _n)$ of the stopping time $\tau $, we have that

\[ \lim _{n \to \infty } X_{\tau _n} = X_\tau \text{ a.s.} \]

Proof ▶

This follows directly as $X$ is right continuous and $\tau _n \downarrow \tau $ a.s.

Lemma 7.69

✓

Let $\tau $ be a stopping time bounded by $t \in T$ and $(\tau _n)$ be a discrete approximation sequence of $\tau $. Then, the sequence of stopping times $\tau _n \wedge t$ is also a discrete approximation sequence of $\tau $.

Proof ▶

Lemma 7.70

✓

Let $\tau $ be a stopping time bounded by $t \in T$ and $(\tau _n)$ be a discrete approximation sequence of $\tau $. Then, for any martingale $X$, the sequence of stopped values $(X_{\tau _n \wedge t})$ is uniformly integrable.

Proof ▶

Follows directly by Lemma 7.49 and Lemma 7.69.

Lemma 7.71

✓

Let $\tau $ be a stopping time bounded by $t \in T$ and $(\tau _n)$ be a discrete approximation sequence of $\tau $. Then, for any right continuous martingale $X$, $X_{\tau } \in L^1$ and $X_{\tau _n \wedge t} \to X_{\tau }$ in $L^1$ as $n \to \infty $.

Proof ▶

By Lemma 7.68, as $X$ is right continuous we have that $X_{\tau _n \wedge t} \to X_{\tau }$ a.s. and so, also in probability. Moreover, by Lemma 7.70, the sequence $(X_{\tau _n \wedge t})$ is uniformly integrable. Thus, by Lemma 7.57 and the Vitali convergence theorem (Lemma 7.59), it follows that $X_{\tau } \in L^1$ and $X_{\tau _n \wedge t} \to X_{\tau }$ in $L^1$ as $n \to \infty $.

Lemma 7.72

$T = \mathbb {R}_+$ is an approximable time index. In particular, for any stopping time $\tau $ on $\overline{\mathbb {R}_+}$, defining $\tau _n = 2^{-n} \lceil 2^n \tau \rceil $, we have that $(\tau _n)$ is a discrete approximation sequence of $\tau $.

Proof ▶

Clearly $\tau _n \downarrow \tau $ as $n \to \infty $ and so it remains to show that each $\tau _n$ is a stopping time. Indeed,

\[ \{ \tau _n \le t\} = \{ \tau \le 2^{-n} \lfloor 2^n t\rfloor \} \in \mathcal{F}_{2^{-n} \lfloor 2^n t\rfloor } \subseteq \mathcal{F}_t \]

where the last inclusion follows as $2^{-n} \lfloor 2^n t\rfloor \le t$.

Lemma 7.73

$T = \mathbb {N}$ is an approximable time index.

Proof ▶

Immediate as we can take $\tau _n = \tau $ for all $n$.

Lemma 7.74 Optional sampling (continuous time)

Let $X$ be a right-continuous $\mathcal{F}$-martingale on an approximable time index. Then, for any stopping times $\sigma , \tau $ with $\tau $ bounded, we have that $X_{\sigma \wedge \tau } = P[X_{\tau } \mid \mathcal{F}_{\sigma }]$ almost surely.

Proof ▶

Fixing $A \in \mathcal{F}_{\sigma }$, we need to show that $P[X_{\tau } \mathbb {I}_A] = P[X_{\sigma \wedge \tau } \mathbb {I}_A]$.

Let $(\tau _n), (\sigma _n)$ be discrete approximation sequences of $\tau $ and $\sigma $ respectively. As $\tau _n, \sigma _n$ take values in a countable set, we have by the discrete time optional sampling theorem (Lemma 7.63) that

\[ X_{\sigma _n \wedge \tau _n} = P[X_{\tau _n} \mid \mathcal{F}_{\sigma _n}] \]

and so, as $\mathcal{F}_{\sigma } \subseteq \mathcal{F}_{\sigma _n}$ by Lemma 7.18, we have that $P[X_{\sigma _n \wedge \tau _n} \mathbb {I}_A] = P[X_{\tau _n} \mathbb {I}_A]$. On the other hand, by Lemma 7.48, the families $\{ X_{\tau _n}\} $ and $\{ X_{\sigma _n \wedge \tau _n}\} $ are uniformly integrable. Thus, as $X$ is right-continuous, $(X_{\sigma _n \wedge \tau _n}, X_{\tau _n}) \to (X_{\sigma \wedge \tau }, X_{\tau })$ a.s. We have $P[X_{\tau } \mathbb {I}_A] = P[X_{\sigma \wedge \tau } \mathbb {I}_A]$ by Lemma 7.71 as desired.

Lemma 7.75

Let $X$ be a right-continuous $\mathcal{F}$-submartingale on an approximable time index. Then, for any stopping times $\sigma , \tau $ with $\tau $ bounded, we have that $X_{\sigma \wedge \tau } \le P[X_{\tau } \mid \mathcal{F}_{\sigma }]$ almost surely.

Proof ▶

Fixing $A \in \mathcal{F}_{\sigma }$, we need to show that $P[X_{\tau } \mathbb {I}_A] \le P[X_{\sigma \wedge \tau } \mathbb {I}_A]$.

\[ X_{\sigma _n \wedge \tau _n} \le P[X_{\tau _n} \mid \mathcal{F}_{\sigma _n}] \]

and so, as $\mathcal{F}_{\sigma } \subseteq \mathcal{F}_{\sigma _n}$ by Lemma 7.18, we have that $P[X_{\sigma _n \wedge \tau _n} \mathbb {I}_A] \le P[X_{\tau _n} \mathbb {I}_A]$. On the other hand, by Lemma 7.56, the families $\{ X_{\tau _n}\} $ and $\{ X_{\sigma _n \wedge \tau _n}\} $ are uniformly integrable. Thus, as $X$ is right-continuous, $(X_{\sigma _n \wedge \tau _n}, X_{\tau _n}) \to (X_{\sigma \wedge \tau }, X_{\tau })$ a.s. We have $P[X_{\tau } \mathbb {I}_A] \le P[X_{\sigma \wedge \tau } \mathbb {I}_A]$ by Lemma 7.71 as desired.

7.6 Martingale convergence

Definition 7.76

✓

Let $X : T \to \Omega \to E$ be a stochastic process, let $\mathcal{F}$ be a filtration on $\Omega $ indexed by $T$ and let $P$ be a measure on $\Omega $. If there exists a function $Y : \Omega \to E$ which is measurable with respect to $\mathcal{F}_\infty $ such that for $P$-almost surely, $X_t$ converges to $Y$ as $t$ goes to infinity, then we say that $Y$ is the limit of $X$. We denote it by $X_\infty $.

In Mathlib, we have results about convergence of martingales to their limit in discrete time.

Theorem 7.77

Let $X$ be an uniformly integrable cadlag martingale with respect to the filtration $\mathcal{F}$. Then there exists a limit process $X_\infty $ measurable with respect to $\mathcal{F}_\infty $ such that $X_t$ converges to $X_\infty $ almost surely as $t$ goes to infinity. Furthermore, $X_t = P[X_\infty \mid \mathcal{F}_t]$ almost surely.

Proof ▶

7.7 Doob’s Lp inequality

In this section, we prove Doob’s Lp inequality.

Lemma 7.78 Doob’s maximal inequality for $\mathbb {N}$

✓

Let $X : \mathbb {N} \rightarrow \Omega \rightarrow \mathbb {R}$ be a non-negative sub-martingale. Then for every $n \in \mathbb {N}$ and $\lambda {\gt} 0$,

\begin{align*} \mathbb {P}\left(\sup _{i \le n}X_i\geq \lambda \right) \le \frac{\mathbb {E}\left[X_n \mathbb {I}_{\sup _{i \le n}X_i \ge \lambda }\right]}{\lambda } \le \frac{\mathbb {E}[X_n]}{\lambda } \: . \end{align*}

Proof ▶

Lemma 7.79 Doob’s maximal Inequality for countable

Let $X : I \rightarrow \Omega \rightarrow \mathbb {R}$ be a non-negative sub-martingale with $I$ countable. Then for every $M \in I,\lambda {\gt} 0$ and $p{\gt}1$ we have

\begin{align*} P\left( \sup _{i\in I, i\leq M}X_i\geq \lambda \right) \le \frac{\mathbb {E}\left[X_M \mathbb {I}_{\sup _{i \le M}X_i \ge \lambda }\right]}{\lambda } \le \frac{\mathbb {E}[X_M]}{\lambda } \: . \end{align*}

Proof ▶

For any finite subset $J \subset I$ with $M \in J$, we have by Lemma 7.78

\begin{align*} P\left( \sup _{i\in J, i \le M}X_i\geq \lambda \right) \le \frac{\mathbb {E}\left[X_{M} \mathbb {I}_{\sup _{i \in J, i \le M}X_i \ge \lambda }\right]}{\lambda } \: . \end{align*}

Then we build a countable increasing sequence of finite sets $J_n$ with $\sup _{i\in I, i\leq M}X_i = \sup _n\sup _{i\in J_n, i \le M}X_i$ and conclude by monotone convergence.

Lemma 7.80 Doob Lp Inequality for countable

Let $X : I \rightarrow \Omega \rightarrow \mathbb {R}$ be a non-negative sub-martingale. Let $I$ be countable. For every $M\in I,\lambda {\gt} 0$ and $p{\gt}1$ we have

\begin{align*} \mathbb {E}\left[ \sup _{i\in I, i \leq M}X_i^p \right] \leq \left(\frac{p}{p-1}\right)^p\mathbb {E}[X_M^p] \: . \end{align*}

That is, for $\Vert \cdot \Vert _p$ the $L^p$ norm, $\left\Vert \sup _{i \le M} X_i \right\Vert _p \leq \frac{p}{p-1} \left\Vert X_M \right\Vert _p \: .$

Proof ▶

\begin{align*} \mathbb {E}\left[ \sup _{i \le M}X_i^p \right] = p \int _0^\infty \mathbb {P}\left( \sup _{i \le M}X_i \geq \lambda \right) \lambda ^{p-1} d\lambda \end{align*}

By Theorem 7.79 and then Fubini’s theorem, we have then

\begin{align*} \mathbb {E}\left[ \sup _{i \le M}X_i^p \right] & \le p \int _0^\infty \mathbb {E}\left[X_M \mathbb {I}_{\sup _{i \le M}X_i \ge \lambda }\right] \lambda ^{p-2} d\lambda \\ & = p \mathbb {E}\left[X_M \int _0^{\sup _{i \le M}X_i} \lambda ^{p-2} d\lambda \right] \\ & = \frac{p}{p - 1} \mathbb {E}\left[X_M (\sup _{i \le M}X_i)^{p-1}\right] \: . \end{align*}

Then by Hölder’s inequality,

\begin{align*} \mathbb {E}\left[ \sup _{i \le M}X_i^p \right] & \le \frac{p}{p - 1} \left(\mathbb {E}[X_M^p]\right)^{1/p} \left(\mathbb {E}\left[\sup _{i \le M}X_i^p \right]\right)^{(p-1)/p} \: . \end{align*}

We then divide the two sides by $\left(\mathbb {E}\left[\sup _{i \le M}X_i^p \right]\right)^{(p-1)/p}$ and raise to the power $p$ to conclude.

Theorem 7.81 Doob Inequality

Let $X: \mathbb {R}_+ \to \Omega \to \mathbb {R}$ be a right-continuous non-negative sub-martingale. For every $T \in \mathbb {R}_+$ and $\lambda {\gt}0$ we have

\begin{align*} P\left( \sup _{t\in [0,T]}X_t \geq \lambda \right) \leq \frac{\mathbb {E}[X_T \mathbb {I}_{\sup _{i \le T}X_i \ge \lambda }]}{\lambda } \leq \frac{\mathbb {E}[X_T]}{\lambda } \: . \end{align*}

Proof ▶

Since $X$ is right-continuous and $[0,T]$ is a compact interval, we have that

\begin{align*} \sup _{t\in [0,T]}X_t = \sup _{t\in [0,T] \cap \mathbb {Q}}X_t \: . \end{align*}

Then apply Lemma 7.79 with $I = [0,T] \cap \mathbb {Q}$ and $M = T$.

Corollary 7.82 Doob Inequality for normed spaces

Let $X:\mathbb {R}_+ \to \Omega \to E$ be a right-continuous martingale with values in a normed space $E$. For every $T$ and $\lambda {\gt}0$ we have

\[ P\left( \sup _{t\in [0,T]} \lVert X_t \rVert \geq \lambda \right) \leq \frac{\mathbb {E}[\lVert X_T \rVert ]}{\lambda }. \]

Proof ▶

By Corollary 7.14, $\lVert X \rVert $ is a sub-martingale. Then apply Theorem 7.81.

Theorem 7.83 Doob’s Lp inequality in $\mathbb {R}$

Let $X:\mathbb {R} \rightarrow \Omega \rightarrow \mathbb {R}$ be a right-continuous non-negative sub-martingale. For every $T, \lambda {\gt}0$ and $p{\gt}1$ we have

\begin{align*} \mathbb {E}\left[ \sup _{t\in [0,T]}X_t^p \right] \leq \left(\frac{p}{p-1}\right)^p\mathbb {E}[X_T^p] \: . \end{align*}

That is, for $\Vert \cdot \Vert _p$ the $L^p$ norm, $\left\Vert \sup _{t\in [0,T]} X_t \right\Vert _p \leq \frac{p}{p-1} \left\Vert X_T \right\Vert _p \: .$

Proof ▶

Since $X$ is right-continuous and $[0,T]$ is a compact interval, we have that

\begin{align*} \sup _{t\in [0,T]}X_t = \sup _{t\in [0,T] \cap \mathbb {Q}}X_t \: . \end{align*}

Then apply Lemma 7.80 with $I = [0,T] \cap \mathbb {Q}$ and $M = T$.

Corollary 7.84 Doob’s Lp inequality for normed spaces

Let $X : \mathbb {R} \rightarrow \Omega \rightarrow E$ be a right-continuous martingale with values in a normed space $E$. For every $T, \lambda {\gt}0$ and $p{\gt}1$ we have

\begin{align*} \mathbb {E}\left[ \sup _{t\in [0,T]} \lVert X_t \rVert ^p \right] \leq \left(\frac{p}{p-1}\right)^p\mathbb {E}[\lVert X_T \rVert ^p] \: . \end{align*}

That is, for $\Vert \cdot \Vert _p$ the $L^p$ norm, $\left\Vert \sup _{t\in [0,T]} X_t \right\Vert _p \leq \frac{p}{p-1} \left\Vert X_T \right\Vert _p \: .$

Proof ▶

By Corollary 7.14, $\lVert X \rVert $ is a sub-martingale. Then apply Theorem 7.83.

Lemma 7.85 Stopped Submartingale

Let $X:\mathbb {R}_+ \to \Omega \to \mathbb {R}$ be a cadlag submartingale and $\tau $ a stopping time. Then the stopped process $X^\tau $ is a submartingale.

Proof ▶

Lemma 7.86 Doob Inequality for stopping times

Let $X:\mathbb {R}\times \Omega \rightarrow \mathbb {R}$ be a right-continuous non-negative sub-martingale. For every $\lambda {\gt}0$ and $p{\gt}1$ and $\tau $ stopping time a.s. bounded by $T{\gt}0$, we have

\[ P\left( \sup _{t\in [0,\tau ]}X_t\geq \lambda \right)\leq \frac{\mathbb {E}[X_\tau ]}{\lambda }. \]

Proof ▶

Corollary 7.87 Doob Inequality for stopping times in normed spaces

Let $X:\mathbb {R}\times \Omega \rightarrow E$ be a right-continuous martingale with values in a normed space $E$. For every $\lambda {\gt}0$ and $p{\gt}1$ and $\tau $ stopping time a.s. bounded by $T{\gt}0$, we have

\[ P\left( \sup _{t\in [0,\tau ]}\lVert X_t \rVert \geq \lambda \right)\leq \frac{\mathbb {E}[\lVert X_\tau \rVert ]}{\lambda }. \]

Proof ▶

By Corollary 7.14, $\lVert X \rVert $ is a sub-martingale. Then apply Theorem 7.86.

Lemma 7.88 Doob’s Lp Inequality for stopping times

\[ \mathbb {E}\left[ \sup _{t\in [0,\tau ]}X_t^p \right]\leq \left(\frac{p}{p-1}\right)^p\mathbb {E}[X_\tau ^p]. \]

Proof ▶

8.1.3 Pascucci.

Corollary 7.89 Doob’s Lp Inequality for stopping times in normed spaces

\[ \mathbb {E}\left[ \sup _{t\in [0,\tau ]}\lVert X_t \rVert ^p \right]\leq \left(\frac{p}{p-1}\right)^p\mathbb {E}[\lVert X_\tau \rVert ^p]. \]

Proof ▶

By Corollary 7.14, $\lVert X \rVert $ is a sub-martingale. Then apply Theorem 7.88.