Almost Sure

15 November 16

Optional Processes

The optional sigma-algebra, ${\mathcal{O}}$, was defined earlier in these notes as the sigma-algebra generated by the adapted and right-continuous processes. Then, a stochastic process is optional if it is ${\mathcal{O}}$-measurable. However, beyond the definition, very little use was made of this concept. While right-continuous adapted processes are optional by construction, and were used throughout the development of stochastic calculus, there was no need to make use of the general definition. On the other hand, optional processes are central to the theory of optional section and projection. So, I will now look at such processes in more detail, starting with the following alternative, but equivalent, ways of defining the optional sigma-algebra. Throughout this post we work with respect to a complete filtered probability space ${(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\in{\mathbb R}_+},{\mathbb P})}$, and all stochastic processes will be assumed to be either real-valued or to take values in the extended reals ${\bar{\mathbb R}={\mathbb R}\cup\{\pm\infty\}}$.

Theorem 1 The following collections of sets and processes each generate the same sigma-algebra on ${{\mathbb R}_+\times\Omega}$.

1. {${[\tau,\infty)}$: ${\tau}$ is a stopping time}.
2. ${Z1_{[\tau,\infty)}}$ as ${\tau}$ ranges over the stopping times and Z over the ${\mathcal{F}_\tau}$-measurable random variables.
4. The right-continuous adapted processes.

The optional-sigma algebra was previously defined to be generated by the right-continuous adapted processes. However, any of the four collections of sets and processes stated in Theorem 1 can equivalently be used, and the definitions given in the literature do vary. So, I will restate the definition making use of this equivalence.

Definition 2 The optional sigma-algebra, ${\mathcal{O}}$, is the sigma-algebra on ${{\mathbb R}_+\times\Omega}$ generated by any of the collections of sets/processes in Theorem 1.

A stochastic process is optional iff it is ${\mathcal{O}}$-measurable.

Before giving the proof of Theorem 1 below, I will first show that certain basic constructions automatically lead to optional processes. We know that optional processes are always progressively measurable. In the previous post, it was shown that the running maximum of a progressive process is itself progressive. In fact, it is optional.

Lemma 3 If X is a progressively measurable process then ${X^*_t\equiv\sup_{s\le t}X_s}$ is optional.

Proof: First, as in the proof of Lemma 5 in the previous post, ${X^*_{t-}}$ is left-continuous and adapted. In particular, it is predictable and hence is optional. As the intervals ${(K,\infty]}$ for real K generate the Borel sigma-algebra on ${\bar{\mathbb R}}$, it is enough to show that ${(X^*)^{-1}((K,\infty])}$ is optional.

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle X^*((K,\infty])\equiv\{t\colon X^*_t > K\}= A\cup B,\smallskip\\ &\displaystyle A=\{t\colon X^*_{t-} > K\},\smallskip\\ &\displaystyle B= \{t\colon X^*_t > K \ge X^*_{t-}\}. \end{array}$

As ${X^*_{t-}}$ is optional, we know that ${A\in\mathcal{O}}$. Next, suppose that ${(t,\omega)}$ is in B for some ${t\in{\mathbb R}_+}$ and ${\omega\in\Omega}$. Then,

$\displaystyle X^*_s\le X^*_{t-}\le K$

for ${s < t}$ and

$\displaystyle X^*_{s-}\ge X^*_t > K$

for ${s > t}$. In either case, ${(s,\omega)\not\in B}$. So, we see that there can be at most a single time lying in B for each ${\omega\in\Omega}$ and, hence, B is the graph ${[\tau]}$ of a map ${\tau\colon\Omega\rightarrow\bar{\mathbb R}_+}$. As ${X^*_t}$ and ${X^*_{t-}}$ are progressive, B is also progressive. By Lemma 7 from the previous post, this means that ${\tau}$ is a stopping time. In particular, using the first statement of Theorem 1,

 $\displaystyle [\tau]=[\tau,\infty)\setminus\bigcup_{n=1}^\infty[\tau+1/n,\infty)\in\mathcal{O}.$ (1)

So, ${(X^*)^{-1}((K,\infty])\in\mathcal{O}}$ as required. ⬜

It was also shown that the limit supremum and left/right limit supremums of progressive processes are again progressive. Actually, the stronger property of optionality holds for most of these.

Lemma 4 If X is a progressively measurable process then

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle t&\displaystyle\mapsto\limsup_{s\uparrow t}X_s,\smallskip\\ \displaystyle t&\displaystyle\mapsto\limsup_{s\uparrow\uparrow t}X_s \end{array}$

are optional. If, furthermore, the filtration ${\mathcal{F}_\cdot}$ is right-continuous then,

$\displaystyle t\mapsto\limsup_{s\rightarrow t}X_s$

is optional.

Proof: For each positive integer n, choose a sequence of times ${0=t^n_0 < t^n_1 < \cdots}$ increasing to infinity, and such that ${\sup_k(t^n_k-t^n_{k-1})}$ goes to zero as n goes to infinity. For example, ${t^n_k=k/n}$. Define processes ${Y^n}$ by ${Y^n_0=X_0}$ and

$\displaystyle Y^n_t=\sup\left\{X_s\colon s\in[t^n_{k-1},t)\right\}$

for ${t\in(t^n_{k-1},t^n_k]}$. This is left-continuous and, by Lemma 4 of the previous post, is adapted. So, ${Y^n}$ and, hence, the limit

$\displaystyle \limsup_{s\uparrow\uparrow t}X_s=\lim_{n\rightarrow\infty}Y^n_t$

Next, define ${Z^n_0=X_0}$ and

$\displaystyle Z^n_t=\sup\left\{X_s\colon s\in[t^n_{k-1},t]\right\}$

for ${t\in(t^n_{k-1},t^n_k]}$. Applying Lemma 3 over each of the intervals ${(t^n_{k-1},t^n_k]}$ shows that this optional. So, the limit

$\displaystyle \limsup_{s\uparrow t}X_s=\lim_{n\rightarrow\infty}Z^n_t$

is also optional.

Finally, define ${W^n_0=\limsup_{s\downarrow0}X_s}$ and

$\displaystyle W^n_t=\lim_{s\downarrow\downarrow t}\sup\left\{X_u\colon u\in[t^n_{k-1},s)\right\}$

for ${t\in(t^n_{k-1},t^n_k]}$. By Lemma 4 of the previous post, ${W^n_t}$ is ${\mathcal{F}_s}$-measurable for each ${s > t}$. So, ${W^n_t}$ is ${\mathcal{F}_{t+}}$-measurable and, if the filtration is right-continuous, W is adapted. As ${W^n}$ is right-continuous over each of the intervals ${(t^n_{k-1},t^n_k]}$, this implies that it is optional. So,

$\displaystyle \limsup_{s\rightarrow t}X_s=\lim_{n\rightarrow\infty}W^n_t$

is optional. ⬜

Notable by their absence in the statement of Lemma 4 are the right-limits ${\limsup_{s\downarrow t}X_s}$ and ${\limsup_{s\downarrow\downarrow t}X_s}$. In fact, these need not be optional even when the filtration is right-continuous and X is itself optional.

Next, we saw in the previous post that stopping times can be characterised by the progressive measurability of their graphs. Optional measurability can equivalently be used. The term `optional time’ is sometimes used as a synonym for stopping time.

Lemma 5 A map ${\tau\colon\Omega\rightarrow\bar{\mathbb R}_+}$ is a stopping time if and only if its graph ${[\tau]}$ is optional.

Proof: If ${\tau}$ is a stopping time then (1) shows that ${[\tau]\in\mathcal{O}}$. Conversely, if ${[\tau]}$ is optional then, in particular, it is progressive. So, Lemma 7 of the previous post shows that ${\tau}$ is a stopping time. ⬜

Proof of Theorem 1

Let ${\mathcal{O}_k}$, ${k=1,2,3,4}$, be the sigma-algebra generated by each of the respective collections of sets/processes of the Theorem 1. The inclusions ${\mathcal{O}_1\subseteq\mathcal{O}_2\subseteq\mathcal{O}_3\subseteq\mathcal{O}_4}$ are immediate. The first inclusion follows by taking ${Z=1}$ in the second statement. The second inclusion holds because the processes in statement 2 are cadlag and adapted, and the third inclusion holds since cadlag processes are, by definition, right-continuous. It only remains to show that ${\mathcal{O}_4\subseteq\mathcal{O}_1}$. I will break this down into the inclusions ${\mathcal{O}_4\subseteq\mathcal{O}_2}$ and ${\mathcal{O}_2\subseteq\mathcal{O}_1}$.

Proof of ${\mathcal{O}_2\subseteq\mathcal{O}_1}$: We need to show that ${Z1_{[\tau,\infty)}}$ is ${\mathcal{O}_1}$-measurable for stopping times ${\tau}$ and ${\mathcal{F}_\tau}$-measurable Z. By the monotone class theorem, it is enough to consider ${Z=1_A}$ for ${A\in\mathcal{F}_\tau}$. Then,

$\displaystyle Z1_{[\tau,\infty)}=1_{[\tau_A,\infty)}$

where ${\tau_A(\omega)}$ equals ${\tau(\omega)}$ when ${\omega\in A}$ and ${\infty}$ otherwise. As this is a stopping time, ${Z1_{[\tau,\infty)}}$ is ${\mathcal{O}_1}$-measurable. ⬜

To complete the proof of Theorem 1, it only remains to show that right-continuous adapted processes are ${\mathcal{O}_2}$-measurable. Although it is not strictly necessary, I will first prove the result for cadlag adapted processes. There are a few reasons for doing this. Firstly, the proof is more straightforward and, in any case, we are often interested in processes which are known to be cadlag rather than merely right-continuous. Secondly, we have not made use of completeness of the filtration yet, and completeness is not actually required for the equivalence of the first three sigma-algebras of the theorem. It is required, though, for the proof that right-continuous adapted processes are ${\mathcal{O}_2}$-measurable. For this reason, in some situations, it may be more appropriate to use either of the first three definitions of the optional sigma-algebra given by Theorem 1 rather than defining it as generated by the right-continuous adapted processes.

Proof of ${\mathcal{O}_3\subseteq\mathcal{O}_2}$: We show that any cadlag adapted process X is ${\mathcal{O}_2}$-measurable. Fixing ${\epsilon > 0}$, it is enough to construct an ${\mathcal{O}_2}$-measurable process Y such that ${\lvert X-Y\rvert < \epsilon}$. By letting ${\epsilon}$ tend to zero, this will show that X is ${\mathcal{O}_2}$-measurable.

Define an increasing sequence of random times ${\{\tau_n\}}$ by ${\tau_0=0}$ and, for each ${n\ge0}$,

 $\displaystyle \tau_{n+1}=\inf\left\{t\ge\tau_n\colon\lvert X_t-X_{\tau_n}\rvert\ge\epsilon\right\}.$ (2)

Then, define the process Y as,

$\displaystyle Y=\sum_{n=1}^\infty X_{\tau_n}1_{[\tau_n,\tau_{n+1})}.$

By the debut theorem for right-continuous processes, ${\tau_n}$ are stopping times. Then, as ${X_{\tau_n}}$ is ${\mathcal{F}_{\tau_n}}$-measurable, writing

$\displaystyle X_{\tau_n}1_{[\tau_n,\tau_{n+1})}=X_{\tau_n}1_{[\tau_n,\infty)}-X_{\tau_n}1_{[\tau_{n+1},\infty)}$

expresses Y as a sum of processes in the form of the second statement of the theorem. So, Y is ${\mathcal{O}_2}$-measurable.

Next, the times ${\tau_n}$ increase to a limit ${\tau}$ and, by construction, ${\lvert X-Y\rvert < \epsilon}$ on the interval ${[0,\tau)}$. So, the result follows so long as ${\tau = \infty}$. On the event ${\{\tau <\infty\}}$, the condition that X has left limits implies that ${X_{\tau_n}}$ converges to a limit ${X_{\tau-}}$ as n goes to infinity. This contradicts the inequality ${\lvert X_{\tau_{n+1}}-X_{\tau_n}\rvert\ge\epsilon}$ and, hence, ${\tau=\infty}$ as required. ⬜

The proof above only made use of completeness of the filtration at one point, when the debut theorem was invoked to show that ${\tau_n}$ are stopping times. However, this was just for convenience and, if we replace (2) by

$\displaystyle \tau_{n+1}=\inf\left\{t\ge\tau_n\colon\lvert X_t-X_{\tau_n}\rvert\ge\epsilon{\rm\ or\ }\lvert X_{t-}-X_{\tau_n}\rvert\ge\epsilon\right\}$

then it can be seen directly that ${\tau_n \le t}$ if and only if

$\displaystyle \sup_{s\in S}\lvert X_s-X_{\tau_n}\rvert\ge\epsilon,$

for any countable ${S\subset[0,t]}$ containing t. As this is ${\mathcal{F}_t}$-measurable, this alternative definition also gives a sequence of stopping times, and the proof does not require completeness of the filtered probability space.

I now complete the proof of Theorem 1 by showing that right-continuous adapted processes are ${\mathcal{O}_2}$-measurable.

Proof of ${\mathcal{O}_4\subseteq\mathcal{O}_2}$: We show that any right-continuous adapted process X is ${\mathcal{O}_2}$-measurable. As above, it is enough to find an ${\mathcal{O}_2}$-measurable process Y satisfying ${\lvert X-Y\rvert < \epsilon}$, for any given ${\epsilon > 0}$.

Define ${\mathcal{T}}$ to be the set of all stopping times such that there exists an ${\mathcal{O}_2}$-measurable process Y with ${\lvert1_{[0,\tau)}X-Y\rvert < \epsilon}$. For any ${\sigma,\tau\in\mathcal{T}}$, there are ${\mathcal{O}_2}$-measurable Y and Z satisfying

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle\lvert 1_{[0,\sigma)}X-Y\rvert < \epsilon,\smallskip\\ &\displaystyle\lvert 1_{[0,\tau)}X-Z\rvert < \epsilon. \end{array}$

Then

$\displaystyle U = 1_{[0,\sigma)}Y+1_{[\sigma\wedge\tau,\tau)}Z$

is ${\mathcal{O}_2}$-measurable and satisfies

$\displaystyle \lvert 1_{[0,\sigma\vee\tau)}X-U\rvert < \epsilon.$

Hence, the maximum ${\sigma\vee\tau}$ is again in ${\mathcal{T}}$. Similarly, for an increasing sequence ${\{\tau_n\}}$ in ${\mathcal{T}}$, there is a sequence ${\{Y^n\}}$ of ${\mathcal{O}_2}$-measurable processes satisfying

$\displaystyle \lvert 1_{[0,\tau_n)}X-Y^n\rvert < \epsilon.$

Then, with ${\tau_0=0}$,

$\displaystyle Y=\sum_{n=1}^\infty1_{[\tau_{n-1},\tau_n)}Y^n$

is an ${\mathcal{O}_2}$-measurable process with

$\displaystyle \lvert1_{[0,\lim_n\tau_n)}X-Y\rvert < \epsilon,$

and the limit of ${\tau_n}$ as n goes to infinity is again in ${\mathcal{T}}$.

This all shows that the supremum of any countable subset of ${\mathcal{T}}$ is in ${\mathcal{T}}$ and, in particular, its essential supremum, say ${\tau^*}$, is again in ${\mathcal{T}}$. That is, there is an ${\mathcal{O}_2}$-measurable process Y with

 $\displaystyle \lvert1_{[0,\tau^*)}X-Y\rvert < \epsilon.$ (3)

Now, defining the stopping time

$\displaystyle \sigma = \inf\left\{t\ge\tau^*\colon\lvert X_t-X_{\tau^*}\rvert\ge\epsilon\right\}$

which, by right-continuity, is strictly greater than ${\tau^*}$ whenever ${\tau^*}$ is finite. Then

$\displaystyle Z=1_{[0,\tau^*)}Y+X_{\tau^*}1_{[\tau^*,\sigma)}$

is ${\mathcal{O}_2}$-measurable and satisfies

$\displaystyle \lvert 1_{[0,\sigma)}X-Z\rvert < \epsilon.$

This shows that ${\sigma\ge\tau^*}$ is in ${\mathcal{T}}$. By the definition of the essential supremum, ${\sigma=\tau^*}$ almost surely and, hence, ${\tau^*}$ is almost surely infinite. So, (3) shows that X is ${\mathcal{O}_2}$-measurable up to evanescence and, by completeness of the filtration, is then ${\mathcal{O}_2}$-measurable. ⬜

Processes With Right Limits

Occasionally, we may want to consider processes which have right limits but are not necessarily right-continuous everywhere. For example, when the underlying filtered probability space is not right-continuous, then not all martingales will have a cadlag version. However, we were able to show that there is always a version with left and right limits, and which has right limits everywhere outside of a fixed countable set of times. Although they are not cadlag, such processes are still optional. I start by proving the following result relating optional processes with respect to the right-continuous filtration ${\mathcal{F}_+}$ to optional processes under the original filtration.

Lemma 6 Let X be an optional process with respect to the right-continuous filtration ${\{\mathcal{F}_{t+}\}_{t\in{\mathbb R}_+}}$. Then, there exists a countable set ${S\subseteq{\mathbb R}_+}$ such that

 $\displaystyle 1_{S^c}X$ (4)

is optional.

Proof: Let us first consider processes of the form ${X=1_{[\tau,\infty)}}$ for an ${\mathcal{F}_+}$-stopping time ${\tau}$. Set,

$\displaystyle S=\left\{t\in{\mathbb R}_+\colon{\mathbb P}(\tau=t) > 0\right\}.$

As ${{\mathbb P}(\tau=t)}$ is strictly positive on S and sums to a finite value, S is countable. We can also show that (4) is optional. As ${Y\equiv1_{(\tau,\infty)}}$ is left-continuous and adapted, it is optional. Also, the random time

$\displaystyle \sigma=\begin{cases} \tau,&\tau\not\in S,\\ \infty,&\tau\in S \end{cases}$

is an ${\mathcal{F}_+}$-stopping time satisfying ${{\mathbb P}(\sigma=t)=0}$ for all times t. This means that ${\{\sigma\le t\}}$ is equal to ${\{\sigma < t\}\in\mathcal{F}_t}$ up to a zero probability set and, by completeness of the filtered probability space, is a stopping time. Hence,

$\displaystyle 1_{S^c}X=(1_{S^c}Y)\vee(1_{S^c}1_{[\sigma,\infty)})$

is optional.

Let ${\mathcal{S}}$ denote the set of all ${\mathcal{F}_+}$-optional processes for which there exists a countable ${S\subseteq{\mathbb R}_+}$ making (4) optional. From what we have just shown, ${1_{[\tau,\infty)}\in\mathcal{S}}$ for all ${\mathcal{F}_+}$-stopping times ${\tau}$. It is also clear that ${\mathcal{S}}$ is closed under taking finite linear combinations, and under taking limits of sequences. Consider a sequence ${X^n\in\mathcal{S}}$, so that there are countable sets ${S_n}$ for which ${1_{S_n^c}X^n}$ are optional. Taking ${S=\bigcup_nS_n}$ then ${1_{S^c}Y}$ is optional for Y a finite linear combination or limit of the sequence ${X^n}$. So, by the functional monotone class theorem, ${\mathcal{S}}$ includes all ${\mathcal{F}_+}$-optional processes. ⬜

Finally, I show that the existence of right limits together with right-continuity outside of a countable set of times is sufficient to conclude that a process is optional. In particular, this includes the modifications of martingales which we previously constructed without requiring right-continuity of the filtration.

Lemma 7 Let X be an adapted process with right-limits everywhere, and which is right-continuous everywhere outside of a countable set ${S\subset{\mathbb R}_+}$. Then, X is optional.

Proof: The process ${X_{t+}=\lim_{s\downarrow\downarrow t}X_s}$ is right-continuous and is adapted with respect to the right-continuous filtration ${\mathcal{F}_{+}}$. Hence, it is ${\mathcal{F}_{+}}$-optional and, by Lemma 6, there exists a countable ${A\subset{\mathbb R}_+}$ such that ${1_{A^c}X_{+}}$ is optional. By enlarging A, if necessary, to include the set of times at which X is not right-continuous, we see that

$\displaystyle 1_{A^c}X=1_{A^c}X_+$

is optional. Therefore, as ${X_t}$ is ${\mathcal{F}_t}$-measurable for each ${t\in A}$,

$\displaystyle X = 1_{A^c}X+\sum_{t\in A}1_{[t]}X_t$

is optional. ⬜