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I have come across a statement I am not sure how to prove.

Assume you have a filtered probability space $(\Omega, \mathcal{F},\{\mathcal{F}_t\},P)$. A stochastic process $X_t$ is progressively measurable if for every $t\ge 0$, the function $f(\omega,s) $from $\Omega\times[0,t]$ to $\mathbb{R}$ given by $f(\omega,s)=X_s(\omega)$, is $\mathcal{F}_t\otimes\mathcal{B}([0,t])$ measurable.

Prog is the smallest sigma-algebra on $\Omega\times[0,\infty)$ such that all progressively measurable processes are Prog-measurable.

Obviously we have that every progressively measurable process is Prog-measurable. However, assume you have a function $g: \Omega\times[0,\infty)\rightarrow \mathbb{R}$, and assume that it is Prog-measurable, then it is stated that the stochastic process $Y_t$ defined by $Y_t(\omega)=g(\omega,t)$ is progressively measurable.

Do you know how to show this?

Attempt:

My only idea is using something like the Doob-Dynkin lemma. The Doob-Dynkin lemma states that if a function $f$ is $\sigma(h)$ measurable then there exists a borel function k, such that $f=k(h).$ The problem I have is that Prog is not generated by a single progressively measurable process it is generated by an infinite amount, so I don't think I can use the Doob-Dynkin lemma directly? But if I can prove that if $g$ is Prog-measurable, then it is some kind of limit or some kind of Borel function of the progressively measurable processes, then it will also be progressively measurable. Is there a way to do this?

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2 Answers 2

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This is a consequence of the functional form of the monotone class theorem as found, for example, at https://almostsuremath.com/2019/10/27/the-functional-monotone-class-theorem/ (see Theorem 1 there).

The key observation is that Prog is generated by an algebra of bounded process that contains the constant processes and is stable under bounded pointwise convergence.

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  • $\begingroup$ Thank you, do you mind checking if this is the argument: Let $H$ be the vector space of bounded progressively measurable functions, $H_+$ is closed under increasing limits, $H$ is also closed under multiplication. $H$ contains the constant function. Then $H$ contains all the functions which are measurable with respect to $\sigma(H)=Prog$. The argument is a little unusual because usually the space which is closed under multiplication is smaller than $H$. $\endgroup$
    – user394334
    Jul 17, 2021 at 16:18
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    $\begingroup$ Yes, this is what I had in mind. As you see, the conclusion is quite general, depending only on certain linear "closure" properties. That is, the same argument works to reach the same conclusion for optional processes, predictable process, ... $\endgroup$ Jul 17, 2021 at 17:17
  • $\begingroup$ Thank you, but what do you mean that we can use the same argument for optional and predictable processes? Left- or right-continuity isn't necessarily preserved under pointwise limits? $\endgroup$
    – user394334
    Jul 17, 2021 at 19:44
  • $\begingroup$ @user394334: Ack! Good point. Strike that comment. $\endgroup$ Jul 18, 2021 at 16:36
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There is no need for the monotone class theorem here. It is sometimes useful to believe that since $\sigma$-algebras are collections of sets, anything that is true about them should follow from operations on measurable sets without reference to measurable functions. (Remember that a measurable function is something like a concatenation of measurable sets.) Hence here it is helpful to replace "if $g$ is $\text{Prog}$-measurable, then..." with the corresponding statement for sets "if $A$ is $\text{Prog}$-measurable, then..."

In this case, it boils down to: $A \in \text{Prog}$ only if the process $\chi_{A}$ is progressively measurable. (Note that, by the definition of $\text{Prog}$, we already know that the "if" version of this statement is true.)

Define $\mathcal{P}$ by \begin{equation*} \mathcal{P} = \{A \subseteq \Omega \times [0,\infty) \, \mid \, \chi_{A} \, \, \text{is progressively measurable}\}. \end{equation*} It is quick to show that $\mathcal{P}$ is a $\sigma$-algebra since $\chi_{\bigcup_{n = 1}^{\infty} A_{n}} = \sum_{n = 1}^{\infty} \chi_{A_{n}}$ if $\{A_{n}\}_{n \in \mathbb{N}}$ is pairwise disjoint, and $\chi_{A^{c}} = 1 - \chi_{A}$.

Since every progressively measurable process is $\text{Prog}$-measurable (by definition), the inclusion $\mathcal{P} \subseteq \text{Prog}$ is immediate. We need to show that $\text{Prog} \subseteq \mathcal{P}$. Since $\text{Prog}$ is on the left-hand side, this suggests using the minimality property of $\text{Prog}$. That is, we only need to show that if $X$ is progressively measurable, then $X$ is $\mathcal{P}$-measurable.

Here is where we get back to the point: $\sigma$-algebras are about sets. Fix $c \in \mathbb{R}$. To show that $X$ is $\mathcal{P}$-measurable, we only need to show that $A(c) = \{X \leq c\} \in \mathcal{P}$ holds. Looking at the definition of $\mathcal{P}$ suggests studying the "process" $\chi_{A(c)}$.

Fix $t \geq 0$. If we define $f^{A(c),t} : \Omega \times [0,t] \to \mathbb{R}$ by $f^{A(c),t}(\omega,s) = \chi_{A(c)}(\omega,s)$, then the progressive measurability of $X$ implies that \begin{equation*} \{f^{A(c),t} = 1\} = \{(\omega,s) \in \Omega \times [0,t] \, \mid \, X^{\omega}_{s} \leq c\} \in \mathcal{F}_{t} \otimes \mathcal{B}([0,t]). \end{equation*} Taking the complement, $\{f^{A(c),t} = 0\} \in \mathcal{F}_{t} \otimes \mathcal{B}([0,t])$. Since $f^{A(c),t}$ takes values in $\{0,1\}$, this proves it is $\mathcal{F}_{t} \otimes [0,t]$-measurable.

We showed that if $X$ is progressively measurable, then $\chi_{A(c)}$ is progressively measurable for all $c \in \mathbb{R}$ --- this fact is really why $\text{Prog}$ completely characterizes progressively measurable processes. Hence, by definition of $\mathcal{P}$, we find that $X$ is $\mathcal{P}$-measurable if it is progressively measurable. But $\text{Prog}$ is the minimal $\sigma$-algebra with this property so $\text{Prog} \subseteq \mathcal{P}$.

Just to reiterate: $X$ is progressively measurable if and only if the "derived sets" $\{A(c)\}_{c \in \mathbb{R}}$ given by $A(c) = \{X \leq c\}$ are progressively measurable. Thus, progressive measurability, at some level, is a question about subsets of $\Omega \times [0,\infty)$ --- so the fact that there is a $\sigma$-algebra $\text{Prog}$ lying behind it is no surprise. In particular, note that $X$ is progressively measurable if and only if $X$ is $\text{Prog}$-measurable. (Further, the definition of $\mathcal{P}$ above gives a succinct, explicit definition of $\text{Prog}$ that does not involve the sometimes-unwieldy "minimal $\sigma$-algebra" concept.)

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