# Is a predictable process adapted?

Let us consider a measurable space $(\Omega, \mathcal{F})$, with a filtration $(\mathcal{F}_t)_t$ of sub $\sigma$-algebras of $\mathcal{F}$. The predictable $\sigma$-algebra $\mathcal{P}$ is the $\sigma$-algebra generated by all processes of the form $$X : [0,\infty) \times \Omega \rightarrow \mathbb{R}$$ such that

• $X(t,\cdot)$ is $\mathcal{F}_t$-measurable for all $t>0$ (i.e. $X$ is adapted to the filtration $(\mathcal{F}_t)_t$),

• $X(\cdot, \omega)$ is left-continuous for all $\omega \in \Omega$.

My question is:

Is every $\mathcal{P}$-measurable process necessarily adapted? If so, is it necessarily adapted to the coarser filtration $(\mathcal{F}_{t-})_t$?

• The answers are yes and yes. The question is not adapted to MO. // Upvoters: why the upvote?
– Did
Jan 17, 2014 at 12:56
• Thanks! Even though I saw more basic questions on probability, I agree that my question is not adapted to MO and I do apologize. Would you be so kind to give me a hint on how to prove it? Thanks again.
– Tom
Jan 17, 2014 at 18:05
– Did
Jan 17, 2014 at 19:28
• Sorry to bother again.. does anybody have by chance a hint on this one?
– Tom
Aug 16, 2015 at 18:39
• No idea; but it's a good question though! May 22, 2016 at 3:13

There is also the concept of an optional process:

Definition 1: The optional $\sigma$-algebra $O$ is generated by all adapted càdlàg processes (continue à droite et limit à gauche/continuous from the right and limit from the left). A stochastic process $X=(X_t)_{t\geq 0}$ is called optional, if $X$ is measurable w.r.t. $O\subseteq \mathcal{P}(\Omega\times \mathbb{R}_{\geq 0})$.

It is possible to show that $P\subseteq O$ ($P$ denotes the predictable $\sigma$-algebra). Thus, every predictable process is optional. For optional process, we have the following result:

Theorem 2: If $X$ is optional, then $X$ is $\mathcal{F}\otimes\mathcal{B}(\mathbb{R}_{\geq 0})-\mathcal{B}(\mathbb{R})$ measurable and the random variable $$X_T 1_{T<\infty}:\Omega\rightarrow \mathbb{R}$$ $$\omega\mapsto X_{T(\omega)}1_{T(\omega)<\infty}$$ is $\mathcal{F}_T-\mathcal{B}(\mathbb{R})$-measurable for any stopping time $T$.

Applying this to the stopping time $T\equiv t$ for $t\geq 0$ gives $X_t=X_T 1_{T<\infty}$ is $\mathcal{F}_T$-measurable. Since $\mathcal{F}_T=\mathcal{F}_t$, we see that a predictable process is adapted to the filtration $(\mathcal{F}_t)_{t\geq 0}$

Regarding your second question, we have the following statement:

Theorem 3: Let $T$ be a stopping time and $X$ a predictable process. Then $X_T 1_{\{T<\infty\}}$ is $\mathcal{F}_{T -}$ measurable.

Applying this to the stopping time $T\equiv t$ yields $X_t=X_t 1_{t<\infty}$ is $\mathcal{F}_{t-}$ measurable for all $t\geq 0$ and a predictable process $X$.

The proof for theorem 3 can be found in the first edition of "Foundations of modern probability theory" by Olav Kallenberg in lemma 22.3.

• Hi peer - thx! Two comments: 1. in the first definition you meant "The optional σ-algebra O", not "The predictable σ-algebra O", right? 2. This would answer yes to my first question, not to the second, right?
– Tom
Nov 18, 2016 at 17:02
• @Tom Yes, sorry for that. I fixed it. I will think about your second question.
– peer
Nov 18, 2016 at 22:05
– peer
Jan 12, 2017 at 12:31

I wanted to note another method for proving the first part. I shall only sketch the argument.

1. Show that the indicator function of every predictable rectangle is progressively measurable. This is very easy to show. Note that the the predictable rectangles form a $\pi$-system which generate the predictable $\sigma$-algebra.
2. Now use the Monotone-Class theorem (see David Williams, Page 37) to show that every bounded predictable process is progressively measurable. (Again very straight-forward)
3. Now, if $X$ is predictable $X \wedge n$ is bounded predictable and therefore progressively measurable from the previous step. Now just take limit $n \to \infty$ to conclude that $X$ is progressively measurable.
4. Now use the fact that a progressively measurable process is measurable and adapted. (This too is easy to show).

Denote

$$\mathcal{P} = \sigma(\mathcal{I}) = \sigma(\{ (s,t]\times E, 0 \leq s < t, E \in \mathcal{F}_s \} )$$

Consider the map

$$f: (\Omega, \mathcal{F}_{t-}) \to ((0,\infty) \times \Omega , \sigma(\mathcal{I}))$$

$$\omega \mapsto (t,\omega)$$

This map $$f$$ is $$\mathcal{F}_{t-}-\mathcal{P}$$ measurable. For this, notice that $$f^{-1}(\mathcal{I}) \subset \mathcal{F}_{t-}$$, since for any element in $$\mathcal{I}$$:

$$f^{-1}( (s,r] \times E ) = \begin{cases} E \quad \text{if } s r \end{cases} \in \mathcal{F}_s \subset \mathcal{F}_{t-}$$

Hence $$f^{-1}(\sigma(\mathcal{I})) \subset \mathcal{F}_{t-}$$ (see Proving measurability of a function only by checking generating sets). Finally as by the assumption $$X : ((0,\infty) \times \Omega, \mathcal{P}) \to (\mathbb{R}, \mathscr{B}(\mathbb{R}))$$ a measurable map, the composition of the two measurable maps $$X_t = X \circ f : (\Omega, \mathcal{F}_{t-}) \to (\mathbb{R}, \mathscr{B}(\mathbb{R}))$$ is $$\mathcal{F}_{t-}$$ measurable.