Filtrations and Adapted Processes

George Lowther Filtrations and Processes, Stochastic Calculus Notes 8 November 0915 September 20

In the previous post I started by introducing the concept of a stochastic process, and their modifications. It is necessary to introduce a further concept, to represent the information available at each time. A filtration ${\{\mathcal{F}_t\}_{t\ge 0}}$ on a probability space ${(\Omega,\mathcal{F},{\mathbb P})}$ is a collection of sub-sigma-algebras of ${\mathcal{F}}$ satisfying ${\mathcal{F}_s\subseteq\mathcal{F}_t}$ whenever ${s\le t}$ . The idea is that ${\mathcal{F}_t}$ represents the set of events observable by time ${t}$ . The probability space taken together with the filtration ${(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\ge 0},{\mathbb P})}$ is called a filtered probability space.

Given a filtration, its right and left limits at any time and the limit at infinity are as follows

$\displaystyle \mathcal{F}_{t+}=\bigcap_{s>t}\mathcal{F}_s,\ \mathcal{F}_{t-}=\sigma\Big(\bigcup_{s<t}\mathcal{F}_s\Big),\ \mathcal{F}_{\infty}=\sigma\Big(\bigcup_{t\in{\mathbb R}_+}\mathcal{F}_t\Big).$

Here, ${\sigma(\cdot)}$ denotes the sigma-algebra generated by a collection of sets. The left limit as defined here only really makes sense at positive times. Throughout these notes, I define the left limit at time zero as ${\mathcal{F}_{0-}\equiv\mathcal{F}_0}$ . The filtration is said to be right-continuous if ${\mathcal{F}_t=\mathcal{F}_{t+}}$ .

A probability space ${(\Omega,\mathcal{F},{\mathbb P})}$ is complete if ${\mathcal{F}}$ contains all subsets of zero probability elements of ${\mathcal{F}}$ . Any probability space can be extended to a complete probability space (its completion) in a unique way by enlarging the sigma-algebra to consist of all sets ${A\subset\Omega}$ such that ${B\subseteq A\subseteq C}$ for ${B,C\in\mathcal{F}}$ satisfying ${{\mathbb P}(C\setminus B)=0}$ . Similarly, a filtered probability space is said to be complete if the underlying probability space is complete and ${\mathcal{F}_0}$ contains all zero probability sets.

Often, in stochastic process theory, filtered probability spaces are assumed to satisfy the usual conditions, meaning that it is complete and the filtration is right-continuous. Note that any filtered probability space can be completed simply by completing the underlying probability space and then adding all zero probability sets to each ${\mathcal{F}_t}$ . Furthermore, replacing ${\mathcal{F}_t}$ by ${\mathcal{F}_{t+}}$ , any filtration can be enlarged to a right-continuous one. By these constructions, any filtered probability space can be enlarged in a minimal way to one satisfying the usual conditions.

Throughout these notes I assume a complete filtered probability space, although many of the results can be extended to the non-complete case without much difficulty. However, for the sake of a bit more generality, I don’t assume that filtrations are right-continuous.

One reason for using filtrations is to define adapted processes. A stochastic process process ${X}$ is adapted if ${X_t}$ is an ${\mathcal{F}_t}$ -measurable random variable for each time ${t\ge 0}$ . This is just saying that the value ${X_t}$ is observable by time ${t}$ . Conversely, the filtration generated by any process ${X}$ is the smallest filtration with respect to which it is adapted. This is given by ${\mathcal{F}^X_t=\sigma\left(X_s\colon s\le t\right)}$ , and referred to as the natural filtration of ${X}$ .

As mentioned in the previous post, it is often necessary to impose measurability constraints on a process ${X}$ considered as a map ${{\mathbb R}_+\times\Omega\rightarrow{\mathbb R}}$ . Right-continuous and left-continuous processes are automatically jointly measurable. When considering more general processes, it is useful to combine the measurability concept with adaptedness. This can be done in either of the following three ways, in order of increasing generality (see Lemma 4 below).

Definition 1

The predictable sigma-algebra on ${{\mathbb R}_+\times\Omega}$ , denoted by ${\mathcal{P}}$ , is generated by the left-continuous and adapted processes. A stochastic process is said to be predictable if it is ${\mathcal{P}}$ -measurable. Alternatively, the predictable processes are sometimes called previsible.

The optional sigma-algebra on ${{\mathbb R}_+\times\Omega}$ , denoted by ${\mathcal{O}}$ , is generated by the right-continuous and adapted processes. A stochastic process is said to be optional if it is ${\mathcal{O}}$ -measurable.

A process ${X}$ is progressively measurable, or just progressive, if for each ${t\ge 0}$ , the map
$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle [0,t]\times\Omega\rightarrow{\mathbb R},\smallskip\\ &\displaystyle (s,\omega)\mapsto X_s(\omega) \end{array}$

is ${\mathcal{B}([0,t])\otimes\mathcal{F}_t}$ -measurable.

The most important of these definitions, at least in these notes, is that of predictable processes. While adapted right-continuous processes will be used extensively, there is not much need to generalize to optional processes. Similarly, progressive measurability isn’t used a lot except in the context of adapted right-continuous (and therefore optional) processes. On the other hand, predictable processes are extensively used as integrands for stochastic integrals and in the Doob-Meyer decomposition, and are often not restricted to the adapted and left-continuous case.

Given any set of real-valued functions on a set, which is closed under multiplication, the set of functions measurable with respect to the generated sigma-algebra can be identitified as follows. They form the smallest set of real-valued functions containing the generating set and which is closed under taking linear combinations and increasing limits. So, for example, the predictable processes form the smallest set containing the adapted left-continuous processes which is closed under linear combinations and such that the limit of an increasing sequence of predictable processes is predictable.

Another way of defining predictable processes is in terms of continuous adapted processes.

Lemma 2 The predictable sigma-algebra is generated by the continuous and adapted processes.

Proof: Clearly every continuous adapted process is left-continuous and, therefore, is predictable. Conversely, if ${X}$ is an adapted left-continuous process then it can be written as a limit of the continuous processes

$\displaystyle X^n_t = n\int_{t-1/n}^t1_{\{|X_{s\vee 0}|\le n\}}X_{s\vee 0}\,ds.$

Continuity of ${X^n}$ follows from the fact that ${t\mapsto\int_{t-1/n}^tf(s)\,ds}$ is continuous for bounded and measurable ${f\colon{\mathbb R}\rightarrow{\mathbb R}}$ . In fact, as ${\lvert f\rvert}$ is bounded by ${n^2}$ , it has Lipschitz constant ${2n^2}$ . The limit ${X^n\rightarrow X}$ follows from left-continuity of ${X}$ , which is therefore in the sigma-algebra generated by the continuous adapted processes ${X^n}$ . ⬜

A further method of defining the predictable sigma-algebra is in terms of simple sets generating it. The following is sometimes used.

Lemma 3 The predictable sigma-algebra is generated by the sets of the form

$\displaystyle \left\{(s,t]\times A\colon t>s\ge 0, A\in\mathcal{F}_s\right\}\cup\left\{\{0\}\times A\colon A\in\mathcal{F}_0\right\}.$ (1)

Proof: If ${S}$ is any of the sets in the collection (1) then the process ${X=1_S}$ defined by ${X_t(\omega)=1_{\{(t,\omega)\in S\}}}$ is adapted and left-continuous, and therefore predictable. So, ${S\in\mathcal{P}}$ .

Conversely, let X be left-continuous and adapted. Then it is the limit of the piecewise constant functions

$\displaystyle X^n_t=X_01_{\{t=0\}}+\sum_{k=1}^\infty X_{(k-1)/n}1_{\{(k-1)/n<t\le k/n\}}$

as n goes to infinity. Each of the summands on the right hand side is easily seen to be measurable with respect to the sigma-algebra generated by the collection (1). So, ${X=\lim_{n\rightarrow\infty}X^n}$ is also measurable. ⬜

In these notes, I refer to the collection of finite unions of sets in the collection (1) as the elementary or elementary predictable sets. Writing these as ${\mathcal{E}}$ then ${\mathcal{P}=\sigma(\mathcal{E})}$ .

Finally, the different forms of measurability can be listed in order of generality, starting with the predictable processes, up to the much larger class of jointly measurable adapted processes.

Lemma 4 Each of the following properties of a stochastic process implies the next

predictable.

optional.

progressive.

adapted and jointly measurable.

Proof: As the predictable sigma-algebra is generated by the continuous adapted processes, which are also optional by definition, it follows that all predictable processes are optional.

Now, if ${X}$ is a right-continuous and adapted process and ${T\ge 0}$ , then the process ${Y_t=X_{t\wedge T}}$ is right-continuous and ${\mathcal{F}_T}$ -measurable at all times. By the joint measurability of right-continuous processes, ${Y}$ is ${\mathcal{B}({\mathbb R})\otimes \mathcal{F}_T}$ -measurable. As this holds for all times ${T}$ , ${X}$ is progressively measurable.

Finally, consider a progressively measurable process ${X}$ . From the definitions, it is jointly measurable. Furthermore, for any time ${t\ge 0}$ , ${(s,\omega)\mapsto X_s(\omega)}$ restricted to ${[0,t]\times\Omega}$ is ${\mathcal{B}([0,t])\otimes\mathcal{F}_t}$ -measurable. Therefore, ${X_s}$ is ${\mathcal{F}_t}$ -measurable for all ${s\le t}$ and, in particular, ${X}$ is adapted. ⬜

Published by George Lowther

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Published 8 November 0915 September 20

87 thoughts on “Filtrations and Adapted Processes”

Student says:

29 September 10 at 2:13 AM

I have been trying to find clarification on a question relating to filtrations and measurability for awhile, but can’t find discussions of it. I figure someone knowledgeable about stochastic processes would know this. If you have a random variable X that is measurable with respect to a sigma algebra generated by random variable Y, then there exists a function g such that X=g(Y). But what if the sigma algebra is generated by an uncountable number of random variables, such as where F is the natural filtration of stochastic process Y. What can be said in this case for r.v. X measurable with respect to F_t? Is there any functional representation?

Thanks for any help.

Reply
1. George Lowther says:
  
  30 September 10 at 10:58 PM
  
  Yes there is! Assuming that X is a real-valued random variable, then it can be expressed as a function of Y.
  If Y:Ω → E is a measurable map from the underlying probability space to a measurable space (E,ℰ) then any σ(Y)-measurable map X:Ω → ℝ can be written as X = g(Y) for a measurable g:E → ℝ. This is an application of the functional monotone class theorem. (i) Show that every X = 1_A for A ∈ σ(Y) is of the form g(Y). (ii) Show that the set of random variables of the form g(Y) is closed under linear combinations and taking increasing limits. (iii) appeal to the monotone class theorem.
  If Y_t is a (real-valued) process, it can be considered as a map to the space ℝ^ℝ₊ of real valued functions.
  
  Reply
Su says:

4 February 11 at 5:00 AM

Hi,

According to Lemma 2, will it be true to say that all brownian motions and martingales that are continuous are predictable processes?

Reply
1. George Lowther says:
  
  4 February 11 at 9:06 PM
  
  Yes, that’s correct. Assuming you are using an appropriate filtration, so that your processes are adapted, then they are predictable.
  
  Reply
  1. Su says:
    
    6 February 11 at 1:30 AM
    
    Thanks for your reply George.
    One more thing that’s bothering me, in the context of finding an equivalent martingale measure. Correct if I am wrong below:
    For discounted asset price process, Dt (assuming a constant discount rate), there are two reasons why the discounted asset price process could be predictable:
    a) Dt will be a martingale under say Q-probability measure, and is hence predictable (under this probability measure only?).
    b) If we assume say Geometric Brownian Motion Model for the asset price, the random component of the asset price will only be the Brownian Motion, which is predictable. Hence St (and Dt) is a predictable process.
    I have read somewhere that Dt is predictable, which is not very intuitive since asset prices from the market cannot be predictable. Hence my two reasoning above: Predictable only under certain probability measure OR predictable for asset prices which we have assumed a certain model.
    Hope you can understand where I am coming from and really appreciate if you could shed some light on this.
    
    Reply
  2. George Lowther says:
    
    25 February 11 at 4:49 PM
    
    Su,
    
    The reason why processes such as Brownian motion are predictable is just because they are continuous and adapted. It does not have anything to do with the martingale property. In fact, it does not have much to do with the measure either. The property of being predictable is not affected by an equivalent measure change.
    
    I think maybe you are a bit confused about the meaning of being predictable in the precise mathematical sense defined above, which is rather different from the (rather ill-defined) everyday notion of something being predictable. Here, predictability only means that you can (roughly speaking) predict the future value in an infinitesimal sense. Just because a Brownian motion is predictable does not mean that you can tell what value it will take at some future point in time. What you can say is that you can tell when the process is going to reach some value a just prior to it happening. This is simply because of continuity. If $X_t=a$ at some time then, by continuity, $\vert X_s-a\vert$ will be arbitrarily small for times s just before t.
    
    So, you can model asset prices by predictable processes. It simply means that they are continuous. Under Black-Scholes, the process is predictable in this sense. That is because it ignores the possibility that the price can suddenly jump.
    
    And, apologies for the late response. Hope that helps clear things up a bit.
    
    Reply
Su says:

12 March 11 at 6:39 AM

Hi George,

Appreciate the explanation. Definitely cleared things up!

Reply
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yaoliang says:

9 June 11 at 4:56 AM

Nice proof of Lemma 2, although it took me a while to see why it does not get through when X is right continuous :). Adapt.

Reply
yaoliang says:

9 June 11 at 5:05 AM

BTW, would that mean if the filtration is right continuous, then optional sigma field = predictable sigma field?

Reply
1. George Lowther says:
  
  9 June 11 at 10:07 PM
  
  Hi Yaoliang. The proof of Lemma 2 doesn’t work if X is only right-continuous because Xⁿ would not converge to X. I’m not sure why you suggest that optional = predictable when the filtration is right-continuous though. This is not true. The filtration generated by a Poisson process is right-continuous (once you complete it), but Poisson processes are not predictable. However, it is often the case that optional = pedictable. For example, this happens if the filtration is generated by a Brownian motion — see my latest post Predictable Stopping Times.
  
  Reply
  1. yaoliang says:
    
    9 June 11 at 10:56 PM
    
    Thanks, George. Here is what I thought about right continuous processes: change the integral interval of X^n from [t-1/n,t] to [t, t+1/n]. Would this be enough to argue that X^n converges to X? But of course, X is only F_+ measurable, that’s why I thought option=predictable when the filtration is right continuous (then F_+=F, hence X is adapted). I believe my claim is false, but still couldn’t see why the proof of Lemma 2 cannot go through.
    
    Reply
    1. George Lowther says:
      
      9 June 11 at 11:09 PM
      
      Lemma 2 relies on the fact that the processes Xⁿ are both continuous and adapted and, then, the fact that they converge to X means that X is measurable with respect to the sigma algebra generated by the continuous and adapted processes.
      If X was only assumed to be right-continuous then, as you say, you could define Xⁿ by integrating over the range [t,t+1/n] instead. However, Xⁿ would not be adapted. So, you can say that X is measurable with respect to the sigma-algebra generated by the continuous processes. But, it need not be measurable with respect to the sigma-algebra generated by the processes which are simultaneously continuous and adapted. It is true though, that X is measurable with respect to the processes which are continuous and $\mathcal{F}_{\cdot+\epsilon}$ -adapted for any given $\epsilon > 0$ . This is much weaker than being measurable with respect to the continous and $\mathcal{F}_{t+}$ -adapted processes.
2. Vlad says:
  
  21 January 17 at 9:30 PM
  
  Hi, George. You say “The filtration generated by a Poisson process is right-continuous (once you complete it)” – I know that was off-hand, but why is this the case? I’m trying to go through the mechanics of proving that statement, and it doesn’t seem to come to easily – the modulus for the right continuity of the process is stochastic.
  
  Reply
  1. George Lowther says:
    
    28 February 17 at 11:23 AM
    
    Hi. One method to prove what you want using results from these notes, is to apply the statements that, Poisson processes are Feller and Feller processes have right-continuous filtration.
    
    A more direct proof would be to show that $\mathbb{E}[U\vert\mathcal{F}_{t+}]=\mathbb{E}[U\vert\mathcal{F}_{t}]$ for bounded rvs U. For U a function of $X_u-X_v$ , with $u,v > t$ it follows from the independent increments property, and it is immediate for U $\mathcal{F}_t$ measurable. It can be extended to general U using the monotone class lemma.
    
    Reply
    1. Vlad says:
      
      28 February 17 at 3:54 PM
      
      Thanks for the references and response, George. It seems like both approaches (Feller and the “direct proof”) only show $\mathbb{E}[U|\mathcal{F}_t]=\mathbb{E}[U|\mathcal{F}_{t+}]$ almost surely, which is fine if the $\sigma$ -algebras are known to be augmented already. However, in the natural filtration $\mathcal{F}_0$ for a Poisson process is trivial, so it would seem that any approach showing conditional expectations are equivalent would suffer from the augmentation assumption. Is this assumption tight? I’d imagine so.
      
      Something still seems off for the direct proof sketch, even with augmentation, though – why couldn’t we apply it to the augmented natural filtration for the Wiener process, which isn’t RC?
    2. George Lowther says:
      
      28 February 17 at 5:07 PM
      
      By the ‘augmented’ natural filtration, you just mean with the zero probability events added, right? In that case, the argument shows that the augmented natural filtration is right-continuous. The argument does indeed apply to the Wiener process, whose augmented natural filtration is right-continuous.
      
      I was not making any claim about the non-augmented natural filtration although, now I think about it, it will still be right-continuous for the Poissoj process (but, not for the Wiener process).
GuidovM says:

14 June 11 at 11:15 PM

Hi George,
Thanks for this great post, it certainly clears things up. Nevertheless I have some trouble understanding the proof of lemma 4. I don’t completely understand how optional impies progressive. You proved that the generators of the optional sigma algebra, are progressively measurable. But I don’t see how this implies that for any optional process this also holds. Do you assume if the generators of a sigma algebra are B([o,t]) x Ft measurable on the subset [0,t] x omega then the same holds for the functions measurable with respect to this sigma algebra? I hope you understand my problem, and I would really appreciate it if you would explain this to me.

All the best,

GuidovM

Reply
1. George Lowther says:
  
  15 June 11 at 1:35 AM
  
  GuidovM,
  
  Yes I did assume that if the generators of a sigma algebra are B([0,t]) × F_t measurable on the subset [0,t] × Ω then the same holds for the functions measurable with respect to this sigma algebra.
  
  Maybe I did not explain this bit well. You can show that the sets A ⊆ R₊×Ω which are B([0,t])×F_t measurable on the subsets [0,t] × Ω forms a sigma-algebra (just check that it is closed under countable unions and taking complements). This is the progressive sigma-algebra. Furthermore, the definition of progressive processes given in Definition 1 is equivalent to saying that the process is measurable with respect to the progressive sigma-algebra. So, the statement “optional ⇒ progressive” is the same as saying that optional sigma-algebra ⊆ progressive sigma-algebra. To prove this you only need to look at processes generating the optional sigma-algebra.
  
  I should add a statement to Definition 1 defining the progressive sigma-algebra and stating that progressive ⇔ measurable with respect to the progressive sigma-algebra. I’ll do this at some point – thanks for mentioning it.
  
  Hope that clears things up.
  
  Reply
Tigran says:

16 November 11 at 2:51 PM

Hi George,

First of all thanks for your great set of notes. They are among the best I have come across so far.

Quick question:
Is it true that if a function f is F1xF2 measurable and that its section f_y(x) is F1* measurable for all y in Omega2, then f is F1xF2* measurable?
Your proof of “optional => progressive” seems to rely on the latter fact.

Thanks in advance!
– Tigran

Reply
1. Tigran says:
  
  16 November 11 at 3:38 PM
  
  Actually I just realized that your proof does not require what I mentioned.
  Never mind!
  
  Reply
2. George Lowther says:
  
  16 November 11 at 10:46 PM
  
  Hi.
  
  No, I don’t rely on that fact. Actually, that’s not true. Consider for example F1, F2 to be the Borel sigma-algebra on the reals. Let f be the indicator function of a non-measurable subset of the diagonal in R². Then, f_y(x) will be measurable for each y, but f itself is not measurable (I’m not sure what you mean by F1* and F1xF2*. Maybe the completion? That depends on having a measure though).
  
  Reply
Tigran says:

21 November 11 at 3:23 PM

Thanks for the reply. I realized after my initial comment that the proof doesn’t rely on this statement.
I also just realized how unclear my initial statement actually was (and how inconsistent the notation was as well).

What I meant to say is that suppose f(x,y) is measurable w.r.t. some joint sigma-algebra (F1xF2) and that its section in y (f_y(x)) is measurable w.r.t. a sigma-algebra smaller than F1 (this smaller sigma-algebra is denoted F1*) for each y.
Note: F1* does not depend on y.
The question is: Is f(x,y) measurable w.r.t. F1*xF2?
I have a feeling the answer in no.

Reply
1. George Lowther says:
  
  21 November 11 at 7:32 PM
  
  The answer is indeed no. Consider any probability space (Ω,F1,P) on which we have defined a uniformly distributed random variable T: Ω → [0,1]. Let F1* consist of the sets in F1 with probability 0 or 1. Let F2 be the Borel sigma-algebra on [0,1].
  
  Now, look at the stochastic process Ω×[0,1] → R given by (ω,t) ↦ X_t(ω) = 1_{T(ω)=t}. This is F1×F2 measurable but not F1*×F2 measurable. However, X_t is almost surely zero at each time t, so is F1* measurable.
  
  Reply
  1. Tigran says:
    
    26 November 11 at 2:55 PM
    
    Thanks again for your answer George. It’s a nice counter-example.
    The only bit that I have yet to understand completely is: why is I_{T(omega}=t} not F1*xF2-measurable.
    It’s probably a trivial detail that I am missing and your input would be great to help me understand this.
    
    P.S.: Once again my hat is off to the clarity of your notes. They are extremely helpful in getting a better intuition of the technicalities in stochastic analysis.
    
    Reply
    1. George Lowther says:
      
      26 November 11 at 3:11 PM
      
      It is true that this process is not F1*xF2 measurable. However, it is not a trivial detail. Far from it. You can prove it by the measurable projection theorem. If it was measurable then this theorem would tell you that the set of ω ∈ Ω such that there exists a t ≤ 1/2 with X_t(ω) = 1 is measurable with respect to the completion of F1*, so has measure 0 or 1. However, this set is just {T ≤ 1/2}, which has probability 1/2. So X is not F1*xF2 measurable.
    2. George Lowther says:
      
      26 November 11 at 3:32 PM
      
      …actually the measurable projection theorem is very similar to the generalised form of the Debut theorem which says that the first time that a jointly measurable process hits a measurable set is itself measurable wrt the completion of the probability space (and is a stopping time if the process is progressive and the filtration is right continuous). In this case, the first time X hits 1 is T, which is not F1* measurable. So X is not F1*xF2 measurable.
Ilya says:

17 February 12 at 4:44 PM

Dear George, thanks for very nice notes – the explanation is highly helpful. I didn’t get the definition of a predictable sigma-algebra: that is generated by left-continuous and adapted processes. W.r.t. which filtration processes generating predictable sigma-algebra should be adapted?
Thanks in advance,
Ilya.

Reply
1. George Lowther says:
  
  22 February 12 at 1:20 AM
  
  The definition of the predictable sigma-algebra does depend on what underlying filtration is used to define adapted processes. Usually, we are given one sigma-algebra $\{\mathcal{F}_t\}_{t\ge0}$ (representing all observable events at time t). Then, we work exclusively with respect to this sigma-algebra, which is used in the definition of adapted processes (and predictable processes, optional processes, stopping times, etc). If we have multiple sigma-algebras and it is not clear which one is being used from the context, then it should be stated. With multiple sigma-algebras, we have multiple predictable sigma-algebras (and multiple optional sigma algebras, etc). Where this happens in these notes I say something along the lines of “the $\mathcal{F}_\cdot$ -predictable sigma algebra” or “predictable with respect to $\mathcal{F}_\cdot$ ”).
  
  Reply
Chilorio says:

4 January 13 at 6:22 AM

Hi!

I don’t quite understand why it is important to assume the usual conditions. What exactly do we miss if don’t assume them? Do stochastic integral stop being adapted? Are we unable to make Doob-Meyer decompositions? Martingales stop having a cadlag version? When we make a Girsanov transformation, the new Brownian motion generates a filtration which does not coincide with the filtration of the original Bm?

Thanks for these great notes!

Reply
1. MathSwineT says:
  
  18 January 13 at 7:43 AM
  
  Cadlag versions and stochastic integrals would both fail to be adapted. Check: compensated Poisson process X, pick its left-continuous version Y and the filtration generated by Y (= the predictable).
  
  Reply
Mario says:

4 September 13 at 7:09 AM

George, I would like to ask a question about this very useful thread. In Lemma 2, when you define the process $X_t^n = \int_{t-1/n}^t etc… I can see that this definition works when $X_t $ is a step process or a simple process. But for general $X_t $ how do you handle them inside the integral?…. Are you saying that one should prove it using the functional monotone class theorem? In this case which version do you use as there is more than one fMonClassTheorem? Thank you in advance if you could throw a few more details in there.

Reply
Chi Dong says:

24 June 14 at 12:29 AM

Hi, I have a quite related question. Suppose X and Y are two continuous processes. For each t, Xt is adapted with respect to {Ys:s X(t,w) and (t, w) –> Y(t,w), is the joint sigma-field generated by X contained in that generated by Y? This kind of joint measurability question is interesting and difficult.

Reply
1. Chi Dong says:
  
  24 June 14 at 12:31 AM
  
  Sorry, there seems to be some system error. My question is like this: Suppose X and Y are two continuous processes. For each t, Xt is adapted with respect to {Ys:s X(t,w) and (t, w) –> Y(t,w), is the joint sigma-field generated by X contained in that generated by Y?
  
  Reply
  1. Chi Dong says:
    
    24 June 14 at 12:32 AM
    
    Sorry, still error. Suppose X and Y are two continuous processes. For each t, Xt is adapted with respect to {Ys:s< = t}.
    
    Reply
  2. Chi Dong says:
    
    24 June 14 at 12:32 AM
    
    Considering X and Y as functions (t,w) –> X(t,w) and (t, w) –> Y(t,w), is the joint sigma-field generated by X contained in that generated by Y?
    
    Reply
vinayak chuni says:

26 October 14 at 11:39 PM

Hi George Lowther
I could not understand what do you mean by the following? I would be grateful if you could explain it
“Given any set of real-valued functions on a set, which is closed under multiplication, the set of functions measurable with respect to the generated sigma-algebra can be identitified as follows. They form the smallest set of real-valued functions containing the generating set and which is closed under taking linear combinations and increasing limits. So, for example, the predictable processes form the smallest set containing the adapted left-continuous processes which is closed under linear combinations and such that the limit of an increasing sequence of predictable processes is predictable.”

Reply
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Anonymous says:

10 August 16 at 12:37 AM

In the proof of lemma 2, do you really need the functional monotone class theorem? Doesn’t continuity follow directly by choosing $\delta = \min(\epsilon/(2n^2) , 1/(2n))$

Reply
1. George Lowther says:
  
  10 August 16 at 10:02 PM
  
  I think you are correct — but why not just $\delta = \epsilon/(2n^2)$ ?
  
  Reply
  1. George Lowther says:
    
    10 August 16 at 10:43 PM
    
    Actually, I see why you used the min, but it is unnecessary. I updated the post. Thanks!
    
    Reply
yan says:

28 February 17 at 9:57 PM

Hi George,

I have a question regarding adaptedness and filtration: is adaptedness a distributional property?
More precisely, Suppose $(\Omega,\mathcal F,\mathbb F,\mathbb P)$ is a probability space supporting a Brownian motion $\{W_t\}_{t=0}^T$ , where the filtration is the augmented brownian filtration $\{F_t^W\}$ and $\mathcal F=F_T^W$ . Let $\{S_t\}_{0 \leq t \leq T}$ be a jointly measurable process which is a martingale in its own natural filtration. I want to understand if there is condition in terms of the joint law of $(W,S)$ (on the canonical path space $\mathcal C([0,T])\times \mathbb D([0,T])$ ) which is sufficient for $\{S_t\}_{0 \leq t \leq T}$ to be $\mathbb F$ -adapted.
My first guess is that if $W$ is a martingale w.r.t. $\{F_t^W \vee F_t^S \}_{0\leq t\leq T}$ , then $\{S_t\}_{0 \leq t \leq T}$ is $\mathbb F$ -adapted. Namely I want to show that enlarge the Brownian filtration by including future information will destroy the martingale property of the brownian motion. I am wondering if this is true or not. Thanks!

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1. George Lowther says:
  
  1 March 17 at 1:28 PM
  
  Are you asking if any strict enlargement of the Brownian filtration (but keeping the same underlying probability space) will necessarily destroy the martingale property of the underlying Brownian motion?
  
  If so, then yes that is true. By the martingale representation theorem, all martingale would have to retain the martingale property, in which case the result follows.
  
  Reply
  1. yan says:
    
    1 March 17 at 3:05 PM
    
    Thank you for your prompt response. As a follow-up question, I am wondering if this is still true in the discrete time case where $W$ is a random walk with jumps that are i.i.d normal random variables(say,with mean zero). I am not aware of any martingale representation in discrete time, except the binomial-tree model. Thanks again!
    
    Reply
    1. George Lowther says:
      
      1 March 17 at 3:44 PM
      
      In the discrete time case you mention, it is possible to enlarge the filtration while keeping the martingale property. For example, add the absolute sizes of all the jumps to $F_0$ .
    2. yan says:
      
      1 March 17 at 4:34 PM
      
      I agree with you that the martingale property can be preserved by enlarging the filtration. However, in this case it seems that we may lose the independent increment property(i.e., $W_{n+1}-W_n$ is no longer independent of $\mathcal F_n$ ). Going back to my original question which involve the process $S$ . Is there a condition on the joint distribution of $(S,W)$ which will guarantee that $S$ is adapted to $\mathcal F^W$ ? Namely, can we find two processes $S_1$ and $S_2$ , one being adapted and one not, such that $(S_1,W)$ has the same distribution as $(S_2, W)$ ? As you pointed out this is not possible in the continuous time case.
      Thank you and looking forward to your reply!
    3. George Lowther says:
      
      2 March 17 at 9:51 AM
      
      On your question about the distribution of $(S,W)$ , I would consider the statement that S is W adapted to be a statement about this distribution, so I am not sure what further to say about it.
  2. yan says:
    
    2 March 17 at 3:38 PM
    
    what I am looking for is a condition on the joint distribution which can be expressed in terms of some equations. For example, that $W$ is $\mathcal F^{W,S}$ martingale is equivalent to saying $\mathbb E((W_{n+1}-W_n)f(W_1,...,W_n,S_1,...S_n))=0$ for all f. Unfortunately, this is not sufficient to give adaptedness as you pointed out. I am just curious if one can formulate condition of this kind which is sufficient. I hope this is a question that worth thinking. thank you!
    
    Reply
    1. George Lowther says:
      
      2 March 17 at 7:18 PM
      
      I think it should be enough to state that, for each $s < t$ , the distribution of $W_t$ conditional on $\mathcal F_s$ Is preserved. In the case of independent increments, it is enough to say that is preserved.
yan says:

28 February 17 at 10:01 PM

Sorry that I miss the initial space in very latex code. I re-typed the comment here:

Suppose $(\Omega,\mathcal F,\mathbb F,\mathbb P)$ is a probability space supporting a Brownian motion $\{W_t\}_{t=0}^T$ , where the filtration is the augmented brownian filtration $\{F_t^W\}$ and $latex\mathcal F=F_T^W$. Let $\{S_t\}_{0 \leq t \leq T}$ be a jointly measurable process which is a martingale in its own natural filtration. I want to understand if there is condition in terms of the joint law of $ latex(W,S)$ (on the canonical path space $ latex\mathcal C([0,T])\times \mathbb D([0,T])$) which is sufficient for $ latex\{S_t\}_{0 \leq t \leq T}$ to be $ latex\mathbb F$-adapted.
My first guess is that if latex$ W$ is a martingale w.r.t. the possibly larger filtration $ latex\{F_t^W \vee F_t^S \}_{0\leq t\leq T}$, then $ latex\{S_t\}_{0 \leq t \leq T}$ is $ latex\mathbb F$-adapted. Namely I want to show that enlarge the Brownian filtration by including future information will destroy the martingale property of the brownian motion. I am wondering if this is true or not. Thanks!

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yan says:

28 February 17 at 10:05 PM

So sorry i miss place the space again. Please ignore the previous comment. this time i should have got it right.

Suppose $(\Omega,\mathcal F,\mathbb F,\mathbb P)$ is a probability space supporting a Brownian motion $\{W_t\}_{t=0}^T$ , where the filtration is the augmented brownian filtration $\{F_t^W\}$ and $\mathcal F=F_T^W$ . Let $\{S_t\}_{0 \leq t \leq T}$ be a jointly measurable process which is a martingale in its own natural filtration. I want to understand if there is condition in terms of the joint law of $(W,S)$ (on the canonical path space $\mathcal C([0,T])\times \mathbb D([0,T])$ ) which is sufficient for $\{S_t\}_{0 \leq t \leq T}$ to be $\mathbb F$ -adapted.
My first guess is that if $W$ is a martingale w.r.t. $\{F_t^W \vee F_t^S \}_{0\leq t\leq T}$ , then $\{S_t\}_{0 \leq t \leq T}$ is $\mathbb F$ -adapted. Namely I want to show that enlarge the Brownian filtration by including future information will destroy the martingale property of the brownian motion. I am wondering if this is true or not. Thanks!

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Anonymous says:

17 July 17 at 5:54 AM

Hi!

You define a predictable sigma-algebra on {{\mathbb R}_+\times\Omega}, denoted by {\mathcal{P}} to be the sigma algebra generated by the left-continuous and adapted processes. You also write that a adapted process is one which in turn depends on the sigma algebra i.e that each r.v is measuable. This seems circular to me, do you mean that we take a left contious processes and then generate it’s natural filtration? Or have I gotten something wrong?

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Santo says:

12 January 18 at 6:10 PM

Assume that $L$ is a process of FV that is adapted to the filtration generated by a Wiener Process $W$ defined in a probability space $(\Omega, \mathcal{F}, P)$ and consider a, say, continuous function $f:[0,T]\to\mathbb{R}$ (I have f(t)=exp(-rt) with $r>0$ in mind). Let us define the stochastic process $X_t:=\int_0^t f(t)dL_t$. Is it the case that the expectation of $X_T$ with respect to $P$ is always well defined? Many thanks!

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sageandonions says:

25 April 18 at 6:45 AM

Hello. It occurs to me that the definition of natural filtration of a rv $X$ as the smallest filtration to which it is adapted is not obviously well-defined, because there is no total order on the set of filtrations. We can define a partial order by saying that one filtration $\mathscr F^1$ is less than or equal to another $\mathscr F^2$ iff $\forall t\geq 0:\ \mathscr F^1_t\subseteq \mathscr F^2_t$. It is imaginable that a rv would have multiple filtrations that are minimum elements of the poset of filtrations under that order. It seems to me that the simplest way to define a natural filtration is by the formula given in the sentence that follows the definition above, ie $\mathscr F^X_t \triangleq \sigma(X_s\ :\ s\leq t)$ . We would then wish to prove that filtration $\mathscr F^X$ is smaller than any filtration to which $X$ is adapted, under the above-defined order. That, together with the observation that $X$ is adapted to that filtration, would entail that filtration $F^X$ is a unique minimum element of the subset of all filtrations to which $X$ is adapted.

I hope this LaTeX works. It’s the first time I’ve tried using it in a WordPress environment 🙂

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1. TheBridge says:
  
  26 April 18 at 1:28 PM
  
  Hi in my opinion what you do in the second part of your post is exactly what it means in the definition of this blog for a filtration to be minimal (and hence the natural filtration w.r.t. the process X), and being minimal has nothing to do with a total order on the set of all filtrations. So even if what you say in the second part makes sense it is redundant with the notions used here and there is no problem of definition. Best regards
  
  Reply
  1. sageandonions says:
    
    27 April 18 at 12:16 AM
    
    It doesn’t require a total order but it does require a partial order and the existence of a minimal element of the set of adapted filtrations. Perhaps I missed it but I couldn’t see any definition of an order relation on filtrations above, which is why I suggested one (there is an implicit order relation used on sigma algebras, but not on filtrations). Having done that, it remains to be proved that the set of filtrations adapted to a process X has a minimal element under that order relation. I imagine one would do that by taking the filtration $(\mathscr F_t)_{t\in\mathbb R_+}$ such that $\mathscr F_t = \sigma(\{X_s\ :\ s\leq t\})$ and proving that it is a minimal element under that order relation.
    
    Reply
    1. TheBridge says:
      
      27 April 18 at 9:53 AM
      
      So far so good, you make explicit what is implicit in the post (but very clear from the context IMO). Nonetheless your precisions are exactly what the definition given here is about and I wouldn’t go as far as qualify the definition of natural filtration as not well defined (introducing pre-order is a bit overkill here). Best regards
xyz says:

7 December 18 at 6:49 PM

In the proof of Lemma 2, you reason as “In fact, as $\lvert f\rvert$ is bounded by $n^2$ , it has Lipschitz constant $2n^2$ .” I doubt that “boundedness” implies “Lipschitz”, the counterexample being $x^{1/2}$ on $[0,1]$ .

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1. George Lowther says:
  
  7 December 18 at 9:21 PM
  
  What I meant is that as f is bounded, its integral is Lipschitz.
  
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Mriganka Basu Roy Chowdhury says:

3 February 20 at 7:32 AM

Thanks for the helpful post! In Lemma 2, either I am missing something, or you don’t mention anywhere that the integral is adapted. It is clear though because adapted left continuous processes are progressive, and progressive processes have adapted integrals.

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1. George Lowther says:
  
  4 February 20 at 9:58 AM
  
  I stated that $X^n$ is adapted, but did not explicitly prove this, because I thought that it is clear. As you note, integrals of progressive processes are adapted. Alternatively, Riemann sum evaluation of the integral shows that it is adapted.
  
  Reply
Gabe says:

18 April 20 at 8:40 PM

Hi George,

I have a related question, that I was not yet able to answer. If we take $\mathbb{F}$ to be the filtration generated by a standard Brownian motion $W$ in $(\Omega, \mathcal{F}, \mathbb{P})$ and we define $\widetilde{W}_t=W_t+\int_0^t\Theta_u du$ , can we find an $\mathbb{F}$ -adapted $\theta$ such that the filtration generated by $\widetilde{W}$ is smaller than $\mathbb{F}$ ?

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1. George Lowther says:
  
  19 April 20 at 5:53 PM
  
  Hi. Nice question! Here is a partial answer:
  
  Let us start with a standard Brownian motion B generating filtration $\mathcal G_\cdot$ . Next, let S be any set which is not in $\mathcal G_0$ . For example, $S=\{B_1 > 0\}$ . Next, let $\mathcal F_t$ be the filtration generated by $\mathcal G_t$ and S. By Jacod’s countable extension, B is still a semimartingale under this enlarged filtration. As its quadratic variation is $[B]_t=t$ , we have $B_t=W_t+A_t$ for some Brownian motion W under the enlarged filtration and continuous FV process A. Furthermore, since $H\cdot A=H\cdot B-H\cdot W=0$ for any predictable process $H$ satisfying $\int \lvert H_t\rvert dt=0$ (a.s.), it follows that A is absolutely continuous and, hence, $A_t=\int_0^t\Theta_sds$ for a predictable $\Theta$ .
  
  So, in summary, we have a Brownian motion W and predictable $\Theta$ adapted to filtration $\mathcal F_t$ , such that $B=W+\int\Theta_s ds$ is a Brownian motion adapted to the strictly smaller filtration $\mathcal G_t$ . All that is missing is to show that $\mathcal F_t$ is generated by W. However, I do not think that this is the case, and S will not be in the natural filtration of W at any time before 1.
  
  Reply
2. George Lowther says:
  
  20 April 20 at 10:10 PM
  
  Hi. I have an answer to your question, which is yes, we can find such a $\Theta$ ! See this post.
  
  Reply
Matan says:

9 May 20 at 3:57 PM

Just a typo: I think that on the RHS of the main equality in the proof of lemma 3 the n’s in the superscript should be deleted.

I enjoy this site.

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1. George Lowther says:
  
  10 May 20 at 10:03 AM
  
  You are correct – I fixed it now. Thanks!
  
  Reply
Philipp Wacker says:

12 August 20 at 10:17 PM

Very small typo: “are automatically jointly measurability” [George: Fixed, Thanks!]

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RozaTh says:

8 October 21 at 3:09 AM

Hi, quick question. In the proof of Lemma 2, why $|f|$ is bounded by $n^2$ ? If $f = 1_{\{|X_{s\vee 0}|\le n\}}X_{s\vee 0}$ shouldn’t the bound be $n$ ?

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1. George Lowther says:
  
  25 February 22 at 1:29 PM
  
  We have $f=n1_{\{\lvert X_{s\vee 0}\rvert\le n\}}X_{s\vee0}$ . You are missing the factor of n.
  
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Julian says:

19 January 22 at 1:39 PM

Hi George, I have a question regarding your definition of optional sigma algebra. You define it as the sigma algebra generated by right-continuous adapted processes. But other references I found all define it as being generated by cadlag adapted processes. May I ask why you drop the left hand limit condition here? Does it not affect anything?

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1. George Lowther says:
  
  20 February 22 at 5:01 PM
  
  I dropped the left-hand limit condition here simply because it is easier in context (right-continuous processes generate the optional processes and left-continuous generate the predictable ones). Requiring the processes to have left-limits is a bit unintuitive and makes no practical difference. They generate the same sigma algebra in the case that the filtered probability space is complete, which I am assuming. See my post on optional processes.
  
  If the filtered probability space is not complete, then the two definitions could differ, but only by evanescent sets.
  
  Reply
H W says:

2 June 22 at 6:24 AM

Sorry, this is a bit late…You state that, “On the other hand, predictable processes are extensively used as integrands for stochastic integrals and in the Doob-Meyer decomposition, and are often not restricted to the adapted and left-continuous case.”. Does it mean that there are predictable processes that are not (merely) left-continuous and adapted? If so, can you please give an example?

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1. George Lowther says:
  
  14 June 22 at 2:39 PM
  
  Yes. For a rather dumb example, just take a deterministic measurable process. Eg, $X_t = 1_{\{t=1\}}$ .
  For slightly more interesting examples, take any progressively measurable process X and translate it to $Y_t = X_{t-h}$ for fixed positive h, so that its values are known a time h in advance.
  
  Reply
facet says:

8 March 23 at 7:39 AM

Hey George, this is late but I just found your site is very helpful. I’m get a bit confused from your proof in lemma 4. So you said optional algebra is generated by right continuous and adapted process, is this mean every optional process is right continuous and adapted? If not, what’s the logic behind section “optional to progressive”?

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1. George Lowther says:
  
  8 March 23 at 8:09 PM
  
  No, it means that optional processes are measurable with respect to the sigma-algebra generated by the right-continuous and adapted processes. This includes pointwise limits of sequences of such processes, limits of sequences of these limits and so on.
  
  Reply
  1. faceet says:
    
    9 March 23 at 6:52 AM
    
    Thank you for the clarification! So the argument is that an optional process is a limit of right-continuous and adapted processes, which are also progressive. Then the claim follows from the fact that a limit of progressive processes is still progressive. Is that right?
    
    Reply
    1. George Lowther says:
      
      26 March 23 at 12:08 PM
      
      almost…an optional process is not necessarily a limit of right-continuous processes. It could be a limit of limits, a limit of limits of limits, and so on. Either way, the fact that right-continuous adapted processes are progressive is enough to imply that optional processes are progressive
Harry says:

17 May 23 at 11:06 PM

In the proof of a process being optional implies it being progressive, should it be $\mathcal{B}([0,T])\otimes \mathcal{F}_t$ instead of $\mathcal{B}([0,\infty])\otimes \mathcal{F}_t$ ? Thank you in advance.

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1. George Lowther says:
  
  18 June 23 at 9:56 AM
  
  You mean in the proof of Lemma 4? Here, Y is defined for all times t >= 0, so is $\mathcal B(\mathbb R)\otimes\mathcal F_t$ -measurable. This immediately implies that restricted to the interval [0,T] (on which it equals X) it is $\mathcal B([0,T])\otimes\mathcal F_t$ -measurable.
  
  Reply
Anonymous says:

26 December 23 at 5:42 PM

In Lemma 2, you define the process $X_n^t$. You showed that this process in continuous, and converges to $X_t$. But don’t you also need to show that $X_n^t$ is adapted, i.e. $X_n^t \in \mathcal{F}_t$? It’s not clear, since the definition of $X_n^t$ involves taking an integral of $X$.

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1. Anonymous says:
  
  26 December 23 at 6:12 PM
  
  Actually I found Prop 1.13 in Karatzas and Shreve which says that every left (or right) continuous adapted process is progressively measurable. This is enough to show that the $X_n^t$ are adapted.
  
  Apologies for the formatting, I’m not sure how to get the latex to show up properly.
  
  Reply
  1. George Lowther says:
    
    4 January 24 at 10:35 AM
    
    X^n_t is an integral of X_s over s <= t which, by Riemann sum approximation, is a limit of linear combinations of X_s over s <= t (and linear combinations+limits preserves measurability), so is F_t-measurable.
    
    Reply
harfaw says:

3 January 24 at 10:06 AM

In Lemma 2 you conclude that $X$ is measurable w.r.t. $\sigma(X^n,n\ge1)$ whenever $X^n \rightarrow X$ . It’s not clear to me why this is true.

I can see that for any Bore set $B$ we have that $\cap_n \{ (t, \omega) | X_t^n(\omega) \in B \} \subset \{ (t, \omega) | X(\omega) \in B \}$ , if $X^n \rightarrow X$ means point-wise convergence on $\mathbb{R}_+ \times \Omega$ .

But you’ve used the other inclusion, i.e. $\cap_n \{ (t, \omega) | X_t^n(\omega) \in B \} \supset \{ (t, \omega) | X(\omega) \in B \}$ , in order to conclude that $X$ is measurable w.r.t. $\sigma(X^n,n\ge1)$ . Why is this inclusion true?

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1. George Lowther says:
  
  4 January 24 at 10:30 AM
  
  Hi. X^n are adapted and converge to X pointwise, so X is also adapted. This is just a standard result that the pointwise limit of a set of random variables measurable wrt a sigma-algebra is itself measurable.
  
  Note in the following post, I mention that the supremum is measurable. So
  lim_n X^n = inf_{m}sup_{n > m}X^n
  is also measurable.
  
  Reply

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