I start these notes on stochastic calculus with the definition of a continuous time stochastic process. Very simply, a stochastic process is a collection of random variables defined on a probability space . That is, for each time , is a measurable function from to the real numbers.

Stochastic processes may also take values in any measurable space but, in these notes, I concentrate on real valued processes. I am also restricting to the case where the time index runs through the non-negative real numbers , although everything can easily be generalized to other subsets of the reals.

A stochastic process can be viewed in either of the following three ways.

- As a collection of random variables, one for each time .
- As a path
one for each . These are referred to as the

*sample paths*of the process. - As a function from the product space

As is often the case in probability theory, we are not interested in events which occur with zero probability. Stochastic process theory is no different, and two processes are said to be *indistinguishable* if there is an event of probability one such that for all and all . This is the same as saying that they almost surely (i.e., with probability one) have the same sample paths. Alternative language which is often used is that and are *equivalent up to evanescence*. In general, when discussing any properties of a stochastic process, it is common to only care about what holds up to evanescence. For example, if a processes has continuous sample paths with probability one, then it is referred to as a continuous process and we don’t care if it actually has discontinuous paths on some event of zero probability.

It is important to realize that even if we have two processes satisfying almost surely, at each time , then it does not follow that they are indistinguishable. As an example, consider a random variable uniformly distributed over the interval , and define the process . For each time , . However, is not indistinguishable from the zero process, as the sample path always has one point at which takes the value . The problem here is the uncountability of nonnegative real numbers used for the time index. By countable additivity of measures, if almost surely, for each , then we can infer that the sample paths of and are almost surely identical on any given countable set of times , but cannot extend this to uncountable sets.

This necessitates a further definition. A process is a *version* or *modification* of if, for each time , . Alternative language sometimes used is that and are *stochastically equivalent*.

Whenever a stochastic process is defined in terms of its values at each individual time , or in terms of its joint distributions at finite times, then replacing it by any other version will still satisfy the definition. It is therefore important to choose a good version. Right-continuous (or left-continuous) versions are often used when possible, as this then defines the process up to evanescence.

Lemma 1Let and be right-continuous processes (resp. left-continuous processes) such that almost surely, at each time . Then, they are indistinguishable.

*Proof:* By countable additivity, the set has probability one. By right-continuity (resp. left-continuity) of the sample paths, it follows that and have the same paths for all . ⬜

As an example, the definition of a standard Brownian motion consists of two parts. First, and, for , is normal with mean 0 and variance , independently of , which defines its finite distributions. The additional condition that it has continuous sample paths is required in order to select the correct version up to evanescence.

Viewing a stochastic process in the third sense mentioned above, as a function on the product space , it is often necessary to impose a measurability condition. The process is said to be *jointly measurable* if it is measurable with respect to the product sigma-algebra . Here, denotes the Borel sigma-algebra. Another benefit of choosing left or right-continuous versions is that it ensures joint measurability.

Lemma 2All right-continuous and left-continuous processes are jointly measurable.

*Proof:* If is right-continuous, then it is the limit of the following sequence of processes

Clearly are jointly measurable functions and, therefore, is also jointly measurable.

The proof for left-continuous processes follows along the same lines, but using in place of . ⬜

Finally, it is often useful to sample a stochastic process at a random time. The value of a process at the measurable time is . This need not even be a measurable quantity unless a good version of the process is used. Fortunately, joint measurability of is enough.

Lemma 3If is jointly measurable and is a random time then is measurable.

*Proof:* Using the functional monotone class theorem, it is only necessary to prove the result for processes with and . However, is clearly measurable. ⬜

Dear Almost sure,

It is in fact a nicely written note.

In Lemma 2 above, I am wondering if you need to add an additional assumption.

For instance, Let is non-measurable random variable w.r.t .

Setting process , then is not jointly measurable function, although it is continuous.

Best

songqsh

Comment by songqsh — 27 April 10 @ 1:21 AM |

Thanks for your comment. If X is a stochastic process, then it is implicit that X_t is a random variable (i.e., it is measurable) for each time t. This rules out your example.

I’ll come back and try to clarify the statement when I have some time.

Comment by George Lowther — 27 April 10 @ 3:58 AM |

Thank you for quick response. It clarifies my question.

Comment by songqsh — 27 April 10 @ 6:30 AM |

Dear Almost Sure,

I have recently discovered your blog and wanted to congratulate you warmly for this work.

Regarding Lemma 3, as you have proven Joint Measurability is sufficient condition for to be well defined random variable.

Is there any reason why you didn’t mention progressive measurability, which is, as far as I know, the sharpest condition for this random variable to make sense.

Best Regards and “encore bravo”

Comment by TheBridge — 16 June 10 @ 10:31 AM |

TheBridge – Welcome

Actually, progressive measurability implies joint measurability. So, joint measurability is a weaker condition, giving a stronger result.

Besides, progressive measurability only makes sense when you have a filtration and is only really needed when is a stopping time. In that case, you get the stronger conclusions that the stopped process is progressive (hence jointly measurable and adapted) and that is -measurable.

See also my later posts about this (Filtrations and adapted processes, Stopping times and the debut theorem and Sigma algebras at a stopping time).

Regards,

George

Comment by George Lowther — 17 June 10 @ 12:56 PM |

Dear Dr.Lowther,

thank you for these notes. However, I have a question. What should we require from a function f so that the process f(X_t) is progressively measurable if X_t is progressively measurable? Is it a well-known result?

Thank you in advance.

Sincerely,

Alex.

Comment by Alex — 13 March 11 @ 2:08 AM |

As compositions of measurable functions are measurable, you just need

fto be Borel measurable.Comment by George Lowther — 13 March 11 @ 2:49 AM |

Dear Dr.Lowther,

as usual I use your notes to undertstand concepts that are not well-written in books. I’m trying to write a complete proof of Lemma 3 and I have a question. If we look at the Functional Monotone Class Theorem on PlanetMath then using the same notation K – the space of simple functions, but what is H then?

Thank you.

Sincerely,

Alex.

Comment by Alex — 27 August 11 @ 5:37 AM |

Hi. Try taking

Hto be the space of all processesXsuch that is measurable. Then,Hcontains the simple functions and is closed under taking linear combinations and taking pointwise limits of sequences. The monotone class theorem implies thatHcontains all jointly measurable processes. Strictly speaking, directly applying the monotone class theorem tells you thatHcontains all bounded and jointly measurable processes, but extending to unbounded ones is not difficult (by taking limits of bounded processes or looking at ).Comment by George Lowther — 27 August 11 @ 1:28 PM |

Dear Almost Sure

I found this nice site accidentally. I really appreciate.

In your simple proof of Lemma 1, you used a countable set for t. Is here a good place to say something about continuity issues like separable process?

Thanks a lot.

Yeev

Comment by Des Yeev — 22 November 11 @ 9:44 PM |

Hi.

I was really only looking at processes with nice modifications here, such has right-continuous processes. As you need to have measurable quantities, and we deal with countably additive measures, it is natural to try and look at quantities which are defined over a finite set of times. Using right-continuity allows you to express quantities in terms of the restriction to a countable index set. This also seems to be the idea behind the notion of separable processes (Springer Online Reference), although it is not really something that I have really looked at before. Certainly, the right-continuous processes are separable relative to the closed subsets of R. Is there much of a gain in formalizing the more general concept?

George

Comment by George Lowther — 22 November 11 @ 11:39 PM |

I still learning about stochastic theory, this are being helpful a lot to me.

Comment by Marcelo Fernandes — 31 May 12 @ 6:24 AM |

Glad these notes are helpful, and thanks for the comment!

Comment by George Lowther — 6 June 12 @ 2:24 AM |

Hi, this is probably obvious but just occurred to me recently, while looking at a problem concerning indistinguishability but given two any two processes how do we know that the set

{X_t=Y_t, for all t>=0} is measurable? It is obviously the intersection of the measurable sets {X_t=Y_t} for all t>= but this is of course an uncountable intersection. Would appreciate any feedback. By the way your blog is excellent and handier than Revuz and Yor. Also are you are interested in non-linear expectations, BSDEs and risk measures and thinking of adding some notes of these topics to your blog?

Comment by Marco — 11 July 12 @ 8:56 PM |

The set {X

_{t}= Y_{t}} need not be measurable. However, the property of processes being indistinguishable still makes sense. You can take measurability of the set (w.r.t. the completion of the probability space) as being part of the definition. So, if the set is not measurable, then the processes are not indistinguishable. Also, as long as both X and Y are jointly measurable then the set will be measurable.I haven’t really thought much about the topics you mention for this blog. However, I’m not sure what I might add in the future, so maybe, at some point.

Comment by George Lowther — 13 July 12 @ 12:05 AM |

Dear Almost Sure,

I am looking for conditions that guarantee the existence of a Stochastic Process of given FDDs that is measurable with respect to the product of the sigma-algebra generated by its cylindrical sets and the Borel algebra on the real line. Continuity in probability suffices for the existence of a measurable SP but not necessarily with the respect to this product of sigma-algebras.

I am particularly interested in the Normal case. Given an autcovariance function that is continuous at the origin, does there always exists a centered stationary Normal stochastic process of this autocovariance function that is measurable with respect to the product of the cylindrical sigma-algebra and the Borel algebra on the real line? If not, are you away of any sufficient conditions that guarantee this?

Comment by Amos Lapidoth — 20 July 12 @ 9:47 AM |

Hi,

Sorry, I don’t have any references for this. I would guess that some kind of uniform continuity in probability should be enough (but I don’t have a reference for that statement). Consider choosing a sequence ε

_{n}of positive reals tending to zero fast enough that X(t_{n}) → X(t) almost surely, for any sequence of times with |t_{n}– t| ≤ ε. Then approximate X by the piecewise constant process X^{n}(t) = X(ε_{n}[t/ε_{n}]). Then takes limits as n goes to infinity to get a jointly measurable modification of X.Comment by George Lowther — 6 September 12 @ 2:02 AM |

Reblogged this on Being simple.

Comment by Marcelo de Almeida — 23 August 13 @ 5:39 AM |

Dear George, does it also makes sense to say that two real-valued random variables are indistinguishable if they are almost surely equal?

Comment by Dimas Abreu Dutra — 5 February 14 @ 8:13 PM |

Hi George – I am new to Stochastic Calculus. I have basic knowledge of probability and statistics. However, I find it difficult to understand term like probability space, measure etc. Are there any pre-requisites?

Comment by Quasar — 3 November 14 @ 11:23 AM |

[…] $[0,T]$, and $W$ the vector space of continuous adapted processes of bounded variation, mod indistinguishability, with pointwise multiplication and the partial order $X le Y$ iff $X_t le Y_t$ for all $t$ almost […]

Pingback by Cauchy-Schwarz inequality for bilinear forms valued in an abstract vector space - MathHub — 15 March 16 @ 11:21 PM |

Hi George, I have two questions and need your help. First in Lemma 1, X and Y are right-continuous process, and for each t, X almost surely equals Y, but t is uncountable, so why the set A has the probability 1 and what’s the function of right-continuous version here? Second, in Lemma 2, I don’t understand the meaning of ‘measurable w.r.t. product sigma-algebra’, can you explain it for me? looking forward to your reply, and thank you very much!

Comment by Anonymous — 2 December 16 @ 4:57 AM |

1) The point of restricting the statement to right-continuous (or left-continuous) processes is that, for X to equal Y everywhere, it is only necessary to show that this holds on the rationals. Taking right-limits extends the equality to irrational t. And, the rationals are countable, so the processes will be equal on the rationals, almost surely.

2) We have two sigma algebras. The Borel sigma algebra on the positive reals , and the sigma algebra on the underlying probability space. A (real) stochastic process is a map . measurable w.r.t. the product sigma algebra means measurable with respect to in the domain and in the codomain.

Comment by George Lowther — 9 December 16 @ 2:40 AM |