A mode of convergence on the space of processes which occurs often in the study of stochastic calculus, is that of *uniform convergence on compacts in probability* or *ucp convergence* for short.

First, a sequence of (non-random) functions converges uniformly on compacts to a limit if it converges uniformly on each bounded interval . That is,

(1) |

as .

If stochastic processes are used rather than deterministic functions, then convergence in probability can be used to arrive at the following definition.

Definition 1A sequence of jointly measurable stochastic processes converges to the limit uniformly on compacts in probability if

as for each .

The notation is sometimes used, and is said to converge ucp to . Note that this definition does not make sense for arbitrary stochastic processes, as the supremum is over the uncountable index set and need not be measurable. However, for right or left continuous processes, the supremum can be restricted to the countable set of rational times, which will be measurable. In fact, for jointly measurable processes, it can be shown that the supremum is measurable with respect to the completion of the probability space, so ucp convergence makes sense. However, that is not needed for these notes.

For each time , the following pseudometric can be defined on the space of locally bounded deterministic processes

Then, uniformly on compacts if for each . This shows that uniform convergence on compacts is in fact given by the following metric

Similarly, a sequence of stochastic processes converges ucp to if in probability. By bounded convergence, this is equivalent to tending to zero. So, ucp convergence is given by the following metric.

(2) |

If a sequence of processes converges ucp, then it is always possible to pass to a subsequence for which uniform convergence on compacts holds with probability one on the sample paths. This is a simple application of the Borel-Cantelli lemma, but allows properties of ucp convergence to be inferred from corresponding properties of uniform convergence on compacts.

Theorem 2The space of cadlag (resp. continuous) adapted processes is complete under ucp convergence.

Furthermore, if then there is a subsequence whose sample paths almost surely converge to those of uniformly on compacts.

*Proof:* Let be a Cauchy sequence under ucp convergence, so that as . Then, there is a subsequence satisfying for all . In this case,

so that is almost surely finite. Restricting to a set of probability one if necessary, we suppose that this is always finite. Then, the sample paths of are Cauchy convergent under uniform convergence on compacts, and there is a limit . As , this will be measurable, so is a stochastic process and is adapted whenever are adapted. If the processes have left limits, then it is clear that and, therefore, the left limits are also Cauchy under uniform convergence on compacts.

As limits can be commuted with uniform convergence of sequences, if the processes are cadlag then,

So, is cadlag and under uniform convergence on compacts. In particular, is continuous whenever are.

To complete the proof, it just remains to show that the original sequence does indeed converge ucp to . As it follows that this converges in probability, so . Then,

as required. ⬜

A consequence of Doob’s martingale inequalities is that convergence of martingales implies ucp convergence.

Lemma 3Let be a sequence of cadlag martingales, and be a process such that as .

Then, is a martingale and has a cadlag version which is the ucp limit of .

*Proof:* Clearly, by -convergence, is a martingale. Furthermore, Doob’s martingale inequality shows that

as for all . Consequently, the sequence is Cauchy under ucp convergence and has a cadlag limit . As (convergence in probability), this limit is a cadlag version of . ⬜

In particular, the space of continuous martingales is complete under convergence.

Corollary 4Let be a sequence of continuous martingales such that as for all and some process . Then, is a martingale and has a continuous version.

*Proof:* By the previous lemma, is a martingale and has a cadlag version such that . Then, by Theorem 2, this limit is continuous. ⬜

#### The semimartingale topology

An even stronger topology than ucp convergence is the *semimartingale topology*. A sequence of cadlag and adapted processes converges to under this topology if, for every sequence of elementary predictable processes with and every , the limit

holds, in probability. Then, a sequence of cadlag adapted processes converges to a limit under the semimartingale topology if converges to zero. As with ucp convergence, this can be described by a metric. For each set

Then, if . It follows that the semimartingale topology is defined by the metric

(3) |

The semimartingale topology is indeed stronger than ucp convergence.

Theorem 5If are cadlag adapted processes with converging to in the semimartingale topology then .

*Proof:* If is a cadlag adapted process and , consider the stopping times

for each positive integer . On the event that , by right-continuity we have for sufficiently large . Hence,

for any real . However, the processes are elementary and . So,

It follows that if in the semimartingale topology then

as , and ucp convergence holds. ⬜

Lemma 3 above states that convergence of martingales implies ucp convergence. In fact, the stronger property of semimartingale convergence holds, although I do not give a proof here (the reader should be able to prove this fact using the martingale inequality established in the later post on martingales as integrators).

Completeness of ucp convergence can be used to also prove completeness of the semimartingale topology.

Lemma 6The space of cadlag and adapted processes is complete under the semimartingale topology.

*Proof:* Let be Cauchy under the semimartingale topology. By Theorem 5 this is also Cauchy under ucp convergence, so has a limit which is cadlag and adapted.

As in probability for each time , it follows that in probability, for all elementary processes . So, setting ,

Taking the supremum over all such elementary processes gives

which, by Cauchy convergence, goes to zero as . ⬜

Finally, the semimartingale topology is *not* a vector topology on the space of cadlag adapted processes. For that to be the case, we would need for all such processes and sequences of real numbers . However, this says that in probability for all elementary processes . Equivalently, for each , the set

is bounded in probability. The processes for which this holds are precisely the *semimartingales*, although I do not prove that here. On these processes, semimartingale convergence is indeed a vector topology.

Hi just to mention that in the second paragraph of the proof of lemma 6 I believe you mean :

“So, setting ” and not “So, setting ”

Regards

Fixed. Thanks for pointing that out.

Hi,

Would you be so kind to give a reference or sketch the poof of your statement:

” Lemma 3 above states that convergence of martingales implies ucp convergence. In fact, the stronger property of semimartingale convergence holds, although I do not prove that fact here. ”

Thanks for your efforts!

It follows from equation (8) in my post “Martingales are integrators”. If is a sequence of martingales with then this shows that for elementary processes bounded by 1, tends to 0 in probability.

I had a question: also the space of càglàd processes is complete under the ucp topology? In other words, if we use the metric defined above in (2), do we obtain that the spec of càglàd processes is complete with this metric?

We do. This is because caglad processes are closed under pathwise uniform convergence.

Why do you write instead of for the metric?

That was just to emphasise that it is translation invariant, so is a function of the difference, .

Hello George, thank you very much for the post! I have one short question: Theorem 2 (and in fact, probably all of the post (?)) seems to hold also for processes with values in, say, general Banach spaces, right?

You probably want separability of the Banach space to avoid measurability issues but, otherwise, yes the results should hold in Banach spaces.

Hello George.

Thank you for your wonderful post.

But I am still struggling to get a reason why the first inequality holds. If you tell me this it could help me. Thank you.

the first inequality as in Theorem 5 *

There was a mistake, I fixed. Hopefully it is clear now. Thanks!

I wonder where that “uniform convergence on compacts” metric comes from?