Continuous local martingales are a particularly well behaved subset of the class of all local martingales, and the results of the previous two posts become much simpler in this case. First, the continuous local martingale property is always preserved by stochastic integration.

Theorem 1IfXis a continuous local martingale and isX-integrable, then is a continuous local martingale.

*Proof:* As *X* is continuous, will also be continuous and, therefore, locally bounded. Then, by preservation of the local martingale property, *Y* is a local martingale. ⬜

Next, the quadratic variation of a continuous local martingale *X* provides us with a necessary and sufficient condition for *X*-integrability.

Theorem 2LetXbe a continuous local martingale. Then, a predictable process isX-integrable if and only if

for all .

*Proof:* If is *X*-integrable then the quadratic variation is finite. Conversely, suppose that *V* is finite at all times. As *X* and, therefore, [*X*] are continuous, *V* will be continuous. So, it is locally bounded and as previously shown, is X-integrable. ⬜

In particular, for a Brownian motion *B*, a predictable process is *B*-integrable if and only if, almost surely,

for all . Then, is a continuous local martingale.

Quadratic variations also provide us with information about the sample paths of continuous local martingales.

Theorem 3LetXbe a continuous local martingale. Then,

Xis constant on the same intervals for which [X] is constant.Xhas infinite variation over all intervals on which [X] is non-constant.

*Proof:* Consider a bounded interval (*s*,*t*) for any , and set for *k*=0,1,…,*n*. By the definition of quadratic variation, using convergence in probability,

where *V* is the variation of *X* over the interval (s,t). By continuity, tends uniformly to zero as *n* goes to infinity, so and [*X*] is constant over (*s*,*t*) whenever the variation *V* is finite. This proves the second statement of the theorem, which also implies that [*X*] is constant on all intervals for which *X* is constant.

It only remains to show that whenever . Applying this also to the countable set of rational times *u* in (*s*,*t*) will then show that *X* is constant on this interval whenever [*X*] is.

The process is a local martingale constant up until *s*, with quadratic variation for . Then is a stopping time with respect to the right-continuous filtration and, by stopping, is a local martingale with zero quadratic variation . Then, as previously shown, is a martingale and, therefore, . This shows that almost surely. Finally, on the set , we have and, hence, . ⬜

Theorem 3 has the following immediate consequence.

Corollary 4Any continuous FV local martingale is constant.

*Proof:* By the second statement of Theorem 3, the quadratic variation [*X*] is constant. Then, by the first statement, *X* is constant. ⬜

The quadratic covariation also tells us exactly when *X* converges at infinity.

Theorem 5LetXbe a continuous local martingale. Then, with probability one, the following both hold.

- exists and is finite whenever .
- and whenever .

*Proof:* By martingale convergence, with probability one either exists and is finite or and are both infinite. It just remains to be shown that, with probability one, exists if and only if is finite..

Let . Then, is a local martingale with quadratic variation bounded by *n*. So, and is an -bounded martingale which, therefore, almost surely converges at infinity. In particular, on the set

we have outside of a set of zero probability. Therefore, almost surely exists on

For the converse statement, set . Then, is a local martingale bounded by *n* and . Hence, is almost surely finite and is finite on the set

outside of a set of zero probability. Therefore, is almost surely finite on the set

⬜

Theorems 3 and 5 are easily understood once it is known that all local martingales are random time-changes of standard Brownian motion, as will be covered in a later post.

The topology of uniform convergence on compacts in probability (ucp convergence) was introduced in a previous post, along with the stronger semimartingale topology. On the space of continuous local martingales, these two topologies are actually equivalent, and can be expressed in terms of the quadratic variation. Recalling that semimartingale convergence implies ucp convergence and that quadratic variation is a continuous map under the semimartingale topology, it is immediate that the first and third statements below follow from the second. However, the other implications are specific to continuous local martingales.

Lemma 6Let andMbe continuous local martingales. Then, asngoes to infinity, the following are equivalent.

- converges ucp to
M.- converges to
Min the semimartingale topology.- in probability, for each .

*Proof:* As semimartingale convergence implies ucp convergence, the first statement follows immediately from the second. So, suppose that . Write and let be the first time at which . Ucp convergence implies that tends to infinity in probability, so to prove the third statement it is enough to show that tends to zero in probability. By continuity, the stopped process is uniformly bounded by 1, so is a square integrable martingale, and Ito’s isometry gives

as *n* goes to infinity. The limit here follows from the fact that is bounded by 1 and tends to zero in probability. So, we have shown that tends to zero in the norm and, hence, in probability.

Now suppose that the third statement holds. This immmediately gives in probability. Letting be the first time at which and be elementary predictable processes, Ito’s isometry gives

So, in particular, in probability. Finally, as whenever , which has probability one in the limit , this shows that tends to zero in probability and tends to zero in the semimartingale topology. ⬜

Applying the previous result to stochastic integrals with respect to a continuous local martingale gives a particularly strong extension of the dominated convergence theorem in this case. Note that this reduces convergence of the stochastic integral to convergence in probability of Lebesgue-Stieltjes integrals with respect to .

Theorem 7LetXbe a continuous local martingale and , beX-integrable processes. Then, the following are equivalent.

- converges ucp to .
- converges to in the semimartingale topology.
- in probability, for each .

*Proof:* This follows from applying Lemma 6 to the continuous local martingales and . ⬜

Theorem 7 also provides an alternative route to constructing the stochastic integral with respect to continuous local martingales. Although, in these notes, we first proved that continuous local martingales are semimartingales and used this to imply the existence of the quadratic variation, it is possible to construct the quadratic variation more directly. Once this is done, the space of *X*-integrable processes can be defined to be the predictable processes such that is almost surely finite for all times *t*. Define the topology on so that if and only if in probability as for each *t*, and use ucp convergence for the topology on the integrals . Then, Theorem 7 says that is the unique continuous extension from the elementary integrands to all of .

Hi, nice blog!

In Theorem 5, what do you mean by [X]_{\infty} < \infty. That the quadratic variation is uniformly bounded by a constant C?

Otherwise your arguments would probably not work? If [X] would only be a.s. finite, then an "n" such that [X] \leq n a.s. would not exist (example: normal random variable takes on only finite values and is therefore a.s. finite but there is no upper bound to the values it takes)…

Is that the way to understand your statement?

Cheers

Chris

Comment by Chris — 12 April 10 @ 8:54 PM |

No, all that matters is that [X]_\infty < n for some n. n is not fixed, so uniform boundedness is not needed. You don't even need [X]_\infty to be almost surely finite. Even if it is only finite on a set of probability p < 1, X will converge on that set (up to a zero probability set).

Comment by George Lowther — 12 April 10 @ 9:38 PM |

Ok, still not sure whether I can follow you.

When [X]_\infty < n for some n and this holds for all \omega in the sample space then it is uniformly bounded by a constant.

When you mean [X]_\infty(\omega) < n(\omega), that is for every element \omega in the sample space you find an n such that [X]_\infty is bounded by n on that given \omega only then the subsequent estimate E[(X^{\tau_n}_t)^2] \leq n does not hold anymore. That estimate would only hold if you have on the right of the estimate something like the ess sup n(\omega), so basically the essential supremum of n over all elements in the sample space. However nothing gurantees you, that this supremum even exists….

Comment by Chris — 13 April 10 @ 12:01 PM |

I think you’re still misunderstanding my argument. There is no need to be thinking about essential supremums. I’ll try modifying the argument and proof to make it clearer. I’m not able to do this right now as I’m away from my computer (just replying by mobile). Hopefully will have time later tonight.

For now, the theorem could be stated more precisely as follows.

then, outside of a set of probability zero, exists and is finite on A, and outside of A.

Comment by George Lowther — 13 April 10 @ 2:10 PM |

Hey,

a lot clearer now thanks….

Comment by Chris — 17 April 10 @ 11:12 AM |

Update:I have added a couple of extra results to this post. Lemma 6 shows that ucp convergence, semimartingale convergence and convergence in probability of quadratic variations all coincide. Theorem 7 uses this to give a much stronger version of the dominated convergence theorem for continuous local martingales.Comment by George Lowther — 13 May 11 @ 11:39 PM |

Hi George,

I have been looking for an answer to the following question on continuous local martingales:

Can one construct a non-zero (in sense of having P>0 of being nonzero) continuous local martingale which is identically equal to $0$ P-a.e. at a fixed time $T$? It has to do with an option hedging problem I am working on.

Thanks in advance!

Tigran

Comment by Anonymous — 28 May 12 @ 9:27 AM |

Yes! Check out my example of a continuous local martingale which is not a proper martingale from my notes.

Comment by George Lowther — 28 May 12 @ 12:25 PM |

[…] with respect to $M$, iff $mathbb{E}left[int_0^t frac{1}{X^2_s}d[M]_sright]0$ (see e.g. Thm 2 in https://almostsure.wordpress.com/2010/04/01/continuous-local-martingales/). Taking e.g. $M_t$ as Brownian motion we have $[M]_t = t$. Now it is known that the second inverse […]

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