Martingale inequalities are an important subject in the study of stochastic processes. The subject of this post is Doob’s inequalities which bound the distribution of the maximum value of a martingale in terms of its terminal distribution, and is a consequence of the optional sampling theorem. We work with respect to a filtered probability space . The absolute maximum process of a martingale is denoted by . For any real number , the -norm of a random variable is

Then, Doob’s inequalities bound the distribution of the maximum of a martingale by the -norm of its terminal value, and bound the -norm of its maximum by the -norm of its terminal value for all .

Theorem 1Let be a cadlag martingale and . Then

- for every ,
- for every ,

We can define a topology on the space of cadlag martingales so that a sequence of martingales converges to a limit if as . It is clear that for any process and, consequently Doob’s second inequality above shows that this topology is equivalent to the seemingly much stronger condition that , for any .

The second statement above does not generalize to and, in fact, there exist martingales such that the maximum has infinite expectation at all positive times.

Doob’s martingale inequalities are a consequence of the following inequalities applied to the submartingale .

Theorem 2Let be a nonnegative cadlag submartingale. Then,

- for each .
- for each .

*Proof:* By completing the filtration if necessary, without loss of generality we assume that it is complete. Consider the first time at which the process reaches the positive value ,

which, by the debut theorem, is a stopping time. Then whenever and, in particular, . Optional sampling gives

Take expectations

Bounding the right hand side above by gives the first inequality. Multiply by and integrate up to some limit to get,

Setting so that , Hölder’s inequality states that

Substituting into the previous inequality,

Finally, cancel from both sides and take the limit to get the second inequality in the statement of the theorem. ⬜

This article was really helpful, thx.

PS: in the last line the constant p/(p-1) is missing…

Comment by Mike — 15 February 12 @ 12:12 PM |

Fixed. Thanks.

Comment by George Lowther — 22 February 12 @ 1:00 AM |

Hi George,

Just a small comment:

I think one needs to first consider a finite time grid to be able to assume that {X_t^* >= K} = {\tau <= t} (second line in proof of theorem 2) and the use MCT to get the result for countable time and continuous time in the cadlag case, like you mentioned in the planetmath proof.

Best,

Tigran

Comment by Anonymous — 1 March 12 @ 11:06 AM |

Hi. You can do it that way, but it is not necessary to start by restricting to the finite case. I already proved the Debut theorem and optional sampling for continuous-time cadlag processes. As we assume the process is cadlag, these can be applied directly to the continuous time case.

Comment by George Lowther — 1 March 12 @ 11:50 PM |

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