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 extend to to give a bound for in terms of . Instead, we have the rather weaker third statement above, which requires to be integrable in order to give a nontrivial upper bound. In fact, there exist martingales *X* such that the maximum has infinite expectation at all positive times even though, by definition, is finite.

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 .
- .

I briefly note that the third inequality looks a bit odd, as it is not dimensionally consistent. This means that, unlike the other two, applying it to for positive *a* gives a slightly different inequality. Specifically, applying it to and dividing through by *a* gives

where *c* is equal to . The optimal value occurs at giving

This is dimensionally consistent — replacing *X* by scales both sides by *a*. Furthermore, as is bounded above by 1, it is stronger in this form than as stated in Theorem 2. However, the versions stated above are simpler, and the important thing is that they have the same coefficient for the dominant term .

To prove Theorem 2, we start with the following submartingale inequality from which each of Doob’s inequalities will follow.

Lemma 3Let be a nonnegative cadlag submartingale. Then, for each ,

(1)

*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 a positive value ,

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

Taking expectations

and (1) follows by letting *L* increase to *K*. ⬜

We use (1) to prove Doob’s inequalities.

*Proof of Theorem 2:* Bounding the right hand side of (1) by gives the first inequality. Next, for , multiply by and integrate up to a 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.

Now, multiply both sides of (1) by and integrating over a range for any ,

using the notation . As we will show in a moment, the inequality

(2) |

holds for nonnegative *x* and *y*. Adding to both sides of the previous inequality and applying (2) with and ,

Subtract from both sides and take the limit to get the third inequality of the theorem.

It only remains to show that (2) holds for all nonnegative reals *x* and *y*. Moving all the terms to the same side, the inequality is equivalent to

By differentiating with respect to *x*, the minimum of the left hand side occurs at , at which it takes the value , proving (2) ⬜

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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