One of the most fundamental and useful results in the theory of martingales is Doob’s maximal inequality. Use to denote the running (absolute) maximum of a process X. Then, Doob’s maximal inequality states that, for any cadlag martingale or nonnegative submartingale X and real ,
with . Here, denotes the standard Lp-norm, .
An obvious question to ask is whether it is possible to do any better. That is, can the constant in (1) be replaced by a smaller number. This is especially pertinent in the case of small p, since diverges to infinity as p approaches 1. The purpose of this post is to show, by means of an example, that the answer is no. The constant in Doob’s inequality is optimal. We will construct an example as follows.
Example 1 For any and constant there exists a strictly positive cadlag -integrable martingale with .
For X as in the example, we have . So, supposing that (1) holds with any other constant in place of , we must have . By choosing as close to as we like, this means that and is indeed optimal in (1).
A natural place to start in trying to construct examples for which (1) is close to equality, is with the methods of the previous post. There, it was shown how to construct martingales with specified terminal distribution and with the maximum possible law for the maximum. In that post, was used to denote the running maximum of X rather than the absolute maximum but, as I will just construct nonnegative examples here, this is of no consequence.
As in the construction of cadlag martingales given in the previous post, start by choosing a (nonnegative) integrable and decreasing function . Its running average is
For a random variable U, uniformly distributed on and defined with respect to a probability space , we construct a cadlag process as
This is a martingale under its natural filtration, and has the terminal and maximum values,
Now, it is simple enough to guess at a form for h. Choosing , this is decreasing for and integrable for . For X to be -integrable, we need to be integrable, which is true whenever . The running average is
So, and, . The requirement of the example is satisfied so long as or, equivalently, . The condition that implies that , so X is an -integrable martingale.
The example constructed above is of a non-continuous martingale. It is not much more difficult to construct a continuous example. As described in the previous post, we can let B be a standard Brownian motion with initial value and stopped at the first time for which . For the example above, this is the first time at which . This gives a uniformly integrable martingale defined over with . Applying a deterministic time change gives a martingale defined over .
The post on Doob’s inequalities also included the inequality
However, it is easy to see that this is optimal. For any and , consider a martingale such that with probability and with probability . Then, (2) is an equality.