It is a consequence of Doob’s maximal inequality that any -integrable martingale has a maximum, up to a finite time, which is also -integrable for any . Using to denote the running absolute maximum of a cadlag martingale *X*, then is -integrable whenever is. It is natural to ask whether this also holds for . As martingales are integrable by definition, this is just asking whether cadlag martingales necessarily have an integrable maximum. Integrability of the maximum process does have some important consequences in the theory of martingales. By the Burkholder-Davis-Gundy inequality, it is equivalent to the square-root of the quadratic variation, , being integrable. Stochastic integration over bounded integrands preserves the martingale property, so long as the martingale has integrable maximal process. The continuous and purely discontinuous parts of a martingale *X* are themselves local martingales, but are not guaranteed to be proper martingales unless *X* has integrable maximum process.

The aim of this post is to show, by means of some examples, that a cadlag martingale need not have an integrable maximum.

#### Example 1

For the first example, I will construct a martingale along the same lines as that used in the post on the maximum maximum of martingales with known terminal distribution. In that post, was used to denote the maximum rather than the *absolute* maximum of the process but, as I will just look at nonnegative martingales, this is not important.

Choosing a nonnegative integrable , its running average is

Letting *U* be a random variable uniformly distributed on , and defined with respect to a probability space , let be the cadlag process

Under its natural filtration, this is a martingale. Its terminal and maximum values are

The inequality for is an equality so long as is decreasing, although that is not necessary here. Fixing any , consider the example

Then, *h* is integrable and

Note that this example gives a non-continuous martingale. However, by the methods used in the post on the maximum maximum of martingales with known terminal distribution, it is possible to construct continuous examples with the same terminal and maximum law. Letting *B* be a Brownian motion with initial value equal to , we would stop this at the first time at which . Then, applying the deterministic time change gives the required continuous example.

#### Example 2

Another method of constructing examples is to start with a nonnegative cadlag martingale , defined with respect a filtered probability space , such that as . The maximum, , cannot be integrable, except where *M* is identically zero. This is because, otherwise, dominated convergence would give the contradiction

Now suppose that we have a random time independent of *M*, and let be the process

(1) |

Using independence between and *M*,

So, *X* is integrable. It is also straightforward to verify that *X* is a martingale under its natural filtration. Its maximum has expectation,

where I have set . As we know that is not integrable, increases to infinity as so, by choosing such that is not integrable, *X* will have nonintegrable maximum.

To give an explicit example, let be a random time with the exponential distribution, . Then let *M* be the martingale

This drops to 0 at time . The maximum has expectation

So, if is a random time independent of and with , then (1) defines a martingale with nonintegrable maximum.

#### Example 3

Finally, I’ll give an example along the same lines as above, but for a continuous martingale. If *W* is a standard Brownian motion starting at , then,

is a martingale. In fact, it is the Doléans exponential of *W*, and is geometric Brownian motion with zero drift. Using the fact that Brownian motion grows at rate , we know that . The expected value of can be computed,

This can be derived using the reflection principle, or by looking up the distribution of the maximum of Brownian motion with drift. Then, if is any random time independent of *W* with , the process (1) defines a continuous martingale with nonintegrable maximum.

Hi George I must have missed something elementary but in example 1, you say and I quote “The inequality for is an equality so long as is decreasing, although that is not necessary here”. I think it is indeed necessary to take a decreasing , because next line you only get and to me this is not enough to prove that the martingale has a non integrable maximum we only have that its maximum is bounded above by which is not very informative, anyway taking a decreasing solves it for me so no big deal. Best regards

Hah, yes, that inequality is useless. I think the text is correct though – I just wrote the inequality the wrong way round. Fixed, thanks.