Martingales with Non-Integrable Maximum

It is a consequence of Doob’s maximal inequality that any ${L^p}$ -integrable martingale has a maximum, up to a finite time, which is also ${L^p}$ -integrable for any ${p > 1}$ . Using ${X^*_t\equiv\sup_{s\le t}\lvert X_s\rvert}$ to denote the running absolute maximum of a cadlag martingale X, then ${X^*}$ is ${L^p}$ -integrable whenever ${X}$ is. It is natural to ask whether this also holds for ${p=1}$ . 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, ${[X]^{1/2}}$ , 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, ${X^*}$ 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 ${h\colon(0,1)\rightarrow{\mathbb R}}$ , its running average is

$\displaystyle \bar h(t)=\frac1t\int_0^th(s)\,ds.$

Letting U be a random variable uniformly distributed on ${(0,1)}$ , and defined with respect to a probability space ${(\Omega,\mathcal{F},{\mathbb P})}$ , let ${\{X_t\}_{t\in[0,1]}}$ be the cadlag process

$\displaystyle X_t=\begin{cases} \bar h(1-t),&\textrm{for }t < 1-U,\\ h(U),&\textrm{for }t\ge1-U. \end{cases}$

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

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle X_1&\displaystyle=h(U),\smallskip\\ \displaystyle X^*_1&\displaystyle\ge\bar h(U). \end{array}$

The inequality for ${X_1^*}$ is an equality so long as ${\bar h}$ is decreasing, although that is not necessary here. Fixing any ${a > 1}$ , consider the example

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle h(t)&\displaystyle=\frac1{t(\log(a/t))^2},\smallskip\\ \displaystyle \bar h(t)&\displaystyle=\frac1{t\log(a/t)}. \end{array}$

Then, h is integrable and

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle {\mathbb E}[X^*_1]&\displaystyle\ge\int_0^1\bar h(t)\,dt\smallskip\\ &\displaystyle=\left[-\log\log(a/t)\right]_0^1=\infty. \end{array}$

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 ${B_0}$ equal to ${\int_0^1h(t)\,dt=1/\log(a)}$ , we would stop this at the first time ${\tau}$ at which ${\bar h(B_\tau)\le h(B^*_\tau)}$ . Then, applying the deterministic time change ${X_t=B_{\tau\wedge(t/(1-t))}}$ gives the required continuous example.

Example 2

Another method of constructing examples is to start with a nonnegative cadlag martingale ${\{M_t\}_{t\in{\mathbb R}_+}}$ , defined with respect a filtered probability space ${(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\in{\mathbb R}_+},{\mathbb P})}$ , such that ${M_t\rightarrow0}$ as ${t\rightarrow\infty}$ . The maximum, ${M^*_\infty}$ , cannot be integrable, except where M is identically zero. This is because, otherwise, dominated convergence would give the contradiction

$\displaystyle {\mathbb E}[M_t]=\lim_{s\rightarrow\infty}{\mathbb E}[M_s]=0.$

Now suppose that we have a random time ${\tau\colon\Omega\rightarrow{\mathbb R}_+}$ independent of M, and let ${\{X_t\}_{t\in[0,1]}}$ be the process

$\displaystyle X_t=M_{\tau\wedge(t/(1-t))}.$

(1)

Using independence between ${\tau}$ and M,

$\displaystyle {\mathbb E}[\lvert X_t\rvert]={\mathbb E}[M_{\tau\wedge(t/(1-t))}]={\mathbb E}[M_0].$

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

$\displaystyle {\mathbb E}[X^*_1]={\mathbb E}[M_\tau^*]={\mathbb E}[f(\tau)]$

where I have set ${f(t)\equiv{\mathbb E}[M^*_t]}$ . As we know that ${M^*_\infty}$ is not integrable, ${f(t)}$ increases to infinity as ${t\rightarrow\infty}$ so, by choosing ${\tau}$ such that ${f(\tau)}$ is not integrable, X will have nonintegrable maximum.

To give an explicit example, let ${\sigma}$ be a random time with the exponential distribution, ${{\mathbb P}(\sigma > t)=e^{-t}}$ . Then let M be the martingale

$\displaystyle M_t=1_{\{t <\sigma\}}e^t.$

This drops to 0 at time ${\sigma}$ . The maximum has expectation

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle {\mathbb E}[M^*_t]&\displaystyle={\mathbb E}\left[1_{\{t < \sigma\}}e^t + 1_{\{t \ge \sigma\}}e^\sigma\right]\smallskip\\ &\displaystyle={\mathbb P}(\sigma > t)e^t-\int_0^t e^s\,d {\mathbb P}(\sigma > s)\smallskip\\ &\displaystyle=1+t. \end{array}$

So, if ${\tau\ge0}$ is a random time independent of ${\sigma}$ and with ${{\mathbb E}[\tau]=\infty}$ , then (1) defines a martingale ${\{X_t\}_{t\in[0,1]}}$ 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 ${W_0=0}$ , then,

$\displaystyle M_t=\exp\left(W_t-\frac t2\right)$

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 ${o(t)}$ , we know that ${M_\infty=0}$ . The expected value of ${M^*}$ can be computed,

$\displaystyle \begin{array}{rl} \displaystyle {\mathbb E}[M^*_t]&\displaystyle={\mathbb E}\left[M_t\vee1\right]+{\mathbb E}\left[\left(W_t+t/2\right)_+\right]\smallskip\\ &\displaystyle\ge1+t/2. \end{array}$

This can be derived using the reflection principle, or by looking up the distribution of the maximum of Brownian motion with drift. Then, if ${\tau\ge0}$ is any random time independent of W with ${{\mathbb E}[\tau]=\infty}$ , the process (1) defines a continuous martingale ${\{X_t\}_{t\in[0,1]}}$ with nonintegrable maximum.

2 thoughts on “Martingales with Non-Integrable Maximum”

TheBridge says:

3 June 20 at 5:10 PM

Hi George I must have missed something elementary but in example 1, you say and I quote “The inequality for ${X_1^*}$ is an equality so long as ${\bar h}$ is decreasing, although that is not necessary here”. I think it is indeed necessary to take a decreasing ${\bar h}$ , because next line you only get $\mathbb E [X^*_1] \le \int_0^1\bar h(t)dt = \infty$ and to me this is not enough to prove that the martingale $X$ has a non integrable maximum we only have that its maximum is bounded above by $+\infty$ which is not very informative, anyway taking a decreasing ${\bar h}$ solves it for me so no big deal. Best regards

1. George Lowther says:
  
  3 June 20 at 10:06 PM
  
  Hah, yes, that inequality is useless. I think the text is correct though – I just wrote the inequality the wrong way round. Fixed, thanks.

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