Lecture 11. Doob’s martingale maximal inequalities

In this post, we prove some fundamental martingale inequalities that, once again, are due to Joe Doob

Theorem (Doob’s maximal inequalities) Let $(\mathcal{F}_t)_{t \ge 0}$ be a filtration on probability space $(\Omega, \mathcal{F},\mathbb{P})$ and let $(M_t)_{t \ge 0}$ be a continuous martingale with respect to the filtration $(\mathcal{F}_t)_{t \ge 0}$ .

Let $p \ge 1$ and $T>0$ . If $\mathbb{E} (\mid M_T\mid^p) <+\infty$ , then we have
$\mathbb{P} \left( \sup_{0 \le t \le T} \mid M_t \mid \ge \lambda\right) \le \frac{\mathbb{E} \left( \mid M_T \mid^p\right)}{\lambda^p}.$

Let $p > 1$ and $T>0$ . If $\mathbb{E} (\mid M_T\mid^p)<+\infty$ , then we have
$\mathbb{E}\left( \left( \sup_{0 \le t \le T} \mid M_t \mid \right)^p \right) \le \left( \frac{p}{p-1} \right)^p \mathbb{E} (\mid M_T \mid^p).$

Proof:

Let $p \ge 1$ and $T>0$ . If $\mathbb{E} (\mid M_T \mid^p) < +\infty$ then, from Jensen’s inequality the process $(\mid M_t \mid^p)_{0 \le t\le T}$ is a submartingale. Let $\lambda > 0$ and

$\tau =\inf \{ s \ge 0 \text{ such that } \mid M_s \mid \ge \lambda \}\wedge T,$

with the usual convention that $\inf \emptyset =+\infty$ . It is seen that $\tau$ is an almost surely bounded stopping time. Therefore, from the Doob’s stopping theorem

$\mathbb{E} (\mid M_{\tau} \mid^p) \le \mathbb{E} (\mid M_{T} \mid^p).$

But from the very definition of $\tau$ ,

$\mid M_{\tau} \mid^p \ge 1_{\sup_{0 \le t \le T} \mid M_t \mid \ge \lambda} \lambda^p+1_{\sup_{0 \le t \le T} \mid M_t \mid < \lambda} \mid M_{T} \mid^p$ .
which implies,
$\mathbb{P} \left( \sup_{0 \le t \le T} \mid M_t \mid \ge \lambda \right) \le \frac{\mathbb{E} \left( \mid M_T \mid^p 1_{\sup_{0 \le t \le T} \mid M_t \mid \ge \lambda} \right)}{\lambda^p}\le \frac{\mathbb{E} \left( \mid M_T \mid^p \right)}{\lambda^p}.$

This concludes the proof of the first part of our statement.

Let now $p > 1$ and $T > 0$ .

Let us first assume that:

$\mathbb{E}\left( \left( \sup_{0 \le t \le T} \mid M_t \mid \right)^p \right) < +\infty$ ,

The previous proof shows that for $\lambda >0$ ,

$\mathbb{P} \left( \sup_{0 \le t \le T} \mid M_t \mid \ge \lambda \right) \le \frac{\mathbb{E} \left( \mid M_T \mid 1_{\sup_{0 \le t \le T} \mid M_t \mid \ge \lambda} \right)}{\lambda}.$

We deduce,

$\int_0^{+\infty} \lambda^{p-1} \mathbb{P} \left( \sup_{0 \le t \le T} \mid M_t \mid \ge \lambda \right) d \lambda \le \int_0^{+\infty} \lambda^{p-2} \mathbb{E} \left( \mid M_T \mid 1_{\sup_{0 \le t \le T} \mid M_t \mid \ge \lambda} \right)d\lambda.$

From Fubini’s theorem,

$\int_0^{+\infty} \lambda^{p-1} \mathbb{P} \left( \sup_{0 \le t \le T} \mid M_t \mid \ge \lambda \right) d \lambda =\int_{\Omega} \left( \int_0^{\sup_{0 \le t \le T} \mid M_t \mid (\omega)} \lambda^{p-1} d\lambda \right) d \mathbb{P} (\omega)$

$=\frac{1}{p} \mathbb{E}\left( \left( \sup_{0 \le t \le T} \mid M_t \mid \right)^p \right).$

Similarly, we obtain

$\int_0^{+\infty} \lambda^{p-2} \mathbb{E} \left( \mid M_T \mid 1_{\sup_{0 \le t \le T} \mid M_t \mid \ge \lambda} \right)d\lambda=\frac{1}{p-1} \mathbb{E} \left( \left(\sup_{0 \le t \le T} \mid M_t \mid\right)^{p-1} \mid M_T \mid \right).$

Hence,

$\mathbb{E}\left( \left( \sup_{0 \le t \le T} \mid M_t \mid \right)^p \right) \le \frac{p}{p-1} \mathbb{E} \left( \left(\sup_{0 \le t \le T} \mid M_t \mid\right)^{p-1} \mid M_T \mid \right).$

By using now Hölder’s inequality we obtain,

$\mathbb{E} \left( \left(\sup_{0 \le t \le T} \mid M_t \mid\right)^{p-1} \mid M_T \mid \right) \le \mathbb{E} \left( \mid M_T \mid^p\right)^{\frac{1}{p}} \mathbb{E} \left( \left( \sup_{0 \le t \le T} \mid M_t \mid \right)^p\right)^{\frac{p-1}{p}},$

which implies

$\mathbb{E}\left( \left( \sup_{0 \le t \le T} \mid M_t \mid \right)^p \right) \le \frac{p}{p-1} \mathbb{E} \left( \mid M_T \mid^p\right)^{\frac{1}{p}} \mathbb{E} \left( \left( \sup_{0 \le t \le T} \mid M_t \mid \right)^p\right)^{\frac{p-1}{p}}.$

As a conclusion if $\mathbb{E}\left( \left( \sup_{0 \le t \le T} \mid M_t \mid \right)^p \right) < +\infty$ , we have:

$\mathbb{E}\left( \left( \sup_{0 \le t \le T} \mid M_t \mid \right)^p \right) \le \left( \frac{p}{p-1} \right)^p \mathbb{E} (\mid M_T \mid^p).$

Now, if $\mathbb{E}\left( \left( \sup_{0 \le t \le T} \mid M_t \mid \right)^p \right) = +\infty$ , we consider for $N \in \mathbb{N}$ , the stopping time $\tau_N=\inf \{t \ge 0, \mid M_t \mid \ge N \} \wedge T$ . By using the above result to the martingale $(M_{t \wedge \tau_N})_{t \ge 0}$ , we obtain

$\mathbb{E}\left( \left( \sup_{0 \le t \le T} \mid M_{t \wedge \tau_N} \mid \right)^p \right) \le \left( \frac{p}{p-1} \right)^p \mathbb{E} (\mid M_T \mid^p),$

from which we may conclude by using the monotone convergence theorem.

4 Responses to Lecture 11. Doob’s martingale maximal inequalities

Danielsen says:

June 29, 2016 at 4:35 pm

Hi, Professor Baudoin. I think in the first line of the proof of the part 2, it should be p>1.

- Fabrice Baudoin says:
  
  June 30, 2016 at 3:24 pm
  
  Thanks. Corrected.
  
Ahmed Taha says:

October 6, 2019 at 5:52 pm

Hi, Professor Baudoin
If I am not mistaken, the last part of the proof actually shows that the supremum is always in L^p.
Is that right?

- Fabrice Baudoin says:
  
  October 10, 2019 at 1:09 am
  
  Yes, that’s right