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.

This entry was posted in Stochastic Calculus lectures. Bookmark the permalink.

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

  1. Danielsen says:

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

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