Doob's martingale inequality

In mathematics, Doob's martingale inequality is a result in the study of stochastic processes. It gives a bound on the probability that a stochastic process exceeds any given value over a given interval of time. As the name suggests, the result is usually given in the case that the process is a non-negative martingale, but the result is also valid for non-negative submartingales.

The inequality is due to the American mathematician Joseph Leo Doob.

Contents

Statement of the inequality

Let X be a submartingale taking non-negative real values, either in discrete or continuous time. That is, for all times s and t with s < t,

\mathbf{E} \big[ X_{t} \big| \mathcal{F}_{s} \big] \geq X_{s}.

(For a continuous-time submartingale, assume further that the process is càdlàg.) Then, for any constant C > 0 and p ≥ 1,

\mathbf{P} \left[ \sup_{0 \leq t \leq T} X_{t} \geq C \right] \leq \frac{\mathbf{E} \big[ X_{T}^{p} \big]}{C^{p}}.

In the above, as is conventional, P denotes the probability measure on the sample space Ω of the stochastic process

X�: [0, T] \times \Omega \to [0, %2B \infty)

and E denotes the expected value with respect to the probability measure P, i.e. the integral

\mathbf{E} \big[ X_{T} \big] = \int_{\Omega} X_{T} (\omega) \, \mathrm{d} \mathbf{P} (\omega)

in the sense of Lebesgue integration. \mathcal{F}_{s} denotes the σ-algebra generated by all the random variables Xi with i ≤ s; the collection of such σ-algebras forms a filtration of the probability space.

Further inequalities

There are further (sub)martingale inequalities also due to Doob. With the same assumptions on X as above, let

S_{t} = \sup_{0 \leq s \leq t} X_{s},

and for p ≥ 1 let

\| X_{t} \|_{p} = \| X_{t} \|_{L^{p} (\Omega, \mathcal{F}, \mathbf{P})} = \left( \mathbf{E} \big[ | X_{t} |^{p} \big] \right)^{1 / p}.

In this notation, Doob's inequality as stated above reads

\mathbf{P} \left[ S_{T} \geq C \right] \leq \frac{\| X_{T} \|_{p}^{p}}{C^{p}}.

The following inequalities also hold: for p = 1,

\| S_{T} \|_{p} \leq \frac{e}{e - 1} \left( 1 %2B \| X_{T} \log X_{T} \|_{p} \right)

and, for p > 1,

\| X_{T} \|_{p} \leq \| S_{T} \|_{p} \leq \frac{p}{p-1} \| X_{T} \|_{p}.

Related inequalities

Doob's inequality for discrete-time martingales implies Kolmogorov's inequality: if X1, X2, ... is a sequence of real-valued independent random variables, each with mean zero, it is clear that

\mathbf{E} \big[ X_{1} %2B \dots %2B X_{n} %2B X_{n %2B 1} \big| X_{1}, \dots, X_{n} \big]
= X_{1} %2B \dots %2B X_{n} %2B \mathbf{E} \big[ X_{n %2B 1} \big| X_{1}, \dots, X_{n} \big]
= X_{1} %2B \cdots %2B X_{n},

so Mn = X1 + ... + Xn is a martingale. Note that Jensen's inequality implies that |M_n| is a nonnegative submartingale if M_n is a martingale. Hence, taking p = 2 in Doob's martingale inequality,

\mathbf{P} \left[ \max_{1 \leq i \leq n} \big| M_{i} \big| \geq \lambda \right] \leq \frac{\mathbf{E} \big[ M_{n}^{2} \big]}{\lambda^{2}},

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let B denote canonical one-dimensional Brownian motion. Then

\mathbf{P} \left[ \sup_{0 \leq t \leq T} B_{t} \geq C \right] \leq \exp \left( - \frac{C^2}{2 T} \right).

The proof is just as follows: since the exponential function is monotonically increasing, for any non-negative λ,

\left\{ \sup_{0 \leq t \leq T} B_{t} \geq C \right\} = \left\{ \sup_{0 \leq t \leq T} \exp ( \lambda B_{t} ) \geq \exp ( \lambda C ) \right\}.

By Doob's inequality, and since the exponential of Brownian motion is a positive submartingale,

\begin{align} & \mathbf{P} \left[ \sup_{0 \leq t \leq T} B_{t} \geq C \right] \\ & = \mathbf{P} \left[ \sup_{0 \leq t \leq T} \exp ( \lambda B_{t} ) \geq \exp ( \lambda C ) \right] \\ & \leq \frac{\mathbf{E} \big[ \exp (\lambda B_{T}) \big]}{\exp (\lambda C))} \\ & = \exp \left( \frac{\lambda^{2} T}{2} - \lambda C \right) \mbox{ since } \mathbf{E} \big[ \exp (\lambda B_{t}) \big] = \exp \left( \frac{\lambda^{2} t}{2} \right). \end{align}

Since the left-hand side does not depend on λ, choose λ to minimize the right-hand side: λ = C / T gives the desired inequality.

References