Doob's martingale inequality

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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 L. Doob.

Statement of the inequality[edit]

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} \left[ X_{t} \big| \mathcal{F}_{s} \right] \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} \left[ X_{T}^{p} \right]}{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, + \infty)

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

\mathbf{E}[X_T] = \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[edit]

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} \left[ | X_{t} |^{p} \right] \right)^{\frac{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 + \| 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[edit]

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

\begin{align}
\mathbf{E} \left[ X_{1} + \dots + X_{n} + X_{n + 1} \big| X_{1}, \dots, X_{n} \right] &= X_{1} + \dots + X_{n} + \mathbf{E} \left[ X_{n + 1} \big| X_{1}, \dots, X_{n} \right] \\
&= X_{1} + \cdots + X_{n},
\end{align}

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

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

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion[edit]

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}{2T} \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} \left[ \exp (\lambda B_{T}) \right ]}{\exp (\lambda C)} \\
& = \exp \left( \tfrac{1}{2}\lambda^{2}T - \lambda C \right) && \mathbf{E} \left[ \exp (\lambda B_{t}) \right] = \exp \left( \tfrac{1}{2}\lambda^{2} t \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[edit]