Incomplete Cholesky factorization

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

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,

$\mathbf{E}\left[X_t\right|{\mathcal{ℱ}}_s]\ge X_s.$

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

$\mathbf{P}[{\sup }_{0\le t\le T}{X}_t\ge C]\le \frac{\mathbf{E}\left[X_T^p\right]}{C^p}.$

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

$X:[0,T]\times \mathrm{\Omega }\to [0,+\mathrm{\infty })$

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

$\mathbf{E}[X_T]=\underset{\mathrm{\Omega }}{\int }{X}_T(\omega )\mathrm{d}\mathbf{P}(\omega )$

in the sense of Lebesgue integration. ${\mathcal{ℱ}}_s$ denotes the σ-algebra generated by all the random variables X_i 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

$S_t={\sup }_{0\le s\le t}{X}_s,$

and for p ≥ 1 let

$\Vert X_t{\Vert }_p=\Vert X_t{\Vert }_{L^p(\mathrm{\Omega },\mathcal{ℱ},\mathbf{P})}={\left(\mathbf{E}\left[\right|X_t{|}^p]\right)}^{\frac{1}{p}}.$

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

$\mathbf{P}[S_T\ge C]\le \frac{\Vert X_T{\Vert }_p^p}{C^p}.$

The following inequalities also hold: for p = 1,

$\Vert S_T{\Vert }_p\le \frac{e}{e-1}(1+\Vert X_T\log X_T{\Vert }_p)$

and, for p > 1,

$\Vert X_T{\Vert }_p\le \Vert S_T{\Vert }_p\le \frac{p}{p-1}\Vert X_T{\Vert }_p.$

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

$\begin{array}{rl}\mathbf{E}[X_1+\dots +X_n+X_{n+1}|X_1,\dots ,X_n]& =X_1+\dots +X_n+\mathbf{E}\left[X_{n+1}\right|X_1,\dots ,X_n]\\ & =X_1+\cdots +X_n,\end{array}$

so M_n = X₁ + ... + X_n 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,

$\mathbf{P}[{\max }_{1\le i\le n}\left|M_i\right|\ge \lambda ]\le \frac{\mathbf{E}\left[M_n^2\right]}{{\lambda }^2},$

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let B denote canonical one-dimensional Brownian motion. Then

$\mathbf{P}[{\sup }_{0\le t\le T}{B}_t\ge C]\le \exp (-\frac{C^2}{2T}).$

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

$\{{\sup }_{0\le t\le T}{B}_t\ge C\}=\{{\sup }_{0\le t\le T}\exp (\lambda B_t)\ge \exp (\lambda C)\}.$

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

$\begin{array}{rl}\mathbf{P}[{\sup }_{0\le t\le T}{B}_t\ge C]& =\mathbf{P}[{\sup }_{0\le t\le T}\exp (\lambda B_t)\ge \exp (\lambda C)]\\ & \le \frac{\mathbf{E}[\exp (\lambda B_T)]}{\exp (\lambda C)}\\ & =\exp (\frac{1}{2}{\lambda }^{2}T-\lambda C)& & \mathbf{E}[\exp (\lambda B_t)]=\exp \left(\frac{1}{2}{\lambda }^{2}t\right)\end{array}$

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

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