# 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 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,

${\displaystyle X_{s}\leq \mathbf {E} \left[X_{t}{\big |}{\mathcal {F}}_{s}\right].}$

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

${\displaystyle \mathbf {P} \left[\sup _{0\leq t\leq T}X_{t}\geq C\right]\leq {\frac {\mathbf {E} \left[X_{T}\right]}{C}}.}$

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

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

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

${\displaystyle \mathbf {E} [X_{T}]=\int _{\Omega }X_{T}(\omega )\,\mathrm {d} \mathbf {P} (\omega )}$

in the sense of Lebesgue integration. ${\displaystyle {\mathcal {F}}_{s}}$ 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

${\displaystyle S_{t}=\sup _{0\leq s\leq t}X_{s},}$

and for p ≥ 1 let

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

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

${\displaystyle \mathbf {P} \left[S_{T}\geq C\right]\leq {\frac {\|X_{T}\|_{1}}{C}}.}$

The following inequalities also hold, : for p = 1,

${\displaystyle \|S_{T}\|_{p}\leq {\frac {e}{e-1}}\left(1+\|X_{T}\log ^{+}X_{T}\|_{p}\right)}$

and, for p > 1,

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

## 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

{\displaystyle {\begin{aligned}\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{aligned}}}

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,

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

which is precisely the statement of Kolmogorov's inequality.

## Application: Brownian motion

Let B denote canonical one-dimensional Brownian motion. Then

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

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

${\displaystyle \left\{\sup _{0\leq t\leq T}B_{t}\geq C\right\}=\left\{\sup _{0\leq t\leq T}\exp(\lambda B_{t})\geq \exp(\lambda C)\right\}.}$

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

{\displaystyle {\begin{aligned}\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{aligned}}}

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.