Doob's martingale inequality

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

${\displaystyle X_{s}\leq \operatorname {E} [X_{t}\mid {\mathcal {F}}_{s}].}$

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

${\displaystyle P\left[\sup _{0\leq t\leq T}X_{t}\geq C\right]\leq {\frac {\operatorname {E} [{\textrm {max}}(X_{T},0)]}{C}}.}$

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

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

and ${\displaystyle \operatorname {E} [X]}$ denotes the expected value with respect to the probability measure P, i.e. the integral

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

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

There are further submartingale 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

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

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

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

The following inequalities also hold :

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

and, for p > 1,

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

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

${\displaystyle {\begin{aligned}\operatorname {E} \left[X_{1}+\cdots +X_{n}+X_{n+1}\mid X_{1},\ldots ,X_{n}\right]&=X_{1}+\cdots +X_{n}+\operatorname {E} \left[X_{n+1}\mid X_{1},\ldots ,X_{n}\right]\\&=X_{1}+\cdots +X_{n},\end{aligned}}}$

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

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

which is precisely the statement of Kolmogorov's inequality.

Let B denote canonical one-dimensional Brownian motion. Then

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

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,

${\displaystyle {\begin{aligned}P\left[\sup _{0\leq t\leq T}B_{t}\geq C\right]&=P\left[\sup _{0\leq t\leq T}\exp(\lambda B_{t})\geq \exp(\lambda C)\right]\\[8pt]&\leq {\frac {\operatorname {E} [\exp(\lambda B_{T})]}{\exp(\lambda C)}}\\[8pt]&=\exp \left({\tfrac {1}{2}}\lambda ^{2}T-\lambda C\right)&&\operatorname {E} \left[\exp(\lambda B_{t})\right]=\exp \left({\tfrac {1}{2}}\lambda ^{2}t\right)\end{aligned}}}$

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.

• Revuz, Daniel; Yor, Marc (1999). Continuous martingales and Brownian motion (Third ed.). Berlin: Springer. ISBN 3-540-64325-7. (Theorem II.1.7)
• Shiryaev, Albert N. (2001) [1994], "Martingale", in Hazewinkel, Michiel (ed.), Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4