Math 2.0

Probability theory and stochastic processes Martingales Example

Doob martingale

Example

Let (Ω,ℱ,P)(\Omega, \mathcal{F}, P) be a probability space with a filtration (ℱn)n≥0(\mathcal{F}_n)_{n \ge 0}, and let XX be a real random variable with E[∣X∣]<∞E[\abs{X}] < \infty. For n≥0n \ge 0 let Mn=E[X∣ℱn],M_n = E[X \mid \mathcal{F}_n], the conditional expectation of XX given ℱn\mathcal{F}_n. Then (Mn)n≥0(M_n)_{n \ge 0} is a martingale, the Doob martingale of XX.

Check of the three clauses:

  1. MnM_n is ℱn\mathcal{F}_n-measurable by the definition of conditional expectation.
  2. E[∣Mn∣]≤E[E[∣X∣∣ℱn]]=E[∣X∣]<∞E[\abs{M_n}] \le E\bigl[E[\abs{X} \mid \mathcal{F}_n]\bigr] = E[\abs{X}] < \infty.
  3. Since ℱn⊆ℱn+1\mathcal{F}_n \subseteq \mathcal{F}_{n+1}, the tower property gives E[Mn+1∣ℱn]=E[E[X∣ℱn+1]∣ℱn]=E[X∣ℱn]=MnE[M_{n+1} \mid \mathcal{F}_n] = E\bigl[E[X \mid \mathcal{F}_{n+1}] \mid \mathcal{F}_n\bigr] = E[X \mid \mathcal{F}_n] = M_n.

Remarks

A Doob martingale is uniformly integrable, and by Lévy's upward theorem it converges almost surely and in L1L^1 to E[X∣ℱ∞]E[X \mid \mathcal{F}_\infty], where ℱ∞\mathcal{F}_\infty is the σ-algebra generated by all ℱn\mathcal{F}_n. In concentration of measure one takes X=f(Y1,…,YN)X = f(Y_1, \dots, Y_N) for independent YiY_i and ℱn=σ(Y1,…,Yn)\mathcal{F}_n = \sigma(Y_1, \dots, Y_n). Then M0=E[X]M_0 = E[X] and MN=XM_N = X, so X−E[X]X - E[X] is a sum of martingale differences, and Azuma's inequality gives McDiarmid's inequality.

Where this leads

1 page builds on this one.

Further reading

  1. Web https://en.wikipedia.org/wiki/Doob_martingale
About this page
TagZ57CG6W9S6
URIm2://Z57CG6W9S6
Kindexample
WritingW0 stub
MSC60G42
Instance ofMartingale
Changed2 November 2026

Markdown · JSON · Edit on GitHub

What this page needs for W1

  • an inventory row: the row of the inventory that the node accounts for (0 of 1)
  • a source with a locator (0 of 1)

The writing levels say how complete the text of a page is. They do not measure its truth.

What links here

Used by