---
tag: Z57CG6W9S6
kind: example
name: Doob martingale
msc: [60G42]
instance_of: [0D0GBNDKDH]
sources:
  - url: https://en.wikipedia.org/wiki/Doob_martingale
---

## Example

Let $(\Omega, \mathcal{F}, P)$ be a [[probability space|0ZYBZ34RT2]] with a [[filtration|HNK1Y62G7V]] $(\mathcal{F}_n)_{n \ge 0}$, and let $X$ be a real random variable with $E[\abs{X}] < \infty$.
For $n \ge 0$ let
$$M_n = E[X \mid \mathcal{F}_n],$$
the [[conditional expectation|0YAMJRK04E]] of $X$ given $\mathcal{F}_n$.
Then $(M_n)_{n \ge 0}$ is a [[martingale|0CDVK9RQBH]], the Doob martingale of $X$.

Check of the three clauses:

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

## Remarks

A Doob martingale is uniformly integrable, and by [[Lévy's upward theorem|W01FD2A8NK]] it converges almost surely and in $L^1$ to $E[X \mid \mathcal{F}_\infty]$, where $\mathcal{F}_\infty$ is the σ-algebra generated by all $\mathcal{F}_n$.
In concentration of measure one takes $X = f(Y_1, \dots, Y_N)$ for independent $Y_i$ and $\mathcal{F}_n = \sigma(Y_1, \dots, Y_n)$.
Then $M_0 = E[X]$ and $M_N = X$, so $X - E[X]$ is a sum of martingale differences, and [[Azuma's inequality|BDA5TTB1F0]] gives [[McDiarmid's inequality|8APFT1RPM4]].

## Explanations

### L1 human

Suppose a random quantity $X$ is revealed bit by bit over time.
At each step, $M_n$ is your best guess of $X$ given what you have seen so far.
Your guess changes as information arrives, but you cannot predict in which direction it will change next.
That is exactly the fair-game property of a martingale.