# martingale convergence theorem

There are several convergence theorems for martingales^{}, which follow from Doob’s upcrossing lemma. The following says that any ${L}^{1}$-bounded martingale ${X}_{n}$ in discrete time converges almost surely.
Note that almost-sure convergence (i.e. convergence with probability one) is quite strong, implying the weaker property of convergence in probability. Here, a martingale ${({X}_{n})}_{n\in \mathbb{N}}$ is understood to be defined with respect to a probability space^{} $(\mathrm{\Omega},\mathcal{F},\mathbb{P})$ and filtration^{} ${({\mathcal{F}}_{n})}_{n\in \mathbb{N}}$.

###### Theorem (Doob’s Forward Convergence Theorem).

Let ${\mathrm{(}{X}_{n}\mathrm{)}}_{n\mathrm{\in}\mathrm{N}}$ be a martingale (or submartingale, or supermartingale) such that $\mathrm{E}\mathit{}\mathrm{[}\mathrm{|}{X}_{n}\mathrm{|}\mathrm{]}$ is bounded over all $n\mathrm{\in}\mathrm{N}$. Then, with probability one, the limit ${X}_{\mathrm{\infty}}\mathrm{=}{\mathrm{lim}}_{n\mathrm{\to}\mathrm{\infty}}\mathit{}{X}_{n}$ exists and is finite.

The condition that ${X}_{n}$ is ${L}^{1}$-bounded is automatically satisfied in many cases. In particular, if $X$ is a non-negative supermartingale then $\mathbb{E}[|{X}_{n}|]=\mathbb{E}[{X}_{n}]\le \mathbb{E}[{X}_{1}]$ for all $n\ge 1$, so $\mathbb{E}[|{X}_{n}|]$ is bounded, giving the following corollary.

###### Corollary.

Let ${\mathrm{(}{X}_{n}\mathrm{)}}_{n\mathrm{\in}\mathrm{N}}$ be a non-negative martingale (or supermartingale). Then, with probability one, the limit ${X}_{\mathrm{\infty}}\mathrm{=}{\mathrm{lim}}_{n\mathrm{\to}\mathrm{\infty}}\mathit{}{X}_{n}$ exists and is finite.

As an example application of the martingale convergence theorem, it is easy to show that a standard random walk^{} started started at $0$ will visit every level with probability one.

###### Corollary.

Let ${\mathrm{(}{X}_{n}\mathrm{)}}_{n\mathrm{\in}\mathrm{N}}$ be a standard random walk. That is, ${X}_{\mathrm{1}}\mathrm{=}\mathrm{0}$ and

$$\mathbb{P}({X}_{n+1}={X}_{n}+1\mid {\mathcal{F}}_{n})=\mathbb{P}({X}_{n+1}={X}_{n}-1\mid {\mathcal{F}}_{n})=1/2.$$ |

Then, for every integer $a$, with probability one ${X}_{n}\mathrm{=}a$ for some $n$.

###### Proof.

Without loss of generality, suppose that $a\le 0$. Let $T:\mathrm{\Omega}\to \mathbb{N}\cup \{\mathrm{\infty}\}$ be the first time $n$ for which ${X}_{n}=a$. It is easy to see that the stopped process ${X}_{n}^{T}$ defined by ${X}_{n}^{T}={X}_{\mathrm{min}(n,T)}$ is a martingale and ${X}^{T}-a$ is non-negative. Therefore, by the martingale convergence theorem, the limit ${X}_{\mathrm{\infty}}^{T}={lim}_{n\to \mathrm{\infty}}{X}_{n}^{T}$ exists and is finite (almost surely). In particular, $|{X}_{n+1}^{T}-{X}_{n}^{T}|$ converges to $0$ and must be less than $1$ for large $n$. However, $|{X}_{n+1}^{T}-{X}_{n}^{T}|=1$ whenever $$, so we have $$ and therefore ${X}_{n}=a$ for some $n$. ∎

Title | martingale convergence theorem |

Canonical name | MartingaleConvergenceTheorem |

Date of creation | 2013-03-22 18:33:47 |

Last modified on | 2013-03-22 18:33:47 |

Owner | gel (22282) |

Last modified by | gel (22282) |

Numerical id | 5 |

Author | gel (22282) |

Entry type | Theorem |

Classification | msc 60G46 |

Classification | msc 60G44 |

Classification | msc 60G42 |

Classification | msc 60F15 |

Related topic | Martingale |

Related topic | MartingaleProofOfKolmogorovsStrongLawForSquareIntegrableVariables |

Related topic | MartingaleProofOfTheRadonNikodymTheorem |

Related topic | UpcrossingsAndDowncrossings |