Martingales

A martingale models a fair game, a process whose best mean-square forecast of any future value is the present value. Two consequences make it the central object of stochastic analysis. Stopping a fair game at a clever time cannot create an edge, and the running maximum of a martingale is controlled by its terminal size.

#The fair-game property

Definition1

An adapted, integrable process $(M_n)_{n\ge 0}$ is a martingale with respect to $(\F_n)$ when $\E[M_{n+1}\mid\F_n]=M_n$ for every $n$ , a submartingale when $\E[M_{n+1}\mid\F_n]\ge M_n$ , and a supermartingale when the reverse inequality holds.

By the tower property a martingale satisfies $\E[M_m\mid\F_n]=M_n$ for all $m\ge n$ , so $\E[M_n]=\E[M_0]$ is constant. If $\varphi$ is convex and $\varphi(M_n)\in L^1$ for every $n$ , then $(\varphi(M_n))$ is a submartingale, since conditional Jensen gives $\E[\varphi(M_{n+1})\mid\F_n]\ge\varphi(\E[M_{n+1}\mid\F_n])=\varphi(M_n)$ . For $\varphi=\abs\cdot$ this is automatic from $M_n\in L^1$ ; for $\varphi=(\cdot)^2$ it requires $M_n\in L^2$ .

#Optional stopping

Theorem2

Let $(M_n)$ be a martingale and $\tau$ a stopping time bounded by $N$ . Then $\E[M_\tau]=\E[M_0]$ .

Proof

Write the stopped value as a telescoping sum weighted by the predictable indicators $\ind\{\tau\ge k\}=1-\ind\{\tau\le k-1\}$ , which are $\F_{k-1}$ -measurable,

M_\tau=M_0+\sum_{k=1}^{N}\ind\{\tau\ge k\}\,(M_k-M_{k-1}). \tag{1}

Taking expectations and conditioning each term on $\F_{k-1}$ ,

\E\big[\ind\{\tau\ge k\}(M_k-M_{k-1})\big]=\E\big[\ind\{\tau\ge k\}\,\E[M_k-M_{k-1}\mid\F_{k-1}]\big]=0, \tag{2}

because $\ind\{\tau\ge k\}$ is $\F_{k-1}$ -measurable and the conditional increment of a martingale vanishes. Hence $\E[M_\tau]=\E[M_0]$ .

#Doob's inequalities

Theorem3

Let $(M_n)$ be a martingale and $M_n^\ast=\max_{0\le k\le n}\abs{M_k}$ . For every $\lambda>0$ ,

\lambda\,\P\big(M_n^\ast\ge\lambda\big)\le\E\big[\abs{M_n}\,\ind\{M_n^\ast\ge\lambda\}\big]\le\E\big[\abs{M_n}\big], \tag{3}

and for $1<p<\infty$ with $M_n\in L^p$ , $\E[(M_n^\ast)^p]\le\big(\tfrac{p}{p-1}\big)^p\,\E[\abs{M_n}^p]$ .

Proof

Decompose $A=\{M_n^\ast\ge\lambda\}=\bigsqcup_{k=0}^n A_k$ with $A_k=\{\abs{M_0}<\lambda,\dots,\abs{M_{k-1}}<\lambda,\ \abs{M_k}\ge\lambda\}\in\F_k$ the event that $\lambda$ is first reached at time $k$ . On $A_k$ we have $\abs{M_k}\ge\lambda$ , and since $\abs M$ is a submartingale $\E[\abs{M_n}\mid\F_k]\ge\abs{M_k}$ , so

\lambda\,\P(A_k)\le\E\big[\abs{M_k}\,\ind_{A_k}\big]\le\E\big[\E[\abs{M_n}\mid\F_k]\,\ind_{A_k}\big]=\E\big[\abs{M_n}\,\ind_{A_k}\big], \tag{4}

using that $A_k\in\F_k$ . Summing over $k$ gives $\lambda\,\P(A)\le\E\big[\abs{M_n}\,\ind_A\big]$ , which is Equation (3). For the $L^p$ bound assume $M_n\in L^p$ , so $M_k\in L^p$ for $k\le n$ by Jensen and $(M_n^\ast)^p\le\sum_{k\le n}\abs{M_k}^p$ gives $\E[(M_n^\ast)^p]<\infty$ . By the layer-cake formula and the tail estimate,

\E[(M_n^\ast)^p]=\int_0^\infty p\lambda^{p-1}\P(M_n^\ast\ge\lambda)\,d\lambda \le\int_0^\infty p\lambda^{p-2}\E\big[\abs{M_n}\,\ind\{M_n^\ast\ge\lambda\}\big]\,d\lambda. \tag{5}

Tonelli evaluates the inner $\lambda$ -integral, $\int_0^{M_n^\ast}p\lambda^{p-2}\,d\lambda=\tfrac{p}{p-1}(M_n^\ast)^{p-1}$ , producing the constant $p/(p-1)$ and the term $\tfrac{p}{p-1}\E[\abs{M_n}(M_n^\ast)^{p-1}]$ . Holder with conjugate exponents $p,\,p/(p-1)$ bounds this by $\tfrac{p}{p-1}\E[\abs{M_n}^p]^{1/p}\E[(M_n^\ast)^p]^{(p-1)/p}$ . When $0<\E[(M_n^\ast)^p]<\infty$ , dividing by $\E[(M_n^\ast)^p]^{(p-1)/p}$ yields the claim. The case $\E[(M_n^\ast)^p]=0$ is trivial.

#Convergence

Theorem4

A martingale bounded in $L^1$ , $\sup_n\E[\abs{M_n}]<\infty$ , converges almost surely to an integrable limit.

Proof

For reals $a<b$ let $U_n([a,b])$ count the upcrossings of $[a,b]$ by $M_0,\dots,M_n$ . The predictable gambling strategy that buys below $a$ and sells above $b$ has gains dominating $(b-a)U_n([a,b])-(M_n-a)^-$ , and since the strategy is a martingale transform its expected gain is zero, yielding Doob's upcrossing bound

(b-a)\,\E\big[U_n([a,b])\big]\le\E\big[(M_n-a)^-\big]\le\abs a+\E[\abs{M_n}]. \tag{6}

The right side is bounded in $n$ , so $\E[U_\infty([a,b])]<\infty$ and $U_\infty([a,b])<\infty$ almost surely. Taking a countable union over rational $a<b$ rules out oscillation, so $M_n$ converges almost surely to a limit $M_\infty$ , which is integrable by Fatou.

#The Doob-Meyer decomposition

Theorem5

(Doob-Meyer.) A right-continuous submartingale of class $\mathrm{D}$ admits a unique decomposition $X_t=M_t+A_t$ with $M$ a martingale and $A$ a predictable increasing process [1], [2].

Applied to $M^2$ for a square-integrable martingale $M$ whose stopped family is uniformly integrable, so that $M^2$ is a submartingale of class $\mathrm{D}$ , the increasing part is the quadratic variation $\langle M\rangle$ , the unique predictable increasing process starting at $0$ that makes $M_t^2-\langle M\rangle_t$ a martingale. The optional stopping of Theorem 2 and the maximal control of Theorem 3 are the two tools that turn this decomposition into a theory of integration against $M$ .

[1]

I. Karatzas and S. E. Shreve, Brownian Motion and Stochastic Calculus, 2nd ed. Springer, 1991.

[2]

D. Revuz and M. Yor, Continuous Martingales and Brownian Motion, 3rd ed. Springer, 1999.

Explore connections

see in the atlas

referenced by (1)

Uniform Integrability and the Vitali Theorem

cite

@misc{martingales,
  author = {Zac Kienzle},
  title  = {Martingales},
  year   = {2026},
  month  = {05},
  url    = {https://zackienzle.com/blog/martingales}
}

#The fair-game property

Definition1

#Optional stopping

Theorem2

Let $(M_n)$ be a martingale and $\tau$ a stopping time bounded by $N$ . Then $\E[M_\tau]=\E[M_0]$ .

Proof

Write the stopped value as a telescoping sum weighted by the predictable indicators $\ind\{\tau\ge k\}=1-\ind\{\tau\le k-1\}$ , which are $\F_{k-1}$ -measurable,

M_\tau=M_0+\sum_{k=1}^{N}\ind\{\tau\ge k\}\,(M_k-M_{k-1}). \tag{1}

Taking expectations and conditioning each term on $\F_{k-1}$ ,

\E\big[\ind\{\tau\ge k\}(M_k-M_{k-1})\big]=\E\big[\ind\{\tau\ge k\}\,\E[M_k-M_{k-1}\mid\F_{k-1}]\big]=0, \tag{2}

because $\ind\{\tau\ge k\}$ is $\F_{k-1}$ -measurable and the conditional increment of a martingale vanishes. Hence $\E[M_\tau]=\E[M_0]$ .

#Doob's inequalities

Theorem3

Let $(M_n)$ be a martingale and $M_n^\ast=\max_{0\le k\le n}\abs{M_k}$ . For every $\lambda>0$ ,

\lambda\,\P\big(M_n^\ast\ge\lambda\big)\le\E\big[\abs{M_n}\,\ind\{M_n^\ast\ge\lambda\}\big]\le\E\big[\abs{M_n}\big], \tag{3}

and for $1<p<\infty$ with $M_n\in L^p$ , $\E[(M_n^\ast)^p]\le\big(\tfrac{p}{p-1}\big)^p\,\E[\abs{M_n}^p]$ .

Proof

\lambda\,\P(A_k)\le\E\big[\abs{M_k}\,\ind_{A_k}\big]\le\E\big[\E[\abs{M_n}\mid\F_k]\,\ind_{A_k}\big]=\E\big[\abs{M_n}\,\ind_{A_k}\big], \tag{4}

\E[(M_n^\ast)^p]=\int_0^\infty p\lambda^{p-1}\P(M_n^\ast\ge\lambda)\,d\lambda \le\int_0^\infty p\lambda^{p-2}\E\big[\abs{M_n}\,\ind\{M_n^\ast\ge\lambda\}\big]\,d\lambda. \tag{5}

#Convergence

Theorem4

A martingale bounded in $L^1$ , $\sup_n\E[\abs{M_n}]<\infty$ , converges almost surely to an integrable limit.

Proof

(b-a)\,\E\big[U_n([a,b])\big]\le\E\big[(M_n-a)^-\big]\le\abs a+\E[\abs{M_n}]. \tag{6}

#The Doob-Meyer decomposition

Theorem5

(Doob-Meyer.) A right-continuous submartingale of class $\mathrm{D}$ admits a unique decomposition $X_t=M_t+A_t$ with $M$ a martingale and $A$ a predictable increasing process [1], [2].

[1]

I. Karatzas and S. E. Shreve, Brownian Motion and Stochastic Calculus, 2nd ed. Springer, 1991.

[2]

D. Revuz and M. Yor, Continuous Martingales and Brownian Motion, 3rd ed. Springer, 1999.

Explore connections

see in the atlas

referenced by (1)

Uniform Integrability and the Vitali Theorem

cite

@misc{martingales,
  author = {Zac Kienzle},
  title  = {Martingales},
  year   = {2026},
  month  = {05},
  url    = {https://zackienzle.com/blog/martingales}
}