Phase 7 - Lesson 7.7

Martingales and Stopping Times (Discrete Time)

Filtrations, the fair-game condition, stopping times, and the optional stopping theorem - the discrete-time skeleton of arbitrage-free pricing.

⏱ 55 min● Advanced🔗 Prereqs: 7.3

↖ Phase 7 hub

Builds on: 7.3 built conditional expectation and sigma-algebras as information.
Leads to: Continuous-time martingales and the martingale representation theorem drive Phases 8-9 and risk-neutral pricing.

Learning Objectives

Click a status chip to cycle: Not started → In progress → Studied → Practiced → Needs review → Mastered.

Define a filtration and adapted process and interpret them as flowing information.
Define a martingale, submartingale, and supermartingale via conditional expectation.
Define a stopping time and verify the condition on concrete rules.
State the optional stopping theorem and apply it to gambling and pricing problems.

Key Vocabulary

Filtration: An increasing family $\F_0\subseteq\F_1\subseteq\cdots$ of sigma-algebras; $\F_n$ is the information available at time $n$.

Adapted process: $(X_n)$ with each $X_n$ $\F_n$-measurable: its value is known once time $n$ is reached.

Martingale: An adapted, integrable $(M_n)$ with $\E[M_{n+1}\mid\F_n]=M_n$; a fair game.

Submartingale / supermartingale: $\E[M_{n+1}\mid\F_n]\ge M_n$ (favorable) / $\le M_n$ (unfavorable).

Stopping time: A random time $\tau$ with $\{\tau\le n\}\in\F_n$ for all $n$: the decision to stop uses only past information.

Optional stopping: Under integrability/boundedness conditions, $\E[M_\tau]=\E[M_0]$ for a martingale.

Intuition & Motivation

Intuition

A filtration is the movie of information: $\F_n$ is everything you could know by time $n$, and it only grows. A process is adapted if you never see the future - its value at time $n$ is knowable from $\F_n$. A martingale is the model of a fair game: your best forecast of tomorrow’s wealth, given everything so far, is exactly today’s wealth. A stopping time is any exit rule that does not peek ahead - ‘sell when the price first hits 110’ qualifies, ‘sell at the peak’ does not. The optional stopping theorem then says: you cannot beat a fair game with any non-anticipating strategy. That single sentence is the discrete-time heart of no-arbitrage pricing.

Filtrations, adaptedness, and martingales

Definition - Filtration and adapted process

A filtration on $(\Omega,\F,\Prob)$ is a nested sequence $\F_0\subseteq\F_1\subseteq\cdots\subseteq\F$ of sub-sigma-algebras. A process $(X_n)_{n\ge0}$ is adapted if $X_n$ is $\F_n$-measurable for every $n$.

Definition - Martingale

An adapted process $(M_n)$ with $\E|M_n|\lt \infty$ is a martingale w.r.t. $(\F_n)$ if

\[\E[M_{n+1}\mid \F_n]=M_n\quad\text{for all }n\ (\text{fair}).\] (7.11)

It is a submartingale if $\E[M_{n+1}\mid\F_n]\ge M_n$ and a supermartingale if $\le M_n$.

Taking expectations in (7.11) and using the tower property gives $\E[M_n]=\E[M_0]$ for all $n$: a martingale has constant mean. The canonical example is a random walk of mean-zero increments; the canonical submartingale is $M_n^2$ when $M_n$ is a martingale (by Jensen).

Stopping times

Definition - Stopping time

A random variable $\tau:\Omega\to\{0,1,2,\dots,\infty\}$ is a stopping time if $\{\tau\le n\}\in\F_n$ for every $n$ (equivalently $\{\tau=n\}\in\F_n$). The event ‘we have stopped by time $n$’ must be decidable from information up to $n$.

First hitting times - $\tau=\min\{n:X_n\ge b\}$ - are stopping times. ‘The last time the price is below $b$’ is not, because deciding it needs the future.

Theorem - Optional stopping (Doob)

Let $(M_n)$ be a martingale and $\tau$ a stopping time. Then $\E[M_\tau]=\E[M_0]$ provided any one holds: (i) $\tau$ is bounded; (ii) $\tau\lt \infty$ a.s. and $(M_{n\wedge\tau})$ is uniformly bounded; or (iii) $\E[\tau]\lt \infty$ and the increments are bounded. For a supermartingale, $\E[M_\tau]\le\E[M_0]$.

Why the conditions matter

Without a regularity condition the theorem fails. In an unbounded symmetric random walk, the strategy ‘stop when first ahead by \$1’ has $\tau\lt \infty$ a.s. yet $\E[M_\tau]=1\ne0=\E[M_0]$ - because $\E[\tau]=\infty$ and wealth is unbounded before stopping. There is no free lunch, only an unbounded bankroll and infinite expected time.

Worked Example - Gambler’s ruin via optional stopping

A fair game: symmetric $\pm1$ random walk $M_n$ started at $M_0=a$, with $0\lt a\lt N$. It is a martingale.

Let $\tau=\min\{n:M_n\in\{0,N\}\}$, the first exit from $(0,N)$. This is a bounded-range stopping time with $\E[\tau]\lt \infty$.

Optional stopping applies: $\E[M_\tau]=\E[M_0]=a$. But $M_\tau\in\{0,N\}$, so $\E[M_\tau]=N\cdot\Prob(\text{hit }N)$.

Solve: $\Prob(\text{reach }N\text{ before }0)=a/N.$ The ruin probability is $1-a/N$.

For the expected duration, apply optional stopping to the martingale $M_n^2-n$: $\E[M_\tau^2-\tau]=a^2$, giving $\E[\tau]=a(N-a)$.

Interactive: simulate a martingale and check the fair-game property

Common Mistakes to Avoid

Calling a look-ahead rule (e.g. ‘sell at the maximum’) a stopping time - it uses future information and is not admissible.
Applying optional stopping without checking a regularity condition; the unbounded random walk is the standard counterexample.
Confusing sub- and supermartingales; a favorable game drifts up ($\ge$), an unfavorable one drifts down ($\le$).
Forgetting integrability $\E|M_n|\lt \infty$ in the definition - without it conditional expectations may not exist.

Quant Practitioner Tips

In finance a discounted asset price is a martingale under the risk-neutral measure - that is precisely no-arbitrage.
To find hitting probabilities or expected durations, hunt for the right martingale ($M_n$, $M_n^2-n$, $e^{\theta M_n}$) and apply optional stopping.
Bounded stopping times always satisfy optional stopping - truncate at a large horizon, then take limits carefully.

Knowledge Check

Q1 Medium

Which random time is a stopping time with respect to the natural filtration of a price process?

The time of the global maximum over all n

The first time the price reaches 120

The time two steps before the crash

The last time before a peak

Q2 Easy

For a martingale $(M_n)$, $\E[M_{n+1}\mid\F_n]$ equals:

$M_0$

$M_n$

$\E[M_n]$

$0$

Q3 Hard

The optional stopping theorem can fail if:

The stopping time is bounded

The martingale has constant mean

The stopping time has infinite expectation and the martingale is unbounded

The filtration is nested

Practical Exercise

A stock modeled as a symmetric $\pm1$ random walk starts at 50. A trader will exit the moment it first hits either 40 or 65. (a) Using optional stopping, find the probability of exiting at 65. (b) Find the expected number of steps until exit.

▶ Show full solution

(a) With $a=50$ measured on a scale where barriers are 40 and 65, shift to $x=a-40=10$ inside $(0,25)$. The martingale $M_n$ gives $\Prob(\text{hit }65)=x/N=10/25=0.4.$

(b) Using the martingale $M_n^2-n$ and optional stopping, $\E[\tau]=x(N-x)=10\times15=150$ steps.

So a 40% chance of the favorable exit and 150 expected steps. Notice the fair game gives an exit probability equal to the fractional distance to the upper barrier - distance, not desire, sets the odds.

After the reveal, answer for yourself: If the walk had a small upward drift, would you still use $M_n$, or an exponential martingale $e^{\theta M_n}$?

Lesson Summary

A filtration encodes flowing information; an adapted process never anticipates it; a martingale is the fair-game model $\E[M_{n+1}\mid\F_n]=M_n$, with sub/supermartingales tilting the drift. Stopping times are non-anticipating exit rules, and the optional stopping theorem - that $\E[M_\tau]=\E[M_0]$ under regularity - formalizes ‘you cannot beat a fair game’, the discrete-time root of arbitrage-free pricing.

Formula Sheet Additions

Martingale property

\[\E[M_{n+1}\mid\F_n]=M_n\]

Fair game: today is the best forecast of tomorrow.

Optional stopping

\[\E[M_\tau]=\E[M_0]\ \ (\text{under regularity})\]

Non-anticipating strategies cannot beat a fair game.

Error Log Checklist

Is my stopping rule non-anticipating ($\{\tau\le n\}\in\F_n$)?
Did I verify a regularity condition before applying optional stopping?
Did I confirm integrability and adaptedness before calling a process a martingale?

Retrieval Practice

Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.

▶ Show retrieval prompts & answers

Q: Define a stopping time and give one example and one non-example.
A: $\tau$ with $\{\tau\le n\}\in\F_n$. Example: first hitting time of a level. Non-example: time of the global maximum (needs the future).

Q: State the optional stopping theorem and one sufficient condition.
A: $\E[M_\tau]=\E[M_0]$ for a martingale, provided e.g. $\tau$ is bounded (or $\E[\tau]\lt \infty$ with bounded increments).

Q: Which martingale yields the expected exit time of a symmetric walk from an interval?
A: $M_n^2-n$; optional stopping gives $\E[\tau]=a(N-a)$.

Flashcards

Click to flip. These feed the site-wide spaced-repetition queue.

Martingale

Adapted, integrable, $\E[M_{n+1}\mid\F_n]=M_n$; constant mean.

Stopping time

Random time with $\{\tau\le n\}\in\F_n$ - no peeking ahead.

Completion Checklist

I can explain the core ideas in my own words
I worked the derivations/examples by hand
I completed the interactive workbench(es)
I passed the knowledge check

Confidence / mastery rating

Personal notes

Source References

This lesson synthesizes and paraphrases concepts from the sources below. No copyrighted text, problem sets, or solutions are reproduced. Return to the originals for full depth.

Probability: Theory and Examples (Rick Durrett, 5th ed.) current - Ch. 4 - Ch. 4: martingales, stopping times, optional stopping, gambler’s ruin.
Stochastic Calculus for Finance II (Steven Shreve, 2004) foundational - Ch. 1-2 - Discrete martingales and risk-neutral pricing intuition.

← 7.6 Next: 8.1 →