Phase 8 - Lesson 8.2

Random Walks, Filtrations, and Information

The simple random walk as the prototype process, its scaling toward Brownian motion, and the filtration formalism that makes ‘information over time’ precise.

⏱ 50 min● Advanced🔗 Prereqs: 7.7, 8.1

↖ Phase 8 hub

Builds on: 7.7 introduced filtrations and martingales; 8.1 gave the Markov property.
Leads to: Rescaling the random walk (Donsker) constructs Brownian motion in 8.3.

Learning Objectives

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Define the simple symmetric random walk and compute its mean and variance growth.
Formalize a filtration as an information flow and identify adapted and predictable processes.
Show the symmetric random walk is a martingale and locate its natural quadratic variation.
Explain the diffusive \(\sqrt n\) scaling that motivates Brownian motion.

Key Vocabulary

Simple random walk: \(S_n=\sum_{i=1}^n \xi_i\) with i.i.d. \(\xi_i=\pm1\); symmetric if each sign has probability 1/2.

Filtration: Increasing sigma-algebras \(\F_0\subseteq\F_1\subseteq\cdots\); \(\F_n=\sigma(\xi_1,\dots,\xi_n)\) is the natural one.

Adapted process: \(X_n\) is \(\F_n\)-measurable - known at time \(n\).

Predictable process: \(H_n\) is \(\F_{n-1}\)-measurable - chosen before the time-\(n\) move (a trading position).

Increments: \(S_n-S_m\) for \(n\gt m\); independent and stationary for a random walk.

Diffusive scaling: \(\mathrm{sd}(S_n)=\sqrt n\): displacement grows like the square root of time.

Intuition & Motivation

Intuition

The simple random walk is the atom of stochastic modeling: add up independent \(\pm1\) shocks. Its mean stays 0 but its spread grows like \(\sqrt n\) - the diffusive law that says a random mover covers distance proportional to the square root of time, not time itself. Two structures organize everything that follows. A filtration is the growing record of what has happened; a process is adapted if it never sees the future and predictable if it is fixed one step in advance - exactly a trading position taken before the price moves. Rescale the walk in space by \(\sqrt n\) and in time by \(n\), and the jagged path smooths into Brownian motion; that limit is the subject of the next lesson.

The simple random walk

Definition - Simple symmetric random walk

Let \(\xi_1,\xi_2,\dots\) be i.i.d. with \(\Prob(\xi_i=+1)=\Prob(\xi_i=-1)=\tfrac12\). The walk is \(S_0=0,\ S_n=\sum_{i=1}^n\xi_i\). Increments are independent and stationary.

Since \(\E[\xi_i]=0\) and \(\Var(\xi_i)=1\): \(\E[S_n]=0\) and \(\Var(S_n)=n\), so \(\mathrm{sd}(S_n)=\sqrt n\). The walk is Markov (future depends only on the current level) and, being a sum of mean-zero independent increments, a martingale.

\[\E[S_{n+1}\mid \F_n]=S_n+\E[\xi_{n+1}]=S_n,\qquad \E[S_n^2-n\mid\F_n\text{-adjusted}]:\ \E[S_n^2]=n.\] (8.2)

Filtrations: information as a growing sigma-algebra

Definition - Natural filtration; adapted vs predictable

The natural filtration is \(\F_n=\sigma(\xi_1,\dots,\xi_n)\), the information from the first \(n\) shocks. A process \((X_n)\) is adapted if \(X_n\) is \(\F_n\)-measurable; it is predictable if \(X_n\) is \(\F_{n-1}\)-measurable.

The distinction is the whole game in trading. Your position \(H_n\) in step \(n\) must be predictable - decided using only \(\F_{n-1}\), before the shock \(\xi_n\) is revealed. Your wealth is then a martingale transform:

\[(H\cdot S)_n=\sum_{k=1}^{n}H_k\,(S_k-S_{k-1})=\sum_{k=1}^n H_k\,\xi_k.\] (8.3)

Theorem - Martingale transform stays a martingale

If \((S_n)\) is a martingale and \((H_n)\) is predictable and bounded, then the transform \((H\cdot S)_n\) is a martingale with \(\E[(H\cdot S)_n]=0\). No predictable betting strategy turns a fair game into a favorable one - the discrete no-arbitrage statement.

The quantity \(\sum_{k\le n}(S_k-S_{k-1})^2=\sum_{k\le n}\xi_k^2=n\) is the walk’s quadratic variation: the increments squared accumulate to \(n\), deterministically. Keep this fact - its continuous analogue \([B]_t=t\) is the crux of stochastic calculus.

Worked Example - Wealth from a predictable strategy is a fair game

Start with a symmetric walk \(S_n\) (fair coin, \(\pm1\)). Choose any predictable stake \(H_n\in\{1,2\}\) using only the past.

Wealth increment is \(H_n\xi_n\). Condition on \(\F_{n-1}\): \(\E[H_n\xi_n\mid\F_{n-1}]=H_n\,\E[\xi_n\mid\F_{n-1}]=H_n\cdot0=0\), since \(H_n\) is known and \(\xi_n\) is independent mean-zero.

Therefore \(\E[(H\cdot S)_n\mid\F_{n-1}]=(H\cdot S)_{n-1}\): the wealth is a martingale.

Hence \(\E[(H\cdot S)_n]=0\) for every predictable strategy - you cannot manufacture edge from a fair coin by timing or sizing.

This is exactly why, later, the discounted price under the risk-neutral measure is a martingale and self-financing strategies cannot arbitrage.

Interactive: random-walk paths and their spread

Common Mistakes to Avoid

Thinking spread grows linearly in time; for a random walk standard deviation grows like \(\sqrt n\), not \(n\).
Letting a position depend on the same step’s shock - that is not predictable and secretly encodes look-ahead (an arbitrage).
Confusing adapted with predictable: adapted knows the present, predictable is fixed one step earlier.
Believing a clever stake sequence can beat a fair game; the martingale-transform theorem forbids it.

Quant Practitioner Tips

Whenever you see ‘decided before the move’, translate it to \(\F_{n-1}\)-measurable = predictable.
The accumulation \(\sum(\Delta S)^2=n\) previews Ito: squared increments, not increments, carry the second-order term.
Diffusive \(\sqrt t\) scaling is why volatility is quoted per \(\sqrt{\text{time}}\) (annualize by \(\sqrt{252}\)).

Knowledge Check

Q1 Easy

For the simple symmetric random walk, \(\mathrm{sd}(S_n)\) grows like:

\(n\)

\(\sqrt n\)

\(\log n\)

constant

Q2 Medium

A trading position \(H_n\) that avoids look-ahead must be:

\(\F_n\)-measurable (adapted)

\(\F_{n-1}\)-measurable (predictable)

independent of the past

constant

Q3 Hard

The martingale-transform theorem implies that a predictable betting strategy on a fair game yields wealth with:

Positive expected gain

Zero expected gain (still a martingale)

Negative expected gain

Undefined expectation

Practical Exercise

Let \(S_n\) be the simple symmetric random walk. (a) Show \(M_n=S_n^2-n\) is a martingale. (b) Interpret the ‘\(-n\)’ term in light of quadratic variation.

▶ Show full solution

(a) \(\E[S_{n+1}^2\mid\F_n]=\E[(S_n+\xi_{n+1})^2\mid\F_n]=S_n^2+2S_n\E[\xi_{n+1}]+\E[\xi_{n+1}^2]=S_n^2+0+1.\) Hence \(\E[S_{n+1}^2-(n+1)\mid\F_n]=S_n^2+1-(n+1)=S_n^2-n=M_n\), so \(M_n\) is a martingale.

(b) The compensator \(n\) equals the accumulated squared increments \(\sum_{k\le n}\xi_k^2=n\), i.e. the quadratic variation of the walk. Subtracting it exactly cancels the systematic growth of \(S_n^2\), turning it into a fair game. In continuous time the same idea makes \(B_t^2-t\) a martingale, because \([B]_t=t\).

After the reveal, answer for yourself: What continuous-time statement does \(S_n^2-n\) being a martingale foreshadow?

Lesson Summary

The simple random walk sums i.i.d. \(\pm1\) shocks; it is Markov, a martingale, and spreads diffusively as \(\sqrt n\). Filtrations formalize accumulating information, distinguishing adapted (known now) from predictable (fixed one step earlier), and the martingale-transform theorem shows no predictable strategy beats a fair game. Its squared increments accumulate deterministically to \(n\), foreshadowing Brownian quadratic variation \([B]_t=t\).

Formula Sheet Additions

Diffusive scaling

\[\Var(S_n)=n,\ \mathrm{sd}(S_n)=\sqrt n\]

Distance grows like the square root of time.

Martingale transform

\[(H\cdot S)_n=\sum_{k\le n}H_k\,\xi_k,\ \E[(H\cdot S)_n]=0\]

Predictable strategies preserve fairness.

Error Log Checklist

Did I use \(\sqrt n\) (not \(n\)) for the standard deviation?
Is my strategy predictable (\(\F_{n-1}\)-measurable), not secretly using \(\xi_n\)?
Did I identify the compensator as accumulated squared increments?

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: Give the mean and standard deviation of \(S_n\) for the symmetric walk.
A: Mean 0; standard deviation \(\sqrt n\) (variance \(n\)).

Q: Distinguish adapted from predictable.
A: Adapted: \(X_n\) is \(\F_n\)-measurable (known now). Predictable: \(H_n\) is \(\F_{n-1}\)-measurable (fixed one step earlier).

Q: Why can't a predictable strategy beat a fair game?
A: The martingale transform of a martingale by a bounded predictable process is again a mean-zero martingale.

Flashcards

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Predictable process

\(H_n\) is \(\F_{n-1}\)-measurable; a position set before the move.

Random walk QV

\(\sum_{k\le n}(\Delta S_k)^2=n\); the compensator making \(S_n^2-n\) a martingale.

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. 4-5 - Ch. 4-5: random walks, martingale transforms, recurrence.
Brownian Motion and Stochastic Calculus (Karatzas & Shreve, 2nd ed., GTM 113) foundational - Ch. 1-2 - Filtrations, adapted/predictable processes, discrete precursors to BM.

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