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

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