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Random Walks & Stochastic

Brownian Motion & Hitting Times

Hard5–6 hrsBrownian MotionHitting TimesReflection PrincipleIto

Overview

Brownian motion is the mathematical model for random noise — it's continuous, nowhere differentiable, and has Gaussian increments. The key practical fact: over time horizon T, Brownian motion moves approximately ±σ√T. Stock prices in Black-Scholes are modeled as geometric Brownian motion — the log is normal. You don't need to derive Itô's lemma; you need to apply it and understand why the σ²/2 drift correction exists.

How to Recognize

  • Continuous-time random process, Gaussian increments
  • First passage time, maximum, or probability of hitting a level
  • Options pricing context — Black-Scholes uses GBM
  • 'Geometric Brownian Motion' or 'log-normal' stock price model

Step-by-Step Approach

  1. 1.BM: B(0)=0, independent increments, B(t)−B(s) ~ N(0, t−s)
  2. 2.Reflection principle: P(max B(s) ≥ a for s≤t) = 2·P(B(t) ≥ a)
  3. 3.GBM: S(t) = S(0)·exp((μ−σ²/2)t + σB(t)) — log returns are normal
  4. 4.Itô's lemma: df(B,t) = f'dB + ½f''dt + f_t dt

Key Formulas

01B(t)N(0,t)B(t) \sim N(0,\,t); B(t)B(s)N(0,ts)B(t)-B(s) \sim N(0,\,t-s), independent
02P(maxstB(s)a)=2Φ ⁣(at)=2P(B(t)a)P(\max_{s\leq t} B(s) \geq a) = 2\,\Phi\!\left(-\frac{a}{\sqrt{t}}\right) = 2P(B(t)\geq a)
03GBM: logS(t)/S(0)N ⁣((μ12σ2)t,  σ2t)\log S(t)/S(0) \sim N\!\left((\mu - \tfrac{1}{2}\sigma^2)t,\;\sigma^2 t\right)
04Itô: dS=μSdt+σSdBtdS = \mu S\,dt + \sigma S\,dB_t

Worked Examples

0Quick Example
P(BM hits level 2 by time 1) = 2·Φ(−2) ≈ 2·0.0228 = 4.56%.

Problem

Standard Brownian motion starts at 0. What is P(it ever reaches level a = 2 by time t = 1)?

Solution

1

P(max B(s) ≥ a for s ≤ t) — probability of hitting level a at or before time t.

2

Reflection principle: any path that reaches a before t and ends above a corresponds to a path that ends above a. Any path that reaches a and ends below a corresponds (by reflection) to a path ending above a.

3

Therefore: P(max ≥ a) = 2·P(B(t) ≥ a).

4

P(B(1) ≥ 2) = 1 − Φ(2) ≈ 1 − 0.9772 = 0.0228.

5

P(max B ≥ 2 by t=1) = 2·0.0228 = 4.56%.

Answer

P(hit level 2 by time 1)=2Φ(2)4.56%P(\text{hit level 2 by time 1}) = 2\Phi(-2) \approx 4.56\%. The reflection principle doubles the terminal probability.

Common Mistakes

  • !

    Using B(t) ~ N(0,1) instead of N(0,t). Standard BM has variance = t, not 1.

  • !

    Forgetting the σ²/2 term in GBM. The log-return has drift μ−σ²/2, not μ.

  • !

    Applying the reflection principle to GBM directly — it only works for standard BM (zero drift). For BM with drift, use Girsanov's theorem.

Practice Problems

Click "Show Answer" to reveal
1

What is P(BM > 3 at time t=4)?

Hint: B(4) ~ N(0,4), so B(4)/2 ~ N(0,1).
2

Stock with μ=15%, σ=25%, S(0)=50. What is E[S(2)]?

Hint: E[S(T)] = S(0)·e^(μT).
3

A stock has S(0)=100, σ=30%, T=1yr, μ=5%. What is P(S < 80)?

Hint: Log return ~ N((0.05−0.045)·1, 0.09). Find P(log(S/100) < log(0.8)).

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