Phase 9 - Lesson 9.3

Stochastic Differential Equations and Geometric Brownian Motion

Solving dS = μS dt + σS dW and why stock models are lognormal.

⏱ 55 min● Advanced🔗 Prereqs: 9.2

Builds on: 9.2 gave Itô's lemma, the tool to solve SDEs.
Leads to: 9.6 prices options assuming GBM under the risk-neutral measure.

Learning Objectives

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

Define an SDE and its strong solution, and state existence–uniqueness conditions.
Derive the closed-form GBM solution $S_t=S_0\exp((\mu-\tfrac12\sigma^2)t+\sigma W_t)$.
Compute the mean, variance, and lognormal distribution of $S_t$.
Implement an Euler–Maruyama simulator and compare it to the exact solution.
Explain why $\E[S_t]=S_0 e^{\mu t}$ despite the $-\tfrac12\sigma^2$ drift of the exponent.

Key Vocabulary

SDE: An equation $dX=a(t,X)dt+b(t,X)dW$ specifying an Itô process by its local drift and diffusion.

Strong solution: A process adapted to the driving Brownian filtration satisfying the SDE for a given $W$ and $X_0$.

Geometric Brownian motion: The solution of $dS=\mu S\,dt+\sigma S\,dW$; an exponential of a Brownian motion with drift.

Lognormal: A variable whose logarithm is normal; $S_t$ under GBM is lognormal.

Euler–Maruyama: The discretization $X_{k+1}=X_k+a\,\Delta t+b\sqrt{\Delta t}\,Z$, $Z\sim\Normal(0,1)$.

Lipschitz / linear-growth: Conditions on $a,b$ guaranteeing a unique strong solution exists.

Intuition & Motivation

Intuition

A stock cannot go negative and its swings scale with its price - a $1000 stock moves in dollars ten times bigger than a $100 stock. Geometric Brownian motion encodes exactly this: proportional drift and proportional volatility. Solving it, the price turns out to be an exponential of Brownian motion, so its logarithm is normal and its level is lognormal.

What an SDE means

The differential form $dS=\mu S\,dt+\sigma S\,dW$ is shorthand for the integral equation $S_t=S_0+\int_0^t\mu S_u\,du+\int_0^t\sigma S_u\,dW_u$, the second integral being an Itô integral.

Existence & uniqueness

If the coefficients $a(t,x),b(t,x)$ are globally Lipschitz in $x$ and grow at most linearly, a unique strong solution exists. GBM's coefficients $a=\mu x,\,b=\sigma x$ are linear, so they qualify.

Worked Example - Deriving the GBM solution

Guess we should work with $Y=\log S$. From 9.2, $dY=(\mu-\tfrac12\sigma^2)dt+\sigma\,dW.$

The right side has constant coefficients, so integrate directly: $Y_t-Y_0=(\mu-\tfrac12\sigma^2)t+\sigma W_t.$

Exponentiate: $S_t=S_0\exp\!\big((\mu-\tfrac12\sigma^2)t+\sigma W_t\big).$

Verify by Itô on $f(t,x)=S_0 e^{(\mu-\sigma^2/2)t+\sigma x}$ with $x=W_t$: you recover $dS=\mu S\,dt+\sigma S\,dW$. Solution confirmed.

Distribution and moments

Since the exponent is normal, $\log S_t\sim\Normal\big(\log S_0+(\mu-\tfrac12\sigma^2)t,\ \sigma^2 t\big)$, so $S_t$ is lognormal. Using $\E[e^{\sigma W_t}]=e^{\sigma^2 t/2}$:

\[\E[S_t]=S_0\,e^{\mu t},\qquad \Var(S_t)=S_0^2 e^{2\mu t}\big(e^{\sigma^2 t}-1\big).\] (9.5)

Key Idea

The $-\tfrac12\sigma^2$ in the exponent and the $+\tfrac12\sigma^2$ from $\E[e^{\sigma W}]$ cancel, so the mean grows at $\mu$ even though the median grows at $\mu-\tfrac12\sigma^2$. Volatility drags the typical path below the mean.

Interactive: sample GBM paths

Common Mistakes to Avoid

Writing $S_t=S_0 e^{\mu t+\sigma W_t}$ (missing $-\tfrac12\sigma^2 t$). That process has mean $S_0e^{(\mu+\sigma^2/2)t}$, not $S_0e^{\mu t}$.
Confusing $\E[S_t]=S_0e^{\mu t}$ (mean) with $S_0e^{(\mu-\sigma^2/2)t}$ (median).
Using Euler–Maruyama with $\Delta t\cdot Z$ instead of $\sqrt{\Delta t}\,Z$ for the noise term.
Thinking GBM can hit zero. It is a strict exponential, always positive.

Quant Practitioner Tips

For simulation prefer the exact log-scheme $\log S_{k+1}=\log S_k+(\mu-\tfrac12\sigma^2)\Delta t+\sigma\sqrt{\Delta t}Z$ - it is unbiased and can't go negative.
Lognormality is why Black–Scholes has closed form; it comes straight from this SDE.
The gap between mean and median growth ($\tfrac12\sigma^2$) is ‘volatility drag’ - central to compounding and leverage.

Knowledge Check

Q1 Medium

The solution of $dS=\mu S\,dt+\sigma S\,dW$ is:

$S_0 e^{\mu t+\sigma W_t}$

$S_0 e^{(\mu-\sigma^2/2)t+\sigma W_t}$

$S_0(1+\mu t+\sigma W_t)$

$S_0 e^{(\mu+\sigma^2/2)t+\sigma W_t}$

Q2 Medium

$\E[S_t]$ for GBM equals:

$S_0 e^{(\mu-\sigma^2/2)t}$

$S_0 e^{\mu t}$

$S_0 e^{\sigma^2 t/2}$

$S_0$

Q3 Easy

In Euler–Maruyama, the noise increment is:

$\sigma S\,\Delta t\,Z$

$\sigma S\,\sqrt{\Delta t}\,Z$

$\sigma S\,Z$

$\sigma S\,(\Delta t)^2 Z$

Practical Exercise

A stock follows GBM with $S_0=100,\ \mu=0.08,\ \sigma=0.3$. (a) Give the distribution of $\log S_1$. (b) Compute $\E[S_1]$ and the median of $S_1$. (c) Which is larger and why?

▶ Show full solution

(a) $\log S_1\sim\Normal(\log100+(0.08-0.045),\,0.09)=\Normal(4.605+0.035,\,0.09)$, i.e. mean $4.640$, variance $0.09$.

(b) Mean: $\E[S_1]=100e^{0.08}\approx108.33$. Median: $e^{4.640}=100e^{0.035}\approx103.56$.

(c) The mean ($108.33$) exceeds the median ($103.56$) because the lognormal is right-skewed: a few large up-moves pull the average above the typical path. The gap is driven by $\sigma$.

After the reveal, answer for yourself: How would halving $\sigma$ change the mean–median gap?

Lesson Summary

An SDE specifies a process by its local drift and diffusion. Geometric Brownian motion, $dS=\mu S\,dt+\sigma S\,dW$, solves to $S_t=S_0\exp((\mu-\tfrac12\sigma^2)t+\sigma W_t)$ - lognormal, always positive, with mean $S_0e^{\mu t}$ and median $S_0e^{(\mu-\sigma^2/2)t}$. Euler–Maruyama discretizes it; the exact log-scheme is unbiased.

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: Write the GBM solution and its mean.
A: $S_t=S_0e^{(\mu-\sigma^2/2)t+\sigma W_t}$, $\E[S_t]=S_0e^{\mu t}$. The exponent's $-\tfrac12\sigma^2$ is cancelled in the mean by $\E[e^{\sigma W_t}]=e^{\sigma^2 t/2}$.

Q: Why is a stock price lognormal under GBM?
A: Because $\log S_t$ is a Brownian motion with drift (a normal variable), so $S_t$ is the exponential of a normal, i.e. lognormal and positive.

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.

Stochastic Calculus for Finance II (Steven Shreve, 2004) foundational - Ch. 4-5 - Ch. 4–5: geometric Brownian motion, its solution, moments, and lognormality.
Monte Carlo Methods in Financial Engineering (Paul Glasserman, 2004) foundational - Ch. 3 - Ch. 3: exact vs. Euler discretization schemes for GBM and diffusion simulation.

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