CONTENTS

Exercise: The Exponential Martingale of Brownian Motion

Problem

Let (Wt)t0(W_t)_{t \ge 0} be a standard Brownian motion and fix σR\sigma \in \mathbb{R}. Define the exponential martingale (also called the Doléans-Dade exponential):
Mt:=exp ⁣(σWt12σ2t)M_t := \exp\!\left(\sigma W_t - \tfrac{1}{2}\sigma^2 t\right)
  1. Show that E[Mt]=1\mathbb{E}[M_t] = 1 for every t0t \ge 0. (Use the MGF of a normal distribution.)
  2. Prove that (Mt)t0(M_t)_{t \ge 0} is a martingale with respect to the natural filtration of WW. That is, for 0st0 \le s \le t:
E[MtFs]=Ms\mathbb{E}[M_t \mid \mathcal{F}_s] = M_s
  1. Financial interpretation: the quantity σ2/2\sigma^2 / 2 is the "drift correction" that appears in the log-return of a geometric Brownian motion. Explain in one paragraph why MtM_t being a martingale is the mathematical content of the statement "discounted asset prices are martingales under the risk-neutral measure" in its simplest possible form.
  2. Extension: What goes wrong if you drop the 12σ2t-\tfrac{1}{2}\sigma^2 t term? Compute E[exp(σWt)]\mathbb{E}[\exp(\sigma W_t)] without the correction, and observe how it depends on tt.

Hint

For parts 1–2, use WtWsN(0,ts)W_t - W_s \sim \mathcal{N}(0, t-s) is independent of Fs\mathcal{F}_s, and the MGF identity E[eλZ]=eλ2/2\mathbb{E}[e^{\lambda Z}] = e^{\lambda^2/2} for ZN(0,1)Z \sim \mathcal{N}(0, 1). Factor MtM_t into a Fs\mathcal{F}_s-measurable piece times a piece involving WtWsW_t - W_s.

Jump to the solution when you're ready.