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Why the Exponential Brownian Martingale Preserves Its Value

Article Quant Q&A · Author: phhhlpfk
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Summary

The note explains the exponential martingale built from standard Brownian motion and clarifies how to express a later Brownian value using an earlier value and an increment. Brownian motion satisfies W(s) = W(t) + [W(s) − W(t)] for s ≥ t; the increment is independent of the information available at time t and has a centered Gaussian distribution with variance s − t. Its exponential moment cancels the time-dependent factor in the process, giving the conditional expectation of the future process as its current value.

The response also notes that Itō’s lemma provides another proof: the drift term vanishes, and the diffusion coefficient is square-integrable over bounded time intervals. The original question’s displayed equality uses W(t) + W(s), which is generally not the correct decomposition; the relevant term is the increment W(s) − W(t). The result assumes the usual Brownian filtration and ordered times, and the discussion is a probability concept rather than a trading strategy.

Key ideas

  • Brownian motion at a later time decomposes into its current value and the increment since that time.
  • The increment is independent of the current Brownian information and has a centered Gaussian distribution.
  • The exponential moment of the increment cancels the deterministic time adjustment in the process.
  • The conditional expectation identity establishes the martingale property.
  • Itō’s lemma supplies an alternative proof under the stated integrability conditions.

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# Standard Brownian Motion and Exponential Martingale calculation


# Standard Brownian Motion and Exponential Martingale calculation












Let $W(t)$ be a standard brownian motion and let $Z(t) = \exp (\lambda W(t) - \frac{1}{2}\lambda^2 t).$ In Xinfeng Zhou's Green Book in the section on Brownian Motion (p.130) he writes as part of the proof that $Z(t)$ is a martingale, that $$\mathbb{E}[Z(t+s)]=\mathbb{E}[\exp(\lambda(W(t)+W(s)) - \frac{1}{2}\lambda^2 (t+s))] $$ My question is how does he go from $Z(t+s) = \exp (\lambda W(t+s) - \frac{1}{2}\lambda^2 (t+s))$ to $W(t)+W(s)$ in the exponent ? Thanks for clearing up my confusion, I am just getting into Brownian Motions and Stochastic Calculus.

## Answer by siou0107 (score 3, accepted)

https://quant.stackexchange.com/a/74516

$Z$ is a martingale, indeed. It is easily proved using Itō's lemma (the drift disappears and the diffusion term is constant, hence square-integrable on any compact). Another way to see it is \begin{align} \mathbb{E} \left[Z \left(s\right) \middle \vert \mathcal{F}_t\right] & = \mathbb{E} \left[Z \left(s\right) \middle \vert W \left(t\right)\right] \\ & =\mathbb{E} \left[e^{\lambda W \left(t\right) - \frac{1}{2} \lambda^2 t + \lambda \left[W \left(s\right) - W \left(t\right)\right] - \frac{1}{2} \lambda^2 \left(s - t\right)}\middle \vert W \left(t\right)\right] \\ & = Z \left(t\right) e^{- \frac{1}{2} \lambda^2 \left(s - t\right)} \mathbb{E} \left[e^{\lambda \left[W \left(s\right) - W \left(t\right)\right]} \middle \vert W \left(t\right) \right] \end{align} From the Markov property of Brownian motion and its centered Gaussian distribution, the conditional expectation is equal to $e^{\frac{1}{2} \lambda^2 \left(s - t\right)}$ and therefore $$ \boxed{\mathbb{E} \left[Z \left(s\right) \middle \vert \mathcal{F}_t\right] = Z \left(t\right)} $$ which is the definition of a martingale.

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