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Girsanov’s Theorem and Risk-Neutral Pricing of Geometric Brownian Motion

Article Quant Q&A · Author: k b
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Summary

The document explains why Girsanov’s theorem matters for geometric Brownian motion in finance. Under the original probability measure, an asset’s drift is its expected return, which generally differs from the risk-free rate. Girsanov’s change of measure shifts the Brownian motion so that the asset’s drift becomes the risk-free rate under a risk-neutral measure. Discounting by the risk-free growth factor then makes the asset price a martingale under that measure.

This matters because derivative values can be calculated as discounted risk-neutral expected payoffs when the required no-arbitrage assumptions hold. The transformation is therefore not merely another solution of the original stochastic differential equation; it supplies the pricing dynamics used for valuation. The explanation is limited to the stated geometric Brownian setting with constant volatility and a risk-free rate. It does not discuss extensions, market incompleteness, or the practical estimation of model inputs.

Key ideas

  • Girsanov’s theorem changes the probability measure and adjusts the Brownian motion driving the asset.
  • Under the risk-neutral measure, the geometric Brownian asset drift equals the risk-free rate.
  • Discounted asset prices are martingales under the risk-neutral measure in this setting.
  • Derivative prices can be obtained from discounted expected payoffs when no-arbitrage assumptions apply.
  • The document’s explanation concerns a simplified geometric Brownian model.

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Full text
# Girsanov Theorem application to Geometric Brownian Motion


# Girsanov Theorem application to Geometric Brownian Motion












I recently read this from a book on mathematical finance

> The important example for finance the (unique) EMM for the geometric Brownian. Let $S_{t}$ be the price of an asset, $${{d{S_t}} \over {{S_t}}} = \mu dt + \sigma d{W_t}$$ and let $r \ge 0$ be the risk-free rate of interest. For the exponential martingale $${Z_t} = \exp \left( { - {t \over 2}{{\left( {{{r - \mu } \over \sigma }} \right)}^2} + {{r - \mu } \over \sigma }{W_t}} \right)$$ the process $W_t^Q \buildrel\textstyle.\over= {{\mu - r} \over \sigma }t + {W_t}$ is Q-brownian motion, and the price process satisfies, $${{d{S_t}} \over {{S_t}}} = rdt + \sigma dW_t^Q$$ Hence, $${S_t} = {S_0}\exp \left( {\left( {r - {1 \over 2}{\sigma ^2}} \right)t + \sigma W_t^Q} \right)$$ and ${S_t}{e^{ - rt}}$ is a Q-martingale

but why is this so important , the result is just the solution to the ${{d{S_t}} \over {{S_t}}} = \mu dt + \sigma d{W_t}$?

## Answer by quallenjäger (score 3)

https://quant.stackexchange.com/a/33370

Denote $B_t=e^{rt}$ the discount factor. Requiring $S_t/B_t$ to be a martingale it would mean the equation $S_0/B_0=E[S_t/B_t]$ hold. Therefore we can calculate the price of an option by discounting the expectation value at the maturity. If $S_t/B_t$ is not a Q-martingale, then we cannot discount the expectation value, which make calculation of $S_0$ extremely difficult.

Another thing to mention: it is not a solution to $dS_t/S_t=\mu*dt+\sigma*dW$. It is a solution to $dS_t/S_t=r*dt+\sigma*dW^Q$. The first dynamic doesn't satisfy risk-neutral-pricing assumption, namely no-arbitrage assumption. The second dynamic is the right dynamic for risk-neutral-pricing. That's why we need girsanov theorem to transform the dynamic.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.