Using Girsanov’s Theorem to Make a Geometric Brownian Price a Martingale
Summary
The answer considers a price following geometric Brownian motion under a probability measure where its drift is nonzero. It explains that the drift can be absorbed into a shifted Brownian motion: under a new measure, the original Brownian motion plus the drift-to-volatility ratio times time is standard Brownian motion. Substituting this shifted process into the price dynamics leaves only the diffusion term, so the price is a martingale under the new measure.
The change of measure is specified with an exponential density, and the result is attributed to Girsanov’s theorem. The answer clarifies that the desired martingale is the price process itself, rather than a proposed time-adjusted function of the price. Its conclusion is conditional on the model and measure-change assumptions: it does not discuss discounting, more general asset dynamics, or conditions needed to ensure the density defines a valid probability measure over a chosen horizon.
Key ideas
- A drifted geometric Brownian price can be written using a Brownian motion with a deterministic time shift.
- Girsanov’s theorem defines a new measure under which that shifted process is standard Brownian motion.
- Under the new measure, the price dynamics contain no drift term and the price is a martingale.
- The result assumes the stated model and does not cover discounted prices or broader validity conditions.
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Full text
# Changing to martingale probability world
# Changing to martingale probability world
This question is really getting me annoyed and I'm struggling to do the final proof, I have no problem obtaining the adjustment to the drift rate necessary to collapse the drift term to make it a martingale measure which I perceive to be = u/sigma, however, I am struggling to find what the fucntion of S is that makes the measure a martingale, from my text books it says that the function should be S_t = S*_t + u/sigma * t ,however when I perform itos lemma on this result I cant seem to get a drift rate of zero, ( I am also assuming that S*_t = S_0), any help willl be greatly aprreciated.
## Answer by Gordon (score 2, accepted)
https://quant.stackexchange.com/a/39693
We assume that, under the probability measure $P$, \begin{align*} dS_t = S_t(\mu dt +\sigma dW_t), \end{align*} where $\{W_t, \, t \ge0\}$ is a standard Brownian motion. Note that, \begin{align*} dS_t=\sigma S_t d\left(\frac{\mu}{\sigma}t+ W_t \right). \end{align*} You need the probability measure $Q$ such that, under $Q$, the process $\{W_t + \frac{\mu}{\sigma}t, \, t\ge 0\}$ is a standard Brownian motion. Towards that, we define the measure $Q$ such that \begin{align*} \frac{dQ}{dP}\big|_t = \exp\left(-\frac{1}{2}\left( \frac{\mu}{\sigma}\right)^2 t-\frac{\mu}{\sigma} W_t\right). \end{align*} Then, by the Girsanov theorem, $Q$ is a probability measure, and the process $\{\widehat{W}_t, \, t\ge 0\}$, where $\widehat{W}_t=W_t + \frac{\mu}{\sigma}t$, is a standard Brownian motion under $Q$. Moreover, since \begin{align*} dS_t =\sigma S_t d\widehat{W}_t, \end{align*} $\{S_t, \, t\ge 0\}$ is a martingale under $Q$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
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