Constructing an Equivalent Martingale Measure for the Bachelier Model
Summary
The document derives an equivalent martingale measure for a Bachelier stock-price model with constant drift and volatility, assuming a zero interest rate. Under the original measure, the stock follows an arithmetic Brownian motion. To make the discounted stock a martingale, the response shifts the Brownian motion’s drift by the amount needed to remove the stock’s expected drift.
It then uses Girsanov’s theorem to express the new measure through a Radon–Nikodym density, an exponential involving the original Brownian motion and the drift-to-volatility ratio. Under this measure, the stock has zero drift and is driven only by the shifted Brownian motion. The response provides the density over a fixed horizon and explains how it follows from the change of measure. Its conclusion assumes the stated constant parameters and zero rate; it does not discuss the call option’s valuation or the conditions and degenerate cases under which the measure exists or is unique.
Key ideas
- The Bachelier model represents the stock price as a drift plus a Brownian shock in arithmetic form.
- With a zero interest rate, the equivalent martingale measure must remove the stock’s drift.
- Girsanov’s theorem expresses that measure through an exponential Radon–Nikodym density.
- Under the changed measure, the shifted Brownian motion drives the stock with zero drift.
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Full text
# EMM for Bachelier model
# EMM for Bachelier model
The stock price is assumed to evolve as $S_{t}=S_{0}+\mu t+\sigma B_{t}$, where $S_{0}>0, \mu>0$ and the process $B_{t}$ is Brownian motion.
The saving account is assumed to be $\beta_{t}=e^{r t}$, with interest rate $r$
A call option with strike $K$ and expiration $T$ pays $C_{T}=\left(S_{T}-K\right)^{+}$ at time $T$.
Assume r = 0. Give the EMM.
My attempt
I am a bit lost when it comes to EMM but this is what I have so far:
- Girsanov's theorem:
$B_{t}$ is a B.M under measure P and C is a constant. Then there exists an EMM q such that:
$\hat{B}_{t}=B_{t}+C_{t} \sim Q$ brownian motion.
$d S_{t}=\mu d t+\sigma d B_{t}$
$r=0 \quad c=\frac{\mu-r}{\sigma}=c=\frac{\mu}{\sigma}$
I am not sure how to continue and give the EMM explicitly.
Edit: After some researching, I have found that the EMM does exist for Bachelier model and is unique by Girsanov's theorem. But I am still a bit lost on how to find it.
## Answer by R. Rayl (score 1, accepted)
https://quant.stackexchange.com/a/63957
It sounds to me like you understand everything apart from how Girsanov's theorem defines the EMM.
Girsanov's theorem tells us that if $B_t$ is standard Brownian motion under $P$, then for any adapted process $\gamma_t$ (satisfying certain conditions) the process $\hat{B}_t$ defined by:
\begin{equation} d\hat{B}_t = \gamma_t dt +dB_t \end{equation} is Brownian motion under another equivalent measure and this equivalent measure (let's call it $Q$) can be defined by its Radon-Nikodym derivative:
\begin{equation} \frac{dQ_T}{dP} = \exp\bigg\{ -\int_0^T \gamma^\top(s) dB_s - \frac{1}{2} \int_0^T \gamma^2(s) ds \bigg\}. \end{equation}
Now, as you have already noticed, what I am calling $\gamma(t)$ should in our case be equal to $\frac{\mu - r}{\sigma} = \frac{\mu}{\sigma}$. Then $d\hat{B}_t = \frac{\mu}{\sigma} dt +dB_t$ and we have: \begin{equation} dS_t = \sigma d\hat{B}_t \end{equation} as desired. So, our EMM, $Q = Q_T \sim P$, is defined by the Radon-Nikodym derivative: \begin{align} \frac{dQ_T}{dP} &= \exp\bigg\{ -\int_0^T \frac{\mu}{\sigma} dB_s - \frac{1}{2} \int_0^T \frac{\mu^2}{\sigma^2} ds \bigg\} \\ &= \exp \Big\{ -\frac{\mu}{\sigma}B_T - \frac{1}{2} \frac{\mu^2}{\sigma^2}T \Big\} \end{align}Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.