# 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.

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}
• Thank you! I am a bit lost when u introduced the Radon-Nikodym derivative. Could you explain how did you get $\frac{d Q_{T}}{d P}=\exp \left\{-\int_{0}^{T} \gamma^{\top}(s) d B_{s}-\frac{1}{2} \int_{0}^{T} \gamma^{2}(s) d s\right\}$ May 13 at 13:58