The Bachelier Model

In this article we look at the Bachelier model.

We assume the dynamics

\[dS(t) = \mu S(t) dt + \sigma dW(t),\]

with $S(0) = S_0$ given and $W(t)$ a standard Wiener process. The drift $\mu S(t) dt$ can be removed with the transformation

\[X(t) = e^{\mu (T - t)} S(t),\]

which brings

\[dX(t) = -\mu e^{\mu(T - t)}S(t) dt + e^{\mu(T - t)}dS(t),\]

that is

\[\begin{aligned} e^{-\mu (T-t)}dX(t) + \mu S(t) dt & = dS(t) \\ & = \mu S(t)dt + \sigma dW(t), \end{aligned}\]

and therefore

\[dX(t) = e^{\mu(T - t)} \sigma dW(t).\]

This equation can be easily integrated, giving

\[X(T) - X(0) = \int_0^Te^{\mu(T - \tau)}\sigma dW(\tau),\]

that is,

\[S_T = e^{\mu T}S_0 + \int_0^Te^{\mu (T - \tau)} \sigma dW(\tau).\]

Using Ito’s isometry, it follows that

\[S_T \sim \mathcal{N}\left( e^{\mu T} S_0, \frac{\sigma^2}{2\mu} \left( e^{2\mu T} - 1 \right) \right),\]

meaning that the distribution of the spot at time $T$ follows a normal distribution with mean $e^{\mu T} S_0$ and variance $\frac{\sigma^2}{2\mu} \left( e^{2\mu T} - 1 \right)$.

At this point we have all the ingredients to compute the price of a call option:

\[\begin{aligned} C & = e^{-rT} \mathbb{E}[(S_T - K)^+] \\ & = e^{-rT} \mathbb{E}\left[(e^{\mu T} S_0 + \sqrt{\frac{\sigma^2}{2\mu} \left( e^{2\mu T} - 1 \right)} Z - K)^+\right] \\ & = e^{-rT} \sqrt{\frac{\sigma^2}{2\mu} \left( e^{2\mu T} - 1 \right)} \mathbb{E} \left[ \left( Z - \frac{K - e^{\mu T}S_0}{\sqrt{\frac{\sigma^2}{2\mu} \left( e^{2\mu T} - 1 \right)}} \right) \right], \end{aligned}\]

where $Z \sim \mathcal{N}(0, 1)$ is a standard normal. Our task reduces to the computation, for $a \in \mathbb{R}$, of

\[\begin{aligned} \mathbb{E}[(Z - a)^+] & = \mathbb{E}[(Z - a) \mathbb{1}_{Z > a}] \\ & = \mathbb{E}[Z\mathbb{1}_{Z > a}] - a \mathbb{P}[Z > a] \\ & = \frac{1}{2 \pi} \int_a^\infty z e^{-z^2/2} dz - a (1 - \Phi(a)), \end{aligned}\]

where $\Phi$ is the cumulative density function of the standard normal distribution. Using $\zeta = z^2/2$, we obtain

\[\begin{aligned} \mathbb{E}[(Z - a)^+] & = \frac{1}{2 \pi} \int_{a^2 /2}^\infty e^{-\zeta} d\zeta - a \Phi(-a) \\ & = - \frac{1}{2 \pi} \left. e^{-\zeta} \right|_{a^2/2}^\infty - a \Phi(-a) \\ & = - \frac{1}{2 \pi} e^{-a^2/2} - a \Phi(-a) \\ & = \varphi(a) - a \Phi(-a) \\ & = \varphi(-a) - a \Phi(-a), \end{aligned}\]

where $\varphi$ is the probability density function of the normal distribution. The call price follows by replacing $a$ with its value above.