I am trying to reach a derivation/proof for how to price a call option when its underlying asset follows a Bachelier process with unknown drift term: $$dS_t=...dt+\sigma dW_t$$

but zero interest rate: $r = 0$.

I try to apply the methods described here: Bachelier model call option pricing formula but I need $r=0$

So I wonder: how will my proces look like when finding the arbitrage free price for the call? Will it necessarily be $dS_t=r S_tdt+\sigma dW_t$ (in this case no drift at all)?

In other words: how to find the right risk neutral measure and use it to derive a call option price at time $t<T$ for $T$ being expiration day of the call?

  • $\begingroup$ I hope my question is clear $\endgroup$
    – Sanjay
    Mar 19, 2018 at 17:42
  • $\begingroup$ Then simply take the results from the question you link to and input value $0$ to $r$. What is the issue? $\endgroup$ Mar 19, 2018 at 19:03
  • 1
    $\begingroup$ More specifically, to replace $\frac{1}{2r}\left(e^{2r(T-t)}-1\right)$ by $(T-t)$. $\endgroup$
    – Gordon
    Mar 19, 2018 at 19:07

1 Answer 1


"how to find the right risk neutral measure [...]?"

Specifically addressing this question, let us work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ filtered with $(\mathcal{F}_t)_{t \geq 0}$. We assume that the asset process under the physical measure $\mathbb{P}$ is: $$\text{d}S_t=\alpha(t,S_t)\text{d}t+\sigma\text{d}W_t$$

where $\alpha(t,S_t)$ is an appropriate$^{*}$ process adapted to the filtration $(\mathcal{F}_t)_{t \geq 0}$.

We know that under the risk-neutral measure $\mathbb{Q}$ the discounted asset price $X_t$ is a martingale. Applying Itô's lemma to $X_t=D_tS_t$, where $D_t =D(0,t)$ is the discount factor from $t$ to $0$: $$\begin{align} \text{d}X_t &= -rD_tS_t\text{d}t+D_t\text{d}S_t \\[3pt] & = -rD_tS_t\text{d}t+D_t\alpha(t,S_t)\text{d}t+\sigma D_t\text{d}W_t \\[3pt] & = (\alpha(t,S_t)-rS_t)D_t\text{d}t+\sigma D_t\text{d}W_t \quad\quad\quad \text{Eq. 1} \end{align}$$

Thus we need to change the drift of $S_t$ from $\alpha(t,S_t)$ to $rS_t$. Now let us define the following process: $$ \tilde{W}_t=W_t+\int_0^t\frac{\alpha(t,S_t)-rS_t}{\sigma}\text{d}t$$

The process $\tilde{W}_t$ is a Brownian Motion under the risk-neutral measure $\mathbb{Q}$ defined by the Doléans-Dade exponential of the adapted process $f(t)=(\alpha(t,S_t)-rS_t)/\sigma$. Then we can make the change of measure$^{\text{*}}$: $$\begin{align} \text{d}S_t & = \alpha(t,S_t)\text{d}t+\sigma\text{d}W_t \\[3pt] & = \alpha(t,S_t)\text{d}t+\sigma\left(\text{d}\tilde{W}_t-\frac{\alpha(t,S_t)-rS_t}{\sigma}\text{d}t\right) \\[3pt] & = rS_t\text{d}t+\sigma\text{d}\tilde{W}_t \end{align}$$

You can then pursue from this last equation in a similar manner as in Bachelier model call option pricing formula but replacing $r$ by $0$ which should simplify things:

$$ S_t=S_0+\sigma\tilde{W}_t \ \sim \ \mathcal{N}(S_0,\sigma^2t)$$

[Edit: why does $\text{Eq. 1}$ imply that the drift of the asset needs to be $rS_t\text{d}t$ under measure $\mathbb{Q}$? We know from risk-neutral theory that the price of a derivative is the expectation of its payoff under measure $\mathbb{Q}$ (see e.g. Proposition 2.9 from $[1]$); we therefore need the distribution of the asset under the risk-neutral measure; we also know that the discounted price of the asset, $X_t=D_tS_t$, is a martingale under $\mathbb{Q}$ (see e.g. Proposition 2.8 from $[1]$). On the other hand, (local) martingales can be characterised as a process without drift, i.e. for a martingale $M_t$: $$M_t=\sigma(t,M_t)\text{d}W_t$$ for some adapted process $\sigma(t,M_t)$, which can be a constant $\sigma$. Thus given $\text{Eq. 1}$, which gives the SDE of $X_t$, in order to make the discounted asset a (local) martingale we need to cancel its drift under $\mathbb{Q}$ by making the drift of the asset $S_t$ to be equal to $rS_t\text{d}t$ under that same measure. In practice, we always work with local martingales that are also martingales hence we do not bother to make all the technical checks. See the technical note from my answer to "Why discounted derivative price is a martingale?" for more details.]

$\text{*:}$ we would need to check Novikov's condition to ensure the change of measure is legit, we assume process $\alpha(t,S_t)$ is such that the condition is fulfilled.


$[1] $ Harrison, M. and Pliska, S. (1980). "Martingales and Stochastic Integrals in the Theory of Continuous Trading", Stochastic Process and their Applications 11, 215-260, North-Holland Publishing Company.

  • $\begingroup$ Actually I knew most of your (really good) answer. At some point u write:"we need to change the drift to rSdt". This is where I struggle. Maybe because I haven't understood Risk Neutral theory totally. Can you add a couple of word for WHY we need to change the drift to exactly " $rS_t dt$" $\endgroup$
    – Sanjay
    Mar 20, 2018 at 21:32
  • $\begingroup$ @Sanjay I updated my answer. $\endgroup$ Mar 20, 2018 at 23:17
  • $\begingroup$ Where did $r$ come from in $Eq. 1$? Are you assuming that $D_t = \exp(-rt)$? $\endgroup$
    – Confounded
    Nov 30, 2018 at 10:30
  • $\begingroup$ Indeed @Confounded. $\endgroup$ Nov 30, 2018 at 11:27

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