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Risk-Neutral Pricing of Bachelier Calls at Zero Interest Rates

Article Quant Q&A · Author: Sanjay

Summary

The answer explains how to move from a Bachelier underlying with an unspecified physical drift to a risk-neutral process for pricing a European call. It starts with the discounted asset price and uses the martingale condition to identify the required drift under the pricing measure. When the interest rate is zero, that drift vanishes, leaving an arithmetic Brownian motion with constant volatility.

The answer then points toward applying the standard Bachelier payoff expectation with the zero-rate process. It outlines a change of measure using a drift adjustment and a Doléans-Dade exponential. The argument assumes the relevant integrability condition, such as Novikov’s condition, holds; it does not work through a full call-price derivation or provide all technical checks needed to ensure the local martingale is a true martingale.

Key ideas

  • Risk-neutral pricing requires the discounted underlying price to be a martingale.
  • Under the risk-neutral measure, the asset drift is the risk-free rate multiplied by its price.
  • With zero rates, the Bachelier process has no drift under the pricing measure.
  • A change of measure can adjust the physical drift, subject to appropriate integrability conditions.

Tags

Full text
# Call option with underlying following a Bachelier process


# Call option with underlying following a Bachelier process












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?

## Answer by Daneel Olivaw (score 4, accepted)

https://quant.stackexchange.com/a/38868

> "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.

References

$[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.

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