Bachelier Call Delta and Risk-Neutral Exercise Probability
Summary
The document examines whether a call option's delta equals its probability of expiring in the money under the Bachelier model. It first corrects a related misconception: Black–Scholes call delta is not generally the exercise probability. For a forward following arithmetic Brownian motion under the risk-neutral measure, the answer derives the call value by integrating the positive payoff over the normal terminal distribution, then differentiates that value with respect to the initial forward.
The resulting forward delta is the discounted normal cumulative probability that the forward exceeds the strike at expiry. Thus, in this setup, delta and exercise probability differ only by the discount factor. The answer also discusses a spot option and applies the chain rule through the forward price, reporting the same normal probability for spot delta under its stated setup. These conclusions rely on the model, European exercise, and risk-neutral probability; they are not a general interpretation of option delta across models or contracts.
Key ideas
- Under the Bachelier model, the forward's terminal value has a normal distribution.
- Differentiating the European call value gives a forward delta equal to a discounted normal cumulative probability.
- The risk-neutral probability of finishing above the strike is the same cumulative probability before discounting.
- The result depends on the model assumptions and does not validate the general claim that delta is exercise probability.
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Full text
# Bachelier option delta = probability of exercise?
# Bachelier option delta = probability of exercise?
Under the Black-Scholes model, the delta of a call option is sometimes interpreted as the probability for the option to end in the money.
If I assume that the underlying follows a normal distribution (Bachelier model), does the same approximation hold for the Bachelier delta?
## Answer by LocalVolatility (score 3)
https://quant.stackexchange.com/a/31997
First, your statement that the delta of a call option under the Black-Scholes model is equal to the exercise probability is not true. This is a common misconception - see for example: this question.
Now regarding your question. Assume that the forward price $F$ for the maturity $T$ under the risk-neutral measure $\mathbb{Q}$ follows
\begin{equation} \mathrm{d}F_t = \sigma \mathrm{d}W_t \end{equation}
for $t \in [0, T]$. Then
\begin{eqnarray} C_0 & = & e^{-r T} \mathbb{E}_{\mathbb{Q}} \left[ \left( F_T - K \right)^+ \right]\\ & = & e^{-r T} \int_{-d}^\infty \left( F_0 + \sigma \sqrt{T} x - K \right) \phi(x) \mathrm{d}x\\ & = & e^{-r T} \left\{ \left( F_0 - K \right) \mathcal{N}(d) + \sigma \sqrt{T} \mathcal{N}'(d) \right\}, \end{eqnarray}
where
\begin{equation} d = \frac{F_0 - K}{\sigma \sqrt{T}}. \end{equation}
Carefully differentiating yields
\begin{eqnarray} \frac{\partial C_0}{\partial F_0} & = & e^{-r T} \mathcal{N}(d), \end{eqnarray}
where we use that
\begin{equation} \frac{\partial}{\partial F_0} \sigma \sqrt{T} \mathcal{N}'(d) = -d \mathcal{N}'(d). \end{equation}
The exercise probability is
\begin{equation} \mathbb{Q} \left\{ F_T > K \right\} = \int_{-d}^\infty \phi(x) \mathrm{d}x = \mathcal{N}(d). \end{equation}
So except for the discounting, the two expressions are the same in case of the Bachelier model.
Edit: The price of a spot asset $S$ doesn't follow an arithmetic Brownian motion under $\mathbb{Q}$. We obtain the delta of a European call option on it as
\begin{equation} \frac{\partial C_0}{\partial S_0} = \frac{\partial C_0}{\partial F_0} \frac{\partial F_0}{\partial S_0} = \mathcal{N}(d). \end{equation}
The exercise probability is still
\begin{equation} \mathbb{Q} \left\{ S_T > K \right\} = \mathcal{N}(d) \end{equation}
since $F_T = S_T$ at maturity.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.