# 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?

• Clarification : In Black-Scholes, the undiscounted "forward" delta is the probability under share measure (foreign measure, in the FX context) of a call finishing in the money. Jan 21 '17 at 0:30

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

$$\mathrm{d}F_t = \sigma \mathrm{d}W_t$$

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

$$d = \frac{F_0 - K}{\sigma \sqrt{T}}.$$

Carefully differentiating yields

\begin{eqnarray} \frac{\partial C_0}{\partial F_0} & = & e^{-r T} \mathcal{N}(d), \end{eqnarray}

where we use that

$$\frac{\partial}{\partial F_0} \sigma \sqrt{T} \mathcal{N}'(d) = -d \mathcal{N}'(d).$$

The exercise probability is

$$\mathbb{Q} \left\{ F_T > K \right\} = \int_{-d}^\infty \phi(x) \mathrm{d}x = \mathcal{N}(d).$$

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

$$\frac{\partial C_0}{\partial S_0} = \frac{\partial C_0}{\partial F_0} \frac{\partial F_0}{\partial S_0} = \mathcal{N}(d).$$

The exercise probability is still

$$\mathbb{Q} \left\{ S_T > K \right\} = \mathcal{N}(d)$$

since $F_T = S_T$ at maturity.

• I dont have the time right now, but I bet I could furnish a model-independent proof of this. Dec 7 '20 at 14:24
• This comment is not helpful whatsoever. Dec 7 '20 at 17:48
• of course it is not helpful. Nonetheless I have a proof that $\partial C /\partial K = -Q(S>K)$ under very general processes. In Bachelier the option price is a function of $S-K$, so we get $\delta=\partial C /\partial S = - \partial C /\partial K = Q$. Otherwise, does $\partial C /\partial K$ even have a "greek" name? @LocalVolatility Feb 27 '21 at 17:52
• Its often called dual delta and the result you mention is extremely well known. Its typically an intermediate step in the derivation if the Breeden-Litzenberger formula. Feb 28 '21 at 23:56
• Thanks! I thought it might be known but didnt know who and when. Looks like I had rederived Breeden-Litzenberger too. I actually trade these implied densities. Mar 2 '21 at 12:02