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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?

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    $\begingroup$ 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. $\endgroup$
    – q.t.f.
    Jan 21, 2017 at 0:30

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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.

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  • $\begingroup$ I dont have the time right now, but I bet I could furnish a model-independent proof of this. $\endgroup$
    – Kostas
    Dec 7, 2020 at 14:24
  • $\begingroup$ This comment is not helpful whatsoever. $\endgroup$ Dec 7, 2020 at 17:48
  • $\begingroup$ 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 $\endgroup$
    – Kostas
    Feb 27, 2021 at 17:52
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    $\begingroup$ 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. $\endgroup$ Feb 28, 2021 at 23:56
  • $\begingroup$ 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. $\endgroup$
    – Kostas
    Mar 2, 2021 at 12:02

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