Deriving Bachelier Call Delta as the Probability of Exercise

Summary

The document derives the delta of a European call when the underlying follows a Bachelier process, with zero drift and zero interest rates. The option value is the expected positive part of the terminal underlying price minus the strike. Since the terminal price changes one-for-one with the current price, differentiating the payoff with respect to the current price yields an indicator that the option finishes in the money.

Taking the conditional expectation of that indicator makes delta equal to the probability of exercise under the model. For a normally distributed terminal price, this probability is the standard normal cumulative distribution evaluated at the current price minus strike, scaled by volatility and the square root of time remaining. The result depends on the stated assumptions; discounting, nonzero drift, or other dynamics would change the derivation or expression. The source provides the analytical argument, without empirical tests or extensions to other option styles.

Key ideas

• Under the stated zero rate and zero drift assumptions, the terminal Bachelier price moves one-for-one with the current price.
• Differentiating the call payoff gives an indicator for finishing above the strike.
• The call delta equals the conditional probability that the option expires in the money.
• Under normal price dynamics, that probability is expressed using the standard normal cumulative distribution.

Tags

# Bachelier model call: computation of delta of a call option


# Bachelier model call: computation of delta of a call option












The price of a call with a stock with Bachellier process as its underlying and zero interest rate is giving by: $$C(t)=(S(t)-K)\Phi(\frac{S(t)-K}{\sigma \sqrt{T-t}})+\sigma \sqrt{T-t} \phi(\frac{S(t)-K}{\sigma \sqrt{T-t}})$$ How do I compute/derive/proof the time $t$ Delta-hedge ratio: hence compute $dC(t)/dS(t)$?

I have tried to reach to the delta by looking at the answers to this question: Bachelier option delta = probability of exercise? but there's some differentiation in there which is not clear for me.

## Answer by Antoine Conze (score 3, accepted)

https://quant.stackexchange.com/a/38912

Zero interest rate and drift so $S(T) = S(t) + \sigma (W(T)-W(t))$ and $\frac{d S(T)}{dS(t)} = 1. $ $$ C(t) = E_t[(S(T) - K)^+] $$ $$ \frac{dC(t)}{dS(t)} = \frac{d}{dS(t)} E_t[(S(T) - K)^+] = E_t[\frac{d}{dS(t)} (S(T) - K)^+] = E_t[\frac{d S(T)}{dS(t)} \text{Indicator}(S(T) > K)]= E_t[\text{Indicator}(S(T) > K)]= \text{Prob}_t[S(T) > K]=\Phi(\frac{S(t)-K}{\sigma \sqrt{T-t}}) $$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.