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.


1 Answer 1


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

  • $\begingroup$ $\frac{d}{dS(t)} E_t[(S(T)-K)\textbf{1}_{S(T) > K}]= E_t[\frac{d}{dS(t)}(S(T)-K)\textbf{1}_{S(T) > K}]$... This is indirect application of Leibniz, right? In that case; we need to ensure contuinuty. As far as I know: $(S(T)-K)\textbf{1}_{S(T) > K}$ is not contnous? $\endgroup$
    – Sanjay
    Mar 21, 2018 at 20:21
  • $\begingroup$ I forgot to tag you before just wanted to make sure if you see the comment $\endgroup$
    – Sanjay
    Mar 21, 2018 at 21:13
  • $\begingroup$ Yes. $x \mapsto (x-K)^+$ is continuous and differentiable. The derivative is discontinuous but it does not matter. $\endgroup$ Mar 22, 2018 at 7:42
  • $\begingroup$ It does not appear correct that $\frac{d}{dS(t)}(S(T)-K)\pmb{1}_{S(T)>K} = \pmb{1}_{S(T)>K}$. $\endgroup$
    – Gordon
    Mar 22, 2018 at 15:15
  • $\begingroup$ $\frac{d}{dS(t)} (S(T)-K)1_{S(T)>K} = \frac{dS(T)}{dS(t)} 1_{S(T)>K}$ $\endgroup$ Mar 22, 2018 at 15:35

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