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Reading: What are some useful approximations to the Black-Scholes formula? I understand that a ATM Call option can be approximated to $$ C(S,t)≈0.4Se^{−r(T−t)}σ \sqrt{T−t}$$ Also, I often hear that an ATM delta is around $\Delta = 0.5$. However, using approximation formaula of an ATM call option price, gives: $$\Delta = 0.4σ \sqrt{T−t}$$ which is significantly lower than the financial considered delta.

This question came to my mind, while I was questionning myself what would be the $\Delta$ of a financial product $F$ paying at $t_1$ an ATM Call option with maturity $T$.

$$ Flow(T) = [S(T) - S(t_1)]_+ , with 0 < t_1 < T $$

Reasonning made me to conclude that in $t_1$, the product price would be an ATM Call option with remaining maturity $ T-t_1$ , so $\Delta(t_1) = 0.4σ \sqrt{T−t_1}$.

But should it be this, or $\Delta = 0.5$ ?

What should be the $ \Delta$ at $t$ with $0 < t < t_1$ ? $\Delta(t_1) = 0.4σ \sqrt{T−t}$ ? Or should it be $0$ for $t < t_1$, and then $\Delta(t_1) = 0.5$ ?

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The difference between the $\Delta=0.5$ and the $\Delta=0.4\sigma\sqrt{T-t}$ is that the latter refers to an option which is always ATM- that is, the strike floats when you bump the stock price to test the delta. This is because when you created the approximation formula for the ATM option you set K=S.

Regarding the second part of your question, I believe it is essentially correct. The delta in the region $0<t<t_1$ is basically the same as the delta at $t_1$. Since, at $t$, you need to hold a position in a forward on the stock deliverable at $t_1$. But the hedge for that is to own the same amount of stock at $t$.

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  • $\begingroup$ Could you explain in which situations it's useful to consider always ATM options ? I understand in such situations $\Delta \approx 0$, since $\sigma \sqrt{T-t} $ is considered small ? - which makes sense since an always ATM options will pay $(S(T) - K)_+ = (S(T) - S(T))_+ = 0 $, thus $\Delta \approx 0$ $\endgroup$
    – user25844
    Nov 29, 2022 at 11:14
  • $\begingroup$ Well there are no such options that are ATM all the way to expiration, since they are worth 0 by definition. But there are lots of applications where the strike is set at some $t_1$ where $t<t_1<T$ $\endgroup$
    – dm63
    Nov 29, 2022 at 11:23

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