Forward ATM Option Delta and the Meaning of a Delta Bump
Summary
The note explains why the familiar delta of an at-the-money call can appear to conflict with a small-vega-style approximation proportional to volatility and the square root of time. The distinction is how “at the money” is maintained: in the approximation, the strike moves with the underlying when it is bumped, whereas conventional delta holds the strike fixed. The two calculations therefore answer different questions.
For a product that pays an at-the-money call at a future date, the answer argues that its delta before that date is approximately the delta of a forward position in the underlying deliverable then. Since hedging that forward requires holding the corresponding amount of stock, the delta does not drop to zero before the option becomes active. The discussion is conceptual and gives no numerical example or full derivation; the approximation is not a substitute for calculating delta under a specified strike and pricing model.
Key ideas
- The usual call delta holds the strike fixed when the underlying price changes.
- An approximation that continually resets the strike to the underlying measures a different sensitivity.
- Before the future option payment date, the product’s delta is linked to the stock position needed to hedge a forward deliverable on that date.
- The ATM approximation depends on its assumptions and should not be confused with a fixed-strike option delta.
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Full text
# Delta of a forward ATM option
# Delta of a forward ATM option
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 $\Delta(t_1) = 0.5$ ?
## Answer by dm63 (score 1, accepted)
https://quant.stackexchange.com/a/73911
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 $t$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.