Why an In-the-Money Call’s Delta Can Rise at High Volatility
Summary
The document explains why the Black–Scholes delta of an in-the-money call can first decline and later increase as volatility rises. Interpreting delta as the risk-neutral probability of exercise, it separates two effects: greater volatility makes the strike closer in standardized distribution terms, while the lognormal model’s convexity adjustment raises the expected stock level. At sufficiently high volatility, the latter effect can dominate for a call.
The explanation relies on the underlying price being bounded below by zero while the call payoff is driven by ordinary price changes rather than log returns. It also notes an asymmetry: the same pattern does not apply to an in-the-money put, since both effects described reduce its moneyness. The discussion is an intuition for the Black–Scholes model, not an empirical result, and it does not provide a derivation or address deviations from model assumptions such as changing volatility or non-lognormal returns.
Key ideas
- An in-the-money call’s delta can decline and then rise as volatility increases.
- Higher volatility reduces standardized moneyness, which initially lowers the call’s exercise probability.
- The lognormal model’s convexity adjustment can eventually outweigh that effect for calls.
- The explanation attributes the asymmetry to the stock price’s zero lower bound and the call payoff structure.
- The described pattern does not apply to in-the-money puts.
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Full text
# Black-Scholes: Delta/probability of exercise increases with volatility
# Black-Scholes: Delta/probability of exercise increases with volatility
The delta for an ITM call option with increasing volatility initially decreases, reaches a global minimum, and then increases.
If we consider delta as a representation of risk-neutral probability of exercise, the first segment of the graph - decreasing with increasing volatility - can be intuitively explained. A lower volatility provides less opportunity for the underlying to move such that the option becomes out of the money.
What is the intuition that for an ITM option, at some point the probability of exercise increases with increasing volatility? Note that this question asks for an intuitive explanation rather than a BS derived equation demonstrating this.
## Answer by hjw (score 1, accepted)
https://quant.stackexchange.com/a/41116
** Edited for more clarity
Its because
- Stock is bounded by 0 so volatility is on log returns $log(\frac{ST}{S0})$
- Option payoff is based on non log returns $\frac{ST}{S0}-1$.
Initially when vol is small and increases, the boundary doesn't affect the realistic range of stock prices. When vol becomes really large, your upside moves, becomes disproportionately larger vs your downside moves bounded by 0.
This is represented as a positive convexity adjustment in black-scholes. So moneyness is affected in 2 different ways due to vol for an ITM Call.
- When vol increases, you are no longer that far from the strike on a distribution basis. So moneyness decreases with the following term in BS getting smaller
$$\frac{log(\frac{S}{K})}{\sigma\sqrt{T-t}}$$
- However convexity adjustment gets more positive making you more ITM as represented by the following term in BS getting larger
$$\frac{\frac{1}{2}\sigma^2(T-t)}{\sigma\sqrt{T-t}}$$
When vol is large, impact of 2. is relatively larger since if you think about it, for large enough vol, you are basically very close to ATM from a distribution perspective and this doesn't change much even if vol increases. Convexity adjustment however continues to increase.
You should find that the phenomenon you described doesn't apply for ITM Puts. Since the both 1. & 2. serves to reduce the moneyness of PutsShown 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.