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Why Call Delta Is Not the Probability of Expiring In the Money

Article Quant Q&A · Author: user9081230912390
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Summary

This note resolves an apparent Black–Scholes paradox: as volatility rises, call delta approaches one, while the put price also rises toward its upper bound. Put–call parity explains the price behavior: for fixed forward price and strike, call and put prices change by the same amount as volatility changes, and the put price is bounded by the discounted strike.

The probability confusion comes from treating delta as the risk-neutral probability of exercise. In Black–Scholes, the risk-neutral probability that a call expires in the money is represented by N(d2), whereas delta is N(d1). Thus a delta near one does not imply that the exercise probability is near one. The explanation is specific to the European option framework and Black–Scholes quantities described; it does not provide a broader treatment of probability measures or other pricing models.

Key ideas

  • Put–call parity makes the European call–put price difference independent of volatility when other inputs are fixed.
  • As volatility rises, call and put prices can both rise while maintaining that fixed difference.
  • A European put price is bounded above by the discounted strike.
  • In Black–Scholes, N(d2) represents the risk-neutral probability of a call expiring in the money.
  • Call delta, N(d1), is not that probability.

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OptionsVolatilityDerivatives pricingStatistics
Full text
# Paradox in option expiry as volatility goes to infinity


# Paradox in option expiry as volatility goes to infinity












As volatility goes to infinity, the delta of a call option goes to 1. The delta approximates the probability that the option expires in the money. So it seems that the probability of expiring in the money is very close to 1.

However, the price of the put option approaches the constant function at the strike price. This is what one would see if the probability of the call expiring out of the money is close to 1.

This seems to suggest that the probability that the call option expires in the money and out of the money is both close to 1.

Put another way, as the volatility increases, the probability of a call expiring in the money increases as well. But so does the price of a put option. So it would seem that the price of a put increases, even though the probability of the put expiring in the money decreases.

How to make sense of this?

## Answer by Quantuple (score 7, accepted)

https://quant.stackexchange.com/a/35645

Where you are right, because of call-put parity $$C-P=DF(F-K)$$ with $F$ representing the forward price, the difference between the European call and a European put price is independent of the volatility.

This suggests that when $C$ increases due to increasing volatility, $P$ should therefore increase by the same amount all other things equal. And indeed the maximum put price is $DF\times K$ at which point the call is worth $DF\times F$.

Where you are wrong is that, under the risk-neutral measure: $\Delta$ (or rather $N(d_1)$ in the BS equation) is not the probability of expiring in the money for a call, $N(d_2)$ is.

See also these questions: Probability of exercise in the Black-Scholes Model and Yet another question about the risk-neutral measure. Why is the risk-neutral probability of an infinitely volatile GBM 0?

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.