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Why At-the-Money Exercise Probability Differs from Black-Scholes D2

Article Quant Q&A · Author: Alex Kofman
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Summary

The document examines a common misunderstanding in Black-Scholes: with spot equal to strike and a zero interest rate, the exercise probability is not generally one half. Under the model, the probability of finishing in the money is represented by the cumulative normal value at D2, but setting spot equal to strike does not make D2 zero when volatility and time to maturity are positive.

The explanation distinguishes the median of the terminal stock price from the median of its logarithm. Under the lognormal model, the logarithm has a negative drift adjustment, so the median terminal price lies below the forward expectation. Thus, a strike equal to today’s spot is not necessarily the median terminal price. The brief answer gives the key intuition but does not derive the full risk-neutral distribution or discuss the effect of a nonzero interest rate.

Key ideas

  • In Black-Scholes, the exercise probability is given by the cumulative normal value at D2.
  • When spot equals strike and the risk-free rate is zero, the volatility term can still make D2 negative.
  • The expected terminal price and its median differ because the stock price is lognormally distributed.
  • A strike at the terminal price median, rather than at the current spot by default, corresponds to a one-half exercise probability.

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# Black-Scholes: If exercise probability is 0.5, should $D_2$=0?


# Black-Scholes: If exercise probability is 0.5, should $D_2$=0?












Let's say we have option strike price equal to current stock price. And we have zero risk-free rate. In this case I assume that probability of exercise is 0.5 because chances that price will go up or down are equal.

As I understand $N(D_2)$ is exactly the probability of exercise and it should be 0.5 . It means that $D_2$ should be 0 in this case. But if we put equal current and strike price and zero rate in its formula it will not be zero:

$D_2=(Ln(S/K) + (r - 0.5*Vol^2)*t)/(Vol*Sqrt(t))$

Here $Ln(S/K)$ and $r$ become zero, but we still have $-0.5*Vol^2*t/(Vol*Sqrt(t))$ that is not zero.

Where is the mistake?

## Answer by Mark Joshi (score 5)

https://quant.stackexchange.com/a/16381

expectations and medians are different things. If the strike is at the median, you get 0.5. However, the median of the log is lower and is

$$ \log S_0 - 0.5 \sigma^2 T $$ The median of $S_T$ is the exp of this.

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.