Deriving the Black–Scholes Strike for a 0.50 Delta Call
Summary
This discussion derives the strike at which a Black–Scholes call has delta 0.50 and clarifies why that strike need not equal the usual forward or spot-based at-the-money strike. Under the Black–Scholes call delta, the condition is that d1 equals zero. Solving the stated d1 expression gives a strike that depends on the spot price, interest rate, volatility, and time to expiry; with zero rates and a one-year expiry, it reduces to the volatility-adjusted expression cited in the question.
The answers distinguish this delta-based strike from the conventional at-the-money definition. At the conventional strike, d1 is generally positive, so call delta exceeds 0.50, though the difference is small for short maturities. The discussion also cautions against interpreting N(d2) as a real-world probability: it is a risk-neutral quantity used in pricing. The derivation relies on the Black–Scholes framework and its assumptions; it does not address alternative models or market-implied skew.
Key ideas
- A Black–Scholes call has delta 0.50 when d1 is zero.
- Solving d1 equal to zero gives a strike adjusted for rates, volatility, and expiry.
- The conventional at-the-money strike generally does not produce exactly 0.50 call delta.
- N(d2) is a risk-neutral pricing probability, not a real-world likelihood.
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# Calculating the 0.50 delta strike
# Calculating the 0.50 delta strike
According to most books the ATM option is the option with a delta of 0.50. However, this is only the case when the distribution is normal. The more positively skewed the distribution, the further the 0.50 delta option is out-of-the-money (for calls). According to the following article, the formula to calculate the 0.50 delta option strike is equal to:
```
S x e^(σ^2/2)
```
I want to know why this is exactly the case. Looking at the delta defintion I have:
```
delta = N(d1) = 0.50
```
Therefore,
```
d1 = 0
```
And
So, how do I get from this well-known formula to the above mentioned formula? Thanks in advance,
## Answer by CABLE (score 4)
https://quant.stackexchange.com/a/54780
The delta you mentioned is the Black-Scholes delta. If you let $r=0$, $T=1$ and solve the equation $d_1=0$, you get what is in the article.
## Answer by Jan Stuller (score 2)
https://quant.stackexchange.com/a/54781
As you point out in your d1 formula:
$$d_1 = \frac{ln \left( \frac{S}{K} \right)+\left(r+0.5\sigma^2 \right)T}{\sigma \sqrt{T}} $$
Therefore, $N(d_1)$ (where $N(.)$ stands for the Standard Normal CDF) is only equal to half when $d_1$ is exactly zero. When an option is ATM, then $S=Ke^{-rT}$. So $N(d_1)$ won't be exactly 0.5, because:
$$d_1 = 0.5\sigma\sqrt(T)$$
For short dated options, $N(d_1)$ of the above will be close to 0.5, whilst for longer-dated options (like 10-year expiry) it will be higher than 0.5.
Indeed, if you set: $S=Ke^\left(-0.5\sigma^2T-rT \right)$, you will set $d_1$ to zero.
People who say that:
(i) $N(d_1)$ for ATM options is exactly half
(ii) ATM option has $N(d_2)$ equal to half because $N(d_2)$ is the probability that the option will end up in the money
Are (in my experience) mostly option traders who lack the technical knowledge to understand how option pricing works. $N(d_2)$ is the risk-neutral probability, so has nothing to do with "likelihood" or "real-world probability" as we humans like to interpret probability. "Risk-neutral" probability is a mathematical construct invented for pricing options.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.