# 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,

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.

• $0=\frac{\ln(S/K)+(0+\sigma^2/2)*1}{\sigma*1}$ gives $0=\ln(S/K)+\sigma^2/2$ gives $-\ln(S/K)=\sigma^2/2$ gives $S/K=e^{-0.5 \sigma^2}$ finally gives $K=S e^{0.5\sigma^2}$ – noob2 Jun 9 '20 at 8:48

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.

• $N(d_1)$ is not the standard risk neutral probability of the event $\{S_T\geq K\}$ though. This is $N(d_2)$. You need a change of numeraire. $N(d_1)$ correspond to the exercise probability under the stock measure which uses $S_t$ or $S_te^{qt}$ as numeraire. – Kevin Jun 9 '20 at 8:37
• @KeChen: you're totally right, careless mistake. – Jan Stuller Jun 9 '20 at 8:53