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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

href="https://www.codecogs.com/eqnedit.php?latex=d1&space;=&space;\frac{ln(\frac{S}{K})+(r+\frac{^{\sigma^{2}}}{2})T}{\sigma\sqrt{T}}

So, how do I get from this well-known formula to the above mentioned formula? Thanks in advance,

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2 Answers 2

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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.

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  • $\begingroup$ $0=\frac{\ln(S/K)+(0+\sigma^2/2)*1}{\sigma*1}$ 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 $K=S e^{0.5\sigma^2}$ $\endgroup$
    – nbbo2
    Jun 9, 2020 at 8:48
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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.

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    $\begingroup$ $N(d_1)$ 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. $\endgroup$
    – Kevin
    Jun 9, 2020 at 8:37
  • $\begingroup$ @KeChen: you're totally right, careless mistake. $\endgroup$
    – Jan Stuller
    Jun 9, 2020 at 8:53

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