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Why Black–Scholes Uses Different Drift Adjustments in d1 and d2

Article Quant Q&A · Author: Anon
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Summary

The document addresses why the Black–Scholes call formula contains different volatility adjustments in d1 and d2 under risk-neutral pricing. It outlines a derivation route: solve the stock price process with its drift set to the risk-free rate, then express the call payoff as the stock value when the option finishes in the money minus the strike value in that same event.

The strike component is associated with the probability term involving d2. The stock-value component requires a separate calculation and leads to d1, rather than reusing d2. This distinction helps explain why the two quantities have opposite half-variance adjustments: they arise from different weighted expectations in the payoff decomposition. The answer is a concise roadmap, not a full derivation; it directs readers to a tutorial for the detailed calculation and does not discuss extensions such as dividends or alternative models.

Key ideas

  • Under risk-neutral pricing, solve the stock process using the risk-free drift.
  • A call payoff can be decomposed into a stock-value term and a strike-value term, each conditional on finishing in the money.
  • The strike component leads to the probability term involving d2.
  • The stock-value expectation is different and leads to d1 rather than d2.
  • The answer sketches the derivation but leaves the detailed calculation to further study.

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OptionsDerivatives pricingStatistics
Full text
# Drift Term in Black Scholes d1, d2


# Drift Term in Black Scholes d1, d2












I am relatively new to quant finance and have been learning the Black-Scholes formula. I have learned that the terminal distribution of $S_T$ is lognormal and that stock returns are expected to follow a geometric Brownian motion (GBM):

$$ dS_t = \mu S_t dt + \sigma S_t dW_t $$

I am a little confused in the derivation of $d_1$ and $d_2$:

$$ d_1 = \frac{\ln(S_0 / K) + (r + \frac{1}{2} \sigma^2)T}{\sigma \sqrt{T}} $$

$$ d_2 = \frac{\ln(S_0 / K) + (r - \frac{1}{2} \sigma^2)T}{\sigma \sqrt{T}} $$

The formulas make sense, except for the drift term $+\frac{1}{2} \sigma^2$ in $d_1$ and $-\frac{1}{2} \sigma^2$ in $d_2$. My understanding is that this drift term arises from the mean of the lognormal distribution and thus the expected value $E[S_T]$. However, since Black-Scholes assumes a risk-neutral martingale measure, it should adjust for this drift term to remove it.

What confuses me is why the drift term is subtracted in $d_2$ and added in $d_1$. Would appreciate any help!

## Answer by Andrea (score 1)

https://quant.stackexchange.com/a/81793

Without repeating too much what is in the link

- solve the SDE (and replace $\mu$ with $r$)

- write call price as $\mathbb{E}[S \, \mathbb{I}_{S>K}]-\mathbb{E}[K \,\mathbb{I}_{S>K}]$

- Compute 2nd term (trivial): it will give you $N(d_2)$ (the easy one)

- At the same time, you will understand why the first cannot be $N(d_2)$ (this is the very important step)

- Find a BS derivation tutorial to compute the $N(d_1)$ of the 1st term

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.