Deriving a Black–Scholes Identity from the Definitions of d₁ and d₂

Summary

The document shows why two exponential terms involving the Black–Scholes variables d₁ and d₂ are equal. It substitutes the standard definitions of those variables, using d₂ = d₁ minus volatility times the square root of time to expiry, into the strike-side term. Expanding the resulting square introduces a cross term that can be replaced using the formula for d₁. The strike and spot ratio then simplifies the expression to the spot-side term, including the dividend discount factor.

This is an algebraic identity, not a pricing strategy or empirical result. It helps verify a step in Black–Scholes derivations, but the explanation assumes the stated definitions and consistent notation for spot, strike, rates, dividends, volatility, and time. It does not discuss numerical implementation, model assumptions, or the broader derivation of the option pricing formula.

Key ideas

• The identity follows by substituting the definitions of d₁ and d₂ into the strike-side exponential term.
• Expanding d₂ squared creates a cross term that is simplified with the formula for d₁.
• The strike and spot ratio converts the remaining expression into the spot-side term.
• The result is an algebraic verification within the stated Black–Scholes notation.

Tags

# Result linked to Black-Scholes evaluation


# Result linked to Black-Scholes evaluation












Why does this

$$Se^{-D(T-t)}e^{-d_1^2/2} - Ee^{-r(T-t)}e^{-d_2^2/2}$$

equal to $0$? (Where $E$ is a strike)

## Answer by Gordon (score 3, accepted)

https://quant.stackexchange.com/a/21991

> Note that this question is similar to Verifying an identity of an equation for Black Scholes formula.

You need to use the fact that \begin{align*} d_1 &= \frac{\ln (S/E) + (r-D)(T-t) + \frac{\sigma^2}{2}(T-t)}{\sigma \sqrt{T-t}}, \\ d_2 &= d_1 - \sigma \sqrt{T-t}. \end{align*} Then, \begin{align*} E e^{-r(T-t)} e^{-d_2^2/2} &=E e^{-r(T-t)} e^{-d_1^2/2 - \frac{\sigma^2}{2}(T-t)+d_1 \sigma \sqrt{T-t}}\\ &=E e^{-r(T-t)} e^{-d_1^2/2 - \frac{\sigma^2}{2}(T-t)+\ln (S/E) + (r-D)(T-t) + \frac{\sigma^2}{2}(T-t)}\\ &=Se^{-D(T-t)} e^{-d_1^2/2}. \end{align*}

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.