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Deriving the Black–Scholes Density Identity for Call Sensitivities

Article Quant Q&A · Author: Alfons Ingomar
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Summary

The question asks why the Black–Scholes quantities involving the stock price and the normal density at d1 equal the discounted strike multiplied by the normal density at d2. The accepted derivation substitutes the standard normal density, uses the relation between d1 and d2, expands the squared term, and applies the definition of d1 to simplify the exponential. This establishes that the two expressions match, so their ratio is one and its logarithm is zero.

The result is an algebraic identity within the Black–Scholes setup, not an independent pricing rule. Its derivation depends on the usual definitions of d1 and d2 and their inputs, including volatility and time to expiry. Another response begins the same expansion but does not complete it, so the document’s useful mathematical explanation comes from the accepted answer. It does not discuss model fit or empirical limits of Black–Scholes.

Key ideas

  • The standard normal density is proportional to the exponential of minus one half the squared argument.
  • The Black–Scholes variables satisfy d2 equals d1 minus volatility times the square root of time to expiry.
  • Expanding the squared d2 term exposes components that cancel the discount factor and strike ratio.
  • The resulting density-weighted expressions are equal, making their logarithmic ratio zero.
  • The identity relies on the Black–Scholes definitions and does not establish empirical validity of the model.

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# Verifying an identity of an equation for Black Scholes formula


# Verifying an identity of an equation for Black Scholes formula












I just started working on the Black Scholes formula with help of the book Financial option valuation by Higham. Apparently you are possible to derive the following function:

$\log(\frac{SN'(d_1)}{e^{-r(T-t)}EN'(d_2)}) = 0$

From the Black scholes formula: $C(S,t)=SN(d_1)-Ee^{-r(T-t)}N(d_2)$

I've been puzzling arround but I'm stuck. This is where I came so far, do you know where I'm going wrong?

$\log(\frac{SN'(d_1)}{e^{-r(T-t)}EN'(d_2)}) = \log(SN'(d_1))-\log(e^{-r(T-t)}EN'(d_2))=0$

## Answer by Gordon (score 3, accepted)

https://quant.stackexchange.com/a/21395

I am trying to fill in what Richard left for the second part: \begin{align*} \exp(-r(T-t))E\, N'(d_2) &= \frac{1}{\sqrt{2\pi}}\exp(-r(T-t))E\, \exp\left(-\frac{1}{2}d_2^2\right) \\ &=\frac{1}{\sqrt{2\pi}}\exp(-r(T-t))E\, \exp\left(-\frac{1}{2}\big(d_1-\sigma\sqrt{T-t}\,\big)^2\right) \\ &=\frac{1}{\sqrt{2\pi}}\exp(-r(T-t))E\\ &\qquad\qquad \exp\left(-\frac{1}{2} d_1^2 -\frac{1}{2}\sigma^2 (T-t) + d_1 \sigma\sqrt{T-t}\right)\\ &=\frac{1}{\sqrt{2\pi}}\exp(-r(T-t))E\\ &\qquad\qquad \exp\left(-\frac{1}{2} d_1^2 +\ln(S/E) + r(T-t)\right)\\ &=\frac{1}{\sqrt{2\pi}} S \, \exp\left(-\frac{1}{2}d_1^2\right)\\ &= SN'(d_1). \end{align*} That is, \begin{align*} \ln\frac{SN'(d_1)}{\exp(-r(T-t))E\, N'(d_2)} = 0. \end{align*}

## Answer by Richi Wa (score 1)

https://quant.stackexchange.com/a/21378

The numerator is $$ S N'(d_1) = S \frac{1}{\sqrt{2 \pi}} \exp(-1/2 d_1^2) = \\ S \frac{1}{\sqrt{2 \pi}} \exp\left(- \frac12 \left(\log(S/E)+ (r + \frac12 \sigma^2(T-t)) \right)^2 / \sigma^2 (T-t) \right) $$ the denominator is: $$ \exp(-r (T-t)) E N'(d_2) = \\ E \frac{1}{\sqrt{2 \pi}} \exp\left(- \frac12 \left(\log(S/E)+(r- \frac12 \sigma^2(T-t)) \right)^2 / \sigma^2 (T-t) -r(T-t) \right). $$ Now what if we extend the square, match exp and log and if then nominator and denominator are equal we get the result.

EDIT: I did not finish the calculation. But the latex above is too much for a comment. Maybe you can do the calculation. It not that clear to see whether the claim is true...

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