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Deriving Black–Scholes Delta by Truncating a Normal Expectation

Article Quant Q&A · Author: rollerboller
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Summary

The document explains the final step in deriving the Black–Scholes call delta from a discounted expectation. Differentiating the payoff with respect to the current stock price introduces the random gross-return factor multiplied by an indicator that the option finishes in the money. Because that return is lognormal, the exercise condition can be rewritten as a threshold condition on a standard normal variable, using the quantity d2.

The remaining expectation is evaluated by integrating the normal density over the threshold region. Completing the square in the exponent shifts the integration limit by volatility times the square root of time, producing d1; the resulting cumulative normal probability is the call delta. The derivation relies on the stated lognormal model and standard normal symmetry when changing the sign of the integration variable. It clarifies an algebraic transformation rather than presenting empirical evidence or addressing model assumptions beyond those given.

Key ideas

  • Differentiating the call payoff produces a return factor multiplied by an indicator for exercise.
  • The lognormal return assumption converts the exercise condition into a normal threshold involving d2.
  • The expectation can be written as an integral over a truncated standard normal distribution.
  • Completing the square shifts the bound to d1, yielding the cumulative normal probability for delta.

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Full text
# Delta derivation from the expectation


# Delta derivation from the expectation












I'm trying to understand the following transformation leading to Delta

$\frac{dC}{dx} = e^{-r\tau} \mathbb{E}[ \frac{\partial}{\partial x}\text{max}(xY-K,0)] = e^{-r\tau} \mathbb{E}[Y \textbf{1}(xY>K)] = e^{-\frac{\sigma^2}{2}\tau}\mathbb{E}[e^{-\sigma\sqrt{\tau}Z} \textbf{1}(Z>-d_2)] = \Phi(d_1)$

I get the first part, but I don't understand the last transformation.

$Y = e^{(r-\frac{\sigma^2}{2})\tau + \sigma \sqrt{\tau}Z}$, Z is Normal(0,1)

x - current stock price

Taken from: http://www.gold-saucer.org/math/diff-int/diff-int.pdf

## Answer by Gordon (score 4)

https://quant.stackexchange.com/a/17691

Since $Y=e^{(r-\frac{\sigma^2}{2})\tau + \sigma \sqrt{\tau}Z}$, then \begin{align*} xY > K \Leftrightarrow Z > -d_2, \end{align*} where \begin{align*} d_2 = \frac{\ln \frac{x}{K} + (r-\frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}. \end{align*} Consequently, \begin{align*} e^{-r\tau}\mathbb{E}\big(Y \mathbb{1}_{\{xY >K\}} \big) &= e^{-\frac{\sigma^2}{2}\tau}\mathbb{E}\big[e^{-\sigma\sqrt{\tau}Z}\mathbb{1}_{\{Z > -d_2\}} \big] \\ &= e^{-\frac{\sigma^2}{2}\tau}\mathbb{E}\big[e^{\sigma\sqrt{\tau}Z}\mathbb{1}_{\{Z < d_2\}} \big]\\ &= \int_{-\infty}^{d_2} e^{-\frac{\sigma^2}{2}\tau + \sigma\sqrt{\tau} x} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{d_2} e^{-\frac{\sigma^2}{2}\tau + \sigma\sqrt{\tau} x -\frac{x^2}{2}}dx\\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{d_2 + \sigma\sqrt{\tau}}e^{-\frac{x^2}{2}}dx\\ &= \Phi(d_1), \end{align*} where $d_1 = d_2 + \sigma\sqrt{\tau}$.

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