Deriving Black–Scholes Gamma by Differentiating an Expectation
Summary
The post works through a derivation of call-option gamma by differentiating an option-price expectation and using the Dirac delta to evaluate the resulting integral. The key step is to recognize the density of the transformed variable as lognormal, apply the delta-function identity to select its value at the current underlying price, and obtain an expression involving the standard normal density evaluated at the Black–Scholes variable d2.
The answer also identifies errors in the source derivation and states the corrected relationship between the d2 density expression and the familiar gamma formula involving d1. It verifies that identity by substituting d1 as d2 plus volatility times the square root of time and simplifying the exponential. The explanation is a mathematical derivation within the Black–Scholes framework; it does not discuss alternative models, numerical methods, or empirical validation. Care is needed to distinguish the normal density from the cumulative normal distribution when reading the formulas.
Key ideas
- Differentiating the option-price expectation twice introduces a Dirac delta term.
- The transformed variable has a lognormal density, which supplies the integrand evaluated by the delta identity.
- The resulting gamma expression uses the normal density at d2 and can be rewritten using d1.
- The derivation corrects typographical errors in the referenced material and assumes the Black–Scholes setup.
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Full text
# Gamma derivation from the expectation
# Gamma derivation from the expectation
I am trying to derive Gamma from the expectation principle (differentiating under expectation sign). I understand these steps
$\frac{d^2 C}{d x^2} = e^{-r\tau} \mathbb{E} [ \frac{\partial}{\partial x}Y 1_{[xY>K]}] = e^{-r\tau} \mathbb{E} [ Y \delta(x-K/Y)] = e^{-r\tau} \mathbb{E} [ K/U \delta(x-U)]$= $Ke^{-r\tau} \mathbb{E} [\frac{\delta(x-U)}{U}]$, where $U=K/Y$
but the last transformations to $Ke^{-r\tau} \int^{\infty}_0 \frac{\delta(x-u)}{u}\frac{\text{exp}(-\frac{1}{2} d_2(u)^2)}{u\sqrt{2\pi}\sigma \sqrt{\tau}}du=\frac{Ke^{-r\tau}}{x^2}\Phi'(d_2(x))=\frac{\Phi(d_1)}{x\sigma \sqrt{\tau}}$ are very confusing for me.
The problem is illustrated in http://www.gold-saucer.org/math/diff-int/diff-int.pdf at page 11.
## Answer by rollerboller (score 2)
https://quant.stackexchange.com/a/17701
The clue was to establish that there are typos in the script. Hence, I should have aimed to prove $\frac{Ke^{-r\tau}}{S^2\sigma\sqrt{\tau}}\Phi'(d_2)=\frac{\Phi'(d_1)}{S\sigma \sqrt{\tau}}$.
Therefore, we have
\begin{equation*} \frac{\partial^2 C}{\partial S^2}=Ke^{-r\tau}\mathbb{E}[\frac{\delta(S-U)}{U}] = Ke^{-r\tau} \int^{\infty}_0 \frac{\delta(S-u)}{u} \overbrace{\frac{\text{exp}(-\frac{(d_2(u))^2}{2})}{u\sqrt{2\pi}\sigma \sqrt{\tau}}}^{\text{log-normal pdf for U}} du \end{equation*}
From the properties of Dirac delta, we know that $$\int^{\infty}_0 \delta(x-u)h(u)du=h(x)$$ Hence, \begin{equation} \frac{\partial^2 C}{\partial S^2}= \frac{Ke^{-r\tau}}{S^2\sigma \sqrt{\tau}}\frac{1}{\sqrt{2\pi}}e^{-\frac{(d_2(S))^2}{2}} = \frac{Ke^{-r\tau}}{S^2\sigma \sqrt{\tau}} \Phi'(d_2) \end{equation}
This can be shown that $\frac{Ke^{-r\tau}}{S^2\sigma\sqrt{\tau}}\Phi'(d_2)=\frac{\Phi'(d_1)}{S\sigma \sqrt{\tau}}$ holds, as \begin{align*} \Phi'(d_1)=\frac{1}{\sqrt{2\pi}}e^{-\frac{\left(d_1\right)^2}{2}}& =\frac{1}{\sqrt{2\pi}}e^{-\frac{\left(d_2+\sigma\sqrt{\tau}\right)^2}{2}} \\ &= \text{exp}(-d_2\sigma\sqrt{\tau}-\frac{\sigma^2\tau}{2})\frac{1}{\sqrt{2\pi}}e^{-\frac{\left(d_2 \right)^2}{2}} \\ &= \frac{K}{S}e^{-r\tau}\Phi'(d_2) \end{align*}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.