# Derivation of Call Delta from Black Scholes Model

How is call delta mathematically derived from Black Scholes Model (without approximation) ? Please help me understand each step mathematically. And how it is approximated to say that delta is the probability of option expiring in the money?

Here's a mathematical derivation of the Black-Scholes delta.

The call option price under the BS model is $$C = S_0 N(d_1) - e^{-rT} K N(d_2) \quad\text{with}\quad d_{1,2} = \frac{\log(S_0\,e^{rT}/K)}{\sigma\sqrt{T}} \pm \frac12 \sigma\sqrt{T},$$ where $$N(x)$$ is the CDF of standard normal.

Using the properties, $$\frac{\partial d_1}{\partial S_0} = \frac{\partial d_2}{\partial S_0} = \frac{1}{S_0\sigma\sqrt{T}}$$ and $$\begin{gather*} d_1^2 - d_2^2 = (A+B)^2 - (A-B)^2 = 4AB = 2\log(S_0\,e^{rT}/K)\\ \quad\text{where}\quad A = \frac{\log(S_0\,e^{rT}/K)}{\sigma\sqrt{T}} \quad\text{and}\quad B = \frac{\sigma\sqrt{T}}{2}, \end{gather*}$$ we differentiate $$C$$ with resect to the spot price $$S_0$$: \begin{align*} D &= \frac{\partial C}{\partial S_0} = \frac{\partial}{\partial S_0}\left( S_0 N(d_1) - e^{-rT} K N(d_2) \right) \\ &= N(d_1) + S_0 n(d_1) \frac{\partial d_1}{\partial S_0} - e^{-rT} K n(d_2) \frac{\partial d_2}{\partial S_0} \\ &= N(d_1) + \frac{n(d_1)}{\sigma\sqrt{T}} \left( 1 - e^{(d_1^2-d_2^2)/2}\frac{K}{S_0e^{rT}} \right) \\ &= N(d_1) + \frac{n(d_1)}{\sigma\sqrt{T}} \left( 1 - \frac{S_0e^{rT}}{K}\cdot\frac{K}{S_0e^{rT}} \right) = N(d_1). \end{align*}

Look here for a detailed derivation of the formula for $$\Delta$$ (be aware that this particular website uses $$r_d$$ to denote the risk-free rate and $$r_f$$ to denote the dividend yield). You can always ask for more specific help regarding a particular step in the derivation.

It is easy to see that $$\mathbb{Q}[\{S_T\geq K\}]= \Phi(d_2)$$. Just replace $$S_T=S_0\exp\left(\left( r-q-\frac{1}{2}\sigma^2\right)T +\sigma\sqrt{T}Z\right)$$ where $$Z\sim N(0,1)$$ and isolate $$Z$$ on the left-hand side. This is the risk-neutral probability of expiring ITM. Note that $$\Delta=\Phi(d_1)=\Phi(d_2+\sigma\sqrt{T})\approx \Phi(d_2)$$. This is since $$\sigma\sqrt{T}$$ is typically very small.

• That derivation is excellent but it starts with a small typo: in BSCallPrice the roles of $r_f$ the risk free rate and $r_d$ the dividend rate are reversed. Oct 6, 2019 at 15:39
• I think it's not a typo because they use $r_f$ constantly and (as far as I checked) consistently to denote the dividend yield and $r_d$ as risk-free interest rate. So I believe they do this intentionally. However, I do agree with you that this notation is odd because $r_f$ should be the risk-free rate (or perhaps the rate in a foreign currcency) but not the dividend yield. Oct 6, 2019 at 16:33
• But I edited my answer and highlighted that notation issue. Thank you for the hint, Alex! Oct 6, 2019 at 16:41
• N represents the cumulative normal distribution, and n represents the normal density. Cumulative is integral of density, so density is then derivative of the cumulative - remember integral is anti-derivative, so derivative of anti--derivative is then the function. For technical approach, please see Leibniz rule (en.wikipedia.org/wiki/Leibniz_integral_rule). As @KeSchn outlined above, if we take derivative of N(s), we get n(s), and if we want to take derivative of N(d_1) where d_1 is a function of s, we will need to add chain rule: n(d1)*d(d1)/dS Oct 6, 2019 at 20:00
• A reminder what N(.) and n(.) look like dwaincsql.files.wordpress.com/2015/05/normal-pdf-cdf-1.png Oct 6, 2019 at 22:42