Deriving Black–Scholes Call Delta and Comparing It with ITM Probability

# Derivation of Call Delta from Black Scholes Model


# 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?

## Answer by Najee (score 7)

https://quant.stackexchange.com/a/71145

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*}

## Answer by Kevin (score 3)

https://quant.stackexchange.com/a/49081

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.