# The Black-Scholes formula

Certainly one of the biggest breakthroughs in quantitative finance, the Black-Scholes (pronounced like “Black-Shoales” not “Black-Skoales”) model was introduced in 1973 and provides a mathematically principled approach to options pricing. While the original models relies on partial differential equations it subsequently found a different interpretation through stochastic processes (martingales) by describing stock prices through a geometric Brownian motion. In this post we want to show that the Black-Scholes formula can be derived using very basic arguments for the special case of European call options.

#### Question:

Can you derive the Black-Scholes formula for the price of a European call option.

The payoff of a European call option at time $T$ with strike price $K$ and asset value $S_T$ is given by $C = \max \{S_T - K, 0\}$. This formula is quite intuitive; if the asset price $S_T$ is larger than $K$ we will execute the option. Otherwise the option has no value and no rational trader would execute a call.

From the above, the value of the option at the time of expiry is straightforward. However, how much should I pay at an earlier time $t < T$ where the final value of $S_T$ is uncertain? Underlying the Black-Scholes model is the assumption that $S_t$ follows a geometric Brownian motion. This can be seen as a process analogue of a log-normal distribution. For us, the only relevant information we need is that the final value $S_T$ follows a log-normal distribution, meaning that $\log S_T \sim \mathcal{N}(\mu, \tau^2)$. We will come back to the question what $\mu$ and $\tau$ are.

We can now calculate the expectation of the payoff $C = \max \{S_T - K, 0\}$ at time $t = 0$:

$\begin{array}{ll} E\left[\max\{S_T - K, 0\}\right] & = \int_{-\infty}^{\infty} \max\{S_0 e^s - K, 0\}\exp\left(-\frac{(s - \mu)^2}{2\tau^2} \right) ds \end{array}$

Note that the integral is $0$ whenever $S_0 e^s < K$. We can use that to rewrite $\displaystyle \begin{array}{ll} & \int_{-\infty}^{\infty} \max\{S_0 e^s - K, 0\}\exp\left(-\frac{(s - \mu)^2}{2\tau^2} \right) ds \\ & = \frac{1}{\sqrt{2\tau^2}}\int_{\log K/S_0}^{\infty} (S_0 e^s - K) \exp\left(-\frac{(s - \mu)^2}{2\tau^2} \right) ds \\ & = S_0\int_{\log K/S_0}^{\infty} \frac{1}{\sqrt{2\tau^2}}\exp\left(s-\frac{(s - \mu)^2}{2\tau^2} \right) ds - K \int_{\log K/S_0}^{\infty} \frac{1}{\sqrt{2\tau^2}}\exp\left(-\frac{(s - \mu)^2}{2\tau^2} \right) ds \end{array}$

Now we can compute

$\begin{array}{ll} \displaystyle \\ \int_{\log K/S_0}^{\infty} \frac{1}{\sqrt{2\tau^2}}\exp\left(-\frac{(s - \mu)^2}{2\tau^2} \right) ds & = 1 - \Phi\left(\frac{\log K/S_0-\mu)}{2\tau^2}\right) \\ & = \Phi\left(\frac{\log S_0/K+\mu)}{2\tau^2}\right) \end{array}$

and using a change of variable $t = \frac{s - \mu}{\tau}$ we can compute

$\begin{array}{ll} & \int_{\log K/S_0}^{\infty} \frac{1}{\sqrt{2\tau^2}}\exp\left(s-\frac{(s - \mu)^2}{2\tau^2} \right) ds \\ & = \quad \int_{\frac{\log K/S_0 - \mu}{\sigma}}^{\infty} \frac{1}{\sqrt{2\tau^2}}\exp\left( \mu + \tau t\right)\exp\left(-\frac{t^2}{2} \right) \tau dt \\ & = \quad \tau\exp\left( \mu + \frac{\tau^2}{2}\right) \int_{\frac{\log K/S_0 - \mu}{\sigma}}^{\infty} \frac{1}{\sqrt{2\tau^2}}\exp\left(-\frac{t^2}{2} + \tau t - \frac{\tau^2}{2}\right) dt \\ & = \quad \exp\left( \mu + \frac{\tau^2}{2}\right) \int_{\frac{\log K/S_0 - \mu}{\sigma}}^{\infty} \frac{1}{\sqrt{2}}\exp\left(-\frac{(t - \tau)^2}{2} \right) dt \\ & = \quad \exp\left( \mu + \frac{\tau^2}{2}\right) \Phi\left( -\frac{\log K/S_0 - \mu}{\sigma} + \tau \right). \end{array}$

This gives us the main part of the famous Black-Scholes formula for a European call:

$E\left[\max\{S_T - K, 0\}\right] = S_0\exp\left( \mu + \frac{\tau^2}{2}\right) \Phi\left( \frac{\log S_0/K + \mu}{\sigma} + \tau \right) + K \Phi\left(\frac{\log S_0/K+\mu)}{2\tau^2}\right).$ Now, $S_T$ is often assumed to have mean $r - \sigma^2T/2$, which is taken to be the risk neutral rate of return and variance $\tau^2 = \sigma^2 T$, i.e. the variance is proportional to the time horizon.

Plugging these values in we obtain

$\begin{array}{ll} E\left[\max\{S_T - K, 0\}\right] & = S_0\exp\left(rT\right) \Phi\left( \frac{\log S_0/K + (r-\sigma^2/2)T}{\sigma\sqrt{T}} + \sigma \sqrt{T} \right) + K \Phi\left(\frac{\log S_0/K+(r-\sigma^2/2)T)}{\sigma \sqrt{T}}\right) \\ & = S_0\exp\left(rT\right) \Phi\left( \frac{\log S_0/K + (r+\sigma^2/2)T}{\sigma\sqrt{T}} \right) + K \Phi\left(\frac{\log S_0/K+(r-\sigma^2/2)T)}{\sigma \sqrt{T}}\right). \end{array}$

While this is the expected value at time $T$, we need to discount this expectation to produce the present value of this expectation. With continuous discounting at rate $r$ we then have

$\begin{array}{ll} PV(C) &= \exp{-rT} E\left[\max\{S_T - K, 0\}\right] \\ & = S_0 \Phi\left( \frac{\log S_0/K + (r+\sigma^2/2)T}{\sigma\sqrt{T}} \right) + \exp{-rT}K \Phi\left(\frac{\log S_0/K+(r-\sigma^2/2)T)}{\sigma \sqrt{T}}\right). \end{array}$ Final note: Often, this formula is written for the expectation at time time $t$ between $0$ and $T$. The calculation is completely equivalent, we just need to substitute $S_0$ with $S_t$ and accordingly the time to maturity from $T$ to $T - t$.