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.


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.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

<span>%d</span> bloggers like this: