# Pricing a Digital Option

This post is based on problems 2.10 and 2.11 in, “Heard on the Street” by Timothy Falcon Crack. I was asked how to price a digital option in a job interview - and had no idea what to do!

# European Call Options

A European call option is the right to buy an asset at the strike price, \(K\), on the option’s expiration date, \(T\). A call is only worth exercising (using) if the underlying price, \(S\), is greater than \(K\) at \(T\), as the payoff from exercising is \(S-K\). The plot below shows the value of a call option, as a function of the underlying asset’s price, with \(K=100\):

Selling a call option with a strike \(K=100\) earns you the call’s price, \(c\), today, but your payoff will be decreasing in the underlying price:

# Digital Call Options

A digital call option with \(K=100\) is similar - it pays off one dollar if \(S\geq100\) at expiration, and pays off zero otherwise:

Suppose you have a model for pricing regular call options. If you’re using Black-Scholes the price of the call, \(c\), is a function of \(K\), \(S\), time to expiration \(T-t\), the volatility of the underlying asset \(\sigma\), and the risk free rate \(r\): \begin{equation} c=F(K,S,T-t,\sigma,r) \end{equation} Now - suppose the model is correct. How can you use \(F(K,\cdot)\) to price the digital option?

# Replicating the Digital Option

The trick is to replicate the digital option’s payoff with regular calls. As a starting point, consider buying a call with \(K=100\) and selling a call with \(K=101\):

This is close to the digital option, but not exactly right. We want to make the slope at 100 steeper, so we need to buy more options. This is because a call’s payoff increases one-for-one with the underlying once the option is in the money, so with one option you are stuck with a slope of one.

Consider buying two calls with \(K=100\) and selling two calls at \(K=100.5\):

As opposed to a slope of 1 between 100 and 101, now we have a slope of two between 100 and 100.5.

Generalizing this idea - consider a number \(\epsilon>0\). To get a slope of \(\frac{1}{\epsilon}\), you buy \(\frac{1}{\epsilon}\) calls at \(K=100\) and you sell \(\frac{1}{\epsilon}\) calls at \(K=100+\epsilon\). Here’s what it looks like for \(\epsilon=\frac{1}{10}\):

Given that the slope is \(\frac{1}{\epsilon}\), to get an infinite slope, we take the limit as \(\epsilon\) goes to zero.

How much will the above portfolio cost? You earn \(\frac{1}{\epsilon}F(100+\epsilon, \cdot)\) from selling the \(K=100+\epsilon\) calls, and pay \(\frac{1}{\epsilon}F(100, \cdot)\) for the \(K=100\) calls. The net cost is: \begin{equation} lim_{\epsilon \rightarrow 0} \frac{F(100+\epsilon,\cdot)-F(100,\cdot)}{\epsilon} \end{equation}

What does this look like? A derivative! It might look more familiar if I re-wrote it as:

\begin{equation} lim_{\epsilon \rightarrow 0} \frac{F(K+\epsilon)-F(K)}{\epsilon} \end{equation}

The price of the digital option is the derivative of \(F\) with respect to the strike price \(K\).

# Conclusion

Many complicated payoffs can be re-created as combinations of vanilla puts and calls. For an overview, see the first few chapters of Sheldon Natenberg’s, “Option Volatility & Pricing”.