Pricing Digital Options with Call Spreads for Smoother Hedging

# Why represent a digital payoff as a call spread


# Why represent a digital payoff as a call spread












Pricing a digital caplet using Hull White model, which pays:

> $1$ if $R>K$, $0$ otherwise.

Why would you represent the payoff as a call spread, i.e.

$$\text{Payoff} = \frac{(R - (K+\epsilon))^+-(R - (K-\epsilon))^+}{2\epsilon}$$

rather than

$$\text{Payoff} = \mathbb{1}_{R>K}?$$

What is the benefit?

## Answer by Daneel Olivaw (score 4, accepted)

https://quant.stackexchange.com/a/60670

An important practical reason is for hedging purposes.

Consider a situation where the option is very close to maturity and the rate $R$ is fluctuating around the strike $K$, such that the option is alternating out-the-moneyness with in-the-moneyness. Compared to a standard call option, where the payoff is continuous in $R$ with a "smooth" transitioning from out-the-moneyness to in-the-moneyness, the payoff of a digital call jumps from $0$ to $1$ when the option gets in-the-money. This makes hedging Greeks such as delta and gamma unstable, and can expose the option's hedger to significant risk and/or costs if he wants to be properly hedged. Generally speaking, this kind of risk is known as pin risk.

Let $D(R)=1_{R>K}$ be the payoff of the digital call. On the other hand, consider the following call spread, which is slightly different to yours (it uses backward differences instead of central differences): $$S(R)=\frac{(R-(K-\varepsilon))^+-(R-K)^+}{\varepsilon}$$ Then the payoff for any $R>K$ is: $$S(R)=\frac{(R-(K-\varepsilon))-(R-K)}{\varepsilon}=1=D(R)$$ However, for $K-\varepsilon<R\leq K$: $$S(R)=\frac{R-(K-\varepsilon)}{\varepsilon}>0=D(R)$$ whereas for $R\leq K-\varepsilon$ we have $D(R)=S(R)=0$. Hence the payoff of this call spread is at least as great as that of the digital call. The advantage is that its Greeks will be smoother and therefore easier to manage $-$ in particular you can merely buy/sell vanilla call options in the market, which are quite liquid. Obviously, the price of $S$ will be higher than that of $D$.

An investment bank might therefore quote the price for a digital call $D$ based on the cost of its overhedge $S$ in order to better manage its risk. It will tend to set the parameter $\varepsilon$ based on the volatility $R$, i.e. the more volatile $R$, the wider $\varepsilon$ to control for any sharp moves in the rate which might leave it exposed to pin risk.

As for the expression in your original post, using central differences and which I will designate with $S^\prime$, it might be used by more aggressive hedgers. Indeed you can easily check that the payoff of $S^\prime$ is lower than or equal to $S$ hence its price should be smaller and thus more competitive. It does still allow for a hedging strategy based on buying/selling vanilla calls, and ensures smooth Greeks.

## Answer by alexbougias (score 2)

https://quant.stackexchange.com/a/60669

First, it is practically more convenient to price any contingent claim in terms of plain vanilla calls/puts, rather than finding a different formula. Second, the European options are traded in the market, thus you can easily calculate the price of a digital option from the observed market prices. In addition, the valuation of the digital will be model-free. Last, the representation of the digital price as the difference between two options with different strikes, provides a tool for retrieving the state-price density of underlying process (and equivalently the option implied return distribution if you combine the full span of options with different strikes).