Pricing Barrier Digital Options in Black–Scholes

# Barrier digital options and pricing


# Barrier digital options and pricing












What do you call options which behave like barrier options but for a digital option?

That is, given $0 < t < T$, then if $S_t > K_t$, the binary option $B(K_T,T)$ comes into play, i.e. which pays out 1 unit if $S_T > K_T$. If however $S_t < K_t$, then we get nothing ,and of coruse if $S_t > K_t$but $S_T < K_T$, then we get nothing as well.

What are these options called, and where can I find their prices? For example, in a Black Scholes setting?

## Answer by LocalVolatility (score 3)

https://quant.stackexchange.com/a/33206

Disclaimer: This answer derives the prices of two different binary options within the Black/Scholes framework. Note that this is not an appropriate valuation model to use for non-European contracts in most real-world markets.

Up-and-In Binary Call

After reading your question for a second time, I agree with Quantuple's comment that you seem to be looking for the solution to an up-and-in binary call option.

Formally, let

\begin{equation} \nu = \inf \left\{ t \in \mathbb{R}_+ : S_t \geq K \right\} \end{equation}

be the first hitting time of $S$ to the strike $K$. The option has a unit payoff conditional on $\nu \leq T$ and $S_T \geq K$, i.e.

\begin{equation} V_T = \mathrm{1} \left\{ S_T \geq K \right\} \mathrm{1} \left\{ \nu \leq T \right\}. \end{equation}

Note however that $S_T \geq K \; \Rightarrow \; \nu \leq T$ and thus $\left\{ S_T \geq K \right\} \subseteq \left\{ \nu \leq T \right\}$. Consequently, we can skip the second indicator and your payoff is just

\begin{equation} V_T = \mathrm{1} \left\{ S_T > K \right\}. \end{equation}

I.e. the price of an up-and-in binary call option is the same of that of a normal binary call option. You thus have the standard result that

\begin{equation} V_0 = e^{-r T} \mathcal{N} \left( d_- \right), \end{equation}

where

\begin{equation} d_- = \frac{1}{\sigma \sqrt{T}} \left( \ln \left( \frac{S_0}{K} \right) + \left( r - \frac{1}{2} \sigma^2 \right) T \right). \end{equation}

Down-and-Out Binary Call

A more interesting case is the down-and-out binary call. This is how I initially understood your question. Now let

\begin{equation} \nu = \inf \left\{ t \in \mathbb{R}_+ : S_t \leq K \right\} \end{equation}

and

\begin{equation} V_T = \mathrm{1} \left\{ S_T \geq K \right\} \mathrm{1} \left\{ \nu > T \right\}. \end{equation}

This option knocks out, should the spot price breach the barrier before maturity. Otherwise it has a digital payoff of one.

Let $\tau = T - t$ be the time-to-maturity. The valuation function $\tilde{V}(S, \tau)$ of this option satisfies the initial boundary value problem

\begin{eqnarray} \mathcal{L} \left\{ \tilde{V} \right\} (S, \tau) & = & 0 \qquad (S, \tau) \in \mathcal{D},\\ \tilde{V}(K, \tau) & = & 0, \qquad \forall \tau \in \mathbb{R}_+\\ \tilde{V}(S, 0) & = & \mathrm{1} \{ S \geq K \}, \end{eqnarray}

where $\mathcal{L}$ is the Black/Scholes forward operator and $\mathcal{D} = \left\{ (S, \tau): S > K, \tau \in \mathbb{R}_+ \right\}$. Using the method of images, see e.g. Buchen (2001), the solution can be shown to be

\begin{equation} \tilde{V}(S, \tau) = \mathcal{B}_K^+(S, \tau) - \stackrel{K}{\mathcal{I}} \left\{ \mathcal{B}_K^+(S, \tau) \right\}, \end{equation}

where

\begin{eqnarray} \mathcal{B}_K^+ (S, \tau) & = & e^{-r \tau} \mathcal{N} \left( d_- \right),\\ d_- & = & \frac{1}{\sigma \sqrt{\tau}} \left( \ln \left( \frac{S}{K} \right) + \left( r - \frac{1}{2} \sigma^2 \right) \tau \right),\\ \stackrel{K}{\mathcal{I}} \left\{ \mathcal{B}_K^+ (S, \tau) \right\} & = & \left( \frac{S}{K} \right)^{2 \alpha} \mathcal{B}_K^+ \left( \frac{K^2}{S}, \tau \right),\\ \alpha & = & \frac{1}{2} - \frac{r}{\sigma^2}. \end{eqnarray}

References

Buchen, Peter W. (2001) "Image Options and the Road to Barriers," Risk Magazine, Vol. 14, No. 9, pp. 127-130