Skip to content
All library documents

Pricing a Piecewise Option Payoff with Vanilla and Digital Options

Article Quant Q&A · Author: Pefok

Summary

The document studies a European payoff that is zero below a lower threshold, declines linearly between two strikes, and pays a fixed amount above the upper strike. It first expresses the price as discounted risk-neutral expectations over the relevant terminal-price regions, using the lognormal Black–Scholes model and standard normal probabilities. The stated integral formula gives one direct pricing route, though the exposition omits the integration steps.

The accepted answer identifies an equivalent portfolio: a put spread combined with cash-or-nothing options at the two boundary strikes. This decomposition makes the payoff easier to price from familiar Black–Scholes instruments and lets the contract’s delta be obtained by combining their Greeks. The discussion assumes the standard Black–Scholes setting and European exercise; it does not address dividends, transaction costs, or numerical validation of the original formula.

Key ideas

  • A piecewise payoff can be valued as the discounted risk-neutral expectation of its terminal cash flow.
  • The payoff can also be decomposed into vanilla puts and cash-or-nothing options at its boundary strikes.
  • A portfolio decomposition allows the contract’s price and delta to be assembled from the component instruments.
  • The derivation relies on the Black–Scholes assumptions and European exercise.

Tags

Full text
# Black and scholes option pricing


# Black and scholes option pricing












I have to solve the following problem in the Black and scholes model: find the price at anty $t\in[0,T)$ for an option whose payoff at the maturity is: \begin{equation} 0 \ \ \ \text{if} \ S_T<K_1\\ K_2-S_T \ \ \ \text{if} \ K_1<S_T<K_2\\ K_2-K_1 \ \ \ \text{if} \ K_2< S_T \end{equation}

SOLUTION I have rewritten the payoff as: \begin{equation} Payoff_T=(K_2-S_T)\textbf{1}_{\{K_1<S_T<K_2\}}+(K_2-K_1)\textbf{1}_{\{S_T>K_2\}} \end{equation} Since $S$ evolves under martingale measure $\mathbb{Q}$ as a geometric Brownian Motion whit dynamic: \begin{equation} S_t=S_se^{(R-\frac{\sigma^2}{2})(t-s)+\sigma Y\sqrt{t-s}} \end{equation} where $Y\sim N(0,1)$ then I whant to compute for which values of $Y$: \begin{equation} K_1<S_T\Rightarrow K_1< S_te^{(R-\frac{\sigma^2}{2})(T-t)+\sigma Y\sqrt{T-t}}\Rightarrow\dfrac{K_1}{S_t}e^{-(R-\frac{\sigma^2}{2})(T-t)}<e^{\sigma Y\sqrt{T-t}}\\ \Rightarrow y_1=\dfrac{\ln(\frac{K_1}{S_t})-(R-\frac{\sigma^2}{2})(T-t)}{\sigma\sqrt{T-t}}<Y \end{equation} similarly I get: \begin{equation} S_T<K_2\Rightarrow Y<\dfrac{\ln(\frac{K_2}{S_t})-(R-\frac{\sigma^2}{2})(T-t)}{\sigma\sqrt{T-t}}=y_2 \end{equation} Now applying the formula for the price in B$\&$S market: \begin{equation} price_t=e^{-R(T-t)}E^{\mathbb{Q}}(Payoff_T|\mathcal{F}_t)\\ =e^{-R(T-t)}\bigg(\int_{y_1}^{y_2}(K_2-S_te^{(R-\frac{\sigma^2}{2})(T-t)+\sigma y\sqrt{T-t}})\frac{1}{\sqrt{2\pi}}e^{-y^2/2}dy+\int_{y_2}^{\infty}(K_2-K_1)\frac{1}{\sqrt{2\pi}}e^{-y^2/2}dy\bigg) \end{equation} Now i omit the computation of this integrals (not difficult) and I have the final formula where I denote with $\Phi(x)=P(X\leq x)$ with $X\sim N(0,1)$: \begin{equation} price_t=e^{-R(T-t)}(K_2-K_1)(1-\Phi(y_2))+K_2e^{-R(T-t)}(\Phi(y_2)-\Phi(y_1))-S_t(\Phi(y_2-\sigma\sqrt{T-t})-\Phi(y_1-\sigma\sqrt{T-t})) \end{equation} At this point my questions are:

- is this computation fine?

- Since the second question of the exercise is to compute the delta of the contract (Derivative w.r.t the underlying S) is it possible to express the payoff in terms of Call/put options for which I know an explicit expression of the delta?

## Answer by Daneel Olivaw (score 5, accepted)

https://quant.stackexchange.com/a/71827

Let us start by considering a bear spread strategy, consisting on long a European put with strike $K_2$ and short another European put with strike $K_1$. Then the payoff of this portfolio at expiry $T$ is: \begin{align} &(K_2-S_T)\textbf{1}_{\{S_T\leq K_2\}}-(K_1-S_T)\textbf{1}_{\{S_T\leq K_1\}} \\ &\qquad=(K_2-S_T)\left(\textbf{1}_{\{K_1< S_T\leq K_2\}}+\textbf{1}_{\{S_T\leq K_1\}}\right) -(K_1-S_T)\textbf{1}_{\{S_T\leq K_1\}} \\ &\qquad=(K_2-S_T)\textbf{1}_{\{K_1< S_T\leq K_2\}} +((K_2-S_T)-(K_1-S_T))\textbf{1}_{\{S_T\leq K_1\}} \\ &\qquad=(K_2-S_T)\textbf{1}_{\{K_1< S_T\leq K_2\}} +(K_2-K_1)\textbf{1}_{\{S_T\leq K_1\}} \end{align} We've matched the first term in your payoff. To match your second, we actually need to subtract $(K_2-K_1)\textbf{1}_{\{S_T\leq K_1\}}$ and add $(K_2-K_1)\textbf{1}_{\{S_T> K_2\}}$: \begin{align} &(K_2-S_T)\textbf{1}_{\{K_1< S_T\leq K_2\}} +(K_2-K_1)\textbf{1}_{\{S_T\leq K_1\}} -(K_2-K_1)\textbf{1}_{\{S_T\leq K_1\}} +(K_2-K_1)\textbf{1}_{\{S_T> K_2\}} \\ \qquad&= (K_2-S_T)\textbf{1}_{\{K_1< S_T\leq K_2\}} +(K_2-K_1)\textbf{1}_{\{S_T> K_2\}} \end{align} Yet, the subtracted term corresponds to the payoff of a cash-or-nothing put option with strike $K_1$ and cash payoff $K_2-K_1$, whereas the added term is equal to the payoff of a cash-or-nothing call option with strike $K_2$ and same cash payment. All these options have known prices and Greeks under the Black-Scholes model.

Therefore, you can price your payoff under a Black-Scholes setting by summing the Black-Scholes prices of 1) a European vanilla put with strike $K_2$, and 2) a European cash-or-nothing call with strike $K_2$ and cash payment $C:=K_2-K_1$, to which you subtract the prices of both 3) a European vanilla put with strike $K_1$, and 4) a European cash-or-nothing put with strike $K_1$ and cash payment $C$.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.