Pricing a Payoff That Takes the Maximum of Two Linear Outcomes

Summary

The document considers a European payoff equal to the larger of a linear upside outcome, the underlying price minus a constant, and a linear downside outcome, another constant minus the underlying price. It proposes representing the payoff as a fixed amount less twice a call payoff struck at the midpoint of the two constants. This transformation reduces the pricing problem to a discounted fixed payment and a standard Black–Scholes call value.

The writer asks whether the proposed expression is correct and whether the Black–Scholes inputs should use the midpoint as strike. The document contains no accepted derivation, validation, or numerical evidence, so it leaves that question unresolved. Any application requires checking the payoff identity and discounting the fixed cash amount consistently; the standard call formula also relies on the usual Black–Scholes assumptions for the underlying and volatility.

Key ideas

• The maximum of the two stated linear payoffs can be expressed using a fixed cash amount and a call payoff at their intersection.
• The strike in that call representation is the midpoint of the two payoff constants.
• Pricing then uses the discounted fixed payment together with a Black–Scholes call value.
• The document poses the derivation as a question and does not confirm or validate the proposed formula.

Tags

# Exotics - Combination of different payoffs using Black-Scholes


# Exotics - Combination of different payoffs using Black-Scholes












I'm currently struggling with the derivation of a formula to price the following exotic option with Black-Scholes.

The option has the maximum payoff of $(S_T-z)$ and $(y - S_T)$, where $S_T$ is the price of the underlying at maturity $T$, $X$ the strike price, and $z$ and $y$ are constants. E.g. $z = 20$ and $y = 40$.

My first approach was to build a formula where I combine the payoff of $max\{(S_T-z);0\}$ and $max\{(y-S_T);0\}$.

$ d_1 =\frac{\ln(\frac{X}{S_0})+(r_f-0.5*\sigma^2)*T}{\sigma*\sqrt{T}} $ and $d_2 = d_1 - \sigma*\sqrt{T}$

$Price = (S_0*\phi{(d_1)}-a*e^{-r_fT}*\phi{(d_2)})+(b*e^{-r_fT}*\phi{(-d_2)}-S_0*\phi{(-d_1)}) $

EDIT: Thanks to the notes I came up with a second approach on how to price this option:

Payoff structure $max(S_T-z;y-S_T) = y - 2*max(S_T-(y+z)/2,0)$, where the price is represented by

$y-S_0*e^{-r_fT} + 2*(S_0*\phi(d_1)-0.5*(y+z)*e^{-r_fT}*\phi(d_2))$

Is this correct and do I have to adjust $d_1$ as well?

So that it would be $d_1 = \frac{\ln(\frac{0.5*(z+y)}{S_0})+(r_f-0.5*\sigma^2)*T}{\sigma*\sqrt{T}}$

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