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Decomposing a Two-Stock Maximum-Payoff Option

Article Quant Q&A · Author: math4biz
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Summary

The document considers an option paying the greater of a scaled price of one stock and a scaled price of another. It suggests expressing a maximum payoff as one underlying value plus a call-like payoff on the difference between the two values. Under geometric Brownian motion, that exchange-option component can be valued with Margrabe’s formula.

The response states that the positive scaling constants do not fundamentally change this decomposition. However, it gives no worked derivation for the original payoff, parameter mapping, or numerical example, so the scaling step is only sketched. Applying the result requires assumptions about the joint dynamics of the stock prices and suitable inputs for the exchange-option formula; those details are not discussed.

Key ideas

  • A maximum of two asset values can be written as one asset value plus the positive part of their difference.
  • Under geometric Brownian motion, the difference-based component is an exchange option that can be valued with Margrabe’s formula.
  • The response asserts that positive payoff scalars do not substantially alter the decomposition, but does not show the full derivation.
  • The document provides no numerical example or discussion of model inputs and assumptions.

Tags

OptionsDerivatives pricingEquities
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# Pricing an option with a certain payoff


# Pricing an option with a certain payoff












Suppose an option with a payoff function

$$ \max((1+k)S_1,kS_2) $$ where $S_1, S_2$ are stock prices and $k>0$ is a constant value.

To value such an option, one would decompose this payoff function into plain vanilla options (and possibly bonds), as they can be valued using the Black-Scholes formula.

How can this payoff function be decomposed?

Thank you very much!

## Answer by Rylan (score 1)

https://quant.stackexchange.com/a/77088

$$\max(S_1, S_2) = S_1 + \max(S_2-S_1, 0)$$ In the case of GBM, the second term in the RHS can be valued using Magrabe's formula. The scalars $k, 1+k$ don't change this significantly.

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.