Decomposing Piecewise Option Payoffs into Calls
Summary
The discussion shows how to represent a capped, tent-shaped payoff as a combination of vanilla call payoffs. The payoff is zero below the lower strike, rises linearly between the two endpoints, and returns to zero above the upper endpoint. Its equivalent call portfolio consists of a long call at the lower strike, twice the short exposure at the midpoint strike, and a long call at the upper strike. This lets the payoff be valued by combining call prices under a pricing model such as Black–Scholes.
The proposed method is to inspect the graph and identify where its slope changes. Each change corresponds to a call position: the call’s strike marks the kink, and its quantity sets the size of the slope adjustment. A second answer verifies the stated identity through algebraic manipulation, while a graphing tool can check the result. The examples illustrate one payoff shape; they do not establish a universal procedure for every payoff or address market frictions and model assumptions in valuation.
Key ideas
- A piecewise linear payoff can be represented as a portfolio of vanilla calls.
- Call strikes correspond to the points where the payoff slope changes.
- Call quantities determine the size and direction of each slope adjustment.
- The example uses a long call at each endpoint and a doubled short call at the midpoint.
- Graphing and algebra can verify the payoff identity before pricing the components.
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# Decomposing option payoffs
# Decomposing option payoffs
Suppose an option payoff function $$max(min(S-1, 2-S), 0)$$ To value such an option, one would decompose this function, for example, as follows: $$max(S-1, 0) - max(2S-3, 0) + max(S-2, 0)$$ Now, it can be valued using the Black-Scholes formula.
How is this decomposition (or any other solution, for that matter) obtained?
I imagine one would start by graphing the payoff function. However, I was not able to come up with any sensible solutions by using this method.
Thank you very much!
## Answer by Kurt G. (score 4)
https://quant.stackexchange.com/a/76756
A good strategy to find the decomposition is to first look at the graph of $$ \max(\min(S-1,2-S),0)\,. $$ It looks like a
- long call with strike $1$ plus
- 2 times short a call with strike $1.5$ plus
- long a call with strike $2$
With Desmos you can check this reasoning.
## Answer by Gordon (score 2)
https://quant.stackexchange.com/a/76758
You can also try with the following: \begin{align*} &\ \max\big(\min\big(S-1,\, 2-S\big),\, 0\big) \\ =&\ \max\big(S-1 + \min\big(0,\, 3-2S\big),\, 0\big)\\ =&\ \max\big(S-1 - \max\big(2S-3, \, 0\big),\, 0\big)\\ =&\ - \max\big(2S-3, \, 0\big) + \max\big(S-1,\, \max\big(2S-3, \, 0\big)\big)\\ =&\ - \max\big(2S-3, \, 0\big) + \max\big(S-1,\, 2S-3, \, 0\big)\\ =&\ - \max\big(2S-3, \, 0\big) + \max\big(\max(S-1,\, 0\big), \, 2S-3\big)\\ =&\ - \max\big(2S-3, \, 0\big) + \max(S-1,\, 0\big) + \max\big(0, \, 2S-3 - \max(S-1,\, 0\big)\big)\\ =&\ - \max\big(2S-3, \, 0\big) + \max(S-1,\, 0\big) + \max\big(0, \, S-2 + S-1 - \max(S-1,\, 0\big)\big)\\ =&\ - \max\big(2S-3, \, 0\big) + \max(S-1,\, 0\big) + \max\big(0, \, S-2\big). \end{align*}Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
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