# Decomposing option payoffs [closed]

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!

• You might find this question interesting. Sep 21 at 21:29
• not sure why this is closed. It is very relevant. Gary Kennedy has a paper around this "A Reduction Algorithm for a Class of Payoff Formulae" papers.ssrn.com/sol3/papers.cfm?abstract_id=1645275 Sep 25 at 19:34

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. • Nice tool, thanks for link Sep 22 at 9:44

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*}