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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!

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2 Answers 2

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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.

enter image description here

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  • $\begingroup$ Nice tool, thanks for link $\endgroup$
    – user35980
    Sep 22, 2023 at 9:44
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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*}

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