The butterfly spread satisfies the inequality

c(X1) - 2c(X2) + c(X3) >= 0

Where call strikes satisfy X1<X2<X3 and X2 - X1 = X3 - X2.

There is a “proof” that was provided here https://quant.stackexchange.com/a/32609/66878. However the proof is missing many steps and serves more as an outline.

Could anyone elaborate on the proof or provide their own in full detail?

Thanks so much, Jordan

  • $\begingroup$ It is the price of the butterfly spread that is non-negative. $\endgroup$
    – nbbo2
    Apr 1, 2023 at 9:44
  • $\begingroup$ Does this answer: quant.stackexchange.com/questions/61031/… fill up the missing gaps for you? $\endgroup$ Apr 1, 2023 at 10:17
  • 2
    $\begingroup$ Make sure you understand the intuition behind the proof: The long butterfly never has a negative payoff, so you cannot receive money for entering into the trade. It would be "free money with no risk" (arbitrage). You only receive money at the start on an option trade that has some possibility of losing money in the future. Once you grasp that the rest are (important) details. $\endgroup$
    – nbbo2
    Apr 1, 2023 at 11:45
  • $\begingroup$ Nbbo2 - thank you for the clarification. You’re right. I wrote (and was thinking) return but clearly this is only true in terms of the value of the position. $\endgroup$
    – Jordan Man
    Apr 1, 2023 at 17:31
  • 1
    $\begingroup$ Prove that $(S-(K+\epsilon))^+ +(S-(K-\epsilon))^+-2(S-K)^+\geq 0 $ for all $S\geq 0$, $K\geq 0$, and $\epsilon\geq 0$, by studying what do you get when $S$ belongs to the $4$ adjacent intervals generated by the 'strikes'. (Note that this is not true for arbitrary sets of 'strikes'.) Then apply the expectation operator on both sides of the inequality (with $S$ now viewed as a random variable with non-negative values) to get the 'prices', and remember its linearity and non-negativity properties (see the wiki or textbooks). $\endgroup$
    – ir7
    Apr 2, 2023 at 1:53


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