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In a situation where $$K_3-K_2=K_2-K_1=h>0$$ and $$K_1\le S_t\le K_3$$ where $$S_T=S_t.e^{[(r-\sigma^2/2)(T-t)+\sigma(W_T-W_t)]}$$ (i.e. Stock process follows GBM under the risk neutral measure).

I know the value of the call under the risk neutral measure is: $$f(S_t)= e^{-r(T-t)}*E((S_T-K_1)^+-2(S_T-K_2)^++(S_T-K_3)^+|\mathcal{F_t})$$ How do we know that the value of the payoff of the butterfly spread using calls is positive for any t<T.

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  • $\begingroup$ It is based on the convexity of the call option payoff. $\endgroup$
    – Gordon
    Commented May 11, 2022 at 14:09
  • $\begingroup$ Can you elaborate or prove it? $\endgroup$
    – user43534
    Commented May 11, 2022 at 14:10
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    $\begingroup$ What do you mean with "payoff ... for any t<T"? The payoff happens exactly at $T$, no? Do you mean value instead of payoff, by any chance? $\endgroup$ Commented May 11, 2022 at 17:58
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    $\begingroup$ Vandalizing questions is not allowed, edit rolled back $\endgroup$
    – Bob Jansen
    Commented May 12, 2022 at 5:19

1 Answer 1

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Note that \begin{align*} K_2 = \frac{K_1+K_3}{2}. \end{align*} Then \begin{align*} &\ \max(S_T-K_1, \, 0) + \max(S_T-K_3, \, 0) \\ =&\ \max\big(S_T-K_1 + \max(S_T-K_3, \, 0), \, \max(S_T-K_3, \, 0)\big)\\ =&\ \max\big(\max(S_T-K_3 + S_T-K_1, \, S_T-K_1), \, \max(S_T-K_3, \, 0)\big)\\ =&\ \max\big(2S_T-(K_1+K_3), S_T-K_1, S_T-K_3, 0\big)\\ \ge&\ \max\big(2S_T-(K_1+K_3), 0\big)\\ =&\ 2\max(S_T-K_2, \, 0). \end{align*}

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  • $\begingroup$ Doesn't strictly positive imply that it is always > $\endgroup$
    – user43534
    Commented May 11, 2022 at 14:58
  • $\begingroup$ Also how is this valid for every $$S_t$$ since $$S_T$$ is also variate on $$r, t and \sigma$$ $\endgroup$
    – user43534
    Commented May 11, 2022 at 14:59
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    $\begingroup$ The above derivation is model-free. $\endgroup$
    – Gordon
    Commented May 11, 2022 at 15:34

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