I am having trouble understanding the transition from the third line to the fourth line.

I see how it works through cases, but is there a general property that allows this?

  • $\begingroup$ You must be kidding. How should anyone know what you mean and answer this 'question'? Have a look on how to ask a good question. $\endgroup$ Mar 10, 2015 at 6:34
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    $\begingroup$ @MarcoBreitig It's true that the conclusion is not very clear, but I've seen much worse here. $\endgroup$
    – SRKX
    Mar 10, 2015 at 7:49

2 Answers 2


Is it true in general - a property of Max and Min: no, it is just true under the assumption stated.

There are just cases - going through the cases is the proof.

What if you write down all possible cases and the fact that $K_1 < K_2$? In fact it is enough to consider:

  1. $S_T < K_1 < K_2$ then $max(K_1,S_T) - max(S_T,K_1,K_2) + K_2 = K_1$

  2. $K_1 < S_T < K_2$ then $max(K_1,S_T) - max(S_T,K_1,K_2) + K_2 = S_T$

  3. $K_1 < K_2 < S_T$ then $max(K_1,S_T) - max(S_T,K_1,K_2) + K_2 = K_2$

which can be summarized as $min(max(K_1,S_T),K_2)$ under the assumption that $K_1 < K_2$.

  • $\begingroup$ You're missing some cases though... $\endgroup$
    – SRKX
    Mar 10, 2015 at 7:52
  • $\begingroup$ If $K_1 < K_2$ is the assumption - which cases am I missing? $S$ can be below $K_1$, between the two or higher. $\endgroup$
    – Richi Wa
    Mar 10, 2015 at 8:23
  • $\begingroup$ Oh you're right I missed $K_1<K_2$ in the picture! $\endgroup$
    – SRKX
    Mar 10, 2015 at 8:30

Note that, \begin{align*} \max(a, b) \pm c &= \max(a\pm c, b\pm c),\\ \min(a, b) \pm c &= \min(a\pm c, b\pm c), \end{align*} and \begin{align*} -\max(a, b) = \min(-a, -b). \end{align*} Then \begin{align*} & \ \max(K_1, S_T)-\max(S_T, K_1, K_2)+K_2 \\ =& \ \max(K_1, S_T)-\max(\max(K_1, S_T), K_2)+K_2\\ =& \ \max(K_1, S_T)+\min\big(-\max(K_1, S_T), -K_2\big)+K_2\\ =& \ \min\big(0, \max(K_1, S_T)-K_2\big) + K_2\\ =& \ \min\big(K_2, \max(K_1, S_T)\big). \end{align*}


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