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Simplifying Max and Min Expressions with Case Analysis and Identities

Article Quant Q&A · Author: Mid
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Summary

The document explains how to simplify a maximum-and-minimum expression involving two ordered strikes and a terminal asset price. One proof divides the possible price outcomes into three regions: below the lower strike, between the strikes, and above the higher strike. The resulting expression is the asset price bounded between the two strikes, or equivalently a nested minimum and maximum.

A second proof uses general identities: adding or subtracting the same value preserves the max or min structure, and negating a maximum converts it to a minimum. These transformations yield the same bounded expression without separately evaluating each case. The result depends on the stated ordering of the strikes; the case analysis and algebra establish an identity, not a pricing model or empirical trading result. The payoff structure may be relevant to option analysis, but the document does not discuss valuation or market behavior.

Key ideas

  • The expression simplifies to the asset price capped by the upper strike and floored by the lower strike.
  • Case analysis proves the identity across prices below, between, and above the strikes.
  • Adding or subtracting a constant can be distributed across a maximum or minimum.
  • Negating a maximum converts it to a minimum of the negated values.
  • The simplification relies on the lower strike being less than the upper strike.

Tags

OptionsDerivatives pricingStatistics
Full text
# What are the properties of Max and Min functions?


# What are the properties of Max and Min functions?












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?

## Answer by Richi Wa (score 1, accepted)

https://quant.stackexchange.com/a/16917

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:

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

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

- $K_1 < K_2 < S_T$ 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 $K_1 < K_2$.

## Answer by Gordon (score 10)

https://quant.stackexchange.com/a/16926

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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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.