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European Call-Put Payoff Bounds and Arbitrage

Article Quant Q&A · Author: Ambitious-Walk3171
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Summary

The document examines a European call with a lower strike and a put with a higher strike, both expiring at the same time. It asks whether a condition linking their premiums to the gap between strikes could create a portfolio with a non-negative payoff and whether no-arbitrage theory rules this out.

The response reasons from the combined options' payoff before premiums: it is at least the difference between the strikes in every outcome. Therefore, the total premiums must be no less than that strike difference to prevent an arbitrage. This is a concise payoff-based bound, not a full treatment of option pricing; it does not discuss transaction costs, exercise conventions beyond the stated European style, or market frictions.

Key ideas

  • A lower-strike call and a higher-strike put with the same expiration can be assessed through their combined payoff.
  • Before premiums, the portfolio payoff is bounded below by the difference between the strikes.
  • No-arbitrage requires the combined premium to be at least that strike difference.
  • The argument is a theoretical bound and does not address market frictions.

Tags

OptionsArbitrageDerivatives pricing
Full text
# Potential arbitrage opportunity or fallacy?


# Potential arbitrage opportunity or fallacy?












Suppose we have two European options with the same expiration: a call priced at $c$ with strike price $K_1$ and a put priced at $p$ with $K_2 (>K_1)$. Further, suppose the zero-points of the two payoff curves intersect on the x-axis, i.e. $K_1+c=K_2-p$. Then, wouldn't the ensuing portfolio always have a non-negative payoff? Is there some theoretical justification to prevent this from happening?

## Answer by dm63 (score 2)

https://quant.stackexchange.com/a/79829

The ‘pure payoff” diagram (excluding option premia) for this structure shows that the payoff is always at least $K_2 - K_1 $ Therefore the total premium $c+p$ must be at least $K_2 -K_1$ so yes, a simple arbitrage argument shows this condition should not occur.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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