European Put Lower Bound Requires Discounting to Maturity

Summary

The document explains an apparent violation of the no-arbitrage lower bound for a European put. For a put with strike K and maturity T, the cited bound is the greater of zero and the present value of the strike minus the current underlying price. A plotted Black–Scholes price seemed to fall below this bound, raising the question of whether the model was inconsistent with the result.

The accepted answer identifies a discounting mistake in the plotted comparison: the strike must be discounted over the full time to maturity. Using a shorter discount period makes the computed lower bound too high and can create a false appearance of a violation. The response gives a corrected expression for the example, but the document does not include the original plot, full parameters, or a broader derivation. Its lesson is a focused check on consistent maturity inputs when comparing option prices with theoretical bounds.

Key ideas

• A European put price must be at least the discounted strike less the underlying price, floored at zero.
• The strike discount factor must use the option's full time to maturity.
• An incorrectly short discount period can make a valid Black–Scholes price appear to violate the lower bound.
• The document presents a correction to one plotted example rather than a general model analysis.

Tags

# Lower bound for European put option prices -- potential contradiction with BS


# Lower bound for European put option prices -- potential contradiction with BS












A classical no-arbitrage argument shows that for a European put with strike $K$ and time to maturity $T$, the price $p$ satisfies $$p \geq \max(0,Ke^{-rT} - S).$$ Is Black-Scholes in contradiction with this result? I've attached a picture of a Mathematica plot of the price of a long-dated put option against stock prices, which shows a violation of the lower bound.

## Answer by Antoine Conze (score 3, accepted)

https://quant.stackexchange.com/a/39557

The discounted intrinsic value in your plot is incorrect because you are not discounting up to 5 years maturity. Correct code should be $\max(0,25 \exp(-0.05 \color{red}{\times 5}) - S)$.

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