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European Put Lower Bound from Put-Stock Bond Arbitrage

Article Quant Q&A · Author: Nick Mugisha
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Summary

The document derives a lower bound for a European put on a non-dividend-paying stock by comparing two portfolios. One portfolio holds the put and one share; the other holds a zero-coupon bond that pays the strike at expiration. At maturity, the put-stock combination is worth at least the strike in every outcome: it is worth the strike when the stock finishes below it and more when the stock finishes above it.

Because the first portfolio’s terminal payoff is never less than the bond’s, absence of arbitrage implies that its current value must be at least as high. Rearranging gives the put’s lower bound as the discounted strike minus the current stock price. The argument relies on the stated setting of European exercise, no dividends, and the bond discounting convention; it offers a payoff comparison rather than a broader discussion of other market frictions or bounds.

Key ideas

  • A European put combined with one share has a terminal value at least equal to the strike.
  • A zero-coupon bond paying the strike at expiration provides the comparison payoff.
  • No-arbitrage pricing requires the put-stock portfolio to cost at least as much as the bond.
  • The resulting lower bound is the discounted strike less the current stock price.

Tags

OptionsDerivatives pricingArbitrage
Full text
# European put options


# European put options












Why is it that for European Puts on Non-Dividend-Paying Stocks, the lower-bound for price is $$p=Ke^{-rT}-S_0?$$

## Answer by Vihaan Shah (score 2)

https://quant.stackexchange.com/a/43655

Let's consider the following 2 portfolios:

Portfolio A: one European put option plus one share,

Portfolio B: a zero coupon bond paying off $K$ at time $T$.

If $S_T<K$ then the option in portfolio A is exercised at $T$ and the portfolio is worth $K$.

If $S_T>K$, then the put option expires worthless and the portfolio is worth $S_T$ at this time. Hence, portfolio A is worth $max(S_T, K)$ at $T$. Portfolio B is worth $K$ in time $T$. Hence, portfolio A is always worth as much as, and can sometimes be worth more than more than B at $T$. It follows then that in the absence of arbitrage opportunities portfolio A must be worth at least as much as B today. Hence: $$p+S_0>= Ke^{-rT}$$ or $$p>=Ke^{-rT}-S_0$$

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.