Why a Put Priced Above Its Strike Creates an Arbitrage

Summary

The document explains the arbitrage implied when a European put's price exceeds its strike, assuming the underlying asset price cannot fall below zero. A put's payoff at maturity is capped at the strike, so a seller's maximum payment is no greater than that amount. With a zero risk-free rate, selling the overpriced put brings in more cash initially than the maximum possible maturity liability. Investing or retaining the strike-sized amount therefore leaves a positive surplus while covering the put payoff.

The answer also gives the positive-rate version of the bound: a put price above the present value of the strike creates the same opportunity. Put–call parity offers a related implication, since under the stated zero-rate condition the put-price inequality also implies a call price above the underlying price. These arguments depend on European-option payoff assumptions, nonnegative underlying value, and the ability to transact and finance at the stated prices; the discussion does not account for frictions or constraints.

Key ideas

• A European put's maturity payoff cannot exceed its strike when the underlying price is nonnegative.
• At a zero interest rate, selling a put priced above its strike produces more initial cash than its maximum payoff liability.
• With positive interest rates, the corresponding upper bound is the discounted strike.
• Put–call parity also implies that the stated overpriced put condition makes the call worth more than the underlying under zero rates.

Tags

# Abritrage when Put Option Greater then Strike Price?


# Abritrage when Put Option Greater then Strike Price?












I am having a tough time conceptualizing this question here: Let $P$= Price of European Option, $S$ = Present Price of Option and $K$ = Strike Price. If $P > K$, why does abritrage exist? Assuming $r = 0$. I really can't figure this out.

I understand that when C(call option) is greater then S abritrage exists because $C - S + K > 0$ and even if you don't execute the option, $C - S > 0$.

Could the same logic be utilized here?

## Answer by ocstl (score 1)

https://quant.stackexchange.com/a/21289

Assuming S is non-negative, the payoff function of a put at maturity is dominated by K:

$P_T = max(K - S_T, 0) = K + max(-S_T, -K) \leq K$

Under the assumption of a zero risk-free rate, one can write a put and invest K until maturity, for a positive cashflow at initiation ($P - K > 0$) and possibility 1 non-negative cashflow at maturity ($P_T \leq K$).

More generally, any put value above $e^{-rT}K$, $T$ being the time to maturity, yields an arbitrage opportunity, it being the amount necessary to dominate the final payoff function.

## Answer by emcor (score 1)

https://quant.stackexchange.com/a/21784

Let $P>K$ and $r=0$.

Then you can short the put and receive $P$.

At maturity, the maximum payoff you have to pay from the put is $K$.

Therefore you have a sure profit of at least $P-K>0$, which is an arbitrage.

## Answer by Phun (score 0)

https://quant.stackexchange.com/a/21280

First, assume r = 0 then Put-Call parity holds: C - P = S - K which can be rewritten as C = S - K + P > S - K + K = S. That's your answer.

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