# 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?

• Hi Efrain Olivenhain=) Welcome to quant.SE. To obtain fast and helpful answers to your questions you should clarify it a little bit: What makes you assume that arbitrage exist if $P>K$? Do you already have some ideas on how to solve this question? As a hint: What can you say about the boundaries of option prices? – muffin1974 Oct 18 '15 at 11:42
• What is the difference between "Price of European Option" and "Present Price of Option "? – Gordon Dec 17 '15 at 19:34

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.

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.

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.