Put–Call Parity and Arbitrage Bounds for a Higher-Strike Put

# Options Arbitrage strategy


# Options Arbitrage strategy












Supposed we have 1-year call option with a strike of 90 and it costs 10. We also have 1-year put on the same stock with a strike of 100. The risk free rate is 5% per annum. the stock is currently trading at 80. For which price of a put option we can find an arbitrage strategy, for 20 or 30? And what the arbitrage strategy would be? I thought about shorting the put and the stock, and buying the call. According to my calculations it will always yield a non-negative profit, but I am not sure whether I am on the right track.

## Answer by dm63 (score 3)

https://quant.stackexchange.com/a/31831

We would like to obtain bounds on P(100). First calculate P(90) from put-call parity:

C(90)-P(90)= stock - PV(strike) =S- 90/1.05 = 80-85.71= -5.71

So P(90)= C(90)+5.71= 15.71

Now observe the following inequalities:

P(100) > P(90) > P(100) -10/1.05

The first of these states that the right to sell for 100 is always better than the right to sell for 90. The second states that the value of the 100-90 put spread can never be greater than the PV of its maximum payoff (10). So from these we obtain :

P(100) > 15.71 and P(100) < 25.24

As the arbitrage bounds for the 100 put. This all assumes the stock does not pay dividends.

## Answer by Ami44 (score 2)

https://quant.stackexchange.com/a/31827

At time of payout you have $$C(K=90) \leqslant C(K = 100) + 10$$ and $$P(K=90) \geqslant P(K=100) - 10$$ From that using put-call parity you can deduct $$ S - 100 * e^{-rT} \leqslant C(K=90) - P(K=100) \leqslant S - 90 * e^{-rT}$$ If this inequality is violated, than you have an arbitrage opportunity. Take advantage of it by buying the stuff that is underpriced and short the overprice things. E.g. assume the first inequality does not hold. Than $C(K=90)$ is too cheap or $P(K=100)$ to expensive. Buy the Call, Sell the Put and the Stock. The payout at T is:

For $S \leqslant 90$: $-100+S-S=-100$

For $90 \leqslant S \leqslant 100$: $S-90 - 100 + S - S = S - 190$

For $100 \leqslant S$: $S - 90-S=-90$

For all three outcomes the payout is greater than -100. Our assumption was that $$C(K=90)-P(K=100)-S < -100 * e^{rT}$$ Which means we got more than 100 (discounted) in cash for our position at time of inception. Since we made more money from selling the shorts than we loose by buying the call and through the payout we make a sure profit.