# Single-period market with probability space [closed]

Let $$C^E$$, $$P^E$$, $$C^A$$, and $$P^A$$ denote prices of a European call option, a European put option, an American call option and an American put option, respectively. All of them with expiry time $$T$$ and the same strike price $$K$$. we assume $$r\geq 0$$ to be the continuously compounded interest rate. I want to show that if it holds that,

$$C^E-P^E-S(0)+Ke^{-rT}<0,$$

then we can make a sure risk-less profit.

Furthermore, I am interested in showing that,

$$C^A-P^A-S(0)+Ke^{-rT}>0,$$

then we can also make a sure risk-less profit.

Anybody have an idea?

• Are you familiar with put call parity? en.wikipedia.org/wiki/Put%E2%80%93call_parity – Bob Jansen Oct 27 '20 at 11:29
• No, I looked at it briefly now, how can that be used for my case? – Jonathan Kiersch Oct 27 '20 at 11:46
• I recommend you work harder on this. It seems you have the right background knowledge to do so. As it stands, this is too basic. – Bob Jansen Oct 27 '20 at 18:52

In both cases, you should argue along the lines of options arbitrage. If you think an asset (or a portfolio) is relatively cheap (as in: arbitrageable) then you simply buy low, sell high.

In your first case, it seems that the call is too cheap, as in:

$$C^E So let's buy the call, sell a put, short a unit of stock, and borrow some money:

At maturity $$T$$, either your call or the put are in the money but in any case, your future payoff is always zero. On the other hand, you paid less for the call than what you got for the put/stock/borrowing position, i.e. you have a risk-less arbitrage profit.

You should be able to follow the same line of thought for the American option situation, knowing that $$C^E=C^A$$ (without dividends), and $$P^A>P^E$$.

HTH?

Edit: Maybe slide 3ff of this source can help?

• It helps with the European case, however, I fail to understand your advise for the american one(???) – Jonathan Kiersch Oct 27 '20 at 14:12
• I have added a reference on European / American PCP properties. – Kermittfrog Oct 27 '20 at 14:24
• I still don't see what to do here. Here is what I have $$C^A<P^A+S-Ke^{-rT}$$, but how does that help, I need to have $$C^A-P^A-S+Ke^{-rT}>0$$ right? – Jonathan Kiersch Oct 27 '20 at 14:31