# Confusion about price at maturity of a call option allowing for arbitrage

I have been reading the book Tomas Bjork's Arbitrage Theory in Continuous Time and could not understand how there could be arbitrage if the price of a contingent claim is not $$X$$.

To give some context, $$X$$ represents a European call option and $$Z$$ is a random variable representing how to stock will move at time 1. So if the stock price is $$S$$ at $$t=0$$, the stock price at $$t=1$$ can be written as $$sZ$$ where $$Z=u$$ would be the stock moving up and $$Z=d$$ would be the stock moving down.

I would like to figure it out myself so maybe just a hint on how this could be would be very helpful.

• I would like to check if I understood the answer correctly if you don't mind. Suppose the stock went up and now the price is $su$ and the strike is $K$ but the price of the option $p$ is less than $su-K$. Then I can buy the option at price $p$, exercise the option and sell the stock. With that, I will end up with $su-K-p$ profit without risk. Similarly if $p$ is greater than $su-K$, then I can sell the option at $p$ buy a share at $su$ and sell it at strike price $K$. $-p<K-su$ so I made a profit of $p-(su-K)$.
• Exactly right. This shows that if $p \neq su-K$ you can make a profit one way or the other, so we must have $p=su-K$ Commented Apr 27 at 4:08