# Call option arbitrage opportunity

I am having trouble wrapping my head around some text provided to us by our lecturer (unfortunately he is currently unavailable). If we let $c$ be the price of a European call option, $S_0$ the current price of an asset (say a stock), $X$ the strike price, $T$ the time to maturity, and $r$ the (static) interest rate. We ignore dividends.

Suppose that $$c = 3, S_0 = 20, X = 18,\\ T = 1, r = 10\%$$ Is there an arbitrage opportunity?

• buy the call, short the stock
• proceeds: $-3+20 =17$; grows to $17e^{0.1} = 18.79 > 18$
• yes!

I don't understand how you can deduce the existence of an arbitrage opportunity from $18.79 > 18$.

• If the stock never moved.. Haha yes.... The risk premium over and above the intrinsic implies a distro of stock returns ... So all joking aside... No. No arb there .. There's risk from hedging, cost of borrowing.. Etc Apr 15, 2013 at 7:15
• The simplest way to show arbitrage opportunities here is the lower bound of the call price which is call > value of underlying asset - PV of strike which should force the call option price to be above about 3.71. Anything below that presents an arbitrage opportunity. Obviously the OP has made clear that he made a lot of simplifying assumptions. Thus, we are talking about a theoretical arbitrage opportunity absent of any transaction costs, liquidity issues, dividends,... Apr 15, 2013 at 8:00
• @Freddy I understand that since $c = 3 < 20 - 18e^{-0.1} = 3.71$ this presents an arbitrage opportunity since by definition a sub-optimal price reflects potential achievable value. Perhaps my original post should have asked how exactly the above quoted arbitrage strategy works. Am I right in saying that if you buy the call and short the stock, you end up with $-3 + 20 = 17$, which you can then invest to get $18.79$ in a year's time; at maturity, you buy the stock for $18$, return it, and pocket $0.79$ instead of getting $-1$, or rather $0$, were you not to engage in arbitrage? Apr 15, 2013 at 9:06
• Not exactly: You buy back the stock at whatever price it is traded at, you exercise the call if the stock price is above your strike at expiry and you get back your investment of 17 plus interest. Apr 15, 2013 at 9:13
• Let $S_1 = 20$, i.e. the stock price stays the same, how do you "get back" your investment if you have to return the stock to whoever lent it to you for you to have gone short with it in the first place? At maturity you're left with 18.79, a call option, and an obligation to return stock. You exercise the call option buying 20 worth of stock for 18, you return this stock. You now have 0.79 left. Alternatively, you buy 20 worth of stock for 18, sell it for 20, pocket 2, you now have 20.79 but you must still return stock, you buy it for 20, return it, and you are again left with 0.79, no? Apr 15, 2013 at 11:02

The option is a contract that gives you the right to buy the stock in one year for 18. Today people are trading the stock for 20, so you can sell the stock short for 20 today, meaning, someone gives you 20 cash today in return for a stock IOU, where you are obligated to deliver the stock to them on a later date.

So you get 20 cash upfront but you need to spend 3 of it to buy the option. After one year this net 17 reinvested becomes 18.79. This is more cash than you need to buy back the stock using your option if the price ends up > 18. So, worst case scenario is the stock ends the year > 18, you exercise the option and buy the stock for 18 from the option seller, and you net 18.79-18.0=.79 cash.

Of course the stock could end the year below 18, and then you wouldn't use the option to buy back the stock, you would just buy the stock at whatever price it trades in the market. So either you make .79 or you make more. Arbitrage.