# Arbitrage strategies in Rubinstein's binomial tree one-step

Suppose that the current stock price is $S_0=20$ and the call option price with no arbitrage is $c=0.633$. Knowing that the expiry stock price can be $S_T=22$ with call option price $1$ or $S_T=18$ with call option price $0$, which strategies can be realized if the current call option is $>0.633$ and $<0.633$?

I thought that if it was $c=0.62$, now, I could:

1. buy the call at $c$
2. sell $∆=0.25$ shares at $S_0$
3. invest $S_0\Delta-c$ at the risk free rate $r=$ 12% for $3$ months.

But what actions do I have to take at expiry? Considering that I have to return the shares and I can cash the profit $(S_0\Delta-c)e^{rT}$.

And if it was $c=0.65$?

• I don't understand the question. Please be clearer by what you mean by "But at expiry?" – user217285 Dec 17 '17 at 10:07
• It means "3 months from now". – Francesco Totti Dec 17 '17 at 10:13
• I don't understand your question either. What do you mean by "which points in order I have to set?" – LocalVolatility Dec 17 '17 at 12:08
• Sorry for my bad english, it's difficult for me. It means "what i could do at 3 months from now". – Francesco Totti Dec 17 '17 at 14:41

Let's first check that 0.633 is the call option price. The risk-neutral probability $p^*$ of an up-tick in the stock is computed by assuming the stock earns a risk-free rate of return: $$20 = e^{-0.03}(22p^* + 18(1-p^*)) \qquad \Rightarrow \qquad p^* = 0.65227.$$ The price of the option is the risk-neutral expected payoff, discounted at the risk-free rate. In this case, the option has payoff 1 in an up-tick and payoff 0 in a down-tick, so $$c = e^{-0.03}p^* = 0.633.$$ The delta-hedge ratio needed for a replicating portfolio is the ratio of the change in the option price to the change in the stock price: $$\Delta = \frac{1 - 0}{22-18} = 0.25.$$ If the option is trading at $C < 0.633$, then you should be able to capture the $0.633 - C$ spread by employing the buy-cheap sell-expensive strategy:

1. Buy the option at $C$
2. Sell $\Delta = 0.25$ shares of the stock

If $C = 0.62$, then selling $\Delta$ shares of the stock and buying the option gives you an initial cash position of $$0.25*20 - 0.62 = 4.38,$$ which grows to $4.38e^{0.03} = 4.5133$ at expiry. We now consider two cases:

1. $S_T = 22$: the option has payoff 1, and the short position has payoff $-0.25*22 = -5.50$, so the total payoff is $1 - 5.50 + 4.5133 = 0.0133$.
2. $S_T = 18$: the option has payoff 0, and the short position has payoff $-0.25*18 = -4.50$, so the total payoff is $0 - 4.50 + 4.5133 = 0.0133$.

If the option is trading at $C > 0.633$, then again employ the buy-cheat sell-expensive strategy. The main point is that you can completely replicate the payoff of the option using just a portfolio of stock and cash; by the law of one price, the cash required to set up the replicating portfolio is the price of the option.

• Thanks for the good reply. I've tried to set up the arbitrage when $C=0,65$: 1. Loan of $S_0∆=20*0,25=5$ at $r$ for $3$ months: i will return $5,1523$ at $3$ months 2. Sell the call at $0,65$ 3. Buy $∆=0,25$ shares. If at $3$ months $S_T=22$ my counterpart exercise the call and the value is $-1$, i have to repay the debit and...then? – Francesco Totti Dec 18 '17 at 8:08
• You seem to be missing the point of the exercise; it should be 100% mechanical. If an option is trading at a price greater than the price of a replicating portfolio, then sell the option and buy the replicating portfolio. You know the payoffs of the option and the stock at the end, so you can compute your net profit as well. Don't get wrapped up in the numbers and "loan"/"borrow" terminology; again this is 100% mechanical, and you can use the template I've provided above. – user217285 Dec 18 '17 at 22:48