Question

A stock is presently selling at a price of $$\$ 50$$ per share. After one time period, its selling price will (in present value dollars) be either $$\$ 150$$ or $$\$ 25 .$$ An option to purchase \(y\) units of the stock at time 1 can be purchased at cost \(c y\). (a) What should \(c\) be in order for there to be no sure win? (b) If \(c=4\), explain how you could guarantee a sure win. (c) If \(c=10\), explain how you could guarantee a sure win. (d) Use the arbitrage theorem to verify your answer to part (a).

Short answer

The cost 'c' should be \(\frac{500}{3}\) for there to be no sure win. To guarantee a sure win for \(c=4\), buy the option and invest \(y\) units of stock, and if the stock price goes down to \(25\), sell the option, resulting in a loss of \(29y\). For \(c=10\), buy the option and invest \(y\) units of stock, and if the stock price goes up to \(150\), sell the option, resulting in a profit of \(90y\). Our answer for part (a) is verified using the arbitrage theorem.

Step-by-step solution

(a) Calculating the cost 'c' for no sure win

To calculate the cost 'c', we need to ensure that in either price scenarios of the stock at time 1, there is no guaranteed win. Let's evaluate both scenarios and set up an equation accordingly: 1. If the stock price goes up to \(150 per share, the profit on each share by buying the option would be \)(150-50)y\(, and the profit by not buying the option would be \)0$. 2. If the stock price goes down to \(25 per share, the profit on each share by buying the option would be \)(25-50)y\(, and the profit by not buying the option would be \)0$. Since we want no sure win, we need the following condition to be satisfied: $$\text{Profit without option} = \text{Profit with option} -c$$ Combining both scenarios, we get: $$(150-50)y - cy = 0$$ and $$(25-50)y - cy = 0$$

Solving for 'c'

Now let's solve both equations for 'c': $$c = (150-50)y$$ and $$c = (25-50)y$$ Equating both: $$(150-50)y = (25-50)y$$ Solving for 'y': $$y = \frac{150-25}{50-(-25)} = \frac{125}{75} = \frac{5}{3}$$ Substituting 'y' back into any equation, we get: $$c = (150-50)\times \frac{5}{3}$$ $$c = \frac{500}{3}$$ So, the cost 'c' should be \(\frac{500}{3}\) for there to be no sure win.

(b) Guaranteeing a sure win for c = 4

If \(c=4\), we can check the profits with the option for both scenarios: 1. If the stock price goes up to \(150, the profit with the option would be \)(150-50)y - 4y = 100y - 4y = 96y$. 2. If the stock price goes down to \(25, the profit with the option would be \)(25-50)y - 4y = -25y - 4y = -29y$. To guarantee a sure win, buy the option and invest \(y\) units of stock. If the stock price goes up to \(150, we have a profit of \)96y\(. If the stock price goes down to \)25, sell the option. In this case, the loss is \(29y\).

(c) Guaranteeing a sure win for c = 10

If \(c=10\), we can check the profits with the option for both scenarios: 1. If the stock price goes up to \(150, the profit with the option would be \)(150-50)y - 10y = 100y - 10y = 90y$. 2. If the stock price goes down to \(25, the profit with the option would be \)(25-50)y - 10y = -25y - 10y = -35y$. To guarantee a sure win, buy the option and invest \(y\) units of stock. If the stock price goes up to \(150, sell the option. In this case, we have a profit of \)90y\(. If the stock price goes down to \)25, sell the option. In this case, the loss is \(35y\).

(d) Verifying part (a) using the arbitrage theorem

The arbitrage theorem states that there is no sure win if and only if no-arbitrage prices exist. We need to check if we have no-arbitrage prices for \(c=\frac{500}{3}\). Let \(p\) be the probability of price going up to \(150, and (1-p)\) be the probability of price going down to $25. According to the theorem, the present price of the asset(50) should be the expected future price: $$50 = p(150) + (1-p)(25)$$ Also, for options, the price should be: $$\frac{500}{3} = p(100) + (1-p)(-25)$$ Now, we need to solve these two equations to check if there's a solution. From the first equation, we get: $$p = \frac{50-25}{150-25} = \frac{1}{3}$$ Then, substituting the value of \(p\) into the second equation: $$\frac{500}{3} = \frac{1}{3}(100) + (1-\frac{1}{3})(-25)$$ $$\frac{500}{3} = \frac{500}{3}$$ Since we have a solution, it means our answer in part (a) is correct as per the arbitrage theorem.