# Pricing a call option – an example

The example in this post illustrates how to price a call option using the one-period binomial option pricing model. The next post will present an example on pricing a put option. The two posts are designed to facilitate the discussion on the binomial option pricing (given in a series of subsequent posts). Links to practice problems are found at the bottom of the post.

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The example

The following gives the information about the stock:

• The stock of XYZ company is currently selling for $50 per share. The price per share 1 year from now is expected to increase to$65 or to decrease to $40. The stock pays no dividends. Consider a call option with the following specifics: • The underlying asset of the call option is the XYZ stock. • The strike price is$55.
• The option will expire in one year.
• The option is assumed to be a European option, i.e. it can be exercised only at expiration.

The annual risk-free interest rate is 2%. There is a benefit to the buyer of the option described above. If the price of the stock goes up to $65 at the end of the 1-year period, the owner of the option has the right to exercise the option, i.e., buying one share at the strike price of$55 and then selling it at the market price of $65, producing a payoff of$10. If the price of the stock goes down to $40 at the end of the 1-year period, the buyer of the option has the right to not exercise the option. The call option owner buys the stock only when he makes money. What would be the fair price of having this privilege? What is the fair price of this call option? ___________________________________________________________________________________ Pricing the call option In this example, the current stock price is$50 and the stock price can be only one of the two possible values at the end of the option contract period (either $65 or$40). The following diagram shows the future state of the stock prices.

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Figure 1 – Stock Price

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The assumption of the 2-state stock prices in 1 year simplifies the analysis of the call option. The value of the call option at the end of 1 year is either $10 (=65-55) or zero. Note that when the share price at the end of the 1-year contract period is less than the strike price of$55, the call option expires worthless. The following diagram shows the value of the call option.

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Figure 2 – Call Option Payoff

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In the above diagram, the value of the call option at the end of 1-year is either $10 or$0. The value of the option at time 0 is $C$, which is the premium of the call option in this example. Our job here is to calculate $C$. The key to finding the value of the option is to compare the payoff of the call to that of a portfolio consisting of the following investments:

Portfolio A

• Buy 0.4 shares of XYZ.
• Borrow $15.683 at the risk-free rate. The idea for setting up this portfolio is given below. For the time being, we take the 0.4 shares and the borrowed amount of$15.683 as a given. Note that $15.683 is the present value of$16 at the risk-free rate of 2%. Let’s calculate the value of Portfolio A at time 0 and at time 1 (1 year from now). The following diagram shows the calculation.

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Figure 3 – Portfolio A Payoff

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Note that the payoff of the call option is identical to the payoff of Portfolio A. Thus the call option in this example and Portfolio A must have the same cost. Since Portfolio A costs $4.317, the price of the option must be$4.317. The Portfolio A of 0.4 shares of stock and $15.683 in borrowing is a synthetic call since it mimics the call option described in the example. Portfolio A is called a replicating portfolio because it replicates the payoff of the call option in question. ___________________________________________________________________________________ Arbitrage opportunities In deriving the cost of the call option of$4.137, we rely on the idea that if two investments have the same payoff, they must have the same cost. This idea is called the law of one price, which is a commonsensical idea and is also an important principle in derivative pricing. If the law of one price is violated, in particular if the price of the call option discussed in this example is not $4.317, there would be arbitrage opportunities that can be exploited to gain risk-free profit. What if the law of one price is violated? For example, what if the option were selling for a higher price (say$4.50)? If the price of the replicating portfolio is less than the price of the option, then we can “buy low and sell high” (i.e. buy the replicating portfolio and sell call option) and obtain a risk-free profit of $0.183. The arbitrage is to buy the synthetic call (Portfolio A) at$4.317 and sell the call option at $4.50. The following table shows the Year 1 cash flows of this arbitrage opportunity. $\text{ }$ Table 1 – Arbitrage opportunity when call option is overpriced $\left[\begin{array}{llll} \text{Year 1 Cash Flows} & \text{ } & \text{Share Price = } \ 40 & \text{Share Price = } \ 65 \\ \text{ } & \text{ } \\ \text{Long synthetic call} & \text{ } & \text{ } & \text{ } \\ \ \ \ \ \text{Hold 0.4 shares} & \text{ } & + \ 16 & + \ 26 \\ \ \ \ \ \text{Repay borrowed amount of } \ 15.683 & \text{ } & - \ 16 & - \ 16 \\ \text{ } & \text{ } \\ \text{Short call } & \text{ } & \ \ \ 0 & - \ 10 \\ \text{ } & \text{ } \\ \text{Total payoff} & \text{ } & \text{ } \ \ 0 & \ \ \ 0 \end{array}\right]$ $\text{ }$ The above table shows that buying a synthetic call (holding 0.4 shares and borrow$15.683) and selling a call will have no loss at the end of 1 year. Yet, the time 0 cash flow is $0.183 (=4.50 – 4.317), and is thus a risk-less profit. If the option is underpriced, then we can still buy low and sell high (in this case, buy call option and sell the replicating portfolio) and obtain risk-free arbitrage profit. For example, let’s say you observe a call option price of$4.00. Then the arbitrage opportunity is to buy the call option at $4.00 and sell a synthetic call (Portfolio A) at$4.317. The time 0 payoff is $0.317, which is a risk-less arbitrage profit. The following table shows the Year 1 cash flows. $\text{ }$ Table 2 – Arbitrage opportunity when call option is underpriced $\left[\begin{array}{llll} \text{Year 1 Cash Flows} & \text{ } & \text{Share Price = } \ 40 & \text{Share Price = } \ 65 \\ \text{ } & \text{ } \\ \text{Short synthetic call} & \text{ } & \text{ } & \text{ } \\ \ \ \ \ \text{Short 0.4 shares} & \text{ } & - \ 16 & - \ 26 \\ \ \ \ \ \text{Receive the amount of } \ 15.683 & \text{ } & + \ 16 & + \ 16 \\ \text{ } & \text{ } \\ \text{Long call } & \text{ } & \ \ \ 0 & + \ 10 \\ \text{ } & \text{ } \\ \text{Total payoff} & \text{ } & \text{ } \ \ 0 & \ \ \ 0 \end{array}\right]$ $\text{ }$ ___________________________________________________________________________________ To complete the picture The call option price of$4.317 is derived by showing that the replicating portfolio has the same payoff as the call option. How do we know that the replicating portfolio consists of holding 0.4 shares and the borrowing of $15.683? In general, the replicating portfolio of a European call option consists of $\Delta$ shares of the stock and the amount $B$ in lending at time 0 (borrowing if negative). By equating the payoff of the replicating portfolio and the payoff of the call option in this example, we have the following equations: $\text{ }$ $\displaystyle \begin{array}{ccc} \displaystyle 40 \ \Delta + B \ e^{0.02} & = & 0 \\ \displaystyle 65 \ \Delta + B \ e^{0.02} & = & 40 \end{array}$ $\text{ }$ Solving these two equations, we obtain $\Delta=\frac{10}{25}=0.4$ and $B=-16 \ e^{-0.02}=15.683$. Therefore, the replicating portfolio for the call option in this example consists of 0.4 shares of the stock and$15.683 in borrowing. The net investment for the replicating portfolio is $4.317 (=0.4(50)-15.683). Because there are only two data points in the future stock prices, the option premium is a linear function of $\Delta$ and $B$. The following is the premium of the call (or put) option using the one-period binomial tree $C=\Delta \ S+B$ where $S$ is the stock price at expiration. The above formula gives the cost of the portfolio replicating the payoff of a given option. It works for call option as well as for put option. We will see that for put options, $\Delta$ is negative and $B$ is positive (i.e. shorting stock and lending replicate the payoff of a put). The number $\Delta$ has a special interpretation that will be important in subsequent discussion of option pricing. It can be interpreted as the sensitivity of the option to a change in the stock price. For example, if the stock price changes by$1, then the option price, $\Delta \ S + B$, changes by the amount $\Delta$. In other words, $\Delta$ is the change in the option price per unit change in the stock price.

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Put-call parity

The put-call parity relates the price of a European call with a European put that has the same strike price and the same time to expiration. The following is a put on XYZ stock that is compatible to the call described above.

• The underlying asset of the put option is the XYZ stock.
• The strike price is \$55.
• The option will expire in one year.
• The option is assumed to be a European option, i.e. it can be exercised only at expiration.

By the put-call parity, the following gives the price of the put option.

\displaystyle \begin{aligned} P(55,1)&=C(55,1)-50+55 \ e^{-0.02} \\&=4.316821227-50+55 \ e^{-0.02} \\&=\ 8.2277 \end{aligned}

The next post will calculate the price of the same put using the binomial model.

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Remarks

We would like to comment that even though the example here may seem like an extreme simplification, the example has great value. First of all, this is an excellent introduction to the subject of option pricing theory. Secondly, the one-period example can be extended to a multi-period approach to describe far more realistic pricing scenarios. For example, we can break a year into many subintervals. We then use the 2-state method to describe above to work backward from the stock prices and option values of the last subinterval to derive the value of the replicating portfolio.

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Practice problems

Practice problems can be found in the companion problem blog via the following links:

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$\copyright \ \ 2015 \ \text{Dan Ma}$