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Call Option Value from Two Approaches

Suppose today the stock price is S and in one year time, the stock price could be either S_1 or S_2. You hold an European call option on this stock with an exercise price of X=S, where S_1<X<S_2 for simplicity. So you'll exercise the call when the stock price turns out to be S_2 and leave it unexercised if S_1.

1. Replicating Portfolio Approach

Case 1 Case 2
Stock Price S_1 S_2
Option: 1 Call of cost c
Exercise? No Yes
Payoff (to replicate) 0 S_2-X
Stock: \delta shares of cost \delta S
Payoff \delta S_1 \delta S_2
Borrowing PV(K)
Repay K K

So we have:

\begin{equation} \delta S_1-K=0 \end{equation}
\begin{equation} \delta S_2 -K = S_2-X \end{equation}

Therefore, the call option value is given by the difference between the cost of \delta units of shares and the amount of borrowing:

\begin{align} c_{REP} &= \delta S - PV(K) \newline &= \delta S - Ke^{-r_f} \newline &= \delta S - \delta S_1e^{-r_f} \end{align}

When \delta is defined as \frac{(S_2-X)-0}{S_2-S_1} as in the textbook (at introductory level),

\begin{equation} c_{REP}= \frac{S_2-X}{S_2-S_1}(S - S_1e^{-r_f}) \end{equation}

2. Risk Neutral Approach

Without too much trouble, we can derive the call value using risk neutral approach as

\begin{align} c_{RN} &= \frac{p(S_2-X)+(1-p)\times0}{e^{r_f}}\newline &= \frac{p(S_2-X)+0}{e^{r_f}}\newline &= p(S_2-X) e^{-r_f} \end{align}

We know that

\begin{equation} p\times \frac{S_2}{S} + (1-p)\frac{S_1}{S} = e^{r_f} \end{equation}

so

\begin{align} p &= \frac{e^{r_f}-\frac{S_1}{S}}{\frac{S_2}{S}-\frac{S_1}{S}}\newline &=\frac{Se^{r_f}-S_1}{S_2-S_1} \end{align}

Therefore,

\begin{align} c_{RN} &= p(S_2-X) e^{r_f}\newline &=\frac{Se^{r_f}-S_1}{S_2-S_1}(S_2-X) e^{-r_f}\newline &=\frac{S-S_1e^{-r_f}}{S_2-S_1}(S_2-X) \end{align}

Identical Result from the Two Methods

It's easy to find that

c_{RN} = c_{REP}

Hence, the call option value from replicating portfolio is the same as from risk neutral approach.

Last update: May 25, 2020

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