Call Option Value from Two Approaches

May 25, 2020 | 2 min read

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}

\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.