# 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:

$$$\delta S_1-K=0$$$
$$$\delta S_2 -K = S_2-X$$$

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),

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

## 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

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

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.