How to find the price of the call option?

Question

Assume we have a financial market consisting of a bond $A$, a stock $S$ and call options $f$ on the stock.

Consider only two times $0$ and $T$ and only two possible outcomes for the value of $S$ at time T.

The bond and stock values at times $0$ and $T$ are given below :

$A(0)=100$, $A(T)=120$, $S(0)=50$, $S^u(T)=70$, $S^d(T)=30$.

1. Construct a portfolio $\phi$ consisting only of shares of $S$ and call options (on $S$) $f$ with strike $K=50$ and expiration $T$ such that the value of $\phi$ at time $T$, $V(T)$, is $120$ regardless of whether the stock goes up or down.

2. What is the price of the call option?

---

I am having trouble understanding the question.

With a call option price at time $T$ of $\$20$, if the stock goes up the value of the portfolio would be $70 \times \frac12 − 20 = 15$ and if the stock goes down the value of the portfolio would be $30 \times \frac12 = 15$, so the portfolio would need to contain $4$ shares of the stock and $8$ options to have a value of $\$120$ whether the stock goes up or down, is this correct?

Answer 1

Let $\phi$ be our portfolio of $\Delta$ of a certain stock $S$ whose initial value is $S_0 \stackrel{ie}{=} 50$ and $f_n$ call options (on $S$) $f$ whose strike price is $K \stackrel{ie}{=} 50$ and maturity/expiration is $T$.

Our portfolio's value is given by $V_t(\phi) = (S_t)(\Delta) - (f_t)(f_n)$ for $t=0,T$

where $f_T = (S_T - K)^{+}$

Then we have:

$$V_0(\phi) = (S_0)(\Delta) - (f_0)(f_n)$$

$$\stackrel{ie}{=} 50(\Delta) - (f_0)(f_n)$$

$$V_T(\phi) = S_T\Delta - f_n(S_T - K)^{+}$$

---

Proposition: To have $P(V_T(\phi) = X) = 1$ (Not sure of the range of values for $X$ but this seems to work for $X=15$ and $X=120$), our portfolio must be:

Long: $\Delta = X/30$ shares

Short: $f_n = X/15$ option

---

Proof:

$$V_T^u(\phi) = S^u_T\Delta - f_n(S^u_T - K)^{+} = 70\Delta - 20f_n \stackrel{set}{=} X$$

$$V_T^d(\phi) = S^d_T\Delta - f_n(S^d_T - K)^{+} = 30\Delta \stackrel{set}{=} X$$

QED

---

The value of the portfolio $\phi$ at time $0$ for such $\Delta$ and $f_n$ is given by

$$V_0(\phi) = (S_0)(X/30) - (f_0)(X/15) = (50/30)(X) - (f_0)(X/15)$$

Also

$$V_0(\phi) = V_T(\phi)e^{-rT} \ \text{Why?}$$

$$ = Xe^{-rT}$$

$$ = Xe^{-(\ln(120/100))} \ \text{Why?}$$

Hence

$$f_0 = \frac{(50/30)(X)-(X)e^{-(\ln(120/100))}}{X/15}$$

$$ = 15((50/30)-\frac{1}{e^{\ln(120/100)}})$$

$$ = 15((50/30)-\frac{1}{120/100})$$

$$ = 15(5/6)$$

$$= 12.5$$

Answer 2

Here's the answer my professor just put up, I got the correct number of shares of stocks and the number of call options, but I don't know how he got the price of the call option at time $0$ to be $12.5$ when he says the call option must replicate the bond?

Assume that the portfolio consists of $x$ shares of stock and y shares of call options.
Let us denote by $C(0)$ the price of the call option.
Denote by $C^u(T)$ the value of the call option at time $T$ in the case when the stock goes up.
Similarly, let $C^d(T)$ denote the value of the call option if the stock goes down.
Clearly, $C^d(T)=0$ and $C^u(T)=S^u(T)−K=20$.
The value of the portfolio in the case that stock goes up is: $120=V^u(T)=xS^u(T)+yC^u(T)=70x+20y$.
The value of the portfolio if the stock goes down is $120=V^d(T)=xS^d(T)=30x$. Therefore we obtain $x=4$ and $y=−8$.
Thus the portfolio consists of buying $4$ shares of stock and short-selling $8$ call options.
The value of the portfolio at time $0$ is: $V(0)=4S(0)−8C(0)=200−8C(0)$.

Since the call option replicates the bond, we must have $V(0)=100$, therefore $C(0)=\frac{100}{8}=12.5$