# Shreve I.3: State prices and other option prices

Suppose the stock having current share price $S_0$ may either go up or down, to $S_1(H)=uS_0$ or $S_1(T)=dS_0$. Suppose we have a “stock option” $V_1$ that pays 1 dollar if $H$ occurs and zero otherwise; $V_1(H)=1$ and $V_1(T)=0$. By the binomial asset pricing model (Shreve volume I chapter 1) the price of this option should be
$$V_0 = \frac{1}{1+r} (\tilde p V_1(H) + \tilde q V_1(T)) = \frac{\tilde p}{1+r}$$
[This is the state price of the state H.] Here
$$\tilde p = \frac{1+r-d}{u-d}$$

Suppose we fix $d=1/2$ and let $u\rightarrow\infty$. So we are looking at a stock that will either drop by 50%, or increase enormously. Then $V_0\rightarrow 0$, i.e., the option becomes worthless.

This may seem surprising since the option depends on heads/tails but not explicitly on the stock price associated with H/T. But the way to look at it is this: if there exists a stock that will either drop by 50% or increase enormously, then it better be very unlikely that the stock should increase.

Say $u=1000$ and $r=1/2$. Then to replicate the option we would buy $\Delta_0 = \frac{V_1(H)-V_1(T)}{S_1(H)-S_1(T)} = \frac{1}{999.5 S_0}$ shares. Because of the $S_0$ in the denominator this is just $\epsilon:=\frac{1}{999.5}$ dollars worth of shares.

We would need an initial capital of
$$X_0 = \frac{1+r-0.5}{(1000-0.5)(1+r)} = \frac{1}{(3/2)(999.5)} = 2\epsilon/3$$
which is two-thirds of the amount we invested in the stock. With such initial capital we have set up a portfolio guaranteeing to get exactly the value of the option $V_1(\omega)$ back at time 1.

Case 1: Tails occurs. Then the value of the option is zero, and the value of our portfolio is calculated as follows: we had to borrow $\epsilon/3$ to buy stock, which means we now owe somebody $(1+r)\epsilon/3 = \epsilon/2$. But the stock we bought now has value $\epsilon/2$; so we have 0 overall.

Case 2: Heads occurs. The value of the option is 1; we owe somebody $\epsilon/2$ but we the stock we bought has value $1000\epsilon$; and indeed
$$1000\epsilon – \epsilon/2 = 1$$

Generally the portion of our capital that we buy stock for is
$$\frac{\Delta_0 S_0}{X_0} = \frac{(\Delta V/\Delta S)S_0}{[\tilde p V_1(H)+\tilde q V_1(T)]/(1+r)} = \frac{\Delta V/(u-d)}{[\tilde p V_1(H)+\tilde q V_1(T)]/(1+r)}$$
$$= \frac{\Delta V/(u-d)}{[\tilde p \Delta V + V_1(T)]/(1+r)}$$

$$= \frac{\Delta V/(u-d)}{[((1+r-d)/(u-d)) \Delta V + V_1(T)]/(1+r)}$$

$$= \frac{(1+r)\Delta V}{[((1+r-d)/(u-d)) \Delta V + V_1(T)](u-d)}$$

$$= \frac{(1+r)\Delta V}{(1+r-d)\Delta V + V_1(T)(u-d)}$$

$$= \frac{(1+r)\Delta V}{(1+r-d)V_1(H) – (1+r-u) V_1(T)}$$

$$= \frac{(1+r)\Delta V}{((1+r)-d)V_1(H) + (u-(1+r)) V_1(T)}$$
where $\Delta V = V_1(H)-V_1(T)$. It is worth looking at when this fraction will be more than 1, i.e., replicating the option involves borrowing from the money market. This is when
$$\frac{V_1(H)}{V_1(T)} > \frac{S_1(H)}{S_1(T)}$$
i.e., the option payoff is, in some sense, more dependent on the heads/tails outcome than the stock price is.
For instance, if the option were constant ($V_1(H)=V_1(T)=c$) then we need not borrow to replicate it; we would buy 0 shares of stock and invest the whole initial capital $c/(1+r)$ in the money market.