Exercise 1.1

Assume in the one-period binomial market of Section 1.1 (The stock price $S$ has two possible outcomes in the next period: it can either go up to $uS_0$ or down to $dS_0$.) that both H and T have positive probability of occurring. Show that condition $0<d<1+r<u$ precludes arbitrage. In other words, show that if $X_0 = 0$ and $$X_1=\Delta_0 S_1 + (1+r)(X_0 - \Delta_0 S_0),$$ then we cannot have $X_1$ strictly positive with positive probability unless $X_1$ is strictly negative with positive probability as well, and this is the case regardless of the choice of the number $\Delta_0$.

In simpler terms, if there’s a chance to gain, there must also be a chance to lose, no matter how the portfolio is constructed using the stock through $\Delta_0$.

Proof: $$X_1 = \Delta_0 S_1 + (1+r)(-\Delta_0 S_0).$$ Expanding $S_1$ for the coin toss results:

  • If the toss results in H (stock price goes up): $S_1 = uS_0$, then $$X_1(H) = \Delta_0 u S_0 + (1+r)(-\Delta_0 S_0) = \Delta_0 S_0 (u-(1+r)).$$
  • If the toss results in T (stock price goes down): $S_1 = dS_0$, then $$X_1(T) = \Delta_0 d S_0 + (1+r)(-\Delta_0 S_0) = \Delta_0 S_0 (d-(1+r)).$$

Suppose $S_0>0$. Because $u>1+r$, $u - (1+r)$ is positive. Because $d<1+r$, $d-(1+r)$ is negative.

Therefore, if $\Delta_0>0$, $X_1(H)>0$ and $X_1(T) <0$. If $\Delta_0<0$, $X_1(H)<0$ and $X_1(T) >0$. If $S_0<0$, we can similarly deduce that $X_1$ must take both positive and negative values, depending on whether the stock price goes up or down.

Thus, under the condition $0<d<1+r<u$, we cannot have $X_1$ always non-negative or always non-positive, regardless of the choice of $\Delta_0$

Exercise 1.2

Suppose in the situation of Example 1.1.1 (for the particular one-period model, let $S_0=4, u=2, d=\frac{1}{2}, K=5, r=\frac{1}{4}$) that the option sells for 1.20 at time zero. Consider an agent who begins with wealth $X_0=0$ and at time zero buys $\Delta_0$ shares of stock and $\Gamma_0$ options. The numbers $\Delta_0$ and $\Gamma_0$ can be either positive or negative or zero. This leaves the agent with a cash position of $-4\Delta_0-1.20\Gamma_0$. If this is positive, it is invested in the money market; if it is negative, it represents money borrowed from the money market. At time one, the value of the agent’s portfolio of stock, option, and money market assets is $$X_1 = \Delta_0 S_1 + \Gamma_0 (S_1-5)^+ - \frac{5}{4}(4\Delta_0+1.20\Gamma_0).$$ Assume that both $H$ and $T$ have positive probability of occurring. Show that if there is a positive probability that $X_1$ is positive, then there is a positive probability that $X_1$ is negative. In other words, one cannot find an arbitrage when the time-zero price of the option is 1.20.

Proof: $$X_1 = \Delta_0 S_1 + \Gamma_0 (S_1-5)^+ - \frac{5}{4}(4\Delta_0+1.20\Gamma_0).$$ Expanding $S_1$ for the coin toss results:

  • If the toss results in H (stock price goes up): $S_1 = uS_0$, then $$\begin{aligned} X_1(H) &= \Delta_0 uS_0+ \Gamma_0 (uS_0-5)^+ - \frac{5}{4}(4\Delta_0+1.20\Gamma_0)\\ &= \Delta_0 \times 2 \times 4 + \Gamma_0 (2 \times 4 - 5)^+ - (5\Delta_0 +1.50\Gamma_0)\\ &= 3\Delta_0+1.5\Gamma_0. \end{aligned}$$
  • If the toss results in T (stock price goes down): $S_1 = dS_0$, then $$\begin{aligned} X_1(T) &= \Delta_0 dS_0+ \Gamma_0 (dS_0-5)^+ - \frac{5}{4}(4\Delta_0+1.20\Gamma_0)\\ &= \Delta_0 \times \frac{1}{2} \times 4 + \Gamma_0 (\frac{1}{2} \times 4-5)^+ - (5\Delta_0 +1.50\Gamma_0)\\ &= -(3 \Delta_0 + 1.50 \Gamma_0). \end{aligned}$$

If $\Delta_0$ and $\Gamma_0$ are set to make $X_1(H)$ positive, $X_1(T)$ is definitely negative. Similarly, if $\Delta_0$ and $\Gamma_0$ are set to make $X_1(T)$ positive, $X_1(H)$ is definitely negative.

Thus, with the option priced at 1.20, there’s no arbitrage strategy that guarantees a profit in both states of the stock.

Exercise 1.3

In the one-period binomial model of Section 1.1, suppose we want to determine the price at time zero of the derivative security $V_1=S_1$ (i.e., the derivative security pays off the stock price.) (This can be regarded as a European call with strike price $K=0$). What is the time-zero price $V_0$ given by the risk-neutral pricing formula $V_0 = \frac{1}{1+r}[\tilde{p}V_1(H)+\tilde{q}V_1(T)]$?

Solution:

Since $V_1=S_1$, if the stock goes up $V_1(H) = S_1(H) = u S_0$; If the stock goes down $V_1(T) = S_1(T) = d S_0$.

The risk neutral probabilities $\tilde{p}$ and $\tilde{q}$ can be solved for and are given by the formulas: $$\tilde{p} = \frac{1+r-d}{u-d},$$ and $$\tilde{q} = \frac{u-1-r}{u-d}.$$

Inserting $\tilde{p}$ and $\tilde{q}$ into the risk-neutral pricing formula, we have $$\begin{aligned} V_0 &= \frac{1}{1+r}\left[\frac{1+r-d}{u-d}u S_0 + \frac{u-1-r}{u-d}d S_0\right]\\ &= \frac{1}{1+r}\left[\frac{(1+1)(u-d)}{u-d}S_0\right]\\ &= S_0 \end{aligned}$$

Exercise 1.4

In the proof of Theorem 1.2.2, show under the induction hypothesis that $$ X_{n+1}(\omega_1 \omega_2 \cdots \omega_n T) = V_{n+1}(\omega_1 \omega_2 \cdots \omega_n T). $$

Proof: Using the wealth equation $X_{n+1} = \Delta_n S_{n+1} + (1+r)(X_n - \Delta_n S_n)$ to compute $X_{n+1}(\omega_1 \omega_2 \cdots \omega_n T)$, $$ X_{n+1}(\omega_1 \omega_2 \cdots \omega_n T) = \Delta_n(\omega_1 \omega_2 \cdots \omega_n) d S_{n}(\omega_1 \omega_2 \cdots \omega_n) + (1+r)(X_n(\omega_1 \omega_2 \cdots \omega_n) - \Delta_n(\omega_1 \omega_2 \cdots \omega_n) S_n(\omega_1 \omega_2 \cdots \omega_n)) $$ suppressing $\omega_1 \omega_2 \cdots \omega_n$ and write the equation as $$ X_{n+1}(T) = \Delta_n d S_{n} + (1+r)(X_n - \Delta_n S_n). $$ Substituting $$ \Delta_n = \frac{V_{n+1}(H)-V_{n+1}(T)}{(u-d)S_n} $$ into $X_{n+1}(T)$ and using the induction hypothesis

$X_{n}(\omega_1 \omega_2 \cdots \omega_n) = V_{n}(\omega_1 \omega_2 \cdots \omega_n)$ for all $\omega_1 \omega_2 \cdots \omega_n,$

and the definition $$ V_{n}(\omega_1 \omega_2 \cdots \omega_n) = \frac{1}{1+r}[\tilde{p}V_{n+1}(\omega_1 \omega_2 \cdots \omega_n H) + \tilde{q}V_{n}(\omega_1 \omega_2 \cdots \omega_n T)], $$ we have $$ \begin{aligned} X_{n+1}(T) &= (1+r)X_n + \Delta_nS_n(d-(1+r)) \\ &= (1+r)V_n + \frac{(V_{n+1}(H)-V_{n+1}(T))(d-(1+r)}{u-d} \\ &= (1+r)V_n -\tilde{p}V_{n+1}(H) + \tilde{p}V_{n+1}(T)\\ &= \tilde{p}V_{n+1}(H) + \tilde{q}V_{n+1}(T) -\tilde{p}V_{n+1}(H) + \tilde{p}V_{n+1}(T)\\ &= V_{n+1}(T). \end{aligned} $$

Reinstating the suppressed coin tosses $\omega_1 \omega_2 \cdots \omega_n$, $$ X_{n+1}(\omega_1 \omega_2 \cdots \omega_n T) = V_{n+1}(\omega_1 \omega_2 \cdots \omega_n T). $$

Exercise 1.5

In example 1.2.4, we considered an agent who sold the look-back option for $V_0=1.376$ and bought $\Delta_0 = 0.1733$ shares of stock at time zero. At time one, if the stock goes up, she has a portfolio valued at $V_1(H) = 2.24$. Assume that she now takes a position of $\Delta_1(H)=\frac{V_2(HH) - V_2(HT)}{S_2(HH) - S_2(HT)}$ in the stock. Show that, at time two, if stock goes up agin, she will have a portfolio valued at $V_2(HH) =3.20$, whereas if the stock goes dow, her portfolio will be worth $V_2(HT) =2.40$. Finally, under the assumption that the stock goes up in the first period and down in the second period, assume the agent takes a position of $\Delta_2(HT)=\frac{V_3(HTH) - V_3(HTT)}{S_3(HTH) - S_3(HTT)}$ in the stock. Show that, at time three, if the stock goes up in the third period, she will have a portfolio value at $V_3(HTH)=0$, whereas if the stock goes dow, her portfolio will be worth $V_3(HTT)=6$. In other words, she has hedged her short position in the option.

Reference

  1. Steven E. Shreve, Stochastic Calculus for Finance I, The Binomial Asset Pricing Model