1 Solutions for Chapter 1

Exercise 1.1 We have

$(1 + 0.1/2)^2 = 1.1025$ so the effective rate is $0.1025$.
$(1 + 0.1/4)^4 - 1 \approx 0.1038$.
$\e^{0.1} - 1 \approx 0.1052$.

Exercise 1.2 We have

Solve $D\e^{rt} = 2D$ where $r=0.1$ for $t$, the time it takes to double your money. Answer: 6.93 years.
Solve $(1+r)^t = 4$ where $r=0.05$ for $t$, the time it takes to quadruple your money. Answer: 28.41 years.

Solve $(1+r)^t = \alpha$ for $t$ to get $t = \log \alpha/\log(1+r)$.

Exercise 1.3 Let $a$ be the amount you have to invest at the beginning of each of the next 60 months in order to have 100,000 at the end of 60 months. Then you have to solve \[ a(1+\frac{0.06}{12})^{60} + a(1+\frac{0.06}{12})^{59} + \ldots + a(1+\frac{0.06}{12}) = a \sum_{k=1}^{60} 1.005^k = 100,000. \] Answer: $a=1426.15$.

Exercise 1.4 Denote by $P$ the present value of the first bond described. We have to solve \[ P = -10000 + \sum_{j=1}^{10} \frac{500}{(1+r/2)^j} + \frac{10000}{(1+r/2)^{10}} \] Using the geometric sum $\sum_{k=1}^n x^k = x(x^n - 1)/(x - 1)$ we obtain

$P = 1706.29$,
$P = 0$ (to a very close approximation!),
$P = -736.01$.

For the second bond, the present value is given by \[ -1000 + 30 \sum_{i=1}^{9} \e^{-0.05*i/2} + 1030 \e^{-0.05*5} = 40.94. \]

Exercise 1.5 See HW solution.

Exercise 1.6 See HW solution.

Exercise 1.7 See HW solution.

Exercise 1.8 See HW solution.

Exercise 1.9 We have $T = 1/4$, $K=10$, $C=P=2.5$, $r=0.06$ and $S = 11$. Then $P+S-C = 11 > 10 > K e^{-rT}$ so the put-call option parity formula does not hold and an arbitrage must exist.

To make money from this we must sell what is overpriced and buy what is underpriced. Specifically short sell one share, sell one put and buy one call option. This leaves us with $P+S-C > 0$ which we put in the bank at $t=0$. This initial portfolio has time 0 value of 0. At time $T = 1/4$ the portfolio is worth

$(P+S-C)e^{rT} - K$ if $S(T) > K$ (the put is not used, we exercise the call which costs us $K$ and we return the share to its owner and pocket the cash);
$(P+S-C)e^{rT} - K$ if $S(T) \leq K$ (we throw away our call, the put is used so we pay out $K$ for a share which we return to its owner and again we pocket the cash).

We make $(P+S-C)e^{rT} - K > 0$ (in time $T$ values) whatever the stock price.

Exercise 1.10 At $t=0$, assemble $A$ with one call option and $K\e^{-rT}$ in the bank and $B$ with one share of the stock. These portfolios are worth $\max \{S_T, K\}$ and $S_T$ at time $T$. For there not to be an arbitrage, $B$ must be worth no more than $A$ at time 0, i.e., $S \leq C + K\e^{-rT}$. As the option must be worth something this proves that $C \geq (S - Ke^{-rT})^+$.

Now modify $A$ by banking 0 initially so $A$ is worth $(S_T - K)^+ \leq S_T$ at time $T$. The usual no-arbitrage argument implies $C \leq S$.

For the put option portfolio $A$ has a share and a put option while $B$ has $K\e^{-rT}$ cash and the inequalities are $(K\e^{-rT} - S)^+ \leq P \leq K\e^{-rT}$. For the final part compare the $K_2$ put plus $K_1\e^{-rT}$ cash with the $K_1$ put plus $K_2\e^{-rT}$ cash.

Exercise 1.11 We have to compare the result of exercising at $T_1$ against waiting until $T_2$. If $S_{T_1} \leq K_1$ then exercising at $T_1$ causes a loss. If $S_{T_1} > K_1$ then at $T_1$ you could

$A$: exercise the call at $T_1$ and put $S_{T_1} - K_1$ in the bank;
$B$: sell the stock short, put $S_{T_1}$ in the bank and wait until $T_2$ to exercise the call (as with an American call).

Let $I_i = 1$ if $S_{T_i} > K_i$; $I_i = 0$ if $S_{T_i} \leq K_i$ for $i = 1$, 2. At time $T_2$, $A$ is worth $I_1 \bigl(S_{T_1} - K_1\bigr)\e^{r(T_2 - T_1)}$ while $B$ is worth $I_2\bigl( S_{T_2} - K_2 \bigr) + S_{T_1}\e^{r(T_2 - T_1)} - S_{T_2}$.

$B$ is worth at least as much as $A$ in the following cases:
$I_1 = 0$, as $A$ is worth 0;
$I_1 = 1$, $I_2 = 0$ if $K_2 < K_1\e^{r(T_2 - T_1)}$ (as $S_{T_2} \leq K_2$);
$I_1 = 1$, $I_2 = 1$ as long as $K_1\e^{r(T_2 - T_1)} > K_2$.

These inequalities are equivalent to that stated in the question so in this case, if $S_{T_1} > K_1$ exercising the option at $T_1$ is worse than selling short at $T_1$ and waiting until $T_2$ and so is certainly not optimal.

Exercise 1.12 Inequality (a) is not true. From put-call parity and $C \geq 0$ we have $P + S \geq K\e^{-rT}$. For large $K$ and small $S$ we must have $P > S$. On the other hand (b) is true because of put-call parity and the fact that $C \leq S$.

Exercise 1.13 Suppose for some expiry dates $s<t$ that $P(s)>P(t)$ and let’s look for an arbitrage. Buy a put option with exercise time $t$ and sell a put option with exercise time $s$ and deposit $P(s)-P(t)$ in the bank. If it’s ever optimal to exercise the time $s$ put then also exercise the time $t$ put for a zero cash flow. If the put we sold is not exercised by $s$, the put we hold is still worth at least $0$. Either way we have at least $P(s)-P(t)$ earning interest in the bank. Therefore there exists an arbitrage. To eliminate this we must have $P(s) \leq P(t)$ whenever $s < t$.

Exercise 1.14 Suppose that $P$ generates cash flow $(x,t)$, where $x= (x_1,\dots,x_k)$ and $t = (t_1,\dots,t_k)$.

We want to find $h_t \in \R$, the amount of bond held at time $t$ in the portfolio $P_B$, which generates the same cash flow $(x,t)$. In other words, we need to find $h_t$ satisfying \[ V^{P_B}_t - V^{P_B}_{t-} = B_t \cdot (h_t - h_{t-}) = \begin{cases} -x_i & \text{if $t= t_i$ for some $i$}\\ 0 & \text{otherwise}. \end{cases} \] This is achieved by $h_t = \sum_{i: t_i \leq t} -x_i/B_{t_i}$.
Portfolio $\widetilde P = P - P_B$ satisfies $V^{\widetilde P}_t = V^P_t - V^{P_B}_t$, and therefore $V^{\widetilde P}_t -V^{\widetilde P}_{t-} = (V^P_t - V^P_{t-}) - (V^{P_B}_t - V^{P_B}_{t-})$, which by part (a) is identically zero for all $t$ (including $t=0$).
Suppose that $P$ is also an arbitrage portfolio. The value at time 0 of the cash flow $(x,t)$ is $\sum_{i} \e^{-r t_i} x_i$, so $e^{-rT} V^P_T + \sum_{i} \e^{-r t_i} x_i > 0$ always.

Using part (b), and the fact that $V^{\widetilde P}_{0-} \equiv 0$, we have $V^{\widetilde P}_0 = 0$. But $V^{P_B}_T = h_T B_T = \sum_{i} -x_i \e^{r(T-t_i)}$ by definition, so \[ V^{\widetilde P}_T = V^P_T - V^{P_B}_T > - \e^{rT} \sum_i \e^{-r t_i} x_i + \sum_i x_i \e^{r(T-t_i)} = 0 \] always, as required.