1 Solutions for Chapter 1

Warm-up

Exercise 1.1 The effective interest rate is given by $1 + r^\star = (1+r/n)^n.$ We have $r=0.1$ throughout.

We have $n=2$, and $(1 + 0.1/2)^2 = 1.1025$, so the effective rate is $0.1025.$
When $n=4$, $r^\star= (1 + 0.1/4)^4 - 1 \approx 0.1038.$
Now interest is compounded continuously, so $r^\star$ is given by $\e^{0.1} - 1 \approx 0.1052.$

Exercise 1.2

To find the time it takes to double our money, $t$, we need to solve $D\e^{rt} = 2D$ where $r=0.1.$

Answer: 6.93 years.
This time, we solve $(1+r)^t = 4$ where $r=0.05$ to get the time it takes to quadruple your money.

Answer: 28.41 years.

In general, we solve $(1+r)^t = \alpha$ for $t$ to get $t = \log \alpha/\log(1+r).$

Exercise 1.3 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. Because $P+S-C$ is too high, we short sell one put plus one share minus one call option (in other words, we 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.

Main problems

Exercise 1.4 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 \[ \begin{aligned} 100,000 & = a\left(1+\frac{0.06}{12}\right)^{60} + a\left(1+\frac{0.06}{12}\right)^{59} + \ldots + a\left(1+\frac{0.06}{12}\right) \\ & = a \sum_{k=1}^{60} 1.005^k. \end{aligned} \] Using the geometric series formula, we find $a=1426.15.$

Exercise 1.5 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 1.6 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 1.7 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 1.8 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 1.9 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 1.10 We have two definitions of arbitrage:

There exists a portfolio $P$ such that $\alpha^T V_T - V_0 > 0$ whatever happens to $S_t$;
There exists a self-financing portfolio $P$ such that $V_0=0$ and $V_T \geq 0$ whatever happens to the share prices, with the possibility that $V_T>0$.

We are going to start off with a market in which there is an arbitrage portfolio in the first sense, and show that we can modify it to find an arbitrage portfolio in the second sense. Hopefully you can see that, if the second type of arbitrage portfolio exists, it is also an arbitrage portfolio in the first sense. So a market either contains both types of arbitrage, or neither.

We begin with the cash flow. If $P$ generates cash flow $(x,t)$, where $x= (x_1,\dots,x_k)$ and $t = (t_1,\dots,t_k)$, then the trades involved in $P$ are producing “left-over” cash £$x_1$ at time $t_1$, and so on. Equivalently, we can say that the value of the assets in $P$ decreases by $x_1$ between time $t_1-$ and time $t_1$. (If $x_1$ is negative then we have an increase in the value: we had to inject cash to be able to perform the trade.)

To replicate this, 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 $h_t = \sum_{i: t_i \leq t} -x_i/B_{t_i}.$
We should expect that the cash flow of the new portfolio $\widetilde P = P - P_B$ is always zero, because we’ve taken the portfolio $P$ and subtracted exactly the right number of bonds to replicate its cash flow.

To check this formally: the value process of $\widetilde{P}$ satisfies \[V^{\widetilde P}_t = V^P_t - V^{P_B}_t \] and therefore its cash flow is given by \[ V^{\widetilde P}_t -V^{\widetilde P}_{t-} = (V^P_t - V^P_{t-}) - (V^{P_B}_t - V^{P_B}_{t-}).\]

Since the cash flow of $P$ is $(x,t)$, the first term is equal to \[ \begin{cases} -x_i & \text{if $t= t_i$ for some $i$}\\ 0 & \text{otherwise}, \end{cases}\] while the second term is \[ B_t \cdot (h_t - h_{t-}) = \begin{cases} -x_i & \text{if $t= t_i$ for some $i$}\\ 0 & \text{otherwise}, \end{cases}\] as we defined in part (a). So the cash flow of $\widetilde{P}$ is identically zero for all $t$ (including $t=0$): $\widetilde{P}$ is self-financing.
Suppose that $P$ is also an arbitrage portfolio, in the first sense: $\alpha^T V^P_T - V^P_0 > 0$.

Let’s prove that $\widetilde{P}$ is an arbitrage portfolio in the second sense. We already know from part (b) that it is self-financing.

To see that $\widetilde{V}_0 = 0$, we can use our work from part (b): we know that $\widetilde{V}_{0-} = 0$ (the portfolio was worth nothing before we started), and we also know that $\widetilde{V}_t = \widetilde{V}_{t-}$ for every $t$ including $t=0$.

We know that \[ \begin{aligned} \widetilde{V}_T &= V^P_T - V^{P_B}_T \\ &= V^P_T - h_T B_T \\ &= V^P_T + (1+r)^T \sum_i e^{ -r t_i} x_i. \end{aligned}\] Multiplying through by $\alpha^T$ and using our arbitrage-portfolio assumption, \[ \alpha^T \widetilde{V}_T > V^P_0 + \alpha^T B_T(1+r)^T \sum_i e^{ -r t_i} x_i. \] But the value at time 0 of $P$ is at least the present value of the cash flow, namely $-\sum_{i} \e^{-r t_i} x_i$, so \[ \alpha^T \widetilde{V}_T > 0,\] whatever happens to the share prices, as required.