2 Solutions for Chapter 2

Warm-up

Exercise 2.1 We have

The martingale probabilities are \[q_u=\Q(\mbox{stock price at time 1 is 200})=(1+2r)/3\] and \[q_d = \Q(\text{stock price at time 1 is 50}) = (2-2r)/3.\] So
1. $C = ((200-150)\times q_u + 0\times q_d)/(1+r) = (50 + 100r)/3(1+r)$ and
2. $P = (0\times q_u + (150-50)\times q_d)/(1+r)= 200(1-r)/3(1+r)$.
$P + S = [200 - 200r + 300 + 300r]/3(1+r) = C + 150/(1+r)$ so put-call parity holds.

Exercise 2.2 We have $r=0.03$, $u=\frac{32}{28}=\frac{8}{7}$ and $d = \frac{26}{28} = \frac{13}{14}$ so \[ q_u = \frac{1.03 - \frac{13}{14}}{\frac{8}{7}-\frac{13}{14}} = \frac{71}{150}, \quad q_d = \frac{ \frac{8}{7} - 1.03}{\frac{8}{7} - \frac{13}{14}} = \frac{79}{150}. \] The price of the call option is \[ C = \frac{1}{1.03}( 4 \times q_u + 0 \times q_d ) = \frac{568}{309}, \] and the replicating portfolio is given by $y = \frac{4-0}{32-26} = \frac{2}{3}$ and $x = \frac{568}{309} - 28y = -\frac{5200}{309}$.

The price of the put option is \[ P = \frac{1}{1.03}( 0 \times q_u + 2 \times q_d ) = \frac{316}{309}, \] and the replicating portfolio is given by $y= \frac{0-2}{32-26} = -\frac{1}{3}$ and $x = \frac{316}{309} - 28y = \frac{3200}{309}$.

Main problems

Exercise 2.3

This market has $u=120/100 = 1.2$, $d=60/100 = 0.6$, and $r=0.1$ and therefore $d<1+r<u$ so the market is arbitrage-free.
We want to find amounts $x$ and $y$ so that $xB_T + yS_T = (S_T - 90)^+$, that is \[ \begin{split} 1.1 x &+ 120 y = 30,\\ 1.1 x &+ 60 y = 0. \end{split} \] Solving for $x$ and $y$ gives $x=-300/11$, and $y=1/2$.

Alternatively, we can get these values directly from the formulas \[ \begin{split} x &= \frac{1}{1+r}\frac{u \Phi(S^d_T) - d \Phi(S^u_T)}{u-d} = \frac{1}{1.1} \frac{1.2 \times 0-0.6\times30}{1.2-0.6}=-\frac{300}{11}\\ y &= \frac{\Phi(S^u_T)-\Phi(S^d_T)}{S^u_T - S^d_T}=\frac{30-0}{120-60}= \frac12. \end{split} \] So, the replicating portfolio is $h=(-300/11,1/2)$, and has value $V_0 = xB_0+yS_0 = x+100y = 250/11$, so this is the maximum premium that you would be willing to pay at time 0.
To find the price of a put option we can use the put-call parity $P+S_0 = C+K(1+r)^{-1}$, so $P = 250/11 + 900/11 - 100 = 50/11$.

Alternatively, we can calculate the price directly, like in part (b): \[ \begin{split} 1.1 x &+ 120 y = 0,\\ 1.1 x &+ 60 y = 30, \end{split} \] which has solution $x=600/11, y=-1/2$ and price $V_0 = x+100y = 50/11$.
The risk-neutral or martingale measure $(q_u,q_d)$ satisfies $q_u+q_d=1$ and $uq_u+dq_d = 1+r$. Therefore \[ q_u = \frac{1+r-d}{u-d} = \frac56, \quad q_d =\frac{u-(1+r)}{u-d} = \frac16. \] Since this is the unique solution, and we have already shown that the market is arbitrage free (part (a)), we know that the market is complete. (Either of Theorems 2.3 or 2.5 tell us this.)

At this point, we can check our prices are correct by using the risk-neutral valuation formula.

Call option: $\displaystyle C=\frac{1}{1.1}\left(\frac56\times 30 + \frac16 \times 0\right) = \frac{250}{11}$

Put option: $\displaystyle P=\frac{1}{1.1}\left(\frac56\times 0 + \frac16 \times 30\right) = \frac{50}{11}$ .
The claim $X = (S_T)^2$ has value 14400 when $S_T=120$ and value 3600 when $S_T=60$. We thus need to find the portfolio $h=(x,y)$ that matches these payoffs. We have \[\begin{split} x &= \frac{1}{1.1}\frac{1.2\times 3600 - 0.6\times 14400}{1.2-0.6} = -\frac{72000}{11},\\[5pt] y &= \frac{14400-3600}{120-60} = 180, \end{split} \] and the price is $-72000/11 + 18000 = 126000/11$. This agrees with the price calculated using the martingale probabilities: \[ \Pi(0;X) = \frac{1}{1.1}\left( \frac56 \times 14400 + \frac16 \times 3600 \right) = \frac{126000}{11}. \]

Exercise 2.4 No, the market is not arbitrage-free. We find an arbitrage portfolio as follows. Supposing $h = (x,y,z)$ is an arbitrage portfolio, its value at time 0 is $V_0^h = x+ 10y+100z = 0$, so $x = -10y-100z$. Then, at time $T$, its value is \[ V^h_T = \begin{cases} 1.1x+12y+130z = y+20z& \mbox{with probability $0.3$},\\ 1.1x+7y+60z = -4y-50z& \mbox{with probability $0.7$}. \end{cases} \] We are free to choose $y$ and $z$ so as to make $V^h_T \geq 0$ always and $V^h_T > 0$ with positive probability. We can easily find values $y,z \in \mathbb{R}$ such that $y+20z>0$ and $-4y-50z > 0$. To find a portfolio that returns a guaranteed risk-free profit, we solve $y+20z = -4y-50z$ to get $y = -14z$, and then $V^h_T = 6z$. So taking $z > 0$, $y = -14z$ and $x = -10y-100z = 40z$ gives an arbitrage portfolio.

Exercise 2.5 We assume $r \neq 0$, so that $B_0 = 1$, $B_1 = 1+r$, but the logical argument is still valid in the case $r=0$ (just set $r$ to zero in the expressions that follow). Note that $1/2 < 1+r < 2$, otherwise the market is not arbitrage-free.

The share price at time 0 is $S_0 = 100$ and at time 1 the share price $S_1$ can take the value $50, 100$ or $200.$ By the risk-neutral valuation formula, once we have a martingale measure $\mathbb{Q}$ for $S_1$, we can calculate the price of the call option by \[ C = \frac{1}{1+r}\E_{\mathbb{Q}}[(S_1 - 150)^+]. \]

Write $q_u, q_m, q_d$ for the martingale probabilities of the stock price at time 1 being 200, 100 or 50, respectively. We know that the martingale measure is defined by the identity $S_0 = \frac{1}{1+r}\E_{\mathbb{Q}}[S_1]$, i.e., \[100 = (200 q_u + 100 q_m + 50 q_d)/(1+r).\] Also, the probabilities $(q_u,q_m,q_d)$ form a probability measure so $q_u+q_m+q_d=1$. Since we have 2 equations in 3 unknowns we should expect there to be a range of solutions. Given the probabilities $(q_u,q_m,q_d)$ we find the price of the call option is \[C = (50q_u +0q_m + 0q_d)/(1+r) = 50q_u/(1+r).\]

Hence we have to find the maximum and minimum value of $q_u$ for which these equations holds. That is, we have to solve the following 2 problems:

Maximise $q_u$ such that $150 q_u + 50 q_m = 100r + 50$, $q_u+q_m \leq 1$ and $q_u,q_m \geq 0.$
Minimise $q_u$ such that $150 q_u + 50 q_m = 100r + 50$, $q_u+q_m \leq 1$ and $q_u,q_m \geq 0.$

For those who are also doing Operations Research, these two problems can be solved with, e.g., the two-phase method or the Big M method. However, there are also other methods for solving these problems. You should be able to solve these problems (e.g graphically)!

The two solutions are:

$q_u=(1+2r)/3,~q_m=0$ and $q_d=1-(1+2r)/3$.
1. $q_u=r,~q_m=1-r$ and $q_d=0$, if $r \geq 0$,
2. $q_u=0, q_m=1+2r$ and $q_d=-2r$, if $r < 0$.

Finally, substituting the minimum and maximum value of $q_u$ into the expression for $C$ yields \[50\max(r,0)/(1+r) < C < 50(1+2r)/\big(3(1+r)\big).\] (This is a non-trivial interval, since $-1/2 < r$ implies $0 < (1+2r)/3$ and $r < 1$ implies $r < (1+2r)/3$.)

Exercise 2.6 We have

Let $h = (x,y,z)$ be a non-zero portfolio such that $V^h_0 = x + 10 y + 20 z = 0$. We must show that $V^h_1$ takes both positive and negative values. We have that $x = -10y-20z$, which can be substituted into $V^h_1$. This gives a system of 3 equations in 2 unknowns: \[\begin{equation*} V^h_1 = \begin{cases} 4y-z & \text{with probability}\; 0.5\\ -y + 0.5z & \text{with probability}\; 0.25\\ -y + 0.2z & \text{with probability}\; 0.25 \end{cases}. \end{equation*}\]

Both $y$ and $z$ cannot be 0 for then $x = 0$ as well, which means that $h$ is the zero portfolio.

Now if $y=0$ then we see that $-z$ and $0.5z$ takes positive and negative values due to $z$ being non-zero. Similarly, if $z=0$ then $4y$ and $-y$ takes positive and negative values. So we may assume that neither $y$ nor $z$ are 0. We may also assume that $y > 0$ by replacing $h$ by $-h$ if necessary.

Now consider the quantity $4y-z$ according to whether it is positive, negative or zero. If $4y-z > 0$ then $-y+0.2z < -y + 0.2 (4y) = -0.2y$, which is negative. If $4y-z < 0$ then $-y+0.5z > -y + 0.5(4y) = y$, which is positive. Finally, if $4y-z=0$ then $-y+0.5z = y$ is positive and $-y+0.2z = -0.2y$ is negative. So the equations take both positive and negative values.

All in all we find that $V^h_1$ always takes both positive and negative values. So the market has no arbitrage.
The equation $V^h_1 = \sqrt{S_1^1} + \sqrt{S_1^2}$ give a system of 3 equations in 3 unknowns: \[\begin{align*} 1.1x + 15y + 21z &= \sqrt{15}+\sqrt{21} \\ 1.1x + 10y + 22.5z &= \sqrt{10}+\sqrt{22.5}\\ 1.1x + 10y + 22.2z &= \sqrt{10}+\sqrt{22.2} \end{align*}\] Observe that the matrix \[\begin{bmatrix} 1.1 & 15 & 21\\ 1.1 & 10 & 22.5\\ 1.1 & 10 & 22.2\\ \end{bmatrix} \] is invertible because it has a determinant of $1.65$. So there is an unique $h^{*} = (x,y,z)$ which solves the above system of equations.
Let $\Pi$ be the arbitrage free price of $X$. If $\Pi \neq V^{h^*}_0$ then we can create arbitrage using the law of one price. Indeed, suppose $\Pi > V^{H^{*}}_0$ with out loss of generality. Then we can short sell $X$ and buy the portfolio $h$ to gain $\Pi - V^{h^{*}}_0$ units of money. Then at time one, $X$ and $h^{*}$ have the same value, which allows us to complete the short sell by selling $h^{*}$ and buying $X$. This leads to arbitrage; so $\Pi = V^{h{*}}_0$.

Exercise 2.7 To rule out an arbitrage on the stock alone we require $\min_i s_i < S_0(1+r) < \max_i s_i$. The problem is readily solved using put-call parity (the put will never be exercised so $P = 0$). For a direct argument let $C$ be the cost of the call option and $S_0$ the initial stock price. At $t=0$, buy $x$ call options and $y$ shares (as usual $y < 0$ is possible by short selling). As $K < \min_i s_i$ the call options will be exercised. The portfolio value at time 1 if $S_1 = s_i$ is \[ x(s_i - K) + ys_i = (x + y)s_i - xK \] for each $i$.

To ensure the return does not depend on $s_i$ we take $y = -x$ for a return of $-xK$ at time 1. To rule out an arbitrage, this riskless return should be equal to the initial cost of the portfolio i.e., \[ x(C - S_0) = -x \frac{K}{1+r} \quad\text{so that}\quad C = S_0 - \frac{K}{1+r}\,. \]