2 Solutions for Chapter 2

Exercise 2.1 See HW solution.

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 \[ \frac{1}{1.03}( 4 \times q_u + 0 \times q_d ) = \frac{568}{309}, \] 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 \[ \frac{1}{1.03}( 0 \times q_u + 2 \times q_d ) = \frac{316}{309}, \] and the replicating portfolio is given by \(y= \frac{0-2}{32-26} = -\frac{1}{3}\) and \(x = \frac{316}{309} - 28y = \frac{3200}{309}\).

Exercise 2.3 We have

  1. 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. \(P = (0\times q_u + (150-50)\times q_d)/(1+r)= 200(1-r)/3(1+r)\).

  2. \(P + S = [200 - 200r + 300 + 300r]/3(1+r) = C + 150/(1+r)\) so put-call parity holds.

Exercise 2.4 See HW solution.

Exercise 2.5 See HW solution.

Exercise 2.6 We have

  1. 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 \(h=0\).

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.

  1. 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.

  2. 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., \[ x(C - S_0) = -x \frac{K}{1+r} \quad\text{so that}\quad C = S_0 - \frac{K}{1+r}\,. \]