3 Solutions for Chapter 3

Warm-up

Exercise 3.1 Let \(Y\) be the random variable representing the number of times the price of the stock goes up.

Under the risk-neutral probabilities we have \(Y \sim \text{Bin}(2,q_u)\), with \[ q_u=(1+r-d)/(u-d) = (1+ 2r)/3.\] The stock price at time 2 is \(S_2=100 \cdot 2^Y (1/2)^{2-Y}\). To eliminate any arbitrages we must have \[ C = (1+r)^{-2} \E[(S_2-K)^+].\] Only the outcome where there are two up-steps contributes anything to this expectation so we have \[C=q_u^2(400 - 150)/(1+r)^2 = 250(1+2r)^2/9(1+r)^2. \]

Exercise 3.2 Use the Cox–Ross–Rubinstein formula \[ \Pi_{T-t} = S_{T-t} \sum_{k=k^\star}^t {\textstyle\binom{t}{k}} m_u^k m_d^{t-k} -K(1+r)^{-t} \sum_{k=k^\star}^t {\textstyle\binom{t}{k}} q_u^k q_d^{t-k}, \] where \[ q_u = \frac{1.5-0.7}{1.1-0.7} = \frac12, \quad q_d = \frac{1.5-1.1}{1.5-0.7} = \frac12 \] and \[ m_u = \frac12\cdot\frac{1.5}{1.1}=\frac{15}{22}, \quad m_d = \frac{1}{2}\cdot\frac{0.7}{1.1} = \frac{7}{22}. \]

To calculate the price at time 0 (when \(t=T=3\)), we need to find \(k^\star\) as the smallest integer greater than \(\log(90/(100\times0.7^3))/\log(1.5/0.7) \approx 1.27\), so \(k^\star = 2\) and

\[\Pi_0 = 100 \sum_{k=2}^3 \binom{3}{k} \left(\frac{15}{22}\right)^k \left(\frac{7}{22}\right)^{3-k} - 90(1.1)^{-3} \sum_{k=2}^3 \binom{3}{k} \left(\frac12\right)^k \left(\frac12\right)^{3-k} \approx 42.26. \]

At time 1, when \(S_1 = 70\), we use the formula with \(k^\star\) the smallest integer greater than \(\log(90/(70\times0.7^2))/\log(1.5/0.7) \approx 1.26\) so \(k^\star = 2\) and

\[ \Pi_2 = 70 \left(\frac{15}{22}\right)^2 - 90(1.1)^{-2} \left(\frac12\right)^2 \approx 13.946. \]

Main problems

Exercise 3.3 The martingale probabilities at each node are:

• at \(t=0\), \[q_u = \frac{50 \times 1.1 - 50}{65-50} = \frac13, q_d = \frac{65 - 50 \times 1.1}{65-50} = \frac23,\]

• at \(t=1\) and \(S_1 = 65\), \[q_u =\frac{65 \times 1.1 - 65}{75-65} = \frac{13}{20}, q_d = \frac{75-65 \times 1.1}{75-65} = \frac7{20},\]

• at \(t=1\) and \(S_1 = 50\), \[q_u = \frac{50 \times 1.1 -30}{60-30} = \frac56, q_d =\frac{60 - 50 \times 1.1}{60-30} = \frac16.\]

1. The payoff of the cash-or-nothing option is \[\Phi(S_2) = \begin{cases}66 & \text{if $S_2 \geq 50$},\\ 0 & \text{if $S_2 < 50$}, \end{cases}\] so \[\Pi_1 = \begin{cases} \frac{1}{1.1} ( 66 \times \frac{13}{20} + 66 \times \frac{7}{20} ) = 60 & \text{if $S_1 = 65$},\\ \frac{1}{1.1} ( 66 \times \frac56 + 0 \times \frac16 ) = 50 & \text{if $S_1 = 50$}, \end{cases}\] and \[\Pi_0 = \frac{1}{1.1}\left( 60 \times \frac13 + 50 \times \frac23 \right) = \frac{1600}{33} \approx 48.48.\]

2. The payoff of the asset-or-nothing option is \[\Phi(S_2) = \begin{cases} S_2 & \text{if $S_2 \geq 50$},\\ 0 & \text{if $S_2 < 50$}, \end{cases}\] so \[\Pi_1 = \begin{cases} \frac{1}{1.1} ( 75 \times \frac{13}{20} + 65 \times \frac{7}{20} ) = 65 & \text{if $S_1 = 65$},\\ \frac{1}{1.1} ( 60 \times \frac56 + 0 \times \frac16 ) = \frac{500}{11} & \text{if $S_1 = 50$}, \end{cases}\] and \[\Pi_0 = \frac{1}{1.1}\left( 65 \times \frac13 + \frac{500}{11} \times \frac23 \right) = \frac{17150}{363} \approx 47.25.\]

Exercise 3.4 The gap option has payoff \[\Phi(S_2) = \begin{cases} 35 & \text{if $S_2 = 75$},\\ 25 & \text{if $S_2 = 65$}, \\ 0 & \text{otherwise}, \end{cases}\] so we can use the martingale probabilities from the previous question to find: at \(t=1\), \[\Pi_1 = \begin{cases} \frac{1}{1.1} ( 35 \times \frac{13}{20} + 25 \times \frac{7}{20} ) = 31.5 & \text{if $S_1 = 65$},\\ \frac{1}{1.1} ( 0 \times \frac56 + 0 \times \frac16 ) = 0 & \text{if $S_1 = 50$}, \end{cases}\] and \[ \Pi_0 = \frac{1}{1.1} \left( 31.5 \times \frac13 + 0 \times \frac23 \right) = \frac{105}{11} = 9.5454...\]

Exercise 3.5 If the portfolio is self-financing, we know that \[ x_t B_t + y_t S_t = x_{t+1} B_t + y_{t+1} S_t. \]

The condition that half of the value of the portfolio is assigned to each of the two assets tells us that \(x_{t+1} B_t = y_{t+1}S_t = \frac{1}{2} \left(x_t B_t + y_t S_t\right)\); rearranging this we get \[ \begin{aligned} x_{t+1} & = \frac{1}{2} \left( x_t + y_t \frac{S_t}{B_t} \right) \\ y_{t+1} & = \frac{1}{2} \left( x_t \frac{B_t}{S_t} + y_t\right). \end{aligned}\]

To find the recurrence relation for the value, we use the fact that the portfolio is self-financing: \[ V_{t+1} = x_{t+1}B_{t+1} + y_{t+1}S_{t+1}. \] We use the fact that \(x_{t+1} B_t = y_{t+1}S_t = \frac{1}{2} V_t\) to write \[ \begin{aligned} V_{t+1} &= \frac{1}{2} \frac{V_t}{B_t} B_{t+1} + \frac{1}{2} \frac{V_t}{S_t} S_{t+1} \\ & = \frac{1}{2} V_t \left( \frac{B_{t+1}}{B_t} + \frac{S_{t+1}}{S_t} \right) \\ & = \frac{1}{2} V_t \left( 1 + r + \frac{S_{t+1}}{S_t} \right). \end{aligned} \]

Exercise 3.6 We cannot use the Cox–Ross–Rubinstein formula because \(u\) and \(d\) are not constant over time; for example, at \(t=0\) we have \(u=9/8\), \(d=7/8\) but at \(t=2\), when \(S_2=100\) we have \(u=11/10\) and \(d=9/10\).

However, the martingale probabilities at each node are the same, all having \(q_u = q_d = 1/2\). So, we have \[S_t = S_0 + 10Y - 10(t-Y) = 80 +20Y -10t,\] where \(Y\) is the number of up-moves. We know \(Y \sim \Bin(t,1/2)\) under \(\Q\), so \[ \textstyle\Pi_0 = \E_\Q[(S_3 - 80)^+] =\E_\Q[(20Y-30)^+] = \frac18 \times 30 + \frac38 \times 10 + \frac38 \times 0 + \frac18 \times 0 = \frac{15}{2}. \]

Exercise 3.7 We use \(u=1.4^{1/12} \approx 1.028\) and \(d=0.8^{1/12} \approx 0.982\). Because the interest is compounded monthly, which agrees with the size of the time intervals of our model, we take \(r = 0.04/12 \approx 0.00333\).
We see that \(d < 1+r< u\) so the model is arbitrage free.
To calculate the price at time 0 of a European put option we can use the CCR formula (with 12 periods) and put-call parity. Here, \(d^{12} = 0.8\) so \[k^\star = \lceil \log(70/(60\times 0.8))/\log((1.4/0.8)^{1/12})\rceil = 9,\] we have \[ q_u = \frac{1+r-d}{u-d}\approx 0.464, \qquad \qquad q_d = \frac{u-(1+r)}{u-d}\approx 0.536, \\ m_u = \frac{u q_u}{1+r}\approx 0.476 \qquad \text{and} \qquad m_d= \frac{d q_d}{1+r}\approx 0.524.\] Hence \[ C = 60 \sum_{k=9}^{12} {\textstyle\binom{12}{k}} m_u^k m_d^{12-k} - 70(1+{\textstyle\frac{0.04}{12}})^{-12}\sum_{k=9}^{12} {\textstyle\binom{12}{k}} q_u^k q_d^{12-k} \approx 0.169. \]

So the price of the put option is \(C - 60 + 70(1+\frac{0.04}{12})^{-12} \approx 7.429\).

Exercise 3.8 We work with a \(T\)-period model with \(u\) and \(d\) constant over time, so that at time \(t=0,1,\dots, T\) the stock price \(S_t\) takes one of the \(t+1\) values \(S_0 u^k d^{t-k}\), for \(k=0,1,\dots,t\). Under the risk-neutral measure we know that \(S_T = S_{T-t} u^Y d^{t-Y} = S_{T-t} d^t (u/d)^Y\), where \(Y\sim \Bin(t,q_u)\) is independent of \(S_{T-t}\), and the no-arbitrage price of a European put option at time \(T-t\) is then \[\begin{split} \Pi_{T-t}(\text{put}) &= (1+r)^{-t} \E_\Q[(K-S_T)^+\mid \cF_{T-t}]\\ &= (1+r)^{-t} \E_\Q[(K-S_{T-t}d^t(u/d)^Y)^+\mid \cF_{T-t}]\\ &=(1+r)^{-t} \sum_{k=0}^{k^\star-1} (K - S_{T-t}d^t(u/d)^k )\textstyle{\binom{t}{k}} q_u^k q_d^{t-k}\\ &=K(1+r)^{-t} \sum_{k=0}^{k^\star-1} {\textstyle\binom{t}{k}} q_u^k q_d^{t-k} - S_{T-t} \sum_{k=0}^{k^\star-1} {\textstyle\binom{t}{k}} m_u^k m_d^{t-k},\\ \end{split} \] where, just like in the CCR formula, \(k^\star = \min\{k \in \N : S_{T-t}d^t (u/d)^k \geq K \}\) and \(m_u = uq_u/(1+r), m_d = d q_d/(1+r)\).

Put-call parity follows from the fact that \(q_u + q_d = m_u + m_d = 1\), so \[\begin{multline*} P - C = K(1+r)^{-t} \sum_{k=0}^{k^\star-1} {\textstyle\binom{t}{k}} q_u^k q_d^{t-k} - S_{T-t} \sum_{k=0}^{k^\star-1} {\textstyle\binom{t}{k}} m_u^k m_d^{t-k}\\ -\bigg( S_{T-t} \sum_{k=k^\star}^t {\textstyle\binom{t}{k}} m_u^k m_d^{t-k} - K(1+r)^{-t} \sum_{k=k^\star}^{t} {\textstyle\binom{t}{k}} q_u^k q_d^{t-k}\bigg) = K(1+r)^{-t} - S_{T-t}. \end{multline*}\]