3 Solutions for Chapter 3

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 $Y \sim \mbox{Bin}(2,q_u)$ 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 See HW solution.

Exercise 3.3 The martingale probabilities 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 $t=1$, $S_1 = 65$: $q_u =\frac{65 \times 1.1 - 65}{75-65} = \frac{13}{20}, q_d = \frac{75-65 \times 1.1}{75-65} = \frac7{20}$,

at $t=1$, $S_1 = 50$: $q_u = \frac{50 \times 1.1 -30}{60-30} = \frac56, q_d =\frac{60 - 50 \times 1.1}{60-30} = \frac16$.

$\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 $\Pi_0 = \frac{1}{1.1}( 60 \times \frac13 + 50 \times \frac23 ) = \frac{1600}{33} \approx 48.48$.

$\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 $\Pi_0 = \frac{1}{1.1}( 65 \times \frac13 + \frac{500}{11} \times \frac23 ) = \frac{17150}{363} \approx 47.25$.

Exercise 3.4 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 $Y \sim \Bin(t,1/2)$ under $\Q$, and \[ \textstyle\Pi_0 = \E_\Q[(S_3 - 80)^+] =\E_\Q[(20Y-30)^+] = \frac18 \times 30 + \frac38 \times 10 + \frac38 \times 0 + \frac18 \times 0 = \frac{15}{2}. \]

Exercise 3.5 See HW solution.

Exercise 3.6 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 $k^\star = \lceil \log(70/(60\times 0.8))/\log((1.4/0.8)^{1/12})\rceil = 9$, and \[ C = 60 \sum_{k=9}^{12} {\textstyle\binom{12}{k}} m_u^k m_d^{12-k} - 70(1+{\textstyle\frac{0.04}{12}})^{-12}\sum_{k=9}^{12} {\textstyle\binom{12}{k}} q_u^k q_d^{12-k} \approx 0.169 \] (using $\displaystyle q_u = \frac{1+r-d}{u-d}\approx 0.464, q_d = \frac{u-(1+r)}{u-d}\approx 0.536, m_u = \frac{u q_u}{1+r}\approx 0.476$ and $\displaystyle m_d= \frac{d q_d}{1+r}\approx 0.524$).

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

Exercise 3.7 See HW solution.

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 \[\begin{multline*} P - C = K(1+r)^{-t} \sum_{k=0}^{k^\star-1} {\textstyle\binom{t}{k}} q_u^k q_d^{t-k} - S_{T-t} \sum_{k=0}^{k^\star-1} {\textstyle\binom{t}{k}} m_u^k m_d^{t-k}\\ -\bigg( S_{T-t} \sum_{k=k^\star}^t {\textstyle\binom{t}{k}} m_u^k m_d^{t-k} - K(1+r)^{-t} \sum_{k=k^\star}^{t} {\textstyle\binom{t}{k}} q_u^k q_d^{t-k}\bigg) = K(1+r)^{-t} - S_{T-t} \end{multline*}\]