5 Solutions for Chapter 5

For Exercises 5.1–5.5 we use the martingale probabilities \[q_u = \frac{1.01-0.5}{2-0.5} = \frac{17}{50}, \quad q_d = \frac{2-1.01}{2-0.5} = \frac{33}{50}. \] at all nodes of the tree.

Exercise 5.1 We have

  1. First calculate the payoffs at time 3:

\[ \begin{array}{c|c|c|c|c|c|c|c|c} \omega& \h\h\h & \h\h\T & \h\T\h & \h\T\T & \T\h\h & \T\h\T & \T\T\h &\T\T\T\\ \hline S_3 & 80 & 20 & 20 & 5 & 20 & 5 & 5 &1.25\\ \min\{S_0,S_1,S_2,S_3\} & 10 & 10 & 10 & 5 & 5& 5 & 2.5 & 1.25\\ \hline X & 70 & 10 & 10 & 0 & 15 & 0 & 2.5 & 0 \end{array} \]

Then work backwards through the tree to find the prices at \(t=2,1,0\):

\[\begin{split} \Pi_2(\h\h) &=\textstyle \frac{1}{1.01}(70 \times \frac{17}{50} + 10 \times \frac{33}{50} ) = \frac{3040}{101},\\ \Pi_2(\h\T) &=\textstyle \frac{1}{1.01}(10 \times \frac{17}{50} + 0 \times \frac{33}{50}) =\frac{340}{101}, \\ \Pi_2(\T\h) &=\textstyle \frac{1}{1.01}(15 \times \frac{17}{50} + 0 \times \frac{33}{50}) =\frac{510}{101},\\ \Pi_2(\T\T) &=\textstyle \frac{1}{1.01}(2.5 \times \frac{17}{50} + 0 \times \frac{33}{50}) =\frac{85}{101},\\[15pt] \Pi_1(\h) &=\textstyle \frac{1}{1.01}(\frac{3040}{101} \times \frac{17}{50} + \frac{340}{101} \times \frac{33}{50} ) = \frac{125800}{10201},\\ \Pi_1(\T) &=\textstyle \frac{1}{1.01}(\frac{510}{101} \times \frac{17}{50} + \frac{85}{101} \times \frac{33}{50}) =\frac{22950}{10201}, \\[15pt] \Pi_0 &=\textstyle \frac{1}{1.01}(\frac{125800}{10201} \times \frac{17}{50} + \frac{22950}{10201} \times \frac{33}{50} ) = \frac{5791900}{1030301} \approx 5.622,\\ \end{split} \]

  1. We do the same for \(X=\max\{S_0,S_1,S_2,S_3\} - S_3\):

\[ \begin{array}{c|c|c|c|c|c|c|c|c} \omega& \h\h\h & \h\h\T & \h\T\h & \h\T\T & \T\h\h & \T\h\T & \T\T\h &\T\T\T\\ \hline S_3 & 80 & 20 & 20 & 5 & 20 & 5 & 5 &1.25\\ \max\{S_0,S_1,S_2,S_3\} & 80 & 40 & 20 & 20 & 20& 10 & 10 & 10\\ \hline X & 0 & 20 & 0 & 15 & 0 & 5 & 5 & 8.75 \end{array} \] \[\begin{split} \Pi_2(\h\h) &=\textstyle \frac{1}{1.01}(0 \times \frac{17}{50} + 20 \times \frac{33}{50} ) = \frac{1320}{101},\\ \Pi_2(\h\T) &=\textstyle \frac{1}{1.01}(0 \times \frac{17}{50} + 15 \times \frac{33}{50}) =\frac{990}{101}, \\ \Pi_2(\T\h) &=\textstyle \frac{1}{1.01}(0 \times \frac{17}{50} + 5 \times \frac{33}{50}) =\frac{330}{101},\\ \Pi_2(\T\T) &=\textstyle \frac{1}{1.01}(5 \times \frac{17}{50} + 8.75 \times \frac{33}{50}) =\frac{1495}{202},\\[15pt] \Pi_1(\h) &=\textstyle \frac{1}{1.01}(\frac{1320}{101} \times \frac{17}{50} + \frac{990}{101} \times \frac{33}{50} ) = \frac{110220}{10201},\\ \Pi_1(\T) &=\textstyle \frac{1}{1.01}(\frac{660}{202} \times \frac{17}{50} + \frac{1495}{202} \times \frac{33}{50}) =\frac{60555}{10201}, \\[15pt] \Pi_0 &=\textstyle \frac{1}{1.01}(\frac{110220}{10201} \times \frac{17}{50} + \frac{60555}{10201} \times \frac{33}{50} ) = \frac{7744110}{1030301} \approx 7.516,\\ \end{split} \]

Note that the question asks us to calculate the prices at all times, which is why we work back through the tree like this. If the question had only asked for the price at time 0, we could have used the risk-neutral valuation formula and we would have only needed to do one calculation (this is what we’ll do in the later questions).

Exercise 5.2 See HW solution.

Exercise 5.3 It is possible to calculate the up-and-out put price directly, but it is slightly quicker to instead find the standard put price and subtract the up-and-in put price.

Since the put option is a European option and doesn’t depend on the path taken by the share price (only on the final price), we quickly calculate \[ \textstyle P = \frac{1}{(1.01)^3}(0 \times (\frac{17}{50})^3 + 10\times 3 (\frac{17}{50})^2\,\frac{33}{50}+ 25\times 3 \frac{17}{50}(\frac{33}{50})^2 + 28.75\times (\frac{33}{50})^3 ) \approx 21.025. \] The up-and-in put only pays out if the price is greater than \(L=30\) at some time and finishes below \(K=30\), so the only path with a positive payoff is HHT, with payoff 10, so \[ \textstyle P_{\text{UI}} = \frac{1}{(1.01)^3}\times 10\times (\frac{17}{50})^2\,\frac{33}{50}\approx 0.741 \] and \(P_{\text{UO}} = P - P_{\text{UI}} \approx 20.285.\)

Exercise 5.4 Writing \(A_S\) for \(\frac{1}{4}(S_0+S_1+S_2+S_3)\) we have

\[ \begin{array}{c|c|c|c|c|c|c|c|c} \omega& \h\h\h & \h\h\T & \h\T\h & \h\T\T & \T\h\h & \T\h\T & \T\T\h &\T\T\T\\ \hline A_S & 75/2 & 45/2 & 15 & 45/4 & 45/4 & \leq10 & \leq10 & \leq10\\ \hline C^{\text{Asian}}_T & 55/2 & 25/2 & 5 & 5/4 & 5/4 & 0 & 0 & 0 \end{array} \]

and \(C^{\text{Asian}}_0 = \frac{1}{(1.01)^3}(\frac{55}{2}\times (\frac{17}{50})^3 + (\frac{25}{2} + 5 + \frac{5}{4})\times(\frac{17}{50})^2\,\frac{33}{50} + \frac{5}{4}\times\frac{17}{50}(\frac{33}{50})^2) \approx 2.617.\)

Exercise 5.5 Let \(A_S\) denote \(\frac{1}{2}(S_1+S_3)\), then

\[ \begin{array}{c|c|c|c|c|c|c|c|c} \omega& \h\h\h & \h\h\T & \h\T\h & \h\T\T & \T\h\h & \T\h\T & \T\T\h &\T\T\T\\ \hline A_S & 50 & 20 & 20 & 25/2 & 25/2 & 5 & 5 & 3.125\\ \hline P^{\text{Asian}}_T & 0 & 0 & 0 & 5/2 & 5/2 & 10 & 10 & 11.875 \end{array} \] and \(P^{\text{Asian}}_0 = \frac{1}{(1.01)^3}(\frac{5}{2}\times (\frac{17}{50})^2\frac{33}{50} + (\frac{5}{2} + 10 + 10)\times \frac{17}{50}(\frac{33}{50})^2 + 11.875\times(\frac{33}{50})^3) \approx 6.733.\)

Exercise 5.6 See HW solution.