5 Solutions for Chapter 5

For Exercises 5.1–5.3 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.

Warm-up

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 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 5.3 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 \(P^{\text{Asian}}_0 = \frac{1}{(1.01)^3}(\frac{5}{2}\times (\frac{17}{50})^2\frac{33}{50} + (\frac{5}{2} + 10 + 10)\times \frac{17}{50}(\frac{33}{50})^2 + 11.875\times(\frac{33}{50})^3) \approx 6.733.\)

Main problems

Exercise 5.4 For these barrier options, what matters is whether or not the price has crossed down across the barrier, and this is equivalent to whether \(S_{\min{}}\) is less than or greater than the barrier value.
The table below shows the value of \(S_3\) and \(S_{\min{}}\) for each of the possible 8 paths through the tree (remember that H corresponds to an up-move and T to a down-move):

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

  1. A down-and-out call option means that the option only pays out if the share price stays above \(L=8\) for all times and finishes above \(K=10\). We can see from the table above that the only paths through the tree which give a positive payoff are HHH, HHT and HTH, and the payoffs are 70, 10 and 10 respectively.
    Therefore the no-arbitrage price at time 0 is \[ \Pi_0 = \textstyle\frac{1}{(1.01)^3}(70 \times (\frac{17}{50})^3 + 10\times (\frac{17}{50})^2\,\frac{33}{50} + 10\times (\frac{17}{50})^2\,\frac{33}{50} ) \approx 4.151. \]

  2. Down-and-in call pays out if the share price drops below \(L=8\) at some time, and then finishes above \(K=10\). This only happens if the path is THH and then the payoff is 10, so the no-arbitrage price is \[ \Pi_0 = \textstyle\frac{1}{(1.01)^3}\times10\times (\frac{17}{50})^2\, \frac{33}{50} \approx 0.741. \]

Since the payoff of a standard European call option is the sum of the down-and-out call and the down-and-in call, the price of a European call must be the sum of the two prices from parts (a) and (b), namely 4.892.

You can check by direct calculation of the price of the call option that this is the case.

Exercise 5.5 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 5.6 Your picture should look like the “recombining trees’’ we drew in lectures, with \(t+1\) nodes for each time \(t =0 ,1,2,3\). The martingale probabilities, which vary through the tree are:

  • At \(S_0=100\): \(\qquad q_u = \frac34, q_d = \frac14\),
  • At \(S_1=120\): \(\qquad q_u=\frac45, q_d = \frac15\),
  • At \(S_1=80\): \(\qquad q_u = \frac7{10}, q_d = \frac3{10}\),
  • At \(S_2=140\): \(\qquad q_u = \frac{17}{20}, q_d = \frac{3}{20}\),
  • At \(S_2=100\): \(\qquad q_u = \frac34, q_d = \frac14\),
  • At \(S_2=60\): \(\qquad q_u = \frac{13}{20}, q_d = \frac{7}{20}\).

We use these to price an American put option with strike price \(K=90\) and expiry date \(T=3\). The prices at time 3 are \((90-S_3)^+ = 0,0,10,50\) for share prices \(S_3 = 160,120,80,40\).

  • At \(t=2\):

    • if \(S_2=140\), both the value of waiting and the return from exercising early are 0, so \(\Pi_2 = 0\);

    • if \(S_2 = 100\), the value of waiting is \(\frac{1}{1.1}(0\times \frac34 + 10 \times \frac14) = \frac{25}{11}\), and the return from exercising is 0, so \(\Pi_2= \frac{25}{11}\);

    • if \(S_2 = 60\), the value of waiting is \(\frac{1}{1.1}(10 \times \frac{13}{20}+ 50 \times\frac{7}{20}) = \frac{240}{11}\) and return from exercising is 30, so we exercise early, and \(\Pi_2 = 30\).

  • At \(t=1\):

    • if \(S_1=120\), the value of waiting is \(\frac{1}{1.1}(0\times \frac45 + \frac{25}{11} \times \frac{1}{5}) = \frac{50}{121}\) and return from exercising is 0, so \(\Pi_1 = \frac{50}{121}\);

    • if \(S_1=80\), the value of waiting is \(\frac{1}{1.1}(\frac{25}{11}\times \frac{7}{10} + 30 \times \frac{3}{10})=\frac{1165}{121}\) and the return from exercising is 10, so we *exercise early**, and \(\Pi_1 = 10\).

  • At \(t=0\):

    the value of waiting is \(\frac{1}{1.1}(\frac{50}{121}\times \frac{3}{4} + 10 \times \frac14 ) = \frac{3400}{1331}\), and the return from exercising is 0, so \(\Pi_0 = \frac{3400}{1331}\).

The price of a European put option is more easily calculated as we do not have the choice to exercise early. You should find:

  • At \(t=2\)

    • if \(S_2 = 140\), \(\Pi_2 = \frac{1}{1.1}(0\times \frac{17}{20} + 0 \times \frac{3}{20} ) = 0\),

    • if \(S_2 = 100\), \(\Pi_2 = \frac{1}{1.1}(0\times \frac{3}{4} + 10 \times \frac{1}{4} ) = \frac{25}{11}\),

    • if \(S_2 = 60\), \(\Pi_2 = \frac{1}{1.1}(10\times \frac{13}{20} + 50 \times \frac{7}{20} ) = \frac{240}{11}\).

  • At \(t=1\)

    • if \(S_1 = 120\), \(\Pi_1 = \frac{1}{1.1}( 0 \times \frac{4}{5} + \frac{25}{11} \times \frac{1}{5} ) = \frac{50}{121}\),

    • if \(S_1 = 80\): \(\Pi_1 = \frac{1}{1.1}(\frac{25}{11} \times \frac{7}{10} + \frac{240}{11} \times \frac{3}{10} ) = \frac{895}{121}\).

  • At \(t=0\)

We know \(S_0 = 100\), so \(\Pi_0 = \frac{1}{1.1}(\frac{50}{121} \times \frac{3}{4} + \frac{895}{121} \times \frac{1}{4} ) = \frac{475}{242}\).

We see that the European put option is worth less than the American put option.

By put-call parity the price of the European call is \[ \textstyle C = P + S_0 - K(1+r)^{-T} = \frac{475}{242} + 100 - \frac{90}{(1.1)^3} = \frac{91425}{2662}. \] We calculate the price of an American call using the same method as for the put option, comparing at each node the value of waiting, versus the return from exercising the option early.

  • At \(t=2\):

    • if \(S_2=140\), the value of waiting is \(\frac{1}{1.1}(70 \times \frac{17}{20} + 30 \times \frac{3}{20} ) = \frac{640}{11}\) and the return from exercising early is 50, so it is optimal to wait, and \(\Pi_2 = \frac{640}{11}\);

    • if \(S_2 = 100\), the value of waiting is \(\frac{1}{1.1}(30 \times \frac34 + 0 \times \frac14) = \frac{225}{11}\), and the return from exercising is 10, so it is optimal to wait, and \(\Pi_2= \frac{225}{11}\);

    • if \(S_2 = 60\), both the value of waiting and the return from exercising are 0, so it is optimal to wait (it is no better to exercise early), and \(\Pi_2 = 0\).

  • At \(t=1\):

    • if \(S_1=120\), the value of waiting is \(\frac{1}{1.1}(\frac{640}{11} \times \frac45 + \frac{225}{11}\frac{1}{5}) = \frac{5570}{121}\) and return from exercising is 30, so it is optimal to wait, and \(\Pi_1 = \frac{5570}{121}\);

    • if \(S_1=80\), the value of waiting is \(\frac{1}{1.1}(\frac{225}{11}\times \frac{7}{10} + 0 \times \frac{3}{10})=\frac{1575}{121}\) and the return from exercising is 0, so it is optimal to wait, and \(\Pi_1 = \frac{1575}{121}\).

  • At \(t=0\): the value of waiting is \(\frac{1}{1.1}(\frac{5570}{121}\times \frac{3}{4} + \frac{1575}{121} \times \frac14 ) = \frac{91425}{2662}\), and the return from exercising is 10, so it is optimal to wait, and \(\Pi_0 = \frac{91425}{2662}\).

This shows that it is never better to exercise this American call option early, and its price at time 0 is the same as the European call option.