Assignment 3

Exercise 3.9 From the description in the question we see that for the first period \(u=1.2, d=0.8\), and for the second period \(u=1.1, d=0.9\). For both periods \(r=0.05\). Using this and the start price \(S_0 = 10\) we can calculate the tree of possible share prices, together with the martingale probabilities (first period: \(q_u = \frac{1.05-0.8}{1.2-0.8} = \frac58\); second period: \(q_u = \frac{1.05-0.9}{1.1-0.9} = \frac34\)):

Now we calculate the price of the call option at each node of the tree, starting at \(t=2\) and working back to \(t=0\):

  • At \(t=2\), the call option has value \((S_2 - 8)^+ = 5.2, 2.8, 0.8\) or 0, depending on the value of \(S_2 = 13.2, 10.8, 8.8\) or \(7.2\).
  • At \(t=1\), if the share price is \(S_1 = 12\), then the call option is worth \[ \frac{1}{1.05}[5.2 \times {\textstyle \frac34 + 2.8 \times \frac14}] = \frac{92}{21}, \] and if the share price is \(S_1 = 8\), then the call option is worth \[ \frac{1}{1.05}[0.8 \times {\textstyle \frac34 + 0 \times \frac14}] = \frac{4}{7}.\]
  • At \(t=0\), the call option is worth \[ \frac{1}{1.05}[{\textstyle \frac{92}{21} \times \frac58 + \frac47 \times \frac38}] = \frac{1240}{441}.\]

Therefore the fair price at time 0 for the call option is \(£\frac{1240}{441} \approx £2.81\).

To find the hedging portfolio we either use the equations \[ x_t = \frac{1}{(1+r)B_{t-1}}\frac{u\Pi^d_t - d\Pi^u_t}{u-d}, \quad y_t = \frac{\Pi^u_t- \Pi^d_t}{uS_{t-1}-dS_{t-1}}, \] or (making use of the fact that \(V_t = \Pi_t\) at every node) find \(y_t\) using the above equation, and find \(x_t\) using \(V_{t-1} = x_tB_{t-1} + y_tS_{t-1}\). Either way, we find that \[ x_1 = -\frac{2960}{441}, \quad y_1 = \frac{20}{21}. \] At time 0, this portfolio has value \(-\frac{2960}{441}+10\times\frac{20}{21} = \frac{1240}{441}\) as expected.
When the stock price increases from \(S_0 = 10\) to \(S_1 = 12\) the portfolio \((x_1,y_1)\) becomes worth \(1.05\times -\frac{2960}{441}+12\times\frac{20}{21} = \frac{92}{21}\), again as expected.
We change the amounts of the bond and stock we hold to \((x_2,y_2)\), which we calculate using the formulas as given above, to find \[ x_2 = -\frac{3200}{441}, \quad y_2 = 1. \] We see that this new portfolio has value \(1.05\times -\frac{3200}{441}+12\times1 = \frac{92}{21}\), so the self-financing condition is satisfied.

Finally, we can check that when the share price decreases from \(S_1 = 12\) to \(S_2 = 10.8\), the portfolio \((x_2,y_2)\) becomes worth \((1.05)^2\times -\frac{3200}{441} + 10.8 \times 1 = 2.8\) and so the portfolio matches the return given by the call option.

It is possible to check that the hedging portfolio always matches the return given by the call option, whatever happens to the share price, but remember that the values \((x_2,y_2)\) depend on what happens to the share price over the first period, so that the amounts of the bond and stock we choose to hold from time 1 to time 2 will be different when \(S_1 = 8\).

Exercise 3.10 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 \[ q_u = \frac{1.05-0.8}{1.1-0.8} = \frac56, \quad q_d = \frac{1.1-1.05}{1.1-0.8} = \frac16 \] and \[ m_u = \frac56\cdot\frac{1.1}{1.05}=\frac{55}{63}, \quad m_d = \frac{1}{6}\cdot\frac{0.8}{1.05} = \frac{8}{63}. \]

To calculate the price at time 0 (when \(t=T=4\)), we need to find \(k^\star\) as the smallest integer greater than \(\log(40/(40\times0.8^4))/\log(1.1/0.8) \approx 2.8\), so \(k^\star = 3\) and

\[\Pi_0 = 40 \sum_{k=3}^4 \binom{4}{k} \left(\frac{55}{63}\right)^k \left(\frac{8}{63}\right)^{4-k} - 40(1.05)^{-4} \sum_{k=3}^4 \binom{4}{k} \left(\frac56\right)^k \left(\frac16\right)^{4-k} \approx 8.188. \]

At time 2, when \(S_2 = 35.2\), we use the formula with \(k^\star\) the smallest integer greater than \(\log(40/(35.2\times0.8^2))/\log(1.1/0.8) \approx 1.4\) so \(k^\star = 2\) and

\[ \Pi_2 = 35.2 \left(\frac{55}{63}\right)^2 - 40(1.05)^{-2} \left(\frac56\right)^2 \approx 1.633. \]

Exercise 3.11

  1. Recall (e.g. from Probability I) that the moment generating function of a random variable \(X\) is the function \(m_X(\theta) = \E[\e^{\theta X}]\). So to work out \(m_Y(\theta)\) you can either directly calculate this expectation for a Binomial random variable, or use the hint and some basic properties of the m.g.f/expectation, namely, that if \(X_1, X_2\) are independent random variables then \[ m_{X_1+X_2}(\theta) = \E[\e^{\theta(X_1+X_2)}] = \E[\e^{\theta X_1}\e^{\theta X_2}] = \E[\e^{\theta X_1}] \E[\e^{\theta X_2}] = m_{X_1}(\theta)m_{X_2}(\theta). \] Since \(Y = \sum_{i=1}^T X_i\) with \(X_i\) i.i.d. Bernoulli random variables, we have \[ m_Y(\theta) = \prod_{i=1}^T m_{X_i}(\theta) = (m_{X_1}(\theta))^T, \] and \(m_{X_1}(\theta) = \E[ e^{\theta X_1} ] = (1-q)\e^0 + q \e^\theta\), giving \(m_Y(\theta) = (1-q+q\e^\theta)^T\) as required.
  1. Since \(S_T = S_0u^Y d^{T-Y}\) where \(Y\sim \Bin(T,q)\), we have \[ \E[S_T] = \E[ S_0 u^Y d^{T-Y} ] = S_0 d^T \E\Big[\Big(\frac{u}{d}\Big)^Y\Big] = S_0 d^T \Big(1-q +q \frac{u}{d} \Big)^T, \] where we used part (a) with \(\e^\theta\) set equal to \(u/d\). Simplifying, we have \[ \E[S_T] = S_0(d(1-q) + uq)^T. \]

Finally, using this expression for \(\E_\Q[S_T]\), we see that the \(q\) that solves \(S_0 = \frac{1}{(1+r)^T}\E_\Q[S_T]\) has \((1+r)^TS_0 = S_0(d(1-q) + uq)^T\), or \(1+r = d(1-q) + uq\), so that \[ q= \frac{1+r-d}{u-d} \] and we see that \(q\) is the martingale probability \(q_u\) of an up-step in the binomial model.