6 Solutions for Chapter 6

These are just practice exercises - good luck!

Warm-up

Exercise 6.1 Use \(\Var(S_t) = \E[(S_t)^2] - \E[S_t]^2\) where \(\E[(S_t)^2] = (S_0)^2 \E[e^{2W}]\) and by standard scaling laws for Normal rvs, \(2W \sim \cN(2t(\mu - \sigma^2/2), 4 t \sigma^2)\). Using the formula given in the previous exercise \(\E[(S_t)^2] = (S_0)^2 \exp 2t(\mu + \sigma^2/2)\) and \(\E[S_t]^2 = (S_0)^2 \exp 2t\mu\). Hence \[ \Var\bigl(S_t\bigr) = \bigl( \e^{t\sigma^2} - 1\bigr)\E[S_t]^2 = (S_0)^2 \e^{2t\mu} \bigl( \e^{t\sigma^2} - 1\bigr) \]

Exercise 6.2 We have

  1. Using \(\E[S_t] = S_0 e^{\mu t}\) we obtain \(\E[S_{10}] = 100\e^2 \approx 738.91\) for each volatility.
  2. Using \(W_t \equiv \log \left( S_t/S_0 \right) \sim \cN(t(\mu - \sigma^2/2), t \sigma^2)\) we find \(\P\bigl( S_{10}>100 \bigr) = \P(W_{10} > 0) \approx 0.998\), \(0.829\), \(0.542\) for \(\sigma = 0.2\), 0.4, 0.6 respectively.
  3. Similarly \(\P(S_{10}<110) = \P(W_{10} < \log 1.1) \approx 0.004\), \(0.191\), \(0.478\) respectively.

Exercise 6.3 We have \(t=1/2\), \(S_0=105\), \(K=100\), \(r=0.10\) and \(\sigma=0.30\). We can proceed directly or calculate the price of the call option and then use put-call parity. Substituting into the Black-Scholes formula yields \(d_1 = 0.5718\), \(N(d_1) = 0.7163\) and \(N(d_1-\sigma \sqrt{t})=0.6404\), so that \(C \approx 14.29\) and using the put-call option parity formula we get \(P = C + 100e^{-0.05} - 105 \approx 4.41\).

Main problems

Exercise 6.4 \(\E(\e^{X}) = \int \e^{x}f(x)\,dx\) where \(v\sqrt{2\pi}f(x) = \exp[-(x-m)^2/2v^2]\). Do the integral by completing the square in the exponent, i.e., \[\begin{align*} \int_{-\infty}^\infty \e^x \exp[-(x-m)^2/2v^2] \,\ud x &= \int_{-\infty}^\infty \exp[-(x^2 -2(m+v^2)x+m^2)/2v^2] \,\ud x \\ &= \exp[(2mv^2+v^4)/2v^2] \int_{-\infty}^\infty \exp[-(x -(m+v^2))^2/2v^2] \,\ud x \\ & = \exp(m+v^2/2) \int_{-\infty}^\infty \exp[-y^2/2v^2] \,\ud y \end{align*}\] (where \(y = x-m-v^2\)). After normalisation by \(v\sqrt{2\pi}\) the final integral is that of a Normal density and so equals 1. Hence \(\E(\e^X) = \e^{m + v^2/2}\) as required. We started above with the formula for \(\E(\e^X)\) but it’s OK to make the substitution \(x = m + vz\) right at the start if you say what you are doing.

For the geometric BM write \(S_t/S_0 = \e^W\) with \(W \sim \cN(t(\mu-\sigma^2/2), t\sigma^2)\) and use the initial result with \(m = t(\mu-\sigma^2/2)\) and \(v^2 = t\sigma^2\). Then \(\E[S_t] = S_0 \E[\e^W] = S_0 \e^{t\mu}\).

Exercise 6.5 Under the geometric Brownian motion model \(\log (S_{u+t}/S_{u}) \sim \cN((\mu - \sigma^2/2)t, \sigma^2 t)\) so its standard deviation is \(\sigma\sqrt{t}\) where \(t\) is measured in years.

For (a) the end of day prices, we can take \(t=1/365\) so that \(\mbox{SD}(\log S_d(n)/S_d(n-1)) = 0.33 \times \sqrt{1/365} \approx 0.0173\). Or, we could quite reasonably claim that since prices only fluctuate of the trading days in the year, we should take \(t\) to be one trading day, i.e., \(t = 1/252\) (the number of trading days in a year being typically 252) to get \(\mbox{SD}(\log S_d(n)/S_d(n-1)) = 0.33 \times \sqrt{1/252} \approx 0.0208\).

For (b) the end of month prices, take \(t=1/12\) so that \(\mbox{SD}(\log S_m(n)/S_m(n-1)) \approx 0.0953\) \(t = d/252\) where \(d\) is the number of trading days in the month (which will depend on the specific month, and year).

Exercise 6.6 We have \(K=42\), \(S_0=40\), \(T=1/3\), \(\mu=0.15\) and \(\sigma=0.24\) so \(\mu - \sigma^2/2 = 0.1212\). We have to calculate \(\P(S_{1/3}>K)\). We know that \(W \equiv \log ( S_{1/3}/S_{0} ) \sim \cN(0.0404, 0.0192)\) and consequently \[\P(S_{1/3}>K) = \P(\log S_{1/3}/S_{0} > \log K/S_{0}) = \P(W > \log 42/40) \approx 0.476. \]

Exercise 6.7 The value of \(\mu\) is necessary but not included in the list. When using the risk neutral drift to calculate call option prices \(\mu\) is not needed but otherwise it is.

Exercise 6.8 A call option with strike price \(K = 0\) is worth exactly the same as a share at time \(T\) so, by the Law of One Price, we must have \(C=S_0\). The same result follows by evaluating \(\e^{-rT}\E_r[S_T]\) and using risk neutral pricing. To see that the Black–Scholes formula also implies this, compute its limit as \(K \to 0\).

Start with the Black–Scholes formula. Clearly \(d_1 \to \infty\) as \(T \rightarrow \infty\) though what happens to \(d_1 - \sigma\sqrt{T}\) depends upon the sign of \(r - \sigma^2/2\). As \(N(d_1) \to 1\) and \(\e^{-rT} \to 0\) we find \(\lim_{T\to \infty} C(K, T) = S_0\).

Exercise 6.9 To find the probability that the option is in the money we must use the actual drift of the stock price. We have \(\mu=0.6\), \(\sigma=0.3\) and \(T=1/2\) so that \((\mu-\sigma^2/2)T = 0.555/2 = 0.2775\) and \(\sigma^2 T = 0.045\) and hence \(\log ( S_{1/2}/S_0 ) \sim N(0.2775, 0.045)\). Consequently \[\begin{eqnarray*} \P(S_{1/2}< 0.9S_0) & = &\P(\log S_{1/2}/S_0 < \log 0.9) \\ & = &N((\log 0.9 - 0.2775)/0.21232) = N(-1.805) \approx 0.036 \end{eqnarray*}\]

For there not to be an arbitrage we must make buying the security and the new investment fair wagers under the risk neutral drift. To achieve this their prices must be their expected values under this drift. Hence, as \(r - \sigma^2/2 = 0.04 - 0.045 = -0.005\), \[ p \equiv \P_r(S(1/2)< 0.9S_0) = \Phi((\log 0.9 + 0.0025)/0.21232) = \Phi(-0.485) \approx 0.314 \] so that \(\E_r(\mbox{return}) = \e^{-rT}100p + 0 \approx \e^{-0.02}31.4 = 30.8\) which we can also find by applying the result for a cash-or-nothing put option.