Valuing a claim on $S^a$: This exercise/solution appears to have a mistake

Question

The below exercise and solution was found in "Models for Financial Economics" by Abraham Weishaus. My issues are:

1. In this problem, $S(t)$ does not satisfy the Black-Scholes framework because volatility is time-varying. According to this paper on page 15, it would appear that we would need to modify the volatility of $S(t)$ for use in the Black-Scholes formula such that:

$$\text{Var}\left(\ln S(t) | S(0)\right) = \frac{1}{1 - 0}\int_0^1 (\sigma(t))^2 dt = \frac{1}{1 - 0}\int_0^1 (0.16t)^2 dt = 0.00853333.$$

2. The second issue is that this appears, based on the wording, to be a gap option where the trigger is $S(1) > 60$ and not $S(1)^{0.9} > 60$.

The author mentions in the errata that the $r$ in the fourth expression from the top of the solution should be changed in the exponent to $\alpha$, but does not address these other concerns.

Are these concerns justified, or is the author's solution correct?

Answer 1

Under the Black-Scholes' framework, the underlying stock price price process $\{S_t, \, t\ge 0\}$ satisfies an SDE of the from $$ dS_t = S_t (\alpha dt + \sigma dW_t),$$ where $\{W_t, \, t\ge 0\}$ is a standard Brownian motion, $\alpha$ is the continuously compounded stock return, and $\sigma$ is the constant volatility. Then $$S_t = S_0e^{(\alpha-\frac{1}{2}\sigma^2)t + \sigma W_t}.$$ Consequently, $$\ln S_t = \ln S_0 + \big(\alpha-\frac{1}{2}\sigma^2\big)t + \sigma W_t,$$ and $$\text{Var}( \ln S_t \mid S_0) = \sigma^2 t.$$ From the given information, $\alpha=0.15$, $\sigma = 0.4$.

Let $X_t = S_t^a$. Note that \begin{align*} X_t &= X_0e^{(a\alpha-\frac{1}{2}a\sigma^2)t + a\sigma W_t}\\ &=X_0e^{(a\alpha + \frac{1}{2}a(a-1)\sigma^2-\frac{1}{2}a^2\sigma^2)t + a\sigma W_t}\\ &\equiv X_0e^{(\mu-\frac{1}{2}a^2\sigma^2)t + a\sigma W_t}. \end{align*} Here, \begin{align*} \mu = a\alpha + \frac{1}{2}a(a-1)\sigma^2 = 0.1287. \end{align*}

The option payoff is given by \begin{align*} (S_1^a - 30) 1_{S_1 > 60} &= (S_1^a - 30) 1_{S_1^a > 60^a}\\ &= (X_1 - 30) 1_{X_1 > 60^a}\\ &=X_1\, 1_{X_1 > 60^a} - 30\, 1_{X_1 > 60^a}. \end{align*} Therfore, the expected payoff is given by \begin{align*} E(X_1) N(d_1) - 30 N(d_2), \end{align*} where \begin{align*} d_1 = \frac{\ln (X_0 / 60^a) + \mu + \frac{1}{2}a^2 \sigma^2}{a\sigma}, \end{align*} and \begin{align*} d_2 = d_1 - a\sigma. \end{align*}