Exercise 4.11. Let
$$
c(t, x)=x N\left(d_{+}(T-t, x)\right)-K e^{-r(T-t)} N\left(d_{-}(T-t, x)\right)
$$
4.10 Exercises
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denote the price for a European call, expiring at time $T$ with strike price $K$, where
$$
d_{ \pm}(T-t, x)=\frac{1}{\sigma_1 \sqrt{T-t}}\left[\log \frac{x}{K}+\left(r \pm \frac{\sigma_1^2}{2}\right)(T-t)\right] .
$$
This option price assumes the underlying stock is a geometric Brownian motion with volatility $\sigma_1>0$. For this problem, we take this to be the market price of the option.
Suppose, however, that the underlying asset is really a geometric Brownian motion with volatility $\sigma_2>\sigma_1$, i.e.,
$$
d S(t)=\alpha S(t) d t+\sigma_2 S(t) d W(t) .
$$
Consequently, the market price of the call is incorrect.
We set up a portfolio whose value at each time $t$ we denote by $X(t)$. We begin with $X(0)=0$. At each time $t$, the portfolio is long one European call, is short $c_x(t, S(t))$ shares of stock, and thus has a cash position
$$
X(t)-c(t, S(t))+S(t) c_x(t, S(t))
$$
which is invested at the constant interest rate $r$. We also remove cash from this portfolio at a rate $\frac{1}{2}\left(\sigma_2^2-\sigma_1^2\right) S^2(t) c_{x x}(t, S(t))$. Therefore, the differential of the portfolio value is
$$
\begin{aligned}
d X(t)=d c( & t, S(t))-c_x(t, S(t)) d S(t) \\
& +r\left[X(t)-c(t, S(t))+S(t) c_x(t, S(t))\right] d t \\
& -\frac{1}{2}\left(\sigma_2^2-\sigma_1^2\right) S^2(t) c_{x x}(t, S(t)) d t, 0 \leq t \leq T .
\end{aligned}
$$
Show that $X(t)=0$ for all $t \in[0, T]$. In particular, because $c_{x x}(t, S(t))>0$ and $\sigma_2>\sigma_1$, we have an arbitrage opportunity; we can start with zero initial capital, remove cash at a positive rate between times 0 and $T$, and at time $T$ have zero liability. (Hint: Compute $d\left(e^{-r t} X(t)\right)$.)