Is the asset-or-nothing call option in this example valued incorrectly in the Black-Scholes framework?

Question

I understand the solution to the author's example below, but I can't help but notice that the implied volatility is an imaginary number:

The time-$t$ price of an All-or-nothing Asset Call is $S_t e^{-\delta(T - t)}N(d_1)$

We have $38.66 = S_0e^{-\delta\cdot T}N(d_1) = 60e^{-0.02\cdot0.5} N(d_1)$ and so $N(d_1) = 0.650808991$ and $d_1 = 0.38751$. Therefore

$$0.38751 = \frac{\ln(60/50) + (0.1 - 0.02 + 0.5\sigma^2)\cdot0.5}{\sigma\sqrt{0.5}},$$ which gives

$\sigma \in \{0.548022 - 0.767436i, 0.548022 + 0.767436i\}$.

I don't see how an imaginary implied volatility is possible, so was this option priced incorrectly under the Black-Scholes framework?

The problem is from "Models for Financial Economics" by Abraham S. Weishaus.

Answer 1

I agree with your computations. The problem is that the initial price of the asset-or-nothing call of 38.66 can't arise within the Black-Scholes framework. This seems to be an inconsistency/error in the question.

Below you see a plot of asset-or-nothing call price as a function of the implied volatility. Note that "1" means 100% implied volatility.

Let $A_0$ be the initial option price. It is easy to check that

\begin{equation} \lim_{\sigma \downarrow 0} A_0 = \lim_{\sigma \uparrow \infty} A_0 = S_0 e^{-\delta T} = 59.4030 \end{equation}

Furthermore

\begin{equation} \arg \min_{\sigma \in \mathbb{R}_+} A_0 = \sqrt{\frac{1}{T} \left( \ln \left( \frac{S_0}{K} \right) + (r - \delta) T \right)} = 94.30\% \end{equation}

and

\begin{equation} \min_{\sigma \in \mathbb{R}_+} A_0 = 44.4070. \end{equation}

However, you can still apply the model-independent put/call parity for asset-or-nothing options as suggested in the answer.