Is the Black-Scholes model price a bijection on the interval of static arbitrage free prices

Question

Consider some stock with observed price $S$ and a call option on the stock with value $C$, time to maturity $T$ and strike $K$. Assume there is a constant, continuously compounded interest rate $r$. Assume everything is traded. It is well known that to avoid static arbitrages we must have $$ C \in (\max\{S-Ke^{-rT},0\},S). $$ Denote the interval $I$. Consider now the Black-Scholes model and the function $f:\sigma \mapsto C_{BS}(\sigma)$ the function sending its volatility into its corresponding model price. The function is monotone so it must be surjective. Since the model is arbitrage free necessarily the image of $f$ is contained in $I$, but is its image equal to $I$? What is known about this situation?

I guess this must be a studied question, but I've never seen it discussed.

Answer 1

It is. Note that the interval I is open on both ends. Moreover, \begin{align*} C_{BS}(\sigma) = S\Phi(d_1)-Ke^{-rT}\Phi(d_2), \end{align*} where $\Phi$ is the cumulative distribution function of a standard normal random variable, and \begin{align*} d_{1,2} = \frac{\ln\frac{S}{K e^{-rT}} \pm \frac{1}{2}\sigma^2T}{\sigma \sqrt{T}}. \end{align*} Note that \begin{align*} \lim_{\sigma\rightarrow \infty} d_{1} = \infty, \,\,\,\mbox{ and }\,\,\, \lim_{\sigma\rightarrow \infty} d_{2} = -\infty. \end{align*} That is, $$\lim_{\sigma\rightarrow \infty} C_{BS}(\sigma) = S.$$ Moreover, note that \begin{align*} \lim_{\sigma\rightarrow 0+} d_{1, 2} = \begin{cases} \infty, & \mbox{ if }\,\, S>K e^{-rT},\\ -\infty, & \mbox{ otherwise}. \end{cases} \end{align*} That is, \begin{align*} \lim_{\sigma\rightarrow 0+} C_{BS}(\sigma) = \max\left(S-Ke^{-rT}, \, 0\right). \end{align*} consequently, $$C_{BS}: (0, \,\,\, \infty) \mapsto \left(\max\left(S-Ke^{-rT}, \, 0\right), \,\,\, S\right)$$ is a bijection (i.e., one-to-one and onto).