Black-Scholes Call Prices Map Volatility Bijectively to Arbitrage-Free Prices
Summary
The document asks whether a European call’s Black-Scholes price, as a function of volatility, covers every price allowed by static no-arbitrage bounds. With the stock price, strike, maturity, and continuously compounded rate fixed, the admissible call-price interval lies above intrinsic value and below the stock price. The answer shows that the mapping from positive volatility to call price is one-to-one and onto this open interval.
The argument uses the limits of the Black-Scholes formula: as volatility approaches zero, the price approaches the call’s intrinsic-value bound; as volatility grows without bound, it approaches the stock price. Together with the model price’s monotonic increase in volatility, these limits establish that each price strictly inside the bounds corresponds to exactly one positive volatility. The result applies under the stated Black-Scholes assumptions and does not provide a numerical inversion procedure.
Key ideas
- For fixed contract inputs, the Black-Scholes call price increases monotonically with volatility.
- As volatility approaches zero, the model price approaches the call’s intrinsic-value bound.
- As volatility grows without bound, the call price approaches the underlying stock price.
- Every call price strictly within the stated no-arbitrage interval corresponds to exactly one positive volatility.
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Full text
# Is the Black-Scholes model price a bijection on the interval of static arbitrage free prices
# Is the Black-Scholes model price a bijection on the interval of static arbitrage free prices
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 by Gordon (score 4, accepted)
https://quant.stackexchange.com/a/18801
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).Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
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