Implied volatility and inverse pricing

Definition Fix a European call with strike $K > 0$ on a stock with current price $S > 0$, time to maturity $\tau = T - t > 0$ in years, a constant continuously compounded rate $r$ and a constant continuous dividend yield $q$. For $\sigma > 0$ the Black-Scholes price is

where

Write $L = (S e^{-q\tau} - K e^{-r\tau})^{+}$ and $U = S e^{-q\tau}$. The following result is proved in the next two cards. For every price $C$ with $L < C < U$ there is exactly one $\sigma_{\mathrm{imp}} \in (0, \infty)$ with

and this number is the implied volatility of $C$. For a price outside the open interval $(L, U)$ no positive volatility exists. The contract, $S$, $K$, $\tau$, $r$ and $q$, and so the forward $S e^{(r - q)\tau}$ and the discount factor $e^{-r\tau}$, stay fixed during the inversion, and the result is a decimal volatility per square root of a year. The Black-Scholes put price is $P_{\mathrm{BS}}(\sigma) = C_{\mathrm{BS}}(\sigma) - S e^{-q\tau} + K e^{-r\tau}$, so a put price strictly between $(K e^{-r\tau} - S e^{-q\tau})^{+}$ and $K e^{-r\tau}$ also has exactly one implied volatility, and a put and a call with the same strike and maturity whose prices satisfy put-call parity have the same implied volatility. Inverting an American price through this European formula returns a number that also absorbs the early-exercise premium, and a normal implied volatility, quoted in currency units per square root of a year, is a different coordinate that is not interchangeable with this one.

Definition Assume $S > 0$, $K > 0$, $\tau > 0$ and real constants $r$ and $q$. The vega of the call, the derivative of its price with respect to $\sigma$, satisfies

for every $\sigma > 0$. To prove it, note that $d_1 - d_2 = \sigma\sqrt{\tau}$ and $d_1 + d_2 = 2(\ln(S/K) + (r - q)\tau)/(\sigma\sqrt{\tau})$, so $\tfrac12(d_1^2 - d_2^2) = \ln(S/K) + (r - q)\tau$, which gives

By the chain rule the derivative of the price is $S e^{-q\tau}\varphi(d_1)\,\partial_\sigma d_1 - K e^{-r\tau}\varphi(d_2)\,\partial_\sigma d_2$, and the identity gives both terms the same factor,

Every factor is positive, so $C_{\mathrm{BS}}$ is continuous and strictly increasing on $(0, \infty)$. The put has the same vega, because the parity difference $K e^{-r\tau} - S e^{-q\tau}$ does not depend on $\sigma$. Vega tends to $0$ as $\sigma \to \infty$. As $\sigma \downarrow 0$ it tends to $0$ when $K \ne S e^{(r - q)\tau}$ and to $S e^{-q\tau}\sqrt{\tau/(2\pi)}$ when the strike equals the forward. At $\tau = 0$ the price is the payoff $(S - K)^{+}$ and does not depend on $\sigma$ at all. The statement is about the model price as a function of its input, not about how quoted prices respond to realized volatility.

Definition Consider a frictionless market with a zero-coupon bond that costs $e^{-r\tau}$ per unit paid at $T$, the stock, which pays a continuous dividend yield $q$ so that $e^{-q\tau}$ shares held with dividends reinvested grow to one share at $T$, and European options on the stock, with $S_T \ge 0$. If there is no arbitrage, every call and put with strike $K > 0$ and time to maturity $\tau \ge 0$ satisfies

with $L$ and $U$ as in the first card. Each bound compares payoffs at $T$. The payoff $(S_T - K)^{+}$ is at least $0$, at most $S_T$, and at least $S_T - K$, which is the payoff of a portfolio that holds $e^{-q\tau}$ shares with dividends reinvested and owes $K e^{-r\tau}$ in the bond. If the call were cheaper than $S e^{-q\tau} - K e^{-r\tau}$, buying it and selling that portfolio would bring in cash today and leave the payoff $(K - S_T)^{+} \ge 0$. If it cost more than $U$, selling it and buying the shares would bring in cash and leave $\min(S_T, K) \ge 0$. The put bounds follow in the same way or from put-call parity. For $\sigma > 0$ and $\tau > 0$ the Black-Scholes call lies strictly inside, and its limits are the endpoints,

As $\sigma \downarrow 0$ both $d_1$ and $d_2$ tend to $+\infty$ when $S e^{-q\tau} > K e^{-r\tau}$ and to $-\infty$ when $S e^{-q\tau} < K e^{-r\tau}$, and when the two are equal they tend to $0$ and the price tends to $\tfrac12(S e^{-q\tau} - K e^{-r\tau}) = 0$. As $\sigma \to \infty$, $d_1 \to +\infty$ and $d_2 \to -\infty$. Together with the previous card, $C_{\mathrm{BS}}$ maps $(0, \infty)$ one to one onto $(L, U)$, which proves the result stated in the first card. A target equal to $U$ is reached only in the limit of infinite volatility, a target equal to $L$ only by the zero-volatility extension $C_{\mathrm{BS}}(0) = L$, and a target outside $[L, U]$ has no solution in this formula and, if it could be traded, would be an arbitrage.

Definition On $(L, U)$ the implied volatility $\sigma_{\mathrm{imp}}(C)$ is the inverse function of $C_{\mathrm{BS}}$. Since $C_{\mathrm{BS}}$ is continuously differentiable with $\nu > 0$, the inverse function theorem makes $\sigma_{\mathrm{imp}}$ continuously differentiable, and differentiating $C_{\mathrm{BS}}(\sigma_{\mathrm{imp}}(C)) = C$ with the chain rule gives

measured in volatility per currency unit. The first-order estimate of the change in implied volatility after a price change $\Delta C$ is therefore $\Delta\sigma \approx \Delta C/\nu$. For an approximate root $\hat\sigma > 0$ the mean value theorem gives an exact identity, with some $\xi$ between $\hat\sigma$ and $\sigma_{\mathrm{imp}}$,

A small price residual guarantees a small volatility error only where vega is not small. Vega is small near expiry, where $\sqrt{\tau}$ is small, and far from the forward, where $\lvert d_1 \rvert$ is large.

Definition For a target $C \in (L, U)$ define the residual

A bracket is a pair $0 < a < b$ with $F(a) < 0 < F(b)$. Since $F$ is continuous and strictly increasing, the root lies in $(a, b)$ and is unique there. A bracket always exists, because $F(\sigma) \to L - C < 0$ as $\sigma \downarrow 0$ and $F(\sigma) \to U - C > 0$ as $\sigma \to \infty$, so a small $a$ and repeated doubling of $b$ find one. Bisection evaluates $F$ at the midpoint $m = (a + b)/2$, replaces $a$ by $m$ when $F(m) < 0$, replaces $b$ by $m$ when $F(m) > 0$ and stops when $F(m) = 0$. After $n$ steps the bracket still contains $\sigma_{\mathrm{imp}}$ and has width $(b - a)/2^n$, a direct bound on the volatility error that does not depend on vega. Newton’s method uses the update

which converges quickly from a good starting point but can leave $(0, \infty)$ or the bracket, and is unreliable where vega is small. A safeguarded method keeps the bracket, accepts a Newton step only when it lands strictly inside the current bracket, bisects otherwise, and updates the bracket with the sign of $F$ at every new point, so it never loses the root.