8.10 Implied volatility

Existence and uniqueness#

Proof

Continuity is clear. $\mathcal V=Se^{-q\tau}\varphi(d_1)\sqrt\tau>0$, so $c$ is strictly increasing. The limits as $\sigma\downarrow0$ and $\sigma\to\infty$ are the two bounds (Remark 8.7.3). Intermediate value theorem plus strict monotonicity give the bijection. For the put, $p(\sigma)=c(\sigma)-Se^{-q\tau}+Ke^{-r\tau}$ is a vertical shift of $c$, so $p(\sigma)=p^{\text{mkt}}$ iff $c(\sigma)=p^{\text{mkt}}+Se^{-q\tau}-Ke^{-r\tau}$, which by parity is the call price.∎

The last statement holds only if market prices satisfy parity with the same $S$, $q$ and $r$. In practice call and put IVs at the same strike differ slightly, and the gap measures the parity residual of 8.2 divided by vega.

Newton's method and its fallback#

Shape of $c(\sigma)$ Differentiate vega with respect to $\sigma$ (this is vomma or volga). Using $\partial_\sigma d_1=-d_2/\sigma$ (check: $d_1=x/(\sigma\sqrt\tau)+\sigma\sqrt\tau/2$ with $x=\ln(S/K)+(r-q)\tau=\ln(F/K)$),

So $c(\sigma)$ is convex for $\sigma<\sigma^*$ and concave for $\sigma>\sigma^*$, where

Proof

Case $\hat\sigma>\sigma^*$. On $[\sigma^*,\infty)$, $f$ is increasing and concave, and $f(\sigma_0)<0$. Suppose $\sigma^*\le\sigma_k<\hat\sigma$. Then $f(\sigma_k)<0$ and $\sigma_{k+1}>\sigma_k$. Concavity puts the graph below its tangent: $0=f(\hat\sigma)\le f(\sigma_k)+f'(\sigma_k)(\hat\sigma-\sigma_k)$, so $\hat\sigma-\sigma_k\ge-f(\sigma_k)/f'(\sigma_k)=\sigma_{k+1}-\sigma_k$, i.e. $\sigma_{k+1}\le\hat\sigma$. The iterates increase and are bounded by $\hat\sigma$, so they converge to some $\bar\sigma\le\hat\sigma$; passing to the limit in the iteration gives $f(\bar\sigma)=0$, so $\bar\sigma=\hat\sigma$ by uniqueness. Case $\hat\sigma<\sigma^*$: on $(0,\sigma^*]$, $f$ is convex, $f(\sigma_0)>0$, and the mirror argument (graph above tangent) shows the iterates decrease to $\hat\sigma$. Quadratic convergence near $\hat\sigma$ is standard for Newton with $f'(\hat\sigma)=\mathcal V>0$.∎

When $F=K$ exactly, $\sigma^*=0$ and $c$ is concave on all of $(0,\infty)$; start at any small positive $\sigma$. The code uses $\max(\sigma^*,10^{-3})$.

Why a fallback anyway With a careless start Newton fails. For a deep OTM option two days from expiry, starting at $\sigma_0=5\%$ gives $d_1\approx-91$ and a vega of order $10^{-1800}$: the first step is a division by (numerical) zero. Production code wraps Newton in a bracketing method. Brent's method (Brent 1973) combines bisection, the secant method and inverse quadratic interpolation; it keeps a bracket $[a,b]$ with $f(a)f(b)<0$ and guarantees convergence at least as fast as bisection, usually superlinearly. Since $f(10^{-6})<0<f(5)$ for any admissible price at realistic maturities, $[10^{-6},5]$ is a valid bracket. m08_implied_vol.solve_iv tries Newton and falls back to scipy.optimize.brentq; the tests force the fallback with $\sigma_0=5\%$ and check that the Manaster–Koehler start needs no fallback for the same option. For industrial use, Jäckel's "Let's Be Rational" (2015) attains machine precision in two iterations everywhere.

Nifty IVs and the skew#

The smile, previewed Repeating this across strikes for the December expiry on 5 Nov 2026 (Black-76 on $F=24{,}719.10$, using OTM options: puts below the forward, calls above) gives roughly: 24,000 put 13.4%; 24,300 put 12.9%; 24,500 put 12.6%; 24,800 call 12.3%; 25,000 call 12.2%; 25,500 call 12.0%. IV falls as the strike rises: the equity-index skew. Under Black–Scholes this curve would be flat. M09 explains what the skew prices (crash risk, the leverage effect) and how to trade it.