Solution: Newton's Method for Implied Vol — Convergence and Edge Cases

Exercise: Newton's Method for Implied Vol — Convergence and Edge Cases

Part 1: why Newton's method converges

Let $f(\sigma) = C_{\text{BS}}(\sigma) - C_{\text{market}}$ . We want to solve $f(\sigma_{\text{imp}}) = 0$ .

Known facts.

$f$ is continuous on $(0, \infty)$ .
$f$ is strictly increasing ( $f'(\sigma) = \text{vega} > 0$ ).
$f$ is strictly convex on $(0, \sigma^*)$ and strictly concave on $(\sigma^*, \infty)$ .
$f(0^+) < 0 < f(\infty)$ (by the arbitrage bounds; market price is between intrinsic and $S_0$ ).

Hence there is a unique root $\sigma_{\text{imp}} > 0$ .

Case 1: $\sigma^{(0)} > \sigma_{\text{imp}}$ with $\sigma^{(0)} < \sigma^*$ (i.e. the convex region). Newton's tangent-line step from the right of the root on a convex function lands between the root and the current iterate. Formally,

\sigma^{(1)} = \sigma^{(0)} - f(\sigma^{(0)})/f'(\sigma^{(0)}) \in (\sigma_{\text{imp}}, \sigma^{(0)})

, and iterating keeps the sequence decreasing and bounded below by

\sigma_{\text{imp}}

, hence convergent to

\sigma_{\text{imp}}

Case 2: $\sigma^{(0)} < \sigma_{\text{imp}}$ with $\sigma^{(0)} < \sigma^*$ (still convex region but left of root). The first Newton step may overshoot past

\sigma_{\text{imp}}

, landing in the concave region. From the concave region, iterates decrease back toward

\sigma_{\text{imp}}

(mirror argument). So after the first overshoot, convergence is monotone.

Case 3: $\sigma^{(0)} > \sigma_{\text{imp}}$ and $\sigma^{(0)} > \sigma^*$ (concave). Tangent-line step from above on a concave function overshoots downward. May briefly enter the convex region, then converges by Case 1.

Conclusion. After at most one "crossing step," the Newton sequence is monotone and converges to

\sigma_{\text{imp}}

quadratically. Global convergence follows from the continuous-and-strict-monotone structure plus the convex-then-concave shape.

Remark. This is stronger than generic Newton guarantees (which require starting in a neighbourhood of the root). The specific shape of the Black-Scholes price in

\sigma

— monotone with a single inflection — is what makes the inversion so robust in practice.

Part 2: deep OTM convergence

At $\sigma = 0.15$ , $S_0 = 100$ , $K = 120$ , $T = 0.25$ , $r = 0$ :

d_1 = \frac{\ln(100/120) + 0.15^2 \cdot 0.25/2}{0.15 \sqrt{0.25}} = \frac{-0.1823 + 0.00281}{0.075} \approx -2.393

\phi(d_1) = \frac{1}{\sqrt{2\pi}}e^{-(-2.393)^2/2} \approx 0.0229

\text{vega} = S_0 \phi(d_1) \sqrt{T} = 100 \cdot 0.0229 \cdot 0.5 \approx 1.14

Very small vega compared to an ATM case (where vega

\approx 20

at these parameters). Newton's step is large when vega is small — the update

\Delta\sigma = -f/\text{vega}

can overshoot wildly, jumping out of the valid region

\sigma > 0

or past the root by a large margin.

Practical fix. Hybrid solver:

Bisection for the first few iterations to get a tight bracket $[\sigma_{\text{lo}}, \sigma_{\text{hi}}]$ containing the root — guaranteed linear convergence, no sensitivity to vega.
Newton within the bracket, but with a safeguard: if the Newton step falls outside the bracket (would overshoot), fall back to bisection for that step.

This "safeguarded Newton" is the standard in production libraries (e.g. QuantLib's impliedVolatility method). It gives the speed of Newton in the generic case and the robustness of bisection in the deep-OTM / deep-ITM edge cases.

Part 3: price outside arbitrage bounds

Arbitrage-free lower bound for the

K = 95

call:

\max(S_0 - Ke^{-rT}, 0) = \max(100 - 95, 0) = 5

. Reported

C_{\text{market}} = 4.50 < 5

— violates the lower bound.

What Newton does.

C_{\text{BS}}(\sigma) > 5

for any

\sigma > 0

(since the price starts at the intrinsic

5

\sigma \to 0^+

). So

f(\sigma) = C_{\text{BS}}(\sigma) - 4.5 > 0.5 > 0

for any

\sigma > 0

. Newton's step

-f/\text{vega}

is always negative. As

\sigma^{(n)} \to 0^+

, vega

\to 0

and the step blows up. Without a safeguard, iterates go to zero or negative.

Production handling.

Pre-check the market price against arbitrage bounds before inversion. If out of band, skip and flag.
Return a sentinel value (NaN, None, or an explicit error code) rather than a nonsensical vol.
Log the stale quote — a repeated violation means a bad data feed, not a rare corner case.

Attempting to invert a violating price is a common source of vol-surface bugs in production; the bounds check is cheap insurance.

Part 4: Manaster-Koehler start

For $S_0 = 100$ , $K = 120$ , $T = 0.25$ , $r = 0$ :

\sigma^{(0)} = \sqrt{\frac{2}{0.25}|\ln(100/120) + 0|} = \sqrt{8 \cdot 0.1823} \approx \sqrt{1.4586} \approx 1.207

So Manaster-Koehler starts at $\sigma^{(0)} \approx 1.21$ — much higher than the naive $0.2$ guess.

Why this is a good start. Manaster and Koehler chose this formula so that Newton's first step is guaranteed to move toward the root — they proved that initialising from this

\sigma^{(0)}

lies on the concave side of

\sigma^*

for any problem, and the first Newton step from there lands close to the root. Starting from

\sigma^{(0)} = 0.2

with a deep-OTM option puts you in the low-vega region where the first Newton step can overshoot.

Empirically for a deep-OTM call with $C_{\text{market}} = 0.005$ , Manaster-Koehler converges in 3–4 iterations where the naive start can take 15+ or fail to converge. This is why industrial libraries choose the starting point carefully rather than using a constant initial guess.

Takeaways

Newton on Black-Scholes is globally convergent for any market price in the arbitrage-free band, thanks to the convex-then-concave monotone shape of $C_{\text{BS}}(\sigma)$ . Quadratic convergence in the generic case.
Deep OTM/ITM is the numerical Achilles' heel. Vega vanishes at the wings, so Newton steps blow up. Safeguarded Newton (with bisection fallback) is the standard fix.
Always pre-validate arbitrage bounds. Garbage-in-garbage-out: inverting a price outside $(\max(S_0 - Ke^{-rT}, 0),\, S_0)$ produces nonsense or non-convergence. Cheap to check, expensive to debug later.
Manaster-Koehler beats a constant start. Initialising Newton intelligently — at the analytically-derived $\sigma^{(0)}$ — gives convergence guarantees that a naive start can't match.