Derivatives Foundation puzzles, solved step by step

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027You back out implied volatility from an option price with Newton's method. For an at-the-money call priced at 40 on a stock at 1,000 with three months to expiry and rates at zero, starting from 30%, how fast does it converge, when can it fail, and what starting guess do traders use?Option pricing intuitionHardAkuna CapitalNew York · 2025

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Starting from 30%, how many Newton steps until the error is below one part in a million?

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Two steps, because Newton converges quadratically near the root. From 30% the error goes 0.099, 8.2e-05, 4.3e-11, then machine precision: the correct digits roughly double each step. The implied volatility is 20.06%. Newton fails where vega is tiny, far out of the money or close to expiry, because dividing by a near-zero slope throws the next guess to nonsense, and it fails outright if the price sits outside the no-arbitrage bounds. Traders start from price over 0.4 x S x sqrt T, which is 20% here.

Why is Newton so fast on this option?

Picture walking towards a wall in the dark by stepping the full distance your outstretched hand estimates. If the floor is level the estimate is right and you arrive in one step; if it slopes gently you arrive in two. Newton does the same with the pricing function: it fits a straight line at the current guess and jumps to where that line hits the target price. The jump is as good as the line, and at the money the call price is almost a straight line in volatility, so the first jump lands within a hair of the answer. Here the price is roughly S x 0.4 x sigma x sqrt T, which is linear in sigma, with a small concave bend. Starting at 30% gives a price of 59.79 against a target of 40; one step takes sigma to 20.0532%, an error of 8.2e-05, and the next step clears ten digits.

The relationship

\sigma_{n+1} = \sigma_n - \frac{C(\sigma_n) - C_{mkt}}{\partial C / \partial \sigma} \qquad |\sigma_{n+1} - \sigma^*| \approx \frac{|C''|}{2|C'|}\,|\sigma_n - \sigma^*|^2

C(sigma)	the model price at the current volatility guess
C mkt	the market price, 40 here
dC/d sigma	vega, the slope of price in volatility
sigma star	the implied volatility being solved for
C'' over 2C'	the curvature of price in volatility relative to its slope; small at the money, so the squaring bites hard

What it says in wordsEach step divides the price gap by the slope, and once close the error is squared, so the correct digits double every step.

For the at-the-money call the error in volatility falls from 0.10 to 8.2e-05 to 4.3e-11 and reaches machine precision by the third step, while for a far out-of-the-money call started where vega is tiny the first step overshoots to 114% and the method needs many more steps to crawl back.

When does the method fail, and what does the failure look like?

Newton divides by vega, so it breaks where vega is close to zero: far out of the money, close to expiry, or at a very low starting volatility. Take an illustrative call struck at 1,300, 30% above spot, priced at 0.50. Its true implied volatility is 23.0%, but at a starting guess of 15% the model price is 0.005 and vega is only 0.50 per unit of volatility, so the first step jumps to 114% and the method needs 7 more steps to get within 7e-06. Start at 10% and vega is 2.4e-04, so the step divides by almost nothing and the next guess is a volatility of 2,095, which is garbage. The other failure is a price with no solution at all: a call priced below its intrinsic value or above the stock has no volatility that produces it, and Newton loops forever. Check the bounds before you iterate.

What starting guess do traders actually use?

Use the at-the-money approximation: an at-the-money call is worth about 0.4 x S x sigma x sqrt T, so invert it. Here that gives 40 / (0.4 x 1,000 x 0.5) = 20%, within 0.0006 of the true 20.06%; the version with the exact constant, sqrt(2 pi / T) x C / S, gives 20.05%. A guess that close means Newton is finishing a job that is already nearly done, which is why production code rarely needs more than three steps. For options away from the money, a guard is standard: a starting volatility of sqrt(2 |ln(S/K)| / T), 145% for the 1,300 strike, from which the method is known to converge, or a bracketed method such as bisection for the first few steps and Newton only to polish. Say the limitation too: all of this assumes a price that the model can reach, and real screens carry stale or crossed quotes that no solver can fix.

Where candidates lose it

The common loss is describing Newton as halving the error, which is bisection, or saying one step per digit, which is a linear method. The word the interviewer wants is quadratic, with the digits doubling, and the reason: near the root the error is squared.

The second is forgetting the failure cases. A candidate who only praises the speed has not run the method on a far out-of-the-money option, where a tiny vega sends the next guess negative. Name vega as the divisor and the failure explains itself.

What the interviewer asks next

Why is the call price nearly linear in volatility at the money, and where does it stop being so?
What goes wrong if you start Newton above the true volatility for a far out-of-the-money put?
How would you make the solver robust enough for a live surface of ten thousand strikes?
Price a call at 40 with the stock at 1,000: is any price between 0 and 1,000 reachable by some volatility?

Asked at Akuna Capital, Quantitative Research, New York, 2025 (Wall Street Oasis): Convergence time of newton's method