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?Akuna CapitalNew York · 2025
Try it first
Starting from 30%, how many Newton steps until the error is below one part in a million?
Show the worked solution
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 relationshipC(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
042An option lets you decide in six months whether it becomes a call or a put, both struck at 100 and expiring in one year. The stock is at 100, pays no dividend, and rates are zero. Express it as a portfolio of plain options.Exotics tradingStructured products
Try it first
Which portfolio of plain options replicates the chooser?
Show the worked solution
A one-year call struck at 100 plus a six-month put struck at 100. At the decision date you take the larger of the call and the put. Write that as the call plus max(P - C, 0). Put-call parity with zero rates and no dividend says P - C = 100 - S at that date, so the extra piece is max(100 - S, 0) paid at six months: a six-month put. At 20% volatility that is 7.97 + 5.64 = 13.60, against 15.93 for a one-year straddle.
Why is a chooser worth less than a straddle but more than a single option?
Booking a restaurant table for a date six weeks away while keeping the right, at four weeks, to switch it to a different restaurant is worth more than a fixed booking but less than holding two bookings to the end. The chooser is the same: it beats any one plain option because you choose with six months of news in hand, but it is cheaper than a straddle because you must give up one leg at six months. A straddle keeps both the call and the put for the full year; the chooser keeps only whichever looks better at the half-way point, and then lives or dies with it. At 20% volatility the one-year straddle is 15.93, a one-year call alone is 7.97, and the chooser sits in between at 13.60. The question is whether you can find that middle number without a model of the choice itself.
How does put-call parity turn the choice into plain options?
At six months you hold max(C, P), where C and P are the values then of the call and the put with six months left. Split it: max(C, P) = C + max(P - C, 0). Now use parity. With no dividend and zero rates, a put minus a call with the same strike and expiry is worth the strike minus the stock, so P - C = 100 - S at the decision date. The awkward piece max(P - C, 0) becomes max(100 - S, 0), which is exactly the payoff of a put struck at 100 that expires at six months, so the chooser is a one-year call plus a six-month put. The call is always in the portfolio because you either keep it or, by parity, the put you switch into is that call plus a short forward, and the short forward's value is what the six-month put pays you when you switch.
The relationshipt the decision date, six months T the final expiry, one year C t, P t the values at the decision date of the call and the put struck at K K - S t put minus call by put-call parity, with zero rates and no dividend What it says in wordsThe choice is a call you always own plus a put on the decision date that pays the switch.The chooser equals a one-year call plus a six-month put, because at the decision date the larger of call and put is the call plus max(100 - S, 0), and at 20% volatility the two pieces add to 13.60, below the 15.93 one-year straddle. How do you check the answer, and what changes with rates and dividends?
Test the two ends of the decision date. If the choice had to be made today, the chooser is just whichever is dearer now, and at the money with zero rates the call and put cost the same, 7.97; the formula gives a call plus a put expiring today at the money, which is worth zero, so it agrees. If the choice is made at one year, you simply take whichever pays, which is a straddle, and the formula gives a call plus a one-year put, which agrees too. A simulation of 40,000 paths to six months, taking the better of call and put on each, gives 13.61 against 13.60. With rates or dividends the same split works, but parity puts a discounted strike on the put: it is struck at K times the discount factor from six months to a year, adjusted for the dividends paid in that window. The limitation to say: the decomposition relies on European options and on parity holding exactly; an American chooser, or one on a stock with a borrow cost, needs a model.
Where candidates lose it
The common loss is answering a straddle. A straddle keeps both legs for a year; the chooser forces you to give one up at six months, so it must cost less, and the difference is the put's last six months of life.
The second is getting the put's expiry wrong: writing a one-year put alongside the call, or a six-month call alongside the put. The switch happens at six months, so the piece that pays for it expires at six months. Write max(C, P) = C + max(P - C, 0) and parity does the rest.
What the interviewer asks next
- What is the chooser worth if the decision date moves to nine months?
- Rates are 6% instead of zero. What strike does the six-month put carry?
- The stock pays a dividend of 3 at month nine. How does the decomposition change?
- The call and the put have different strikes, 95 and 105. Is there still a closed form?
