Derivatives Foundation puzzles, solved step by step
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- 100
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078Three calls on the same stock and expiry trade at: strike 90 for 14, strike 100 for 8, strike 110 for 1. Is there an arbitrage? Build it.Market makingVolatility trading
Try it first
Which relationship between the three prices should you test first?
Show the worked solution
Yes. Buy the 90 call, sell two 100 calls, buy the 110 call, and you are paid 1 to own a payoff that is never below zero. The butterfly costs 14 - 16 + 1 = -1, so you receive 1 today. At expiry it pays nothing below 90, rises to 10 at 100 and falls to nothing above 110, never negative. Call prices must be convex in strike: the 100 call cannot exceed (14 + 1)/2 = 7.5, and it trades at 8.
Where does the 7.5 come from?
Picture three houses on one street at 90, 100 and 110 square metres, priced at 14, 8 and 1. The middle house should be worth no more than the average of its neighbours if each extra metre is worth less than the last; if it is priced above the average, you sell it and buy the two neighbours. A call's price falls as the strike rises, and it falls at a decreasing rate, so the price at the middle strike must sit on or below the straight line between its neighbours. The line from 14 at 90 to 1 at 110 passes through 7.5 at 100. The market says 8. That half point is the mispricing.
The butterfly of long one 90 call, short two 100 calls and long one 110 call pays nothing below 90, peaks at 10 at a stock price of 100 and pays nothing above 110, and because it costs minus 1 the whole payoff sits at least 1 above the price paid, while on the right the quoted 8 for the 100 call sits above the 7.5 chord between the 90 and 110 calls, which breaks convexity. How do you prove the payoff is never negative?
Walk the stock price up. Below 90 nothing is in the money, payoff zero. Between 90 and 100 only the 90 call pays, so the payoff is the stock price minus 90, rising to 10. Between 100 and 110 the two short 100 calls start paying out, and the slope turns to 1 - 2 = -1, so the payoff falls from 10 back to zero at 110. Above 110 the long 110 call kicks in and the slope is 1 - 2 + 1 = 0: flat at zero. The payoff is zero or positive everywhere, you were paid 1 to hold it, so you have a riskless profit of at least 1 and up to 11.
The relationshipC(K) the price of the call struck at K 7.5 the midpoint of the chord between the outer strikes 8 the quoted middle price, half a point too high What it says in wordsWith equally spaced strikes, the middle call can never cost more than the average of the outer two, because the butterfly that tests it can only pay out zero or more.Check the other bounds too, out loud, so the interviewer sees you are not pattern matching. A call spread can never be worth more than the gap between its strikes: the 90/100 spread costs 6 and the 100/110 spread costs 7, both under 10, so those pass. Prices fall as strike rises: 14, 8, 1, pass. Only convexity fails. The limitation: in a real screen you trade at bids and offers, not mids, and a half point of theoretical edge can vanish inside the spread. The structure is an arbitrage at these prices; whether you can execute it is a separate question.
Where candidates lose it
Candidates test the wrong bound. They check that each call is worth more than its intrinsic value, or that a spread is worth less than the strike gap, find both fine, and declare no arbitrage. Convexity is the bound that catches a dear middle strike, and it is the one most people forget.
The second loss is building the butterfly the wrong way round. If you sell the wings and buy the middle you pay 1 to hold a payoff that is zero or negative. Say which leg is long before you say the price.
What the interviewer asks next
- Reprice the 100 call so there is no arbitrage. What is the highest price it can take?
- If the strikes were 90, 100 and 120, how would you weight the legs?
- Why does the same convexity rule hold for puts?
083A price starts at 100 and moves up or down 1 each day with equal chance for 10 days. What is the probability that it touches 104 at some point in those 10 days?Exotics tradingQuant trading
Try it first
Which quantity is easier to count, and how does it relate to touching?
Show the worked solution
232 of the 1,024 paths touch 104, about 22.7%. A path ending at or above 104 must have touched it: that needs at least 7 up days, 176 paths. A path that touches 104 and ends below it can be reflected in the 104 line after its first touch, and the reflection ends above 104, needing at least 8 up days: 56 paths. The pairing is one to one, so the count is 176 + 56 = 232.
Why is the reflection a fair way to count?
Imagine a walker who reaches a wall at some point and then wanders on. Flip every step after the first touch, up for down, and you get another walker who also touched the wall at the same moment and is now the same distance on the other side of it. Each flipped step is as likely as the original, so the two walks have equal probability. Reflection pairs every path that touches 104 and finishes below it with exactly one path that finishes above 104, so the hard count becomes an easy count of endpoints. Paths that finish at exactly 104 are their own partners and are counted once.
A path that climbs to 104 by day 4 and ends at 100 is mirrored in the 104 line after the first touch into a path that ends at 108, and counting by endpoint gives 176 paths ending at or above 104 plus 56 mirrors of paths ending above it, 232 of 1,024 paths, which is 22.7%. How do the binomial counts fall out?
After 10 days of plus or minus 1, the position is 100 plus ups minus downs, which is 100 + 2 x ups - 10. Ending at or above 104 needs ups of at least 7: C(10,7) + C(10,8) + C(10,9) + C(10,10) = 120 + 45 + 10 + 1 = 176. Ending strictly above 104 means at or above 106, since the endpoint is always even, so ups of at least 8: 45 + 10 + 1 = 56. The probability of touching is (176 + 56)/1,024 = 232/1,024, about 22.7%, against only 176/1,024 = 17.2% for ending at or above 104. The gap is the 56 paths that visited the barrier and came back.
The relationshipS_10 the net move after ten days P(S_10 >= 4) paths ending at or above the barrier, 176 P(S_10 > 4) paths ending strictly above, which are the reflections of touch-and-return paths, 56 What it says in wordsThe chance of touching a level equals the chance of ending at or past it plus the chance of ending strictly past it.Say why a derivatives desk cares. A knock-out or one-touch option pays on exactly this event, and the reflection principle is how the closed-form barrier prices are derived in continuous time. The limitation is in the symmetry: reflection needs the up and down steps to be equally likely and the barrier to sit on the lattice. With a drift, or in continuous time with a non-zero rate, the reflected path is not equally likely and a correction factor appears, which is why barrier formulas carry that extra power of the barrier over the spot.
Where candidates lose it
The usual loss is answering the probability of ending at or above 104, 17.2%, and forgetting the paths that touched and came back. The interviewer asked about touching, and the difference is a quarter of the answer.
The second loss is trying to enumerate paths that stay below 104 by hand. There are 792 of them and no clean way to list them in the room. Reflection exists so you do not have to.
What the interviewer asks next
- What is the probability the price touches 104 and ends at 100?
- Change the odds to 55% up. Why does reflection stop working as stated?
- How does this count turn into the price of a one-touch option?
093A stock at 100 will be 120 with probability 70% or 90 with probability 30% in one period, and interest rates are zero. Price a call struck at 100. Why does the 70% not appear in your answer?Quant trading
Try it first
What is the call worth?
Show the worked solution
The call is worth 20/3, about 6.67, and the 70% does not matter because the call can be copied with stock and cash. Hold 2/3 of a share and borrow 60: at 120 the copy is worth 80 - 60 = 20, at 90 it is worth 60 - 60 = 0, matching the call in both states. The copy costs 66.67 - 60 = 6.67, so the call must too. The real-world odds are already in the stock price, which the copy uses.
How do you copy the call?
If a shop sells a gift box of two items for more than the items cost separately, you buy the items and skip the box; the box's price is pinned by what goes in it. Options work the same way. Find a mix of stock and cash that pays exactly what the call pays in every state, and the call must cost what the mix costs, whatever anyone believes about the odds. The call pays 20 or 0, a swing of 20, while the stock swings from 120 to 90, a swing of 30. So hold 20/30 = 2/3 of a share. At 120 that is 80, which is 60 too much, and at 90 it is 60, also 60 too much: borrow 60 today and repay it in either state.
Two thirds of a share financed with 60 of borrowing pays 20 when the stock goes to 120 and 0 when it goes to 90, exactly like the call, so the call costs what that portfolio costs, 2/3 x 100 - 60 = 6.67, and the real-world 70% chance of an up move, which would give an average payoff of 14, never enters. Where did the 70% go?
It is in the stock price. A stock that goes up 70% of the time to 120 and is still priced at 100 today is one the market demands a return on, because it is risky. The copy buys the stock at that price, so it inherits whatever the market thinks of the odds and the risk. The option is priced relative to the stock, not relative to anyone's forecast, so the real-world probability cancels out of the answer. What does appear is a different probability, q, the one that makes the stock earn the risk-free rate: 100 = q x 120 + (1 - q) x 90, so q = 1/3. Discounting the call's payoff at q gives 1/3 x 20 = 6.67, the same answer by another route.
The relationshipDelta shares held in the copy, the call's swing over the stock's swing B cash borrowed, 60, so the copy pays nothing in the down state q the risk-neutral probability of the up move, not a forecast What it says in wordsHold two thirds of a share, borrow sixty, and you have built the call for 6.67; the risk-neutral probability of a third gives the same number.Then show the arbitrage, since that is what makes the answer binding. If the call traded at 8, sell it and buy the copy for 6.67: you pocket 1.33 today and the two positions cancel in both states. If it traded at 5, do the reverse. The limitation is the one-step world: real prices take many values, so the copy has to be rebalanced as the stock moves, which is where the Black-Scholes model comes from, and where trading costs and jumps make the copy imperfect.
Where candidates lose it
The common loss is answering 14, the call's expected payoff under the stated odds. It is the natural first move and the one the question is built to catch. Expected payoff under real-world odds is not a price unless everyone is indifferent to risk.
The second loss is getting q = 1/3 and then calling it the true chance of an up move. It is not a forecast. Say what it is: the probability that makes the stock's expected return equal the risk-free rate.
What the interviewer asks next
- Price the put struck at 100 in the same tree, and check put-call parity.
- If interest rates were 5% for the period, what is q and what is the call worth?
- The stock's up probability rises to 90% but its price stays at 100. What happens to the call price, and why?
096A game flips a fair coin until the first tail and pays Rs 2^n if the first tail comes on flip n, but the house can pay at most Rs 1,024. What is the fair price, and why does the uncapped game break the idea of a fair price?Market makingProp trading firms
Try it first
How much does each possible flip contribute to the expected payout, below the cap?
Show the worked solution
Rs 11. A first tail on flip n pays 2^n with probability 1/2^n, so each of flips 1 to 10 contributes exactly Rs 1, Rs 10 in all. Reaching flip 11 or later has probability 1/1,024 and pays the cap of Rs 1,024, adding one more rupee. Without the cap every flip keeps adding Rs 1 and the expected value is infinite, yet nobody would pay much to play, which shows expected value alone cannot price a bet on outcomes the payer cannot honour.
Why does every flip add exactly one rupee?
Think of a raffle where each ticket in the next bundle is half as likely to win but the prize is twice as large. Every bundle is worth the same to you. Here the first tail on flip 1 pays Rs 2 half the time, on flip 2 pays Rs 4 a quarter of the time, on flip 3 pays Rs 8 an eighth of the time. The payout doubles exactly as the probability halves, so each flip contributes 2^n x 1/2^n = Rs 1 to the expected value, and the fair price is simply the number of flips before the cap bites, plus the capped tail. Ten flips reach Rs 1,024; anything later is paid at the cap, worth 1,024 x 1/1,024 = Rs 1.
Each flip from 1 to 10 contributes payout times probability of exactly Rs 1, and reaching flip 11 or later adds the capped Rs 1,024 times a 1 in 1,024 chance, another Rs 1, for a fair price of Rs 11, while without the cap the Rs 1 contributions would carry on forever. What goes wrong without the cap?
Remove the cap and the sum is 1 + 1 + 1 + ... with no end, an infinite expected value. Yet the game pays Rs 8 or less seven times in eight, and more than Rs 11 only one time in eight. The infinite value lives entirely in outcomes so rare and so large that no house could pay them, so the cap is not a detail; it is the price. Raising the cap barely moves it: a cap of Rs 1 crore makes the game worth only about Rs 24.19, because each doubling of the cap adds one rupee. That is the market maker's answer to the paradox: price what the counterparty can actually pay.
The relationship2^n the payout if the first tail comes on flip n 1/2^n the chance the first tail comes on flip n 1/1024 the chance the first ten flips are all heads What it says in wordsTen rupees from the flips the cap does not touch, and one more from the capped tail.Mention the other classic resolution, then put it in its place. Economists answer the paradox with diminishing utility: a doubling of wealth is worth less to you than the first rupee, so a risk-averse player pays little. True, but on a desk the binding constraint is usually the counterparty's balance sheet, not your utility. The limitation of the Rs 11 is the same as any fair value: it is an average over many plays. Played once, the price feels steep because 87.5% of the time you get Rs 8 or less.
Where candidates lose it
The common loss is computing an infinite expected value and stopping there, or saying the cap makes the game worth Rs 1,024. The cap removes the infinity but adds only one rupee for the capped tail.
The second loss is getting the boundary wrong: counting 11 flips below the cap, or forgetting the capped tail entirely and saying Rs 10. Write the n = 10 payout, Rs 1,024, next to the cap and the count is obvious.
What the interviewer asks next
- What is the fair price if the house can pay at most Rs 1 crore?
- What is the median payout of the uncapped game?
- Make me a market on this capped game, and say which side you would rather be on if the house's credit were in doubt.