Derivatives Foundation puzzles, solved step by step

Puzzles: 100
Traced to a firm: 66
Topics: 12
Hard: 29

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010In some stock, the 99-strike call trades at 5.60 and the 101-strike call at 4.70, same expiry. Estimate the price of a digital option that pays 1 if the stock finishes above 100 at that expiry.Option payoffs and no-arbitrageCoreExotics tradingStructured products
Try it first
Before any arithmetic: which combination of the two calls has a payoff that looks most like a step at 100?
Show the worked solution
About 0.45. Buying the 99 call and selling the 101 call pays 0 below 99, 2 above 101 and a straight ramp between. Divide that by the width of 2 and the payoff is 0 below 99, 1 above 101 and a ramp through 100: a digital with its edge smoothed over two points. Its cost is (5.60 minus 4.70) over 2, which is 0.45. The narrower the spread, the closer the ramp sits to the step, and the price converges to the digital.
Why does a call spread stand in for a digital?
A light switch is a step: off or on. A dimmer that goes from fully off to fully on over a tiny turn of the knob is, for every practical purpose, the same switch. A call spread over its width is a dimmer: it ramps from 0 to 1 across the two strikes, and as the strikes close in on 100 the ramp becomes the step. So the digital is the limit of a scaled call spread, and a traded call spread gives you a price for it without any model.
The 99 to 101 call spread divided by its width of 2 pays 0 below 99, ramps to 1 at 101 and crosses the digital's step exactly at 100, so the two payoffs differ only inside the narrow band between the strikes and the spread's price, (5.60 minus 4.70) over 2, gives a digital value of 0.45.
The relationship
$D(100) \approx \frac{C(99) - C(101)}{101 - 99} = \frac{5.60 - 4.70}{2} = 0.45 \qquad D(K) = -\frac{\partial C}{\partial K}$
C(K) the price of a call struck at K
D(K) the price of a digital paying 1 above K
(C(99) - C(101)) / 2 the slope of the call price in strike, estimated across 100
What it says in wordsThe digital is minus the slope of the call price with respect to strike, and a centred call spread measures that slope.
Is 0.45 the digital's price or an approximation, and which way is it off?
It is an approximation to the slope at 100 taken from two points either side. Because the spread is centred on 100, the first-order error cancels and what remains is small, of the order of the curvature of the call price between 99 and 101. If the digital were struck at 99 instead, the same spread would overstate it, because the call price is convex in strike and the ramp sits above the step on that side. On a desk you would quote the digital from the tightest spread the market will show you, and hedge it with that spread, so the approximation is also the hedge.
What does 0.45 say about the market, and what is the limitation?
A digital paying 1 above 100 at 0.45, with rates near zero, means the pricing probability of finishing above 100 is about 45%, slightly below one half. That is a risk-neutral probability, not a forecast, and it is pulled down by the skew: with a steeper put skew, out-of-the-money calls are cheaper in volatility terms and the slope in strike is steeper, which moves the digital. Say that a flat-volatility formula would miss this, and that the call spread picks the skew up automatically because it uses the two traded prices.
Where candidates lose it
The common error is to take the difference of the two call prices, 0.90, and present it as the digital. That is the price of a spread that pays 2 above 101, not 1. Divide by the width.
The second loss is reaching for a lognormal formula with a guessed volatility. The question gives you two traded prices precisely so you can price the digital without a model; use them.
What the interviewer asks next
The 99.5 and 100.5 calls are 5.37 and 4.93. What does that pair say about the digital, and why might it differ from 0.45?
How would you hedge a short digital you sold at 0.45, and what goes wrong near expiry?
Price a digital that pays 1 if the stock finishes below 100.
025A stock trades at 500. The one-year 500-strike call is priced at 40 and the 500-strike put at 45, and interest rates are zero. What dividend is the options market pricing in, and what would you check before trading on it?Option payoffs and no-arbitrageCoreEquity derivativesMarket making
Try it first
Before using any formula: the put costs more than the call at the same at-the-money strike. With zero rates, that tells you
Show the worked solution
A dividend of 5 per share. With zero rates, put-call parity says call minus put equals spot minus the dividend minus the strike. Here 40 minus 45 = 500 minus D minus 500, so D = 5, and the implied forward is 495. Before trading against it, check that the options are European or that early exercise is worth nothing, what it costs to borrow the stock, whether the ex-date falls before expiry, and whether all four prices are live and tradable.
Why does a put costing more than a call point to a dividend?
Suppose you agree today to buy a friend's scooter in a year for its fair price, but the friend will keep using it to deliver parcels and pocket the fees until then. You would pay less than today's price, by the fees they collect. Holding a call and selling a put at the same strike is an agreement to buy the stock at the strike, so call minus put must equal the forward price minus the strike, and the forward is spot minus whatever the holder collects before expiry. With zero rates there is no interest to account for, so a put dearer than the call by 5 says the forward is 5 below spot.
The relationship
$C - P = S - D - K \;\Rightarrow\; 40 - 45 = 500 - D - 500 \;\Rightarrow\; D = 5$
C, P the call and put prices at the same strike and expiry
S the spot price, 500
K the strike, 500
D the cash the stock pays out before expiry, here the implied dividend
What it says in wordsWith zero rates, the call minus the put equals the spot less the dividend less the strike, so the dividend is what balances the equation.
Call minus put is minus 5, so the options imply a forward of 495, and since the stock's forward is 500 minus the dividend, the market is pricing a dividend of 5; if the company pays only 3, the remaining 2 has to be explained by something else, such as the cost of borrowing the stock.
What would you check before trading on it?
Say your own estimate of the dividend is 3. Then a reversal looks attractive: short the stock, buy the call, sell the put, collecting 500 + 45 minus 40 = 505; at expiry the options deliver the stock back at 500 and you owe the 3 dividend, 503 in all, a locked-in 2 per share. That 2 is only yours if you can borrow the stock for less than 0.4% of its price over the year; a hard-to-borrow stock shows up in parity exactly as an extra dividend. Then check the rest: American calls can be exercised just before a dividend, the ex-date must fall before expiry, and four bid-offer spreads can eat a gap of 2 on their own.
What is the general lesson the interviewer wants?
That parity is an accounting identity between three things you can trade, and any gap between what it implies and what you believe is a claim about something you have not yet priced. The options market does not quote a dividend; it quotes a forward, and the dividend, the borrow cost and the interest rate are the pieces you split it into. A good answer gives the 5, names the implied forward of 495, and then lists what could make the 5 something other than a dividend. The limitation to say plainly: with non-zero rates, the strike is discounted and the arithmetic shifts, so state the zero-rate assumption before quoting the number.
Where candidates lose it
The common answer is that the put is dearer because the market expects the stock to fall. At one strike and one expiry, direction cannot make the put dearer than parity allows; anyone could sell the put, buy the call and short the stock against it. The gap is a forward, not a view.
The second loss is stopping at 5. The question asks what you would check, and the borrow cost is the one interviewers wait for: an implied dividend above the announced one is often a stock that is expensive to short.
What the interviewer asks next
Interest rates are now 6% a year. Redo the implied dividend, and say which way it moves.
The company announces a dividend of 8. Which trade would you put on with these four prices, and what is the risk?
Why might an American call on this stock be worth more than its European twin, and how would that distort the implied dividend?
044Using only calls, build a payoff that is zero below 90, rises one for one to 10 at 100, falls back to zero at 110, and stays at zero above that.Option payoffs and no-arbitrageCoreEquity derivativesStructured products
Try it first
Which call portfolio gives the tent?
Show the worked solution
Long one 90 call, short two 100 calls, long one 110 call: a call butterfly. Read the slope of the target from left to right: 0, +1, -1, 0. A long call adds +1 to the slope at its strike, so the changes of +1 at 90, -2 at 100 and +1 at 110 give the weights. Check the corners: 0 at 90, 10 at 100, 0 at 110 and above. At 20% volatility and three months it costs about 3.69.
How do you read a payoff picture as a list of calls?
A road that is flat, then climbs, then drops, then is flat again can be described by where the gradient changes and by how much. A payoff made of straight pieces is the same: each call adds one unit of slope from its strike onwards, so the number of calls at a strike is simply the change of slope at that strike. The tent has slope 0 below 90, +1 from 90 to 100, -1 from 100 to 110, and 0 above 110. The changes are +1 at 90, -2 at 100 and +1 at 110. So you buy one 90 call, sell two 100 calls and buy one 110 call. Above 110 the three legs pay (S - 90) - 2(S - 100) + (S - 110) = 0, which confirms the payoff returns to zero and stays there.
The long 90 call, the two short 100 calls and the long 110 call add up to a tent that is zero below 90, peaks at 10 at 100 and returns to zero from 110 onwards, because the slope changes by +1, -2 and +1 at the three strikes.
The relationship
$\text{Payoff}(S) = (S - 90)^+ - 2(S - 100)^+ + (S - 110)^+, \qquad \text{weight at } K = \text{slope just above } K - \text{slope just below } K$
(S - K)+ the payoff of a call struck at K, the larger of S - K and zero
weight at K the number of calls to hold at that strike; negative means sell
S the stock price at expiry
What it says in wordsThe weight on each strike is the jump in slope there, which is how any straight-line payoff is built from calls.
What does the butterfly cost, and what does its price tell you?
Take an illustrative stock at 100 with 20% volatility and three months to expiry. The calls cost 10.71, 3.99 and 0.95, so the butterfly costs 10.71 - 2 x 3.99 + 0.95 = 3.69. The payoff is a tent of height 10 and base 20, and its price is close to the chance of finishing near 100 times the peak. Divide the cost by the peak payoff of 10 and you get 0.369, close to the 0.383 risk-neutral chance that the stock ends between 95 and 105, which is why a butterfly is the market's way of pricing the probability of a narrow range. Shrink the strike gap towards zero and the scaled butterfly becomes the risk-neutral density itself, a result traders use to read the distribution off a strip of call prices.
Where does the no-arbitrage check come in?
The tent never pays less than zero, so it can never cost less than zero. That means C(90) - 2C(100) + C(110) must be at least 0: call prices must be convex in strike. Suppose a screen shows the 90 call at 11.00, the 100 call at 7.00 and the 110 call at 2.50. The butterfly costs 11.00 - 14.00 + 2.50 = -0.50: you are paid 0.50 to hold a payoff that is never below zero. A negative butterfly price is a free lunch, and spotting it on a quote sheet is the reason the question is asked. The limitation in practice is that each leg has a bid and an offer, so the check must use the prices you can actually trade at, buying at offers and selling at bids, and small apparent violations usually vanish once the spread is paid.
Where candidates lose it
The common loss is short one 100 call instead of two. One short call only cancels the slope of the 90 call, which flattens the payoff at 10 forever; it takes two to turn the slope down to -1 and bring it back to zero.
The second is building the tent and stopping. The interviewer usually follows with the price: a butterfly must cost more than zero because it never pays less than zero, and a candidate who connects that to convexity in strike has answered the real question.
What the interviewer asks next
Build the same tent with puts only. Is the cost the same?
Build a payoff that is 0 below 90, rises to 10 at 100 and stays at 10 above.
The 90, 100 and 110 calls trade at 11.00, 7.00 and 2.50. What do you do?
As the gap between the strikes shrinks, what does the scaled butterfly price approach?
050You can buy the 90 call, sell the 110 call, buy the 110 put and sell the 90 put, all European and expiring in one year, for a net 19.20. What does the position pay at expiry, what interest rate does it imply, and when is that attractive?Option payoffs and no-arbitrageHardMarket makingRates derivatives
Try it first
What does the four-leg position pay at expiry?
Show the worked solution
It pays exactly 20 at any stock price, so 19.20 today implies a one-year rate of 20/19.20 - 1, about 4.17%. The 90 call and short 90 put make a long forward at 90; the short 110 call and long 110 put make a short forward at 110. The stock cancels and 110 - 90 = 20 is left. Buying the box is lending at 4.17% and selling it is borrowing at that rate, so it suits a lender whose other return is lower, or a borrower whose funding costs more.
Why does the payoff not depend on the stock?
Agree to buy a scooter from one friend for 90 and to sell it to another for 110, both next year, and you will make 20 whatever scooters cost by then. The box is the same pair of agreements: long the 90 call and short the 90 put is a promise to buy at 90, and short the 110 call and long the 110 put is a promise to sell at 110, so the stock comes in and goes out and only the gap between the strikes, 20, is left. Seen as spreads, the call spread pays from 0 below 90 up to 20 above 110, and the put spread pays the mirror image, 20 below 90 down to 0 above 110. Wherever the stock ends, the two add to 20.
The long call spread rises from 0 to 20 between 90 and 110 while the long put spread falls from 20 to 0 over the same range, so their sum is a flat 20 at every stock price, and paying 19.20 for it is lending at 4.17%.
How do you turn the price into a rate?
You pay 19.20 today and receive 20.00 in a year with no market risk, so it is a deposit. The simple rate is 20 / 19.20 - 1 = 4.17%, and the continuously compounded rate is ln(20 / 19.20) = 4.08%. A box is a zero-coupon bond built from options, and its price is the strike gap times the discount factor, so any box that trades away from the market's interest rate is mispriced. Prices consistent with this rate, for an illustrative stock at 100 with 20% volatility, are a 90 call at 16.11, a 110 call at 5.69, a 110 put at 11.29 and a 90 put at 2.51: 16.11 - 5.69 + 11.29 - 2.51 = 19.20. Volatility does not enter the box price at all, because every volatility effect in the calls is cancelled by the puts.
The relationship
$\underbrace{(S - 90)^+ - (90 - S)^+}_{S - 90} \; - \; \underbrace{\big[(S - 110)^+ - (110 - S)^+\big]}_{S - 110} = 20, \qquad r = \frac{20}{19.20} - 1 = 4.17\%$
(S - K)+ a call payoff struck at K
(K - S)+ a put payoff struck at K
call minus put at the same strike a forward to buy at that strike, paying S - K
r the simple one-year rate implied by paying 19.20 for 20
What it says in wordsA long forward at 90 and a short forward at 110 leave a fixed 20, and its price gives the interest rate.
When is it attractive, and what can go wrong?
Compare 4.17% with your own rates. If cash would otherwise earn 3.5%, buying the box lends at a better rate: 19.20 at 3.5% grows to only 19.87, against 20 from the box. If your funding costs 5%, selling the box borrows more cheaply: you receive 19.20 today and owe 20, while 20 owed at 5% would have raised only 19.05. Whether a box is cheap or dear is never a property of the box alone; it depends on the rate you can otherwise lend or borrow at, which is why boxes are a funding trade. The risks to name are practical. American options can be exercised early, which breaks the box, so use European index options. Four bid-offer spreads and fees can eat the 0.80 of interest. And the counterparty, or the clearing house, must still be there in a year.
Where candidates lose it
The common loss is analysing the four legs as a view on the stock, describing a bull call spread and a bear put spread and forgetting to add them. Add them: the stock cancels and the position is a loan.
The second is calling a box at 19.20 an arbitrage on its own. It is only cheap or dear against a rate, so say which rate you are comparing with, your deposit rate if you buy and your funding rate if you sell.
What the interviewer asks next
The same box trades at 19.80. What rate does that imply, and who would sell it?
Why does an American-style box carry early-exercise risk, and which leg is the danger?
Build a box with strikes 95 and 105. What should it cost at the same rate?
How is a box related to put-call parity at each strike?
059On a stock that pays no dividend, the three-month 100-strike call trades at 6.50 and the six-month 100-strike call trades at 6.10. Is there an arbitrage, and how would you capture it?Option payoffs and no-arbitrageCoreMarket makingVolatility trading
Try it first
Two calls, same strike, the longer one cheaper. What do you do?
Show the worked solution
Yes. Sell the three-month call at 6.50, buy the six-month call at 6.10, and lock in 0.40 today with no risk. On a non-dividend stock, a call with more time to expiry is worth at least as much as the same strike with less, because at three months the six-month call is still worth at least its intrinsic value, which is exactly what the expiring call pays out. The position never owes money, so the 0.40 is a pure profit.
Why must the longer call be worth at least the shorter one?
A ticket that lets you buy a train seat any time in the next six months is worth at least as much as one that expires in three; you can always do with the long ticket exactly what you would have done with the short one, and then keep it. At the three-month expiry the short call pays max(S - 100, 0), and the six-month call at that moment is worth at least max(S - 100, 0) too, because a call on a non-dividend stock is never worth less than its intrinsic value and is usually worth more. So holding the long and being short the short can never leave you with a negative balance at three months, and the 0.40 you took in today is yours.
At-the-money call value rises with time to expiry, so the six-month call must be worth at least the 6.50 the three-month call trades at; the quoted 6.10 sits below that floor, and selling the three-month while buying the six-month locks in 0.40 with a spread that can only pay more.
What happens at three months in each case?
Walk through both branches. If the stock is above 100, the short call is exercised against you, you deliver stock for 100, and you still hold a six-month call worth more than S - 100, so you exercise or sell it and finish ahead; if the stock is at or below 100, the short call expires worthless and you are left holding a live six-month call for free. Either way you keep the 0.40 from today and own something worth zero or more. The limitation is the non-dividend assumption: with a large dividend before six months, early exercise of an American three-month call could matter, and the comparison needs the dividend's present value subtracted.
The relationship
$C(K, T_2) \ge C(K, T_1) \quad \text{for } T_2 > T_1 \text{ on a non-dividend stock} \qquad 6.10 < 6.50 \Rightarrow \text{sell } T_1,\ \text{buy } T_2,\ \text{keep } 0.40$
C(K, T) the price of a call with strike K and time to expiry T
T_1, T_2 three and six months
0.40 the cash locked in today, the smallest profit the trade can make
What it says in wordsA longer-dated call at the same strike can never be worth less than a shorter one, so a longer call quoted cheaper is sold against the shorter one for a riskless profit.
Why would a market maker see this quote and what does it say about the vol surface?
The three-month price of 6.50 corresponds to an implied volatility of about 33% with zero rates, and at that volatility a six-month call would be worth around 9.18. A quote of 6.10 for the six-month implies a term structure of volatility so inverted that no volatility at all could justify it, which is why this is an arbitrage and not merely a view on calendar spreads. Real quotes rarely break the bound outright; what you see instead is a long-dated call a few ticks above the floor on an illiquid name, and the question is then whether the bid-ask spread swallows the edge before you can trade both legs.
Where candidates lose it
Candidates treat it as a volatility question and start talking about the term structure. Implied volatility cannot push a longer call below a shorter one at the same strike, so the answer is a bound, not a view.
The second loss is getting the direction wrong under pressure. Sell what is dear, the three-month at 6.50; buy what is cheap, the six-month at 6.10. Then walk the two branches at the first expiry out loud.
What the interviewer asks next
Does the same bound hold for puts on a non-dividend stock?
The stock pays a large dividend in month four. Can the three-month call now be worth more than the six-month?
What if the two calls had different strikes, say 100 and 105?
Both quotes are mid prices and the bid-ask is 0.30 on each. Is the arbitrage still there?
072A stock at 100 will be at 80, 100 or 130 at expiry, each equally likely in your view. What is the expected payoff of a 100-strike call and of a 100-strike put, and why is the expected payoff not what a market maker would charge?Option payoffs and no-arbitrageWarm upSell-side sales and trading
Try it first
Three equally likely outcomes. Expected call payoff and expected put payoff?
Show the worked solution
Expected payoffs are 10 for the call and 6.67 for the put, and neither is a price. The call pays 0, 0 and 30 across the three outcomes, averaging 10; the put pays 20, 0 and 0, averaging 6.67. A market maker charges the cost of hedging, and with zero rates call minus put must equal stock minus strike, which is 0, so prices of 10 and 6.67 would be an arbitrage against the maker. Hedge-implied odds that keep the stock worth 100 give both options the same price.
Why is the expected payoff under your odds not the price?
A shopkeeper who believes the monsoon will be good does not price umbrellas off that belief; the price is set by what the umbrellas cost to stock and what the shop next door charges. An option maker does not hold the option to expiry hoping for the payoff; they hedge it with the stock, so the price is the cost of that hedge, and the cost of the hedge depends on the stock's current price of 100, not on anyone's forecast of where it goes. Your equal odds imply the stock is expected to be worth 103.33, above today's 100; that optimism is yours to trade, not something the maker will pay you for inside an option price.
Under equal odds on 80, 100 and 130 the call's expected payoff is 10 and the put's is 6.67, but with zero rates call minus put must equal stock minus strike, which is zero, so those two numbers cannot both be prices; hedge-implied odds that keep the stock worth 100, such as 0.40, 0.33 and 0.27, price both options at 8.
What is the one-line test that catches the mistake?
Put-call parity. Buying the call and selling the put with the same strike gives you the stock minus 100 in every outcome, which with zero rates is worth 100 - 100 = 0 today, so the call and the put must have the same price. Expected payoffs of 10 and 6.67 fail that test by 3.33, and a maker who quoted them would be lifted on the put and hit on the call until the prices met. The test needs no probabilities at all, which is the point: parity is enforced by hedging, not by views.
The relationship
$C - P = S - K\,e^{-rT} = 100 - 100 = 0 \qquad \text{but } E[C] - E[P] = 10 - 6.67 = 3.33 = E[S] - K$
C, P the prices of the 100-strike call and put
S - K stock minus strike, the value of a long call and short put in every state
E[S] - K the drift your odds put on the stock, 103.33 - 100
What it says in wordsPrices must satisfy parity, and the gap between the two expected payoffs is exactly the drift your personal odds assign to the stock.
What odds would a maker use, and are they unique here?
Odds that make the stock worth its forward, 100 with zero rates. Any set of probabilities on 80, 100 and 130 with an average of 100 prices the two options consistently; one such set is 0.40, 0.33 and 0.27, which gives the call 0.267 x 30 = 8 and the put 0.40 x 20 = 8, equal as parity demands. The limitation is that with three outcomes and only the stock to hedge with, the set is not unique: the common level of the call and put price is pinned down by parity only up to a range, and in practice it is the market's volatility quote that chooses the point inside it.
Where candidates lose it
The arithmetic is easy and candidates get 10 and 6.67 quickly; the loss comes in the second half, where they say a market maker would add a spread to the expected payoff. The real answer is that the probabilities themselves are wrong for pricing, because the price is the cost of a hedge.
Say parity: call minus put equals stock minus strike, zero here, so the two prices must be equal. That one sentence shows you know why risk-neutral pricing exists.
What the interviewer asks next
Find a set of probabilities on 80, 100 and 130 under which the stock is worth 100, and price both options.
Interest is 5% for the period. What does parity say the call minus the put is worth now?
Why is the set of hedge-implied probabilities not unique with three outcomes and one stock?
The stock will be at 80 or 130 only. Price the call by replication.
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.Option payoffs and no-arbitrageHardMarket 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 relationship
$C(100) \le \tfrac{1}{2}\,[C(90) + C(110)] = \tfrac{1}{2}(14 + 1) = 7.5 < 8$
C(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?

C(K)	the price of a call struck at K
D(K)	the price of a digital paying 1 above K
(C(99) - C(101)) / 2	the slope of the call price in strike, estimated across 100

C, P	the call and put prices at the same strike and expiry
S	the spot price, 500
K	the strike, 500
D	the cash the stock pays out before expiry, here the implied dividend

(S - K)+	the payoff of a call struck at K, the larger of S - K and zero
weight at K	the number of calls to hold at that strike; negative means sell
S	the stock price at expiry

(S - K)+	a call payoff struck at K
(K - S)+	a put payoff struck at K
call minus put at the same strike	a forward to buy at that strike, paying S - K
r	the simple one-year rate implied by paying 19.20 for 20

Try it first

Why does a call spread stand in for a digital?

Is 0.45 the digital's price or an approximation, and which way is it off?

What does 0.45 say about the market, and what is the limitation?

Where candidates lose it

What the interviewer asks next

Try it first

Why does a put costing more than a call point to a dividend?

What would you check before trading on it?

What is the general lesson the interviewer wants?

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How do you read a payoff picture as a list of calls?

What does the butterfly cost, and what does its price tell you?

Where does the no-arbitrage check come in?

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Why does the payoff not depend on the stock?

How do you turn the price into a rate?

When is it attractive, and what can go wrong?

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Why must the longer call be worth at least the shorter one?

What happens at three months in each case?

Why would a market maker see this quote and what does it say about the vol surface?

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Why is the expected payoff under your odds not the price?

What is the one-line test that catches the mistake?

What odds would a maker use, and are they unique here?

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Where does the 7.5 come from?

How do you prove the payoff is never negative?

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