Derivatives Foundation puzzles, solved step by step

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

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004Two stocks each have 30% annual volatility and a correlation of 0.5. What is the volatility of a basket that holds half of each? What if the correlation were zero?Volatility and correlationWarm upEquity derivativesRisk management
Try it first
Before the formula: can a 50/50 basket of two 30% stocks ever be more volatile than 30%?
Show the worked solution
25.98% at a correlation of 0.5, and 21.21% at zero. Basket variance is the sum of the two weighted variances plus twice the weighted covariance: 0.25 x 0.09 + 0.25 x 0.09 + 2 x 0.25 x 0.5 x 0.09 = 0.0675, whose square root is 25.98%. At zero correlation the cross term vanishes, leaving 0.045, whose root is 21.21%. The basket is less volatile than either stock because they do not move in step.
Why is the basket calmer than the stocks inside it?
Two commuters who each arrive late by a random ten minutes rarely arrive late together; the average of their lateness swings less than either does alone. Volatilities do not add; variances do, and the cross term that joins them is scaled by the correlation, so anything below a correlation of 1 cuts the basket's swing below its parts. At a correlation of 1 the two stocks are one stock and you get 30% back. At minus 1 they cancel exactly and the basket is flat.
Basket volatility rises with correlation from 0% at minus 1 through 21.21% at zero and 25.98% at 0.5 to the parts' 30% only at a correlation of 1, so the gap below 30% is the diversification and it exists only because the stocks are imperfectly correlated.
The relationship
$\sigma_B^2 = w^2\sigma^2 + w^2\sigma^2 + 2w^2\rho\,\sigma^2 = 2 \times 0.25 \times 0.09\,(1+\rho) \quad\Rightarrow\quad \sigma_B = 0.30\sqrt{\tfrac{1+\rho}{2}}$
w the weight of each stock, 0.5
sigma each stock's volatility, 0.30
rho the correlation between the two stocks
sigma_B the basket's volatility
What it says in wordsWith equal weights and equal volatilities, the basket's volatility is the single-stock volatility times the square root of (1 plus rho) over 2.
How do you do it in your head?
Use the shortcut in the formula: with two equal stocks the basket volatility is 30% times the square root of (1 plus rho) over 2. At rho 0.5 that is 30% times the root of 0.75, about 0.866, giving 26.0%; at rho 0 it is 30% times the root of 0.5, about 0.707, giving 21.2%. Say the structure first, then the number, so a slip in the arithmetic does not look like a slip in the thinking.
What is the limitation you should name?
The formula treats correlation as a fixed number, and it is not. Correlations between stocks tend to rise in a sell-off, which is exactly when a basket holder wants the diversification, so the 25.98% is a fair-weather figure. On a derivatives desk that is why basket options and dispersion trades are priced with a correlation assumption that is marked, stressed and hedged rather than looked up once. Say that the answer depends on the correlation you assume, and that the assumption is the risk.
Where candidates lose it
The fast wrong answer is 30%, from averaging the two volatilities. Volatility is a square root, and square roots do not average. Add the variances and the covariance, then take the root.
The second loss is forgetting the factor of 2 on the cross term. With it, the correlation 0.5 answer is 25.98%; without it, you get 23.72% and an interviewer who knows the number immediately.
What the interviewer asks next
Three stocks at 30% volatility, all pairwise correlations 0.5, equal weights. What is the basket volatility?
As the number of equally correlated stocks grows large, where does the basket volatility settle, and why?
The basket option is quoted at 24% implied volatility. What correlation is the market pricing?
018A stock trades at Rs 1,000 and its options are priced at 16% implied volatility. You buy an at-the-money option and delta-hedge it every day. Roughly how large a daily move does the stock need to make for you to break even?Option pricing intuitionWarm upVolatility tradingMarket making
Try it first
Answer in your head before reading on: the breakeven daily move is about
Show the worked solution
About 1% a day, Rs 10. With roughly 256 trading days in a year and volatility growing with the square root of time, daily volatility is annual volatility divided by 16, so 16% a year is 1% a day. A delta-hedged long option earns half its gamma times the square of each day's move and pays theta every day; the two cancel when the move equals the implied daily move. Using 252 days gives 1.008%, still Rs 10.
Where does dividing by 16 come from?
Walk randomly on a straight road, one step forward or back each second, and after 100 seconds you are typically about 10 steps from where you began, not 100, because most steps undo each other. Price moves add up the same way. Volatility scales with the square root of time, and the square root of 256 trading days is 16, so an annual volatility divided by 16 is the standard deviation of one day's move. Traders call this the rule of 16. It makes 16% implied volatility the cleanest number on the screen: 1% a day, Rs 10 on this stock.
The relationship
$\sigma_{\text{day}} = \frac{\sigma_{\text{year}}}{\sqrt{256}} = \frac{16\%}{16} = 1\%,\qquad 1\% \times 1{,}000 = \text{Rs } 10$
sigma year the implied volatility, quoted per year
256 trading days in a year, rounded so the square root is a whole number
sigma day the standard deviation of one day's percentage move
What it says in wordsOne day's typical move is the annual volatility divided by the square root of the number of trading days.
Why does that daily move decide whether the hedged option makes money?
Once the delta is hedged, the option's daily P&L is two pieces. Gamma pays you half gamma times the square of the move, in either direction; theta charges you a fixed amount for the day passing. For a one-month at-the-money option here, gamma is 0.0087 per rupee and theta is 0.435 a day. A Rs 10 move earns 0.5 x 0.0087 x 100 = 0.435, exactly the theta, because the option's price was built so that theta pays for a move of one implied standard deviation. A flat day loses 0.435; a Rs 20 day makes 1.31.
A delta-hedged long at-the-money option loses its theta of 0.435 on a flat day, breaks even when the stock moves Rs 10 either way, and makes 1.31 on a Rs 20 move, because the gain grows with the square of the move while theta is fixed, and Rs 10 is 16% divided by 16.
What does the quick answer leave out?
Three things, and naming one earns the follow-up. First, the breakeven is on the average squared move, not the average move. If the stock's moves are normal with a standard deviation of Rs 10, its average absolute move is only about Rs 7.98, so a stock that typically moves Rs 8 a day is already moving enough to break even. Second, gamma changes as the stock drifts away from the strike and as expiry nears, so the Rs 10 holds for an at-the-money option on the day you measure it. Third, hedging once a day adds noise to the P&L even when realised volatility exactly matches implied. The rule of 16 is a desk shortcut, not a pricing model.
Where candidates lose it
The common slip is dividing 16% by the number of trading days, or by 365, and quoting a breakeven of a few paise. Volatility adds in squares, so time enters under a square root; dividing by days is the mistake the question is built to catch.
The second loss is quoting Rs 10 as an average move to expect every day. It is a standard deviation: plenty of days will move Rs 2 and a few will move Rs 25, and the hedged option breaks even only if the average of the squared moves matches 100.
What the interviewer asks next
The same stock's options are priced at 32% volatility. What is the breakeven move, and what is the theta in terms of gamma?
Over a week the stock moves 5, minus 12, 3, 15 and minus 9. Did a delta-hedged long option make or lose money, roughly?
Why might a trader quote 252 days rather than 256, and when does the difference matter?
020One glass holds 100 ml of wine and another holds 100 ml of water. You take a spoonful of wine, tip it into the water and stir. Then you take a spoonful of the mixture and tip it back into the wine glass. Is there now more wine in the water glass, or more water in the wine glass?Games and logicWarm upProp trading firms
Try it first
Decide before any arithmetic: after the two spoonfuls,
Show the worked solution
Exactly the same. Each glass ends with 100 ml, so whatever wine is missing from the wine glass has been replaced, millilitre for millilitre, by water, and the missing wine can only be in the water glass. With a 10 ml spoon and a thorough stir, the return spoon carries back 0.91 ml of wine and 9.09 ml of water, leaving 9.09 ml of water in the wine and 9.09 ml of wine in the water.
Why does the first spoon feel like it settles the question?
Because it is pure wine going one way and a diluted mixture coming back, so it feels as though more wine travelled. Think instead of two cricket teams of eleven who swap some players and still field eleven each. Each glass ends with exactly 100 ml, so every millilitre of wine that left the wine glass and did not come back has been replaced by a millilitre of water: the two foreign amounts must be equal. The number of team A players now in team B is the number of team B players now in team A, however the swaps were done.
With a 10 ml spoon, the wine glass goes from 100 ml of wine to 90 ml and then back to 100 ml holding 9.09 ml of water, while the water glass goes to 110 ml and back to 100 ml holding 9.09 ml of wine, so the two foreign amounts are equal.
What do the millilitres actually look like?
Take a 10 ml spoon. After the first transfer the water glass holds 100 ml of water and 10 ml of wine, 110 ml in all, so a stirred spoonful from it is 10/110 wine. The return spoon carries 0.91 ml of wine and 9.09 ml of water, so 9.09 ml of wine stays behind in the water glass and 9.09 ml of water arrives in the wine glass. The arithmetic confirms the argument, but the argument came first and did not need the spoon size, the stirring or any division.
The relationship
$\text{water in wine} = 10 \times \frac{100}{110} = 9.09,\qquad \text{wine in water} = 10 - 10 \times \frac{10}{110} = 9.09$
10 the spoon, in millilitres
100/110 the share of water in the stirred water glass after the first transfer
10/110 the share of wine in that glass
What it says in wordsThe water carried into the wine glass equals the wine left behind in the water glass, both 9.09 ml for a 10 ml spoon.
Why do the interviewer's variations not change the answer?
Interviewers vary the story: no stirring, five spoonfuls back and forth, a ladle instead of a spoon. As long as both glasses end at their starting volume, the answer is equal, because the argument uses only the totals. The limitation to say out loud: if the return spoon is a different size from the first, the glasses end at different volumes and the amounts differ, so check the volumes before using the shortcut. The desk lesson is the bookkeeper's: in a closed system, look at the totals before tracking every transfer, the same way a net position check catches a booking error faster than replaying every ticket.
Stage Wine glass Water glass
Start 100 wine 100 water
After spoon 1 90 wine 100 water + 10 wine
After spoon 2 90.91 wine + 9.09 water 90.91 water + 9.09 wine
Tracking a 10 ml spoon through both transfers leaves each glass at 100 ml with 9.09 ml of the other liquid, which is what the conservation argument predicted without any arithmetic.
Where candidates lose it
The common answer is more wine in the water, because the first spoon was undiluted. It anchors on one transfer and forgets that the second spoon also took some of that wine back.
The second loss is reaching the right answer by long arithmetic and then failing the follow-up, such as an unstirred glass or several transfers, because there was no argument underneath. Give the volume argument first and use the numbers only as a check.
What the interviewer asks next
The return spoon is 5 ml instead of 10 ml. Which glass now holds more of the other liquid, and by how much?
You repeat the two-spoon swap many times. What do both glasses converge to?
Where on a trading desk does checking a total first save you from tracking every transfer?
031A stock is at 1,000, volatility is 20% and rates are near zero. Estimate the three-month at-the-money call in your head, and the straddle.Option pricing intuitionWarm upMarket makingVolatility trading
Try it first
Say the call price before you reach for a formula.
Show the worked solution
Call about 40, straddle about 80. The rule is 0.4 x S x sigma x sqrt T. Three months is a quarter of a year, so sqrt T is 0.5 and the volatility over the period is 10%; 0.4 x 1,000 x 0.1 = 40. With rates at zero the at-the-money put is worth the same, so the straddle is 80. The full model gives 39.88 for the call, because the exact constant is 1 over sqrt(2 pi), 0.3989, not 0.4.
Where does the 0.4 come from?
A tailor who knows a customer's height is normally distributed around 170 cm with a spread of 10 cm can say how far above 170 the average tall customer stands: about 0.4 of the spread, which is 4 cm, because the mean of the positive half of a normal is sigma over sqrt(2 pi). An at-the-money call pays the positive half of the stock's move, and the average of the positive half of a normal is 0.4 of its standard deviation, so the call is worth 0.4 times the standard deviation of the move over its life. The standard deviation of the move is S x sigma x sqrt T, which is 1,000 x 0.2 x 0.5 = 100 here, and 0.4 of 100 is 40. The exact constant is 1 / sqrt(2 pi) = 0.3989, so the rule gives 40 where the precise version gives 39.89, and the model 39.88.
The relationship
$C_{ATM} \approx 0.4\, S\, \sigma \sqrt{T} = 0.4 \times 1000 \times 0.20 \times 0.5 = 40 \qquad C_{exact} = S\left(2N\!\left(\tfrac{\sigma\sqrt{T}}{2}\right) - 1\right)$
S the stock price, 1,000
sigma sqrt T the volatility scaled to the option's life: 20% x 0.5 = 10%
0.4 the approximation to 1 over root 2 pi, which is 0.3989
N the standard normal distribution function
What it says in wordsAn at-the-money call is about four tenths of one standard deviation of the stock's move over its life.
Across volatilities from 0 to 100% the rule 0.4 x S x sigma x sqrt T sits almost on top of the full model price for a three-month at-the-money call, giving 40 against 39.88 at 20% volatility, and bends below it only at high volatility where the model's price curves.
Why is the straddle just double, and when is it not?
With rates at zero and no dividends, the forward equals the spot, so an at-the-money call and put have the same value by put-call parity, and the straddle is simply two calls, about 80. The straddle costs 8% of the stock for a three-month bet, which is the whole quarter's one-standard-deviation move of 10% times 0.8: that is the number a volatility trader carries in their head. With rates or dividends the forward moves away from spot, at-the-money means at-the-forward, and the call and put split the straddle unevenly, though their sum barely changes. The rule also assumes the volatility is the right one for this strike, which on a real surface with skew it may not be.
How far can you push the rule?
The rule is linear in volatility and time, the model is not. For three months the gap is 0.12 on 40 at 20% volatility, and even at 60% volatility the rule gives 120 against 119.2. Over one year at 20% the rule gives 80 against the model's 79.66. Up to a total move of about 50% the rule is within a couple of percent, which is every interview case and most of the real book; beyond that the model price flattens because a call can never be worth more than the stock. Say the limitation and then use the rule anyway: on a desk the question is never whether 40 is exactly right, it is whether 44 on the screen is rich or cheap.
Where candidates lose it
The common loss is forgetting to scale the volatility to the horizon and quoting 0.4 x 1,000 x 0.2 = 80 for the call, which is the one-year number. Say sqrt T out loud: three months is a half.
The second is giving the call and then stalling on the straddle, or doubling the call without saying why. The put equals the call only because rates are zero and there is no dividend; name parity and the interviewer knows you understand what at the money means.
What the interviewer asks next
Rates are now 8%. Which is worth more at the money, the call or the put, and by roughly how much?
The stock pays a 2% dividend before expiry. What changes?
Quote me the one-month straddle on the same stock, then the one-year.
The market is paying 44 for the call. What volatility is it implying, roughly?
035There are 100 coins on the table. Players take turns removing 1 to 10 coins, and whoever takes the last coin wins. Do you want to go first, and what is your first move?Games and logicWarm upQuant trading
Try it first
Go first or second, and what is the opening?
Show the worked solution
Go first and take 1, leaving 99. Work backwards: whoever faces 11 coins loses, because any take of 1 to 10 leaves 1 to 10 for the other player to finish. The same holds for 22, 33 and every multiple of 11. From 100, taking 1 leaves 99, a multiple of 11; after that, whatever the opponent takes, you take 11 minus it, stepping down 88, 77, 66 and so on to 0, where you take the last coin.
Why work backwards from the last coin?
If you are climbing stairs with a friend and the rule is that the person who steps onto the top stair wins, you do not plan from the bottom; you ask which stair you must leave your friend on so that they cannot reach the top in one go. Games with a fixed last move are solved from the end: find the positions where the player to move loses, then find the positions from which you can push your opponent onto one of them. With 1 to 10 coins allowed, facing 1 to 10 coins is a win, you take them all. Facing 11 is a loss, because every move leaves between 1 and 10. Facing 12 to 21 is a win, since you can reduce to 11. Facing 22 is a loss again. The losing positions repeat every 11.
Every multiple of 11 from 0 to 99 is a losing position for the player who must move, so the first player takes 1 to leave 99 and then answers every take of t with 11 minus t, stepping down through 88, 77 and 66 until the last coin.
How do you find the period without listing every position?
The period is the largest take plus one, 11, because that is the one total a pair of moves can always be made to add up to: whatever your opponent takes between 1 and 10, you can take the balance of 11. The losing positions are the multiples of the largest take plus one, and the winning opening move is the remainder when the pile is divided by that number. 100 divided by 11 is 9 remainder 1, so take 1. If the rule allowed 1 to 7 coins, the period would be 8 and the opening would be 100 mod 8, which is 4. If the pile had been 99 to start with, you would want to go second, because the first player cannot leave a multiple of 11.
The relationship
$\text{losing positions} = \{0, 11, 22, \ldots, 99\}, \qquad \text{opening} = 100 \bmod 11 = 1$
11 the largest allowed take plus one, the amount you can always complete in a pair of moves
100 mod 11 the remainder when 100 is divided by 11; take exactly this many
What it says in wordsTake the remainder on your first move, then keep each pair of moves summing to 11.
What changes if the last coin loses instead of wins?
Then you want to hand your opponent the last coin, so the position you avoid facing is 1 coin, and the losing positions shift up by one: 1, 12, 23 and so on up to 100. Facing 100 in that version you are already lost, so you would want to go second, which shows the interviewer that you re-derive the pattern rather than remember it. The method is the same in every variant: name the terminal position, step back one move at a time to find the first losing position, then find the period. The limitation of the trick is that it needs a game with perfect information and no chance; add a die that sets each turn's maximum and the clean period disappears.
Where candidates lose it
The common loss is taking 10, because a bigger move feels like a stronger start. It leaves 90, which is not a multiple of 11, and a prepared opponent takes 2 to leave 88 and wins from there.
The second is knowing the answer and not the reason. Say why 11 is the period: any take of 1 to 10 can be completed to 11. Without that sentence the interviewer will change the numbers and watch you stall.
What the interviewer asks next
Players may take 1 to 7 coins instead. Do you go first, and what is the opening?
The player who takes the last coin loses. Do you go first?
There are two piles, 100 and 60, and you may take from either pile. Who wins?
Each turn a die sets the maximum take. Is there still a strategy, and what is it?
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.

w	the weight of each stock, 0.5
sigma	each stock's volatility, 0.30
rho	the correlation between the two stocks
sigma_B	the basket's volatility

sigma year	the implied volatility, quoted per year
256	trading days in a year, rounded so the square root is a whole number
sigma day	the standard deviation of one day's percentage move

10	the spoon, in millilitres
100/110	the share of water in the stirred water glass after the first transfer
10/110	the share of wine in that glass

Stage	Wine glass	Water glass
Start	100 wine	100 water
After spoon 1	90 wine	100 water + 10 wine
After spoon 2	90.91 wine + 9.09 water	90.91 water + 9.09 wine

Try it first

Why is the basket calmer than the stocks inside it?

How do you do it in your head?

What is the limitation you should name?

Where candidates lose it

What the interviewer asks next

Try it first

Where does dividing by 16 come from?

Why does that daily move decide whether the hedged option makes money?

What does the quick answer leave out?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the first spoon feel like it settles the question?

What do the millilitres actually look like?

Why do the interviewer's variations not change the answer?

Where candidates lose it

What the interviewer asks next

Try it first

Where does the 0.4 come from?

Why is the straddle just double, and when is it not?

How far can you push the rule?

Where candidates lose it

What the interviewer asks next

Try it first

Why work backwards from the last coin?

How do you find the period without listing every position?

What changes if the last coin loses instead of wins?

Where candidates lose it

What the interviewer asks next

Try it first

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?

Where candidates lose it

What the interviewer asks next