Hedge Funds puzzles, solved step by step

Puzzles: 100
Traced to a firm: 38
Topics: 14
Hard: 30

Showing 11–20 of 100

011An ant starts at one corner of a cube and at each step walks along an edge to a randomly chosen neighbouring corner. What is the expected number of steps until it first reaches the opposite corner?Random walks and Markov chainsCoreQuant and systematic fundsProp and quant trading firms
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Pick your estimate before you set anything up.
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10 steps. Group the eight corners by how many edges separate them from the target: one corner at distance 3, three at 2, three at 1, and the target itself. From 3 the ant must move to 2; from 2 it slips back to 3 one time in three; from 1 it reaches the target one time in three. Solving the three equations gives 10 from the start, 9 from distance 2 and 7 from distance 1.
How do you turn eight corners into four states?
Ask a lost tourist how far they are from the station, not which street they are on. Every corner at the same distance from the target behaves the same way, so the distance is all you need to track. Collapsing the cube by distance turns eight corners into four states, 3, 2, 1 and 0, and the walk becomes a short chain. From distance 3 all three neighbours are at distance 2. From 2, one neighbour is back at 3 and two are at 1. From 1, two neighbours are at 2 and one is the target.
Grouped by distance from the target, the ant moves from 3 to 2 for certain, from 2 forward with probability two thirds, and from 1 home with probability one third, which gives expected times of 10, 9 and 7 steps from distances 3, 2 and 1.
The relationship
$E_3 = 1 + E_2 \qquad E_2 = 1 + \tfrac{1}{3}E_3 + \tfrac{2}{3}E_1 \qquad E_1 = 1 + \tfrac{2}{3}E_2 + \tfrac{1}{3}\cdot 0$
E_k the expected steps still needed from a corner at distance k
1 the step being taken now
What it says in wordsEach expected time is one step plus the average of the expected times from wherever that step lands.
How do you solve the equations quickly?
Substitute from the two ends into the middle. Put E3 = 1 + E2 and E1 = 1 + 2E2/3 into the middle equation and it collapses to E2 = 2 + 7E2/9, so E2 = 9, E3 = 10 and E1 = 7. Then sanity-check the odd-looking one: from distance 1 the ant sits right next to the target yet still needs seven steps on average, because two of its three moves lead away. A number that surprises you is worth one sentence of explanation, not a recalculation.
What is the skill a fund is actually testing?
The same first-passage logic tells you how long a mean-reverting spread takes to reach a target or a stop, and how many moves a process needs to hit a barrier. Whenever many states behave alike, lump them, write one equation per lumped state, and solve: that is the skill, and the cube is only the costume. Candidates who write eight equations, one per corner, get the right answer too, but too slowly for the room.
Where candidates lose it
Answering 3, the length of the shortest path, is the fast mistake. The ant does not know where it is going, and from every corner except the start it is at least as likely to wander as to advance.
The slower mistake is writing eight equations, one per corner. It works, but it takes far longer than the interview allows. Say the symmetry out loud first: corners at the same distance are interchangeable.
What the interviewer asks next
What is the expected number of steps for the ant to return to its starting corner?
What if the ant stays put with probability one half at each step?
On a square, what is the expected time to reach the opposite corner?
012A fund's NAV falls from 100 to 80 in year one and rises to 110 in year two. It charges a 20% performance fee above a high-water mark tracked separately for each investor. Investor A came in at 100 at the start of year one; investor B came in at 80 at the start of year two. On how much gain does each pay the fee in year two?Returns, compounding and feesCoreFund of funds and allocatorsMulti-manager platforms
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In year two, on how much gain per unit does investor A pay the fee?
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Investor A pays the fee on 10 per unit, a fee of 2; investor B pays it on 30, a fee of 6. Each high-water mark is the highest value that investor's own units have reached: 100 for A, who lived through the fall, and 80 for B, who bought at the bottom. The fund made 37.5% in year two for both of them, yet B pays three times A's fee, because every point B made is new profit for B.
Why does the same year produce two different fees?
Two shopkeepers pay a helper a bonus only when monthly sales beat their own best month so far. One had a record month last year and a slump since; the other opened last month. The same good month earns the helper a bonus from the new shop and little or nothing from the old one. A high-water mark is personal: it is the peak value of that investor's own units, so it depends on when they came in. A bought at 100 and watched it fall to 80, so the rise back to 100 only repairs A's loss. B bought at 80, so the whole rise to 110 is new money for B.
The fund falls from 100 to 80 and rises to 110; investor A's high-water mark of 100 means A is charged only on the 10 above it, a fee of 2, while investor B's mark of 80 means B is charged on the full 30, a fee of 6.
Investor Entry NAV High-water mark NAV, end of year 2 Gain charged Fee at 20% NAV after fee
A 100 100 110 10 2.0 108.0
B 80 80 110 30 6.0 104.0
Per unit, investor A is charged on 10 and keeps 108, while investor B is charged on 30 and keeps 104, although both held the same fund through the same year.
How do funds keep the two investors apart?
If the fund kept one NAV and one mark for everyone, either A would be charged on a recovery or B would ride free on 20 points of profit. Funds solve this with per-investor accounting: a separate series of units for each subscription date, or equalisation adjustments, so each investor pays on their own gain and nobody else's. Series are the easier version to explain in the room: B's units are a new series that starts life with a mark of 80. The limitation is worth a line: not every fund does this, so an allocator reads the offering document before assuming it.
What does this do to the manager's incentives?
A manager whose older investors sit below their marks earns no performance fee on them until the loss is repaid. For an allocator, a high-water mark is a fee holiday on the recovery, and it belongs only to the investors who stayed through the loss. For a manager deep under water it can mean years of work for no incentive fee, which is why some funds in that position close and relaunch rather than climb back to their marks, and why allocators ask about it.
Where candidates lose it
Candidates work out one fee for the whole fund, usually 20% of the year's 30 point gain, and apply it to everyone. That overcharges A by 4 per unit, which is precisely what per-investor marks exist to prevent.
The mirror mistake is giving B the benefit of A's mark and charging only the gain above 100. B never lost anything; every point from 80 to 110 is B's profit. The fee follows the investor, not the fund.
What the interviewer asks next
If the NAV had only reached 95 in year two, what would each investor pay?
How would a 5% hurdle rate change A's and B's fees?
Why might a manager well below the high-water mark close the fund and launch a new one?
013A researcher tests 20 unrelated trading signals, each at a 5% significance level, and none of them truly works. What is the chance that at least one of them looks significant?Statistics and estimationWarm upQuant and systematic funds
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Your instinct: the chance of at least one false discovery?
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About 64.2%. A useless signal clears a 5% bar by luck one time in twenty. The chance that all 20 stay insignificant is 0.95 to the twentieth, about 35.8%, so the chance that at least one looks like a discovery is 64.2%. On average the search turns up one false signal, as 20 x 5% suggests, but at least one appears in roughly two searches out of three.
Why does testing more ideas manufacture a winner?
Ask twenty friends to flip a coin five times each. Any one of them flips five heads only once in 32 tries, yet the chance that at least one of the twenty does is 47.0%, and that friend will look gifted. Each test is a lottery ticket for a false discovery, and buying twenty tickets makes a win likely even when nothing works. A signal chosen because it looked best among twenty has not passed a 5% test; it has passed a 64.2% one.
The chance of at least one false positive rises from 5% for one useless signal to 40.1% for ten and 64.2% for twenty, passing even odds at 14 tests, although every individual test is run at 5%.
The relationship
$P(\text{at least one false positive}) = 1 - (1-\alpha)^m = 1 - 0.95^{20} \approx 0.642$
\alpha the significance level of each test, 5%
m the number of independent tests, 20
What it says in wordsThe chance that every test stays quiet shrinks with each test added, so the chance of a false winner grows.
How do you correct for it?
Tighten the bar to match the number of tries. The Bonferroni correction tests each signal at 5% divided by 20, which is 0.25%, and that brings the chance of any false discovery back to 4.9%. The cost is power: a real but modest signal now struggles to get through. The other defence is data the search never touched: choose the best signal on one period, then test it once on another.
What does a quant fund take from this?
Research teams run thousands of tests, and the ones that get presented are the survivors. Count every test, including the ones you ran and forgot, because the significance of the survivor depends on how many were tried. That is why systematic funds keep research logs and hold data back, and why a backtest with a t-statistic of 2 means much less after a large search than after one planned test. Say the limitation: the 64% assumes independent tests; correlated signals give a lower figure, but rarely a comfortable one.
Where candidates lose it
The fast wrong answer adds the probabilities: 20 x 5% = 100%, a certainty. Adding only works for events that cannot happen together; here several false positives can appear at once, so go through the complement.
The quieter error is answering 5%, treating the batch as one test. The interviewer wants you to see that the error rate of the search is not the error rate of each test inside it.
What the interviewer asks next
How many tests at 5% before a false positive is more likely than not?
What significance level per test keeps the family-wide chance at 5% across 100 tests?
Why does out-of-sample testing help, and what can still go wrong with it?
014A corporate bond has a spread duration of 6 and convexity of 50. Its credit spread widens by 50 basis points. Roughly what happens to its price?Valuation, accounting and macro riddlesCoreAQR Capital ManagementGreenwich · 2021
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Which is closest?
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The price falls by about 2.94%. Spread duration of 6 says a 0.50 percentage point widening costs 6 x 0.50% = 3.00%. Convexity of 50 adds back one half x 50 x 0.005 squared, about 0.06%, because the price curve bends upwards. On a bond priced at 100 that is a move to about 97.06. At 50 basis points the convexity term is small; at 300 or 500 it is not.
What do duration and convexity each measure?
Picture a playground slide that curves and flattens towards the bottom. Judge the drop from the steepness at the top and you overstate it, because the slide levels off as you go. Spread durationThe percentage change in a bond price for a one percentage point change in its credit spread, holding the risk-free rate fixed. is the steepness at today's spread; convexity is the flattening, so the straight-line estimate always overstates the loss when spreads widen. Duration gives the first-order move, 6 x 0.50% = 3.00% down; convexity corrects it by a term that depends on the square of the move.
For a 50 basis point widening, duration of 6 gives minus 3.00% and convexity of 50 adds back 0.06%, a fall of 2.94%; the convexity cushion grows with the square of the move, to 2.25 points at 300 basis points and 6.25 at 500.
The relationship
$\frac{\Delta P}{P} \approx -D_s\,\Delta s + \tfrac{1}{2}C\,(\Delta s)^2 = -6(0.005) + \tfrac{1}{2}(50)(0.005)^2 = -3.00\% + 0.0625\% \approx -2.94\%$
D_s spread duration, 6
C convexity, 50
\Delta s the change in spread as a decimal, 50 basis points = 0.005
What it says in wordsThe price moves by the duration term plus a smaller correction that grows with the square of the spread change.
When does the convexity term start to matter?
It grows with the square of the move. At 50 basis points convexity is worth 0.06% against a 3.00% duration loss; at 300 basis points duration says -18% and convexity adds back 2.25%, which is no longer small. That is why a credit desk can run duration-only risk for everyday moves but needs convexity for stress scenarios. One more distinction marks a strong answer: for a fixed-coupon bond spread duration and rate duration are close, but a floating-rate note has almost no rate duration and still carries several years of spread duration.
Say the limitation plainly. Both numbers are local, measured at today's spread, and a distressed bond stops behaving like this long before default, when its price starts tracking the expected recovery instead. For a bond trading near par, as here, the two-term estimate is good to a few hundredths of a per cent for moves of this size.
Where candidates lose it
Candidates give minus 3% and stop, which is fine as a first line but ignores the second number the question handed you. Worse is using convexity with the wrong sign, making the loss bigger: for a plain bond convexity always cushions a spread widening.
The other slip is units. Fifty basis points is 0.005 in the formula; squaring 0.50 instead turns a 0.06% correction into 6.25% and produces a price that rises when spreads widen.
What the interviewer asks next
What if the spread tightens by 50 basis points instead?
Why can a callable bond have negative convexity?
How would you hedge the spread risk of this bond?
Asked at AQR Capital Management, Investment Research, Greenwich, 2021 (Wall Street Oasis): Discussion on credit spreads on fixed income products and duration.
015Under a normal model with 1% daily volatility, how often should a move of 4% or more in either direction happen? You have seen two such days this year. What do you conclude?Continuous probability and distributionsCoreQuant and systematic fundsProp and quant trading firms
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How often does the normal model expect a 4% day?
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The normal model expects a 4% day about once every 63 years, so two in one year says the model is wrong, not that you were unlucky. A 4% move is four standard deviations, and the normal puts about 0.0063% of days beyond that in either direction, 0.016 such days a year. Under the model, two in a year has a chance of roughly 1 in 7,933. Real returns have fatter tails and volatility that clusters.
How rare is four standard deviations under a normal curve?
Adult heights are roughly normal. Someone four standard deviations above the average is so rare that you could meet many thousands of people without seeing one. The normal tail thins faster than exponentially, so each extra standard deviation makes an event far rarer: beyond 2 is about 1 day in 22, beyond 3 about 1 in 370, beyond 4 about 1 in 15,787. With 1% daily volatility, a 4% day is a four standard deviation day.
Under a normal model with 1% daily volatility, moves beyond 4% in either direction cover only 0.0063% of days, about 0.016 days a year or one every 63 years, so seeing two in a single year points to a model with tails that are too thin.
The relationship
$P(|Z| \ge 4) = 2\,(1-\Phi(4)) \approx 6.3\times10^{-5} \qquad 252 \times 6.3\times10^{-5} \approx 0.016 \text{ per year}$
Z the daily return divided by its 1% volatility
\Phi the standard normal cumulative distribution
252 trading days in a year
What it says in wordsTwo thin tails times the number of trading days gives the expected count of 4% days a year.
What do two such days in a year actually tell you?
Work out how surprising the evidence is under the model. With 0.016 expected a year, two or more has a probability of about 1 in 7,933 under the normal model, so either this was an extraordinarily rare year or the model is wrong, and the second is far more likely. Market returns have fatter tails than the normal and volatility that comes in clusters, so a 1% volatility estimated over calm months understates risk once the market turns. The honest conclusion is to re-estimate volatility with recent data and stop quoting tail odds from the normal curve.
What does a risk manager do with this?
Two things. Replace the normal tail with something that respects the data, a fatter-tailed distribution or historical scenarios, and let the volatility estimate react faster, for example by weighting recent days more heavily. Then ask the more useful question for the book: not how likely a 4% day is, but what the book loses if it happens twice in a month. A risk limit calibrated on the normal model is exactly the number this evidence has just discredited.
Where candidates lose it
Candidates compute the rarity correctly and then conclude the year was unlucky. That is the wrong way round: when an event the model calls once in 63 years happens twice in one, the evidence is against the model, not against the market.
The other slip is using one tail. A move of 4% in either direction means both tails, which doubles the probability; one tail alone gives about once in 125 years.
What the interviewer asks next
Under the same model, how often should a 3% day occur?
If volatility is really 1.5%, how often is a 4% day?
How would you estimate volatility so that it reacts quickly to a change of regime?
016How many rolls of a fair die do you expect to need before you have seen all six faces at least once?Counting and combinatoricsCoreQuant and systematic fundsProp and quant trading firms
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Your estimate?
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14.7 rolls. Split the wait into six stages, one per new face. The first roll always shows a new face. With k faces already seen, each roll is new with probability (6 - k)/6, so that stage takes 6/(6 - k) rolls on average. Adding 1 + 1.2 + 1.5 + 2 + 3 + 6 gives 14.7, and the last face alone costs six of those rolls.
Why does the wait get longer as you go?
Collecting a set of six cricket cards from cereal packets feels quick at first: almost every packet brings a new card. By the end you are opening packet after packet for the one card you lack. The chance of something new falls as the collection grows, so the wait for each new face grows too, and the last face dominates. With five faces seen, only one roll in six is any use.
The expected rolls for each new face rise from 1 for the first to 1.2, 1.5, 2, 3 and finally 6 for the last, adding to 14.7 rolls, with the final face alone taking 6.
The relationship
$E[N] = \sum_{k=0}^{5}\frac{6}{6-k} = 6\left(1 + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{4} + \tfrac{1}{5} + \tfrac{1}{6}\right) = 14.7$
k the number of faces already seen
6/(6 - k) the expected rolls to find one more new face
What it says in wordsThe total wait is the sum of six geometric waits, each longer than the last.
Why is each stage 6/(6 - k) rolls?
Each stage is a run of independent tries with a fixed chance of success, and the average length of such a run is one over that chance. If a new face turns up with probability p on each roll, you wait 1/p rolls on average: 6/5 rolls when five faces are still new, 6/1 when only one is. That rule, the mean of a geometric wait, is the one piece of theory the puzzle needs, and it is worth saying before you start adding.
Where does the same shape show up in markets?
Any wait to see every one of a set of outcomes has a long tail. Waiting until every stock on a thin watch list has traded at least once, or until a survey has reached every group in a sample, behaves the same way: most of the time goes on the last few. The general answer is n times the sum 1 + 1/2 + ... + 1/n, which grows like n times the natural log of n; for 100 equally likely items it is about 519 draws, not 100.
Where candidates lose it
Answering 6 assumes no repeats. Candidates usually sense that is wrong but then guess 10 or 12 instead of splitting the wait into stages.
The other slip is adding the probabilities instead of their inverses. The stages are waits, and a wait for an event of probability p lasts 1/p rolls on average; state that rule before you sum.
What the interviewer asks next
How many rolls on average to see every face of a 20-sided die?
What is the expected number of distinct faces seen in six rolls?
How many rolls on average to see a 6 twice?
017You roll a fair die until the first 6 appears and are paid the sum of every roll, including the final 6. What is the expected payout?Expected value and dice gamesHardQuant and systematic fundsProp and quant trading firms
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What is the expected payout?
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21. The number of rolls until the first 6 averages 6, so there are on average 5 non-six rolls plus the 6. A roll known not to be a 6 is equally likely to be 1 to 5, so it averages 3, not 3.5, giving 5 x 3 + 6 = 21. Wald's identity confirms it: 6 expected rolls times 3.5 a roll is also 21, because a stop that looks only at past rolls does not bias the total.
Why is it tempting to get 23.5?
Suppose you keep buying scratch cards until one wins. Every card before the winner is, by definition, a loser, so those cards are worth less than an average card. The stopping rule changes the rolls before the stop: each one is known not to be a 6, so it averages 3, and valuing them at 3.5 overpays by 0.5 a roll, 2.5 in all. Five non-sixes at 3.5 plus a 6 is 23.5, which counts the high side twice: once in the 3.5 and again in the final 6.
In a typical run of 4, 1, 5, 2, 3 and then 6, the five rolls before the stop are non-sixes averaging 3 and the total is 21; 5 x 3 + 6 and 6 x 3.5 both give 21, while 5 x 3.5 + 6 = 23.5 wrongly treats the early rolls as ordinary rolls.
How do the two routes agree?
Route one splits the sum: the expected number of non-six rolls times their average, plus the final 6. The count of rolls is a geometric wait with success chance 1/6, so it averages 6, of which 5 are non-sixes: 5 x 3 + 6 = 21. Route two is Wald's identityFor a stopping rule that uses only rolls already seen, the expected total equals the expected number of rolls times the average roll.: the expected total is the expected number of rolls times the average roll, 6 x 3.5 = 21. The low early rolls and the high final roll balance exactly.
The relationship
$E[S] = E[N]\cdot E[X] = 6 \times 3.5 = 21 \qquad E[S] = (E[N]-1)\times 3 + 6 = 5\times 3 + 6 = 21$
S the total paid
N the number of rolls, including the 6
X a single roll, averaging 3.5 before any conditioning
What it says in wordsCounted either as all rolls at 3.5 or as non-sixes at 3 plus a 6, the expected payout is 21.
Why would a trading firm ask this?
Stopping rules are everywhere on a desk: exit at the first stop-loss hit, rebalance at the first breach of a band. The question checks whether you can tell when a stopping rule biases what you observe, as it does for the early rolls, and when it does not, as for the total. Say the condition too: Wald's identity needs the decision to stop to use only rolls already seen, and the expected number of rolls to be finite. A rule that could peek at the next roll would break it.
Where candidates lose it
The slip is 5 x 3.5 + 6 = 23.5. It treats the rolls before the 6 as ordinary rolls, when the stopping rule guarantees none of them is a 6, which pulls their average down to 3.
The opposite slip is to distrust 6 x 3.5 because stopping at a 6 seems to bias it. It does not: the total is unbiased for any stopping rule that looks only at the past. Give both routes and say why they agree.
What the interviewer asks next
What is the expected payout if the final 6 is not paid?
You stop at the first 5 or 6 instead. What is the expected payout?
If you could choose to stop whenever you like, what would you pay to play?
018One hundred lockers start closed. Person 1 toggles every locker, person 2 toggles every second locker, person 3 every third, and so on up to person 100. Which lockers end open?Logic and brainteasersCoreProp and quant trading firmsLong-short equity funds
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Which lockers end open?
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The ten perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. Locker n is toggled once by each person whose number divides n, so its final state depends on how many divisors n has. Divisors come in pairs, d and n/d, which cancel out. Only a perfect square has an unpaired divisor, its square root, so only squares are toggled an odd number of times and end open.
What decides whether one locker ends open?
A light switch flipped an even number of times ends where it started; flipped an odd number of times, it ends the other way. Each locker is a switch, flipped once for every divisor of its number, so the question is which numbers from 1 to 100 have an odd number of divisors. Locker 12 is touched by persons 1, 2, 3, 4, 6 and 12, six times, and ends closed.
Of the 100 lockers only the ten perfect squares end open, because locker 12 and every non-square has its divisors in pairs, an even number of toggles, while locker 36 and every square has one unpaired divisor, its square root.
Why do only perfect squares have an odd number of divisors?
Pair every divisor d with n divided by d. For 12 the pairs are 1 and 12, 2 and 6, 3 and 4: six divisors, even. The pairing breaks only when a divisor is paired with itself, d = n/d, which happens exactly when n is a perfect square. For 36 the pairs are 1 and 36, 2 and 18, 3 and 12, 4 and 9, with 6 left over: nine divisors, odd, so locker 36 ends open. There are ten squares up to 100, so ten lockers.
The relationship
$d \cdot \frac{n}{d} = n \quad\text{and}\quad d = \frac{n}{d} \iff d^2 = n$
d a divisor of the locker number n
n/d its partner divisor
What it says in wordsDivisors cancel in pairs, and only a perfect square leaves one divisor without a partner.
Why does a fund ask a puzzle like this?
It tests whether you look for structure before you simulate. Walking through a hundred people toggling lockers is hopeless in an interview; turning it into a question about divisors takes one sentence and makes the answer obvious. The move, recasting a process as a property you can count, is the one that turns a messy trading rule into a quantity you can compute. Check it on a small case out loud: with 10 lockers, 1, 4 and 9 end open.
Where candidates lose it
Candidates start simulating: person 1 opens everything, person 2 closes the evens, person 3 toggles multiples of 3, and they lose track by person 5. The interviewer wants you to stop and ask what decides one locker's final state.
The other miss is answering the primes. A prime is touched exactly twice, by person 1 and by the person with its own number, so every prime ends closed.
What the interviewer asks next
Which lockers are toggled exactly three times?
With 1,000 lockers, how many end open?
Which locker under 100 is toggled the most, and how many times?

Investor	Entry NAV	High-water mark	NAV, end of year 2	Gain charged	Fee at 20%	NAV after fee
A	100	100	110	10	2.0	108.0
B	80	80	110	30	6.0	104.0

019Three dealers quote USD/INR at 84.00, EUR/USD at 1.10 and EUR/INR at 93.00, each a single price you can deal at. Is there an arbitrage, which way do you trade it, and what is the profit on EUR 1 million?Market making and trading gamesCoreProp and quant trading firmsVolatility and relative value funds

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Which statement is right?

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Yes: the direct quote is rich by 60 paise, so buy euros through dollars and sell them for rupees at 93.00, making Rs 6 lakh on EUR 1 million before costs. The implied rate is 1.10 x 84.00 = 92.40 rupees per euro. Spend Rs 9.24 crore on USD 1.1 million, turn that into EUR 1 million, and sell the euros for Rs 9.30 crore. You finish in rupees, where you started, with Rs 6 lakh more.

How do you spot the mispricing in one line?

If one stall sells mangoes at Rs 100 a dozen and the stall next to it buys them back at Rs 10 each, you would buy dozens and sell singles all day. Every pair of currencies can be priced two ways, directly or through a third currency, and when the two prices differ you buy on the cheap route and sell on the rich one. Through dollars a euro costs 1.10 dollars at Rs 84.00 each: Rs 92.40. The direct dealer pays Rs 93.00. That 60 paise gap is the whole trade.

A euro costs Rs 92.40 when bought through dollars at 84.00 and 1.10 but fetches Rs 93.00 from the direct dealer, so running rupees to dollars to euros and back to rupees earns 60 paise a euro, Rs 6 lakh on EUR 1 million before costs.

Step	Trade	You pay	You receive
1	Buy USD with rupees at 84.00	Rs 9,24,00,000	USD 1,100,000
2	Buy EUR with dollars at 1.10	USD 1,100,000	EUR 1,000,000
3	Sell EUR for rupees at 93.00	EUR 1,000,000	Rs 9,30,00,000
	Net, in rupees		Rs 6,00,000

Starting and ending in rupees, the three legs turn Rs 9.24 crore into Rs 9.30 crore, a profit of Rs 6 lakh on EUR 1 million, before spreads and dealing costs.

What stops this from being free money in practice?

Three things. Real quotes have a bid and an offer, and the gap survives only if it is wider than the three spreads you cross plus the cost of dealing. A 60 paise gap on a 92 rupee price is about 0.65%, far wider than dealer spreads in major currencies, which is why a gap that size would be traded away almost at once. And the legs must be done together: if one price moves before you finish, you are left holding an open currency position instead of a locked-in profit.

Where candidates lose it

The common slip is running the loop the wrong way round: selling euros through dollars and buying them directly. That locks in a 60 paise loss on every euro. Decide which route is rich before placing any leg, and say it out loud.

The second is quoting the profit in a mix of currencies or on the wrong notional. Start and end in the same currency; starting from rupees makes the answer a clean Rs 6 lakh on EUR 1 million.

What the interviewer asks next

EUR/INR is quoted 92.95 / 93.05, USD/INR 83.99 / 84.01 and EUR/USD 1.09975 / 1.10025. Is there still an arbitrage?
Why do gaps like this almost never appear in major currencies?
If the third leg fails to fill, what position are you left with?

020A book holds Rs 60 crore of a stock with 30% volatility and Rs 40 crore of another with 20% volatility, and the two have a correlation of 0.5. What is the book's volatility in rupees, and what share of the risk comes from each position?Portfolio and risk mathsHardMan GroupBoston · 2022
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What share of the book's risk comes from the Rs 60 crore position?
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The book's volatility is about Rs 23.1 crore a year, and the Rs 60 crore position carries about 74% of it on 60% of the capital. Stand-alone risks are Rs 18 crore and Rs 8 crore. Book variance is 18 squared plus 8 squared plus 2 x 0.5 x 18 x 8, which is 532, so volatility is Rs 23.07 crore. Each position's contribution is its covariance with the book over the book's volatility: Rs 17.17 crore and Rs 5.90 crore, which add back to the total.
Why is risk not shared out like capital?
Two friends share a taxi. One rides twice as far, straight through the traffic jam; the other gets off after a short hop. Splitting the fare by the number of bags each carries would be absurd. Risk belongs to a position in proportion to how much it moves and how much it moves with everything else, not to how much money sits in it. Here the first stock is larger, more volatile and positively correlated with the second, so it carries far more than its 60% of the capital.
Stand-alone risks of Rs 18 crore and Rs 8 crore add to Rs 26 crore, diversification at a 0.5 correlation removes Rs 2.93 crore, and the book's Rs 23.07 crore of volatility splits 74.4% to the first position and 25.6% to the second, against a 60 to 40 split of capital.
The relationship
$\sigma_{book}^2 = a^2 + b^2 + 2\rho ab \qquad RC_A = \frac{a^2 + \rho ab}{\sigma_{book}} \qquad RC_B = \frac{b^2 + \rho ab}{\sigma_{book}}$
a, b stand-alone rupee volatilities: 60 x 30% = 18 and 40 x 20% = 8
\rho the correlation, 0.5
RC a position's contribution to book volatility
What it says in wordsEach position owns its own variance plus half the shared term, and dividing by the book's volatility turns that into rupees of risk.
Position Capital, Rs crore Volatility Stand-alone risk Risk contribution Share of risk
A 60 30% 18.0 17.17 74.4%
B 40 20% 8.0 5.90 25.6%
Book 100 26.0 23.07 100.0%
Rs crore of annual volatility. The stand-alone risks add to Rs 26.0 crore, but the book's volatility is Rs 23.07 crore, of which position A contributes 74.4% and position B 25.6%.
Why do the contributions add up exactly to the total?
Split the variance. The cross term, 2 x 0.5 x 18 x 8 = 144, is shared equally, 72 to each position. So position A owns 324 + 72 = 396 of the 532 of variance and position B owns 64 + 72 = 136, and dividing each by the book's volatility of 23.07 gives rupee contributions that sum exactly to Rs 23.07 crore. The diversification benefit is the gap between the stand-alone total of Rs 26 crore and the book's Rs 23.07 crore.
What does a risk manager do with the split?
Cut where the risk is, not where the money is. Each rupee in position A carries 28.6 paise of marginal risk against 14.7 paise in position B, so trimming Rs 10 crore from A lowers book volatility by roughly Rs 2.9 crore. Recomputing exactly gives Rs 2.84 crore, close to the estimate. The limitation: the split is a snapshot at one correlation, and when correlations move, both the total and the split move with them.
Where candidates lose it
The quick answer shares risk like capital, 60 and 40, or like stand-alone risk, 18 and 8. The first ignores volatility and the second ignores correlation; neither sums to the book's actual Rs 23 crore of risk.
The second slip is adding the stand-alone risks to get the book's risk, Rs 26 crore. Volatilities do not add unless the correlation is exactly 1; variances do, with the cross term included.
What the interviewer asks next
If the correlation fell to zero, how would the risk split between the two positions?
How much of position B would you add to minimise the book's volatility, holding A fixed?
How do transaction costs change which position you trim first?
Asked at Man Group, Investment Management, Boston, 2022 (Wall Street Oasis): How do you understand portfolio risk and transaction cost?

E_k	the expected steps still needed from a corner at distance k
1	the step being taken now

\alpha	the significance level of each test, 5%
m	the number of independent tests, 20

D_s	spread duration, 6
C	convexity, 50
\Delta s	the change in spread as a decimal, 50 basis points = 0.005

Z	the daily return divided by its 1% volatility
\Phi	the standard normal cumulative distribution
252	trading days in a year

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