Hedge Funds puzzles, solved step by step

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

Showing 1–10 of 35 · filtered from 100Clear filters

001A book holds 20 independent positions of 5% each, and each has a 10% chance of going to zero over the year. What is the probability that you lose 15% or more of the book?Betting and sizingCoreMulti-manager platformsProp and quant trading firms
Try it first
Before calculating: roughly how likely is a loss of 15% or more?
Show the worked solution
About 32%. A 15% loss means three or more of the 20 positions go to zero. The count of zeros is binomial with 20 tries at 10%, so the chance of zero, one or two is 12.2% + 27.0% + 28.5% = 67.7%, and the chance of three or more is 32.3%. The book expects two blow-ups a year, so three is not a tail event.
Why is a rare event per name a common event per book?
Think of a wedding with twenty guests, each with a one in ten chance of arriving late. Any single guest is almost certainly on time, but a host who plans for nobody being late is planning badly: on average two will be. When you hold many independent risks, the question is not whether one fails but how many do, and the expected count here is 20 x 10% = 2. A loss of 15% needs three failures, which is one more than an average year.
With 20 positions each carrying a 10% chance of going to zero, the most likely outcomes are one or two zeros; three or more zeros, a loss of 15% or more, happen in 32.3% of years.
How do you count the ways to lose three or more?
Count the outcomes you can live with and subtract. It is faster to add up zero, one and two blow-ups and take them from one than to add up three through twenty. Zero needs all twenty to survive: 0.9 to the twentieth, 12.2%. One needs a single failure, with twenty choices of which name: 20 x 0.1 x 0.9 to the nineteenth, 27.0%. Two has 190 possible pairs: 190 x 0.01 x 0.9 to the eighteenth, 28.5%.
The relationship
$P(X \ge 3) = 1 - \sum_{k=0}^{2} \binom{20}{k} (0.1)^k (0.9)^{20-k} \approx 1 - 0.677 = 0.323$
X the number of positions that go to zero
\binom{20}{k} the number of ways to choose which k names fail
0.1 and 0.9 the chance one name fails, and survives
What it says in wordsThe chance of three or more failures is one minus the chance of zero, one or two.
What does a risk manager take from this?
Sizing each position so a single wipe-out is survivable does not make the book survivable. Five per cent a name feels small, yet a 15% drawdown is roughly a one in three year event on these odds, and a platform with a 10% drawdown limit would see it breached in 60.8% of years, because two zeros, the average outcome, already cost 10%. Say the limitation as well: the positions are assumed independent. In a sell-off failures cluster, and correlation fattens exactly the tail you have just computed.
Where candidates lose it
The fast wrong answer multiplies: 10% cubed is 0.1%, so three blow-ups look like a freak. That ignores the 1,140 different ways to choose which three names fail, and it ignores four, five and more.
The second loss is stopping at exactly three. The question says 15% or more, so either sum three through twenty or, far faster, take the complement of zero, one and two. Say which route you are taking before you start.
What the interviewer asks next
What is the chance of losing 10% or more?
If the 20 names are positively correlated, does the chance of a 15% loss rise or fall, and why?
How would you resize the book so a 15% loss happens less than one year in ten?
010Every stock in a universe has 30% volatility and every pair has a correlation of 0.3. What is the volatility of an equal-weighted portfolio of 10 stocks, of 100 stocks, and of infinitely many?Portfolio and risk mathsCoreMulti-manager platformsQuant and systematic funds
Try it first
Where does the volatility end up with infinitely many stocks?
Show the worked solution
About 18.2% for 10 stocks, 16.6% for 100, and a floor of 16.4% for infinitely many. Portfolio variance is 30% squared times (0.3 + 0.7/n): the 0.7/n part is stock-specific noise that averages away, and the 0.3 part is shared movement that never does. The floor is 30% times root 0.3. Ten stocks capture most of the benefit; the next ninety add little.
Why does diversification stop working?
A choir of a hundred singers each slightly off key sounds more in tune than one singer, because the individual errors cancel. But if the whole choir takes its note from one badly tuned piano, no number of singers fixes it. Stock-specific risk is the individual error and averages away; the shared correlation is the piano, and it stays however many names you add. With every pair at 0.3, the shared part is 30% of each stock's variance.
Equal-weighted portfolio volatility falls from 30% for one stock to 18.2% for ten and 16.6% for a hundred, flattening onto a floor of 16.4% set by the 0.3 correlation that no amount of diversification removes.
The relationship
$\sigma_p^2 = \sigma^2\left(\rho + \frac{1-\rho}{n}\right) \qquad \sigma_\infty = \sigma\sqrt{\rho} = 30\% \times \sqrt{0.3} \approx 16.4\%$
\sigma each stock's volatility, 30%
\rho the correlation between every pair, 0.3
n the number of stocks, equally weighted
What it says in wordsPortfolio variance is a shared part that stays plus a specific part that shrinks with every name added.
How do the three numbers come out?
Plug in. Ten stocks: 0.09 x (0.3 + 0.07) = 0.0333, a volatility of 18.2%. One hundred: 0.09 x 0.307 = 0.0276, 16.6%. Infinitely many: 0.09 x 0.3 = 0.027, 16.4%. Going from one stock to ten cuts risk from 30% to 18.2%; going from ten to a hundred cuts only another 1.6 points. That is why a long book of 30 names, at 17.1%, is not as undiversified as it sounds, and why names added past a point buy almost nothing.
What does this mean for a hedge fund book?
The only way under the floor is to remove the shared factor itself, which is what a short leg or an index hedge does. If the correlation comes from the market, shorting the market against the long book strips out the shared piece and leaves stock-specific risk, which does diversify. The limitation is that correlations are not fixed. In a sell-off they rise, and the floor rises with them: at a correlation of 0.6 it is 23.2%, so a book that looked diversified at 0.3 starts behaving like a concentrated one.
Where candidates lose it
The common miss is saying volatility goes to zero with enough stocks. That holds only if the stocks are uncorrelated; any shared correlation leaves a floor, and the interviewer is testing whether you know it is there.
The second is computing the floor as 30% x 0.3 = 9%, which applies the correlation to volatility instead of variance. Variance floors at 0.3 times 0.09; take the square root at the end, not the start.
What the interviewer asks next
How many stocks do you need to get within one point of the floor?
If correlation rises to 0.6 in a crisis, where is the new floor?
How does a long-short book change this calculation?
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
Try it first
Pick your estimate before you set anything up.
Show the worked solution
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
Try it first
In year two, on how much gain per unit does investor A pay the fee?
Show the worked solution
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?
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?
Show the worked solution
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
Try it first
Your estimate?
Show the worked solution
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?
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
Try it first
Which lockers end open?
Show the worked solution
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

Try it first

Which statement is right?

Show the worked solution

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?

022Two independent, unbiased forecasts of next quarter's GDP growth have error standard deviations of 2% and 3%. How should you combine them, and how accurate is the blend?Statistics and estimationCoreQuant and systematic funds
Try it first
What weight should the 2% forecast get?
Show the worked solution
Weight them 9/13 and 4/13, about 69% and 31%, and the blend's error falls to about 1.66%. For independent unbiased forecasts the best weights are proportional to one over each error variance: 1/4 for the 2% forecast and 1/9 for the 3% one. The blended error variance is 1 over (1/4 + 1/9), which is 36/13, so its standard deviation is 1.66%, better than either forecast alone.
Why does blending two forecasts beat the better one?
Ask two people to guess the weight of a pumpkin at a village fair. One is usually closer, but their mistakes are unrelated, so averaging tends to cancel part of each. Independent errors partly cancel when you average, so even a weaker forecast adds information, provided it gets a smaller weight. Throwing the 3% forecast away leaves you at 2%; blending it in well gets you to 1.66%.
Blending with inverse-variance weights of 9/13 and 4/13 gives an error of 1.66%, lower than the better forecast's 2.00%, while equal weights give 1.80% and weights of 3/5 and 2/5 give 1.70%.
The relationship
$w_A = \frac{1/\sigma_A^2}{1/\sigma_A^2 + 1/\sigma_B^2} = \frac{1/4}{1/4 + 1/9} = \frac{9}{13} \qquad \sigma_{blend} = \sqrt{\frac{1}{1/4 + 1/9}} = \sqrt{\frac{36}{13}} \approx 1.66\%$
\sigma_A, \sigma_B the two forecasts' error standard deviations, 2% and 3%
w_A the weight on the sharper forecast
What it says in wordsEach forecast is weighted by its precision, one over its variance, and the blend's precision is the sum of the two.
Why inverse variance and not inverse error?
The blend's error variance is w squared times 4 plus (1 minus w) squared times 9. Setting its slope to zero gives w = 9/(4 + 9), so the weights follow one over the variance, which penalises the noisier forecast harder than one over the standard deviation would. Check the alternatives: equal weights give an error of 1.80%, weights of 3/5 and 2/5 give 1.70%, and the inverse-variance weights give the minimum, 1.66%.
What would you check before trusting the blend?
Two assumptions carry the answer. The forecasts must be unbiased and their errors independent; if both forecasters lean on the same survey, their errors are correlated and the gain from blending shrinks. With an error correlation of 0.5, the best blend gives the sharper forecast 6/7 of the weight and improves the error only from 2.00% to 1.96%. Ask where each forecast comes from before you average them.
Where candidates lose it
Candidates either average equally, which overweights the noisier forecast, or keep only the better one, which throws information away. Both miss that the right weights come from the variances.
The subtler slip is weighting by one over the standard deviation, 3/5 and 2/5. It is close but not optimal; state the inverse-variance rule and show that the blended error beats 2%.
What the interviewer asks next
What if the two forecast errors have a correlation of 0.5?
How would you estimate each forecaster's error variance in practice?
One forecast is biased upwards by 0.5%. What do you do?
028A stick is broken at two points chosen independently and uniformly at random along its length. What is the probability that the three pieces can form a triangle?Continuous probability and distributionsCoreQuant and systematic fundsProp and quant trading firms
Try it first
Where does the answer land?
Show the worked solution
One in four. Three lengths make a triangle only if no piece is longer than the other two together, which for a stick of length 1 means no piece is longer than one half. Plot the two break points as a dot in the unit square. The dots that keep every piece under half fill two triangles of area 1/8 each, so the chance is 1/4.
What exactly has to be true for a triangle?
Try to make a triangle from three sticks where one is longer than the other two laid end to end: the short ones cannot reach each other across the long one. Three lengths form a triangle only if the longest is shorter than the sum of the other two, and with a total of 1 that means every piece must be under one half. That turns a geometry question into a single condition you can test on any pair of break points.
How do you turn two random breaks into an area?
Call the break points x and y, each uniform between 0 and 1. Every possible outcome is a dot in the unit square, and every dot is equally likely, so probability is area. Take the half where x is below y: the pieces are x, y minus x and 1 minus y, and all three are under half only when x is below 1/2, y is above 1/2 and y minus x is below 1/2. Those three lines cut out a triangle with corners (0, 1/2), (1/2, 1/2) and (1/2, 1), area 1/8. The half where y comes first gives its mirror image, another 1/8.
Each dot in the square is one pair of break points; only the two green triangles, each of area 1/8, keep every piece shorter than half the stick, so a triangle forms with probability 1/4.
The red regions are also worth naming, because each one is a different piece being too long. Three failure regions of area 1/4 each are the first piece, the middle piece and the last piece exceeding half. They cannot overlap, since two pieces cannot both be longer than half, so the answer is 1 minus 3/4, which is 1/4, a second route to the same number.
Mention the variant, because interviewers often switch to it: break the stick once, then break the longer piece at a random point. The answer changes to about 0.386, which is 2 ln 2 minus 1, because the second break is no longer uniform over the whole stick. Knowing the setup changes the answer shows you are reading the question.
Where candidates lose it
The usual loss is guessing from symmetry, 1/2 or 1/3, without writing the condition. Once you say no piece may exceed half, the square and the area come quickly.
The second loss is solving only the case where x is left of y, getting 1/8 and forgetting the mirror half. Either break point is equally likely to be the left one, so the good region always comes in two pieces.
What the interviewer asks next
Break the stick once, then break the longer piece at a random point. What is the chance now?
What is the expected length of the longest piece?
What is the probability that the three pieces form an acute triangle?

X	the number of positions that go to zero
\binom{20}{k}	the number of ways to choose which k names fail
0.1 and 0.9	the chance one name fails, and survives

\sigma	each stock's volatility, 30%
\rho	the correlation between every pair, 0.3
n	the number of stocks, equally weighted

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

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

Try it first

Why is a rare event per name a common event per book?

How do you count the ways to lose three or more?

What does a risk manager take from this?

Where candidates lose it

What the interviewer asks next

Try it first

Why does diversification stop working?

How do the three numbers come out?

What does this mean for a hedge fund book?

Where candidates lose it

What the interviewer asks next

Try it first

How do you turn eight corners into four states?

How do you solve the equations quickly?

What is the skill a fund is actually testing?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the same year produce two different fees?

How do funds keep the two investors apart?

What does this do to the manager's incentives?

Where candidates lose it

What the interviewer asks next

Try it first

How rare is four standard deviations under a normal curve?

What do two such days in a year actually tell you?

What does a risk manager do with this?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the wait get longer as you go?

Why is each stage 6/(6 - k) rolls?

Where does the same shape show up in markets?

Where candidates lose it

What the interviewer asks next

Try it first

What decides whether one locker ends open?

Why do only perfect squares have an odd number of divisors?

Why does a fund ask a puzzle like this?

Where candidates lose it

What the interviewer asks next

Try it first

How do you spot the mispricing in one line?

What stops this from being free money in practice?

Where candidates lose it

What the interviewer asks next

Try it first

Why does blending two forecasts beat the better one?

Why inverse variance and not inverse error?

What would you check before trusting the blend?

Where candidates lose it

What the interviewer asks next

Try it first

What exactly has to be true for a triangle?

How do you turn two random breaks into an area?

Where candidates lose it

What the interviewer asks next