Hedge Funds puzzles, solved step by step

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

Showing 41–50 of 100

041Two orders arrive one after the other. The first arrives after a wait that is exponential with a mean of one minute; the second arrives after a further, independent exponential wait with the same mean. What is the probability that both have arrived within one minute?Continuous probability and distributionsHardCitadelChicago · 2025
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Your estimate:
Show the worked solution
1 minus 2/e, about 26.4%. The total wait is the sum of two independent exponential waits. Convolving the two densities gives t e^-t, a gamma shape that starts at zero because two steps cannot both be instant. Its area from 0 to 1 is 1 minus e^-1 (1 + 1), which is 1 minus 2/e. A second route: it is the chance that a Poisson process with rate 1 produces at least two arrivals in one minute.
Why is this not the chance of one wait, squared?
Squaring would be right if both orders were racing from the same start line. Here they queue: the second clock only starts when the first order lands. It is like two buses where you must take the first to reach the stop for the second. The event is that the sum of the two waits is under one minute, which is stricter than each wait being under one minute. The chance one wait is under a minute is 1 minus 1/e, about 63%; squaring gives 40.0%, which answers a different question.
How do you get the density of the sum?
Add up every way to split the total t between the two waits. The density of a sum of independent waits is the convolutionThe density of a sum of two independent variables, found by integrating one density against the other shifted over every possible split of the total. of their densities, and for two exponentials it is t e^-t. Every split of t into s and t minus s has density e^-s times e^-(t minus s), which is e^-t whatever s is, and there is a length t of possible splits. Integrate t e^-t from 0 to 1 by parts and you get 1 minus 2/e.
The total of two independent one-minute exponential waits has density t e^-t, which starts at zero and peaks at one minute, so only 26.4% of its area, 1 minus 2/e, lies below one minute.
The relationship
$P(X_1 + X_2 \le 1) = \int_0^1 t e^{-t}\,dt = 1 - 2e^{-1} \approx 0.264$
X1, X2 the two independent exponential waits, each with mean one minute
t e^-t the density of their sum, from convolving the two exponential densities
What it says in wordsThe chance the total wait is under a minute is the area under the gamma density up to one minute.
Check it with counting. Exponential waits are the gaps of a Poisson process, so both orders arrive within a minute exactly when the process makes at least two arrivals in that minute. With one arrival expected per minute, the chance of zero is e^-1 and of exactly one is e^-1, so at least two is 1 minus 2/e, the same number. Say both routes and the interviewer will usually skip ahead.
Where candidates lose it
The common loss is squaring the single-wait probability, which answers the question of two independent orders racing in parallel. Read the setup again: one after the other means the waits add.
The second loss is freezing on the convolution integral. If the integral will not come, switch to the Poisson count: at least two arrivals in one minute. Candidates who know one route and not the other are the ones interviewers push hardest.
What the interviewer asks next
What is the probability that three orders in sequence all arrive within two minutes?
Given that both orders arrived within one minute, what is the expected arrival time of the first?
The two waits have means of one and two minutes. What is the density of their sum?
Asked at Citadel, Quant Research Interview, Chicago, 2025 (Wall Street Oasis): if i knew this was about convolutions, i would have answered better.
042A bowl holds 100 noodles. You repeatedly pick two free ends at random and tie them together, until no free ends remain. What is the expected number of loops?Counting and combinatoricsHardD.E. ShawNew York · 2026
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Roughly how many loops?
Show the worked solution
About 3.28 loops. With k strands in the bowl there are 2k free ends. Pick any end; the other end you pick is one of the remaining 2k minus 1, and exactly one of those belongs to the same strand, so this tie closes a loop with chance 1/(2k minus 1). Either way the number of strands falls by one. Adding 1/199 + 1/197 + ... + 1/3 + 1 gives about 3.284.
What does one tie do, whatever happens?
Start with the bookkeeping, because it makes the rest easy. Every tie reduces the number of loose strands by exactly one: either it closes a strand into a loop, or it joins two strands into one longer strand. So there are always exactly 100 ties, and with k strands left there are 2k free ends. The question becomes how many of those 100 ties happen to close a loop.
What is the chance a given tie closes a loop?
Think of a room of dancers holding hands in lines: grab one free hand, then pick a second free hand at random, and a circle forms only if the second hand is at the other end of the same line. With 2k free ends, the second end is one of 2k minus 1, and exactly one of them is the other end of the strand you picked, so the chance is 1/(2k minus 1). Give each tie an indicator that is 1 if it closes a loop; by linearity of expectation the expected number of loops is the sum of the chances, from 1/199 for the first tie up to 1 for the last.
The first tie closes a loop with chance 1 in 199 and the chances stay tiny until the last few ties, 1/5, 1/3 and 1, so the expected number of loops from 100 noodles is only 3.28, about a third of it from the last three ties.
The relationship
$E[\text{loops}] = \sum_{k=1}^{100} \frac{1}{2k-1} = 1 + \tfrac13 + \tfrac15 + \dots + \tfrac{1}{199} \approx 3.28$
k the number of strands in the bowl before a tie
1/(2k-1) the chance that tie joins the two ends of one strand
What it says in wordsAdd each tie's chance of closing a loop to get the expected number of loops.
For a sense check without a calculator: the sum of odd reciprocals up to 1/(2n minus 1) is about half of ln n plus ln 2 plus half of Euler's constant, which for n = 100 gives 3.28. The number of loops grows only like the logarithm of the number of noodles: a million noodles would give only about 7.9 loops.
Where candidates lose it
The first loss is trying to track the lengths of the strands, which quickly becomes impossible. The length of a strand never matters; only the count of strands does.
The second loss is getting 1/(2k minus 1) right but summing it wrong, for example as 100 x 1/199. Write the sum out from the last tie backwards, 1 + 1/3 + 1/5, and the size of the answer becomes obvious.
What the interviewer asks next
What is the variance of the number of loops?
What is the probability that you end with exactly one big loop?
How does the answer grow with the number of noodles, roughly?
Asked at D.E. Shaw, Research, New York, 2026 (Wall Street Oasis): What is the expected number of loops from tying 100 noodles' ends together randomly
043You may roll a fair die up to three times. After each roll you either stop and are paid the face in rupees, or throw that roll away and roll again; if you reach the third roll you must take it. What is the best stopping rule, and what is the game worth?Expected value and dice gamesHardSquarepoint CapitalLondon · 2026
Try it first
On the first roll you get a 4. What do you do?
Show the worked solution
Stop on the first roll only with a 5 or 6, on the second with a 4 or more, and the game is worth 14/3, about Rs 4.67. Work backwards. The last roll is worth 3.5. With two rolls left, keep 4, 5 or 6 and reroll otherwise: worth (4 + 5 + 6)/6 + 3/6 x 3.5 = 4.25. With three rolls left, keep only what beats 4.25, a 5 or 6: worth (5 + 6)/6 + 4/6 x 4.25 = 14/3.
Why start from the last roll?
Deciding whether to take a job offer is easier if you know what your fallback is worth. Each keep-or-reroll decision compares the roll in hand with the value of the rolls still to come, so you need the value of the future first, and the only stage with no future is the last one. On the last roll you must take whatever comes, which is worth 3.5 on average. That single number lets you solve the stage before it, and so on backwards. This is {term('backward induction', 'Solving a sequence of decisions from the last one to the first, using the value of each later stage to make the earlier decision.')}, the core of dynamic programming.
With one roll left the game is worth 3.5, so with two left you keep 4 or more and the game is worth 4.25; with three left you keep only 5 or 6, and the whole game is worth 14/3, about 4.67.
How do the thresholds come out?
With two rolls left, a roll of 4, 5 or 6 beats the 3.5 you expect from rerolling, and 1, 2 or 3 does not. So the two-roll game is worth the average of the kept faces times their chance, plus the chance of rerolling times 3.5: 15/6 + 1.75 = 4.25. On the first roll the fallback is now 4.25, so a 4 is no longer good enough: only 5 or 6 is kept. That gives 11/6 plus 4/6 x 4.25, which is 1.833 plus 2.833, or 14/3.
The relationship
$V_n = \frac{1}{6}\sum_{f=1}^{6} \max\left(f,\, V_{n-1}\right), \quad V_1 = 3.5,\; V_2 = 4.25,\; V_3 = \tfrac{14}{3}$
V_n the value of the game with n rolls left
f the face you just rolled
max(f, V_{n-1}) keep the roll or throw it away, whichever is worth more
What it says in wordsEach stage is worth the average, over the six faces, of the better of keeping the face or playing on.
Add the pattern. More rolls always raise the value, but by less each time: 3.5, 4.25, 4.67, then about 4.94 with four rolls. An extra option is always worth something and never worth more than what it can still improve. That is the same logic as valuing a trade you can exit early: the right to wait is priced by what the future is worth, not by the average outcome.
Where candidates lose it
The common loss is using 3.5 as the threshold at every stage, which keeps a 4 on the first roll. The fallback on the first roll is the two-roll game, 4.25, not a single roll.
The second loss is solving forwards and getting lost. Say you will start from the last roll, compute 3.5, 4.25 and 14/3 in that order, and the thresholds fall out.
What the interviewer asks next
What is the game worth with four rolls, and what is the first-roll threshold?
You must pay Rs 1 for every reroll. How do the thresholds change?
You are paid the square of the face instead. What is the optimal rule?
Asked at Squarepoint Capital, Quantitative Research, London, 2026 (Wall Street Oasis): Dynamic programming questions with focus on probability at the end.
044You have two identical eggs and a 100-storey building. An egg breaks if dropped from some floor or higher and survives from any floor below it. What is the minimum number of drops that guarantees you find that floor?Logic and brainteasersHardProp and quant trading firmsLong-short equity funds
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Minimum guaranteed number of drops:
Show the worked solution
14 drops. With k drops available, the first egg should go from floor k: if it breaks, the second egg checks the k minus 1 floors below one at a time. If it survives, you have k minus 1 drops left, so the next gap is one smaller. k drops therefore cover k + (k minus 1) + ... + 1 = k(k + 1)/2 floors. 13 drops cover 91 floors, 14 cover 105, so 14 is the minimum: drop from 14, 27, 39, 50 and so on.
Why does binary search fail here?
Binary search assumes you can keep testing after a failure. With two eggs, the first break leaves you one egg, and one egg can only be used safely by walking up one floor at a time. Once the first egg breaks, every floor below it that has not been ruled out costs one drop of the second egg, so large jumps with the first egg are expensive. Dropping the first egg at floor 50 and seeing it break could cost 49 more drops. It is like searching for a leak with one spare pipe: once the first one bursts, you test the rest slowly.
How do you balance the worst cases?
Make every worst case take the same number of drops. If you allow k drops in total, the first drop should be from floor k, the next k minus 1 floors higher, the next k minus 2 higher, because each first-egg drop used leaves one fewer drop for the second egg's walk. With k = 14 the first egg goes from 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99 and 100. If it breaks at 27, the second egg tests 15 to 26: 2 plus 12 is 14 drops. The same count holds at every step.
Dropping the first egg from floors 14, 27, 39, 50 and onward, with each gap one floor smaller, makes every worst case exactly 14 drops, because 14 drops can cover up to 14 x 15 / 2 = 105 floors while 13 drops cover only 91.
The relationship
$k + (k-1) + \dots + 1 = \frac{k(k+1)}{2} \ge 100 \;\Rightarrow\; k = 14$
k the number of drops you allow in the worst case
k(k+1)/2 the most floors k drops with two eggs can cover
What it says in wordsThe smallest k whose triangle number reaches 100 is the answer.
Say what an interviewer wants beyond the number. The move is to fix the budget of drops and ask how many floors it can cover, rather than fixing the building and searching for a strategy. That reversal is what makes the problem easy, and it generalises: with three eggs and k drops, the floors covered are the two-egg coverage for each smaller budget, plus one per drop, added up, which is why three eggs need only 9 drops for 100 floors.
Where candidates lose it
The common loss is answering 7 from binary search, which forgets that the second break ends the experiment. The next is 19 from fixed steps of ten, which is safe but not the minimum.
The other loss is reaching 14 by trial and error and not being able to say why 13 fails. Give the k(k + 1)/2 argument: 13 drops cover at most 91 floors.
What the interviewer asks next
What if you have three eggs?
With two eggs, how many floors can you handle with 20 drops?
What is the expected number of drops with your strategy if the breaking floor is uniformly random?
045You make a market on the number of heads in 10 fair coin flips: 4.5 bid, 5.5 offered. A counterparty who has already seen the first three flips lifts your offer. What does the trade tell you, and where do you requote?Market making and trading gamesHardCitadelNew York · 2025
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Given that they bought at 5.5, the fair value is about
Show the worked solution
The lift says they saw at least two heads, so the fair value is now at least 5.75, not 5; requote around 5.75 bid, 6.5 offered. The other seven flips are worth 3.5 heads, so the buyer's value is heads seen plus 3.5. Paying 5.5 only makes sense with two heads (5.5) or three (6.5). Those are 3 to 1 likely, giving 5.75; a buyer who needs a strict edge saw three heads, worth 6.5.
What is the trader's view before they trade?
A friend offers to buy your raffle ticket after the first few numbers are drawn. The offer itself is the warning. The informed trader values the contract at heads already seen plus 3.5, the expected heads in the seven unseen flips, so their value is 3.5, 4.5, 5.5 or 6.5 with chances 1, 3, 3 and 1 in 8. Against your market of 4.5 bid and 5.5 offered, they buy only if their value is at least 5.5, and sell to you at 4.5 only if it is 4.5 or less. The flat 5 you quoted around is right only for someone who has seen nothing.
The informed trader's value is 3.5, 4.5, 5.5 or 6.5 depending on how many heads they saw, so a lift at 5.5 means two or three heads, and weighting those 3 to 1 puts the fair value given the trade at 5.75, above your 5.5 offer.
How do you turn the trade into a new fair value?
Condition on the fact that they traded. Only the two-head and three-head worlds produce a buy at 5.5, and they are 3/8 and 1/8 likely, so given a lift the value is (3 x 5.5 + 1 x 6.5) / 4 = 5.75. If you assume they would not bother trading at zero edge, only the three-head world is left and the value is 6.50. Either way you sold too cheaply: this is adverse selectionThe tendency of a market maker to trade most with the people who know more, so the trades that happen are the ones that lose money for the market maker., and it is the cost every market maker prices into the spread.
Now requote. Your bid should not be below what you now believe the floor is, and your offer should sit where even the best informed buyer has no edge. Something like 5.75 bid, 6.5 offered does both: a buyer who saw three heads is indifferent at 6.5, and you are no longer selling below value. Cut your size too, because you know someone is trading with more information than you, and say you would ask whether they could see the flips before quoting again.
Where candidates lose it
The common loss is staying at 5 because the coin is fair. The coin is fair; the counterparty is not uninformed. The trade itself carries information and you must update on it.
The other loss is overreacting and moving to 8 or 9, as if the trader knew all ten flips. They saw three. Condition on what could have made them trade, weight those worlds, and move by exactly that much.
What the interviewer asks next
The same trader then hits your new bid. What do you conclude?
How wide should your first market have been if you knew one counterparty could see three flips?
What if the trader had seen the first three flips but traded a small size and then a large size?
Asked at Citadel, Quantitative Trading, New York, 2025 (Wall Street Oasis): Superday was more market-making but requires very sold foundation in math and statistics.
046A fund has annual volatility of 12% and its benchmark index has 15%. The correlation between them is 0.9. What is the fund's tracking error?Portfolio and risk mathsCoreMulti-manager platformsQuant and systematic funds
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Your estimate of the tracking error:
Show the worked solution
About 6.7%. Tracking error is the volatility of the fund's return minus the benchmark's. The variance of a difference is the two variances minus twice the covariance: 0.12 squared plus 0.15 squared minus 2 x 0.9 x 0.12 x 0.15, which is 0.0144 + 0.0225 - 0.0324 = 0.0045. The square root is 6.71%. A correlation of 0.9 sounds tight but still leaves a sizeable gap.
What exactly is tracking error measuring?
Two friends walking to the same station take slightly different routes; tracking error is how far apart they typically are, not how fast either walks. Tracking errorThe standard deviation of the difference between a fund return and its benchmark return, usually quoted per year. is the volatility of the return difference, fund minus benchmark, so it depends on both volatilities and on how closely the two move together. The variance of a difference is Var(F) + Var(B) - 2 Cov(F, B), and the covariance is the correlation times the two volatilities.
Drawn as arrows of length 12 and 15 set 25.8 degrees apart, the angle whose cosine is 0.9, the fund and benchmark tips sit 6.7 apart, which is the tracking error; it would be 3.0% only at a correlation of 1 and rises to 13.7% at 0.5.
Why is the answer so much bigger than 3%?
Because the correlation is below 1, the gap is not just the difference in size. The variance of the difference is 144 + 225 - 324 = 45 in squared percent, and the square root of 45 is 6.7%: the 10% of correlation that is missing contributes more than the 3 point difference in volatility. The arrow picture shows it: two arrows 12 and 15 long, set about 26 degrees apart, have tips further apart than 3. That is just the law of cosines.
The relationship
$TE = \sqrt{\sigma_F^2 + \sigma_B^2 - 2\rho\,\sigma_F\sigma_B} = \sqrt{0.0144 + 0.0225 - 0.0324} \approx 6.7\%$
sigma_F, sigma_B the fund and benchmark volatilities, 12% and 15%
rho the correlation between their returns, 0.9
What it says in wordsTracking error is the volatility of the gap between fund and benchmark returns.
Say what it means for an allocator. A fund with 0.9 correlation to its index can still trail or beat it by 6 to 7 points in an ordinary year, so a single year of underperformance tells you very little. The limitation: the calculation assumes the correlation and volatilities are stable, and they tend to shift in stressed markets, which is when tracking error matters most.
Where candidates lose it
The fast wrong answer is 3%, the difference in volatilities, which is only true at a correlation of exactly one. The interviewer chose 0.9 precisely because it sounds close to one.
The second loss is forgetting the factor of 2 on the covariance term, which gives root(207), over 14%. Write the variance of a difference in full before plugging in numbers.
What the interviewer asks next
What correlation would give a tracking error of 5%?
If the fund had a beta of 0.72 to the index, what is its information ratio if it beats the index by 2% a year?
Why might a fund's realised tracking error jump in a sell-off?
047An index rises 10% one day and falls 10% the next. What happens to the index over the two days, and to a fund that delivers exactly twice the index's daily return?Returns, compounding and feesCoreFund of funds and allocatorsMulti-manager platforms
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Over the two days, the 2x fund is
Show the worked solution
The index ends down 1% and the 2x fund down 4%, not 2%. The index goes 100, 110, 99. The fund goes 100, 120, 96, because it doubles each day's move: up 20%, then down 20% of a bigger number. Two days of index returns of plus and minus r cost the index r squared, 1%, but cost the 2x fund 4r squared, 4%. The drag comes from resetting leverage daily, and it grows with volatility.
Why is the index down at all after up 10% and down 10%?
A salary cut of 10% after a 10% raise leaves you below where you started, because the cut is taken on the bigger salary. An up move of r followed by a down move of r multiplies to (1 + r)(1 - r) = 1 - r squared, so the index loses r squared, here 1%. This is the same volatility drag that makes compound returns lower than average returns. It is small for the index because r squared is small.
Why does the 2x fund lose four times as much?
Because it resets its leverage each day. The fund multiplies to (1 + 2r)(1 - 2r) = 1 - 4r squared, so its drag is four times the index's, not two times. With r = 10% that is 4%. A buyer expecting twice the index return over the period expected 2 x (-1%) = -2% and got -4%. The extra 2 points is the price of re-levering after the gain, which buys more exposure at the top, and de-levering after the loss.
Over two days of plus and minus 10%, the index ends at 99, down 1%, while the fund that doubles each daily move ends at 96, down 4%, twice the naive 2% loss, because leverage is reset every day.
The relationship
$(1+Lr)(1-Lr) = 1 - L^2 r^2 \qquad L = 2,\; r = 0.10:\; 1 - 0.04 = 0.96$
L the daily leverage multiple, 2
r the size of each day's index move, 10%
What it says in wordsA leveraged fund's drag in a choppy market grows with the square of the leverage.
Say where the drag does not apply. In a steady trend, daily resetting helps: two days of +10% take the index to 121 and the 2x fund to 144, more than the naive 142. Daily-reset leveraged funds win in trends and lose in chop, so holding one for months is a bet on the path, not just the direction. That is why allocators treat them as trading tools rather than long-term holdings.
Where candidates lose it
The common loss is saying the 2x fund is down 2%, doubling the index's two-day result. The fund doubles each daily return, and the compounding does the rest.
The second loss is saying the index is flat. Up 10% and down 10% is always a loss; say 1 minus r squared and the two answers come together.
What the interviewer asks next
What is the two-day return of a fund delivering minus twice the daily return?
Over a year with 20% index volatility and no trend, roughly how much does a 2x daily fund lag twice the index return?
Two days of +10%: how does the 2x fund compare with twice the index return?
048A strategy's true annual Sharpe ratio is 1.0. How many years of monthly returns do you need before its average return shows a t-statistic of 2? What if the true Sharpe is 0.5?Statistics and estimationCoreViking Global InvestorsNew York · 2014
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Years needed for a Sharpe of 0.5:
Show the worked solution
About 4 years for a Sharpe of 1.0 and about 16 years for a Sharpe of 0.5. The t-statistic of a mean return is the mean over its standard error, which works out to the annual Sharpe ratio times the square root of the number of years, whatever the data frequency. Setting Sharpe x root(years) = 2 gives years = (2 / Sharpe) squared: 4 for 1.0, 16 for 0.5 and just 1 for 2.0.
Why does the t-statistic grow with the square root of time?
A coin that lands heads 55% of the time looks fair after 20 tosses; you need hundreds before the bias shows through the noise. The average return grows in proportion to time, but the noise around it grows only with the square root of time, so the signal-to-noise ratio, the t-statistic, grows with root time. With monthly data, the t-statistic is the monthly Sharpe times root(12 x years), and the monthly Sharpe is the annual Sharpe divided by root 12, so the twelves cancel: t = annual Sharpe x root(years).
Because the t-statistic equals the Sharpe ratio times the square root of years, a Sharpe of 2.0 clears t = 2 after 1 year, a Sharpe of 1.0 after 4 years and a Sharpe of 0.5 only after 16 years.
Why does monthly data not shorten the wait?
More frequent data gives more observations but each is noisier relative to its mean. Sampling the same years more often does not add information about the mean return; only more years do. This is why a {term('t-statistic', 'An estimate divided by its standard error; a value around 2 is the usual threshold for saying an effect is unlikely to be pure noise.')} on the average return depends on the span of the data, not the number of rows. Frequency helps you estimate volatility, not the mean.
The relationship
$t \approx SR_{\text{annual}}\sqrt{Y} \;\Rightarrow\; Y = \left(\frac{2}{SR}\right)^2: \quad SR = 1 \to 4, \quad SR = 0.5 \to 16$
SR the true annual Sharpe ratio
Y years of data
2 the target t-statistic
What it says in wordsThe years needed to prove a strategy grow with the inverse square of its Sharpe ratio.
Say the practical point. Most real strategies have Sharpe ratios well below 1, so their track records are too short to separate skill from luck with any confidence. An allocator looking at a three-year record with a Sharpe of 0.8 sees a t-statistic of about 1.4. The limitation of the rule: it assumes returns are independent and stable over the whole sample, and fat tails or regime changes make the real uncertainty larger.
Where candidates lose it
The common loss is thinking monthly data gives twelve times the evidence, which leads to answers like four months. The twelve cancels, because the monthly Sharpe is smaller by root 12.
The second loss is saying a Sharpe of 0.5 needs twice as long as 1.0. The dependence is on the square: half the Sharpe, four times the data.
What the interviewer asks next
How many years for a Sharpe of 0.3?
You test 20 strategies and pick the best one with t = 2.2. How much do you trust it?
Would daily data change the answer for estimating the Sharpe ratio itself rather than the mean?
Asked at Viking Global Investors, Quantitative Research, New York, 2014 (Wall Street Oasis): how to reject a hypothesis test, what's your structure of your code, what's the sample size
049You short a stock at Rs 100 at 5% of NAV. It rises to Rs 150 while the rest of the book is flat. What is your loss, and what share of NAV is the short position now?Valuation, accounting and macro riddlesCoreLong-short equity fundsGlobal macro funds
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The short's share of NAV at Rs 150 is about
Show the worked solution
You lose 2.5% of NAV, and the short is now about 7.7% of the book. On NAV of 100, the short is worth 5. A 50% rise makes it a liability of 7.5, a loss of 2.5, so NAV falls to 97.5. The position is now 7.5 over 97.5, which is 7.7%. A losing short grows as a share of the book, so risk rises exactly when the trade is going wrong, the opposite of a losing long.
How do the loss and the new weight work out?
Keep NAV at 100 so every number is a percentage. The short is a liability that rises with the price: 5 at Rs 100 becomes 7.5 at Rs 150, a loss of 2.5, which takes NAV from 100 to 97.5. The weight is the liability over the new NAV: 7.5 / 97.5 = 7.69%. Both parts of the fraction move against you: the top grows and the bottom shrinks.
Why is a losing short more dangerous than a losing long?
A losing long is like a debt that shrinks as the thing you own loses value: a 5% long falling to half is worth 2.5 of NAV 97.5, 2.6%, and it can never lose more than the 5 you put in. A losing short does the reverse: the more it loses, the bigger it gets, and there is no ceiling on the price, so there is no cap on the loss. Left alone, a short that doubles is 10 of NAV 95, 10.5%, and one that triples is 16.7%. This is why long-short funds size shorts smaller and cut them faster than longs.
A 5% long whose price halves shrinks to 2.6% of the book and can lose at most 5% of NAV, while a 5% short whose price rises to 150 grows to 7.7% and keeps growing, with no cap on the loss.
The relationship
$w_{\text{short}} = \frac{0.05 \times 1.5}{1 - 0.05 \times 0.5} = \frac{7.5}{97.5} \approx 7.7\%$
0.05 the starting weight, 5% of NAV
1.5 the price relative, 150 over 100
0.05 x 0.5 the loss as a share of starting NAV, 2.5%
What it says in wordsThe short's new weight is its grown liability over the NAV that the loss has reduced.
Add what a desk does about it. Many funds set a stop or a maximum weight for each short and cover part of it as it rises, so that a losing short is brought back towards its original risk. The discipline exists because the position sizes itself up without anyone deciding to. Short squeezes make this worse, since the price can gap up on little news when many holders cover at once.
Where candidates lose it
The common loss is saying the weight is 7.5%, forgetting that NAV fell. It is 7.5 over 97.5, not over 100. A smaller trap is saying 5%, as if the short's size were fixed at the entry value.
The bigger miss is not drawing the lesson. The interviewer wants to hear that shorts grow when they lose and longs shrink, so short risk compounds against you and needs its own limits.
What the interviewer asks next
At what price does the short reach 10% of NAV?
How much would you buy back to bring the short to 5% of NAV at Rs 150?
The stock pays a dividend while you are short. What happens to your P&L?
050A researcher regresses 12-month forward returns on a signal using monthly observations, so consecutive observations overlap by 11 months, and reports a t-statistic of 4.0 from ordinary least squares. Roughly what is the honest t-statistic?Statistics and estimationHardQuant and systematic funds
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The honest t-statistic is closest to
Show the worked solution
Roughly 1.2, not 4.0. Consecutive 12-month returns share 11 months, so 240 monthly rows over 20 years hold only about 20 independent observations. OLS standard errors assume independence and come out too small by roughly the square root of the overlap, root 12, about 3.5. Dividing 4.0 by 3.46 gives about 1.15: the result is no longer significant. A Newey-West or Hansen-Hodrick standard error does this properly.
What does the overlap do to the regression?
Asking twelve friends for restaurant advice sounds like twelve opinions, but if eleven of them only repeat what the first one said, you have heard about one. Each 12-month return shares 11 months with its neighbour, so the rows are mostly the same data counted again, and the regression thinks it has twelve times more independent evidence than it does. The slope estimate is not biased by the overlap. What breaks is the standard error, because the residuals are strongly correlated from one row to the next, and that breaks one of the {term('OLS assumptions', 'The conditions under which ordinary least squares standard errors are correct, including residuals that are uncorrelated across observations.')}.
Monthly observations of 12-month returns share 11 of every 12 months, so 240 rows over 20 years hold only about 20 independent observations, and the reported t-statistic of 4.0 shrinks to about 1.2 once divided by root 12.
Why divide by root 12 and not by 12?
The standard error scales with one over the square root of the number of independent observations. If the effective sample is twelve times smaller, the standard error is root 12, about 3.46, times larger, and the t-statistic is 3.46 times smaller: 4.0 becomes about 1.15. This is a rough correction. The exact factor depends on how persistent the signal is: for a slow-moving signal, such as a valuation ratio, it is close to root 12; for a fast-moving one it can be smaller.
The relationship
$t_{\text{honest}} \approx \frac{t_{\text{OLS}}}{\sqrt{h}} = \frac{4.0}{\sqrt{12}} \approx 1.15$
h the overlap horizon, 12 months
t_OLS the t-statistic from plain OLS standard errors, 4.0
What it says in wordsWith overlapping returns of horizon h, the plain t-statistic is too large by about the square root of h.
Say how you would fix it properly: use Newey-West standard errors with at least 11 lags, or Hansen-Hodrick errors built for exactly this overlap, or run the regression on non-overlapping annual data and accept the smaller sample. Any of those should give a t-statistic well below 4.0, and a researcher who reports only the OLS number has not yet shown the signal works.
Where candidates lose it
The common loss is accepting the 4.0 because the slope looks economically sensible. The overlap does not move the slope; it fakes the precision, and the interviewer wants to see you spot that.
The second loss is overcorrecting, dividing by 12 instead of root 12. Standard errors shrink with the square root of the sample, so the correction is the square root of the overlap.
What the interviewer asks next
How many Newey-West lags would you use here, and why?
Would non-overlapping annual regressions give the same slope but a bigger standard error?
Why do long-horizon return predictability studies often report very high R squared values?

X1, X2	the two independent exponential waits, each with mean one minute
t e^-t	the density of their sum, from convolving the two exponential densities

k	the number of strands in the bowl before a tie
1/(2k-1)	the chance that tie joins the two ends of one strand

V_n	the value of the game with n rolls left
f	the face you just rolled
max(f, V_{n-1})	keep the roll or throw it away, whichever is worth more

k	the number of drops you allow in the worst case
k(k+1)/2	the most floors k drops with two eggs can cover

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