Hedge Funds puzzles, solved step by step

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

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032Two ice cream sellers each choose a spot on a straight 1 km beach. Sunbathers are spread evenly along it and each walks to the nearer seller. Where do the sellers end up, and is that the best outcome for the customers?Logic and brainteasersCoreProp and quant trading firmsLong-short equity funds
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Where do two self-interested sellers settle?
Show the worked solution
Both end up side by side in the middle, and customers are worse off. From the quarter points, a seller who steps inward keeps everyone behind them and wins beach from the rival, so both drift to the centre. There they still split customers half and half, but the average walk doubles from 125 m to 250 m. Competition moves the sellers to the spot that maximises share, not the one that serves customers best.
Why can neither seller stay at the quarter points?
Picture two petrol pumps on a highway. Each wants the drivers on its own side plus as many from the middle as it can reach first. A seller keeps every customer on the far side of them wherever they stand, so moving towards the rival only ever adds customers. From 250 m, A steps to 400 m against B at 750 m: the dividing line moves to the midpoint, 575 m, and A's share rises from 50% to 57.5%. B then responds the same way, and the dance ends only when both stand at 500 m.
Sellers at the quarter points split the beach evenly with an average walk of 125 m; when A edges inward to 400 m it wins 57.5% of the beach, and the process ends with both at the middle, still splitting 50/50 but with the average walk doubled to 250 m.
What does the middle cost the customers?
With both sellers in the centre, a sunbather is on average a quarter of the beach away, 250 m. With sellers at 250 m and 750 m, nobody is more than 250 m away and the average is 125 m. The shares are identical in both setups; the only thing that changed is how far customers walk, so the stable outcome is strictly worse for them and no better for the sellers. This is the {term('Hotelling model', 'A model of competition on a line, set out by Harold Hotelling in 1929, in which rivals crowd towards the centre to win the middle ground.')}, and it is why rival shops cluster and why two parties often converge on the middle voter.
Say the equilibrium idea in one line: a pair of positions is stable when neither player can do better by moving alone, and the middle is the only such pair here. Then say where the model breaks: if customers stop buying when the walk is too long, or if prices can differ, the sellers have a reason to spread out again. Interviewers ask this to see whether you can reason about another player's best response, which is most of trading.
Where candidates lose it
The common loss is answering the quarter points, because that is the sensible arrangement. The question asks where self-interested sellers end up, and the quarter points are not stable.
The second loss is getting to the middle but not saying what it costs. The interviewer wants the contrast: same shares, twice the walking. Name the stable point, then name the welfare cost.
What the interviewer asks next
What happens with three sellers?
If customers refuse to walk more than 300 m, where do the sellers stand?
Where do you see the same pattern in markets or in fund positioning?
034A stock trades at Rs 1,000 and its one-month options are priced at 30% implied volatility. Without a calculator, roughly what does a one-month at-the-money straddle cost, and how far must the stock move for the buyer to break even at expiry?Options and payoffsCoreVolatility and relative value fundsProp and quant trading firms
Try it first
Closest estimate for the straddle:
Show the worked solution
About Rs 69, so the stock must move about 6.9% either way. Scale the 30% annual volatility to one month by the square root of time: 30% x root(1/12) is about 8.7%. An at-the-money straddle is worth about 0.8 times that move times the price: 0.8 x 8.7% x 1,000 is about Rs 69. At expiry the buyer profits only below about Rs 931 or above Rs 1069.
Where does the 0.8 x sigma x root T rule come from?
A straddle pays the size of the move, whichever way it goes, so its value is the expected absolute move. For a normal distribution, the average absolute move is the standard deviation times root(2/pi), which is about 0.8. Over one month the standard deviation of the price is S x sigma x root T = 1,000 x 0.30 x root(1/12), about Rs 87. Times 0.8 gives about Rs 69. Black-Scholes with zero rates gives Rs 69.1, so the shortcut is good to within a rupee for short-dated at-the-money options.
A one month straddle struck at Rs 1,000 costs about Rs 69, so at expiry it loses money anywhere between about Rs 931 and Rs 1069 and profits only on a move of more than about 6.9% either way.
Why is the breakeven move bigger than the typical move traders expect?
Buying a straddle is like buying insurance against a big move in either direction: you pay the average claim up front. The breakeven at expiry is the premium, about 6.9%, while the average absolute move is the same 6.9% by construction, so a buyer at fair implied volatility breaks even only on average. The buyer makes money when realised volatility turns out higher than the 30% priced in, and the seller makes money when it turns out lower. That is why desks talk about straddles as a bet on volatility, not on direction.
The relationship
$\text{Straddle} \approx \sqrt{\tfrac{2}{\pi}}\,\sigma\sqrt{T}\,S \approx 0.8 \times 0.30 \times \sqrt{1/12} \times 1{,}000 \approx 69$
sigma implied volatility, 30% a year
T time to expiry in years, 1/12
S the stock price, Rs 1,000
What it says in wordsAn at-the-money straddle costs about 0.8 times the one standard deviation move over its life.
Say the limits. The rule assumes the option is at the money and short dated, and rates are small; for long-dated or out-of-the-money options it drifts. A single call or put is half the straddle, about 0.4 x sigma x root T x S, which is a useful second number to have ready.
Where candidates lose it
The most common loss is forgetting to scale volatility to one month, which gives an answer near Rs 240 or Rs 300. Implied volatility is quoted per year; divide by root 12 for a month.
The second loss is forgetting the 0.8 and pricing the straddle at the full one standard deviation move, Rs 87. Say where the 0.8 comes from, the average absolute value of a normal, and the answer sounds reasoned rather than memorised.
What the interviewer asks next
What does a one-week straddle cost on the same stock?
Roughly what is the one month at-the-money call worth on its own?
The stock moves 5% by expiry. Did the straddle buyer or seller win, and what does that say about realised volatility?
035Three assets all have the same pairwise correlation, rho. What is the lowest value rho can take? What is the answer for n assets?Portfolio and risk mathsCoreMulti-manager platformsQuant and systematic funds
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Lowest possible common correlation for three assets:
Show the worked solution
Minus one half for three assets, and minus 1/(n minus 1) for n. Give each asset unit variance and add them up. The variance of the sum is 3 plus 6 rho, because there are three variances and six pairwise covariance terms. A variance cannot be negative, so rho is at least -1/2. With n assets the sum's variance is n plus n(n minus 1) rho, which gives rho at least -1/(n minus 1): -0.33 for four, -0.11 for ten.
Why can three assets not all be perfectly opposed?
Three friends cannot all disagree with each other on a yes-or-no question: if Ravi says yes and Meena says no, Arjun agrees with one of them. Perfect opposition is a relationship between two things; with three, two of them must lean the same way. The same limit holds for correlations. A set of numbers in a correlation matrix has to be internally consistent, and equal pairwise correlations become impossible well before -1 once there are three or more assets.
How do you find the exact floor?
Use the one fact that can never fail: a variance is zero or more. Add the three assets with unit variance; the sum has variance 3 from the three diagonal terms plus 6 rho from the six covariance terms, and 3 + 6 rho must be at least zero. That gives rho of at least -1/2. The picture is three arrows 120 degrees apart: each pair has a cosine of -1/2, and the three add to exactly zero, which is the boundary case. For n assets, n + n(n minus 1) rho at least zero gives -1/(n minus 1).
Three unit arrows 120 degrees apart add to zero, the arrangement where every pair has correlation minus one half, and the variance of the sum, 3 + 6 rho, turns negative below that value, which is impossible, so rho cannot fall below minus one half.
The relationship
$\operatorname{Var}\Big(\sum_{i=1}^{n} X_i\Big) = n + n(n-1)\rho \ge 0 \;\Rightarrow\; \rho \ge -\frac{1}{n-1}$
n the number of assets, each with variance 1
rho the common pairwise correlation
n(n-1) the number of ordered pairs, each contributing one covariance of rho
What it says in wordsThe variance of the equal-weighted basket must be non-negative, which caps how negative a shared correlation can be.
Say why a risk desk cares. A correlation matrix that breaks this rule is not a valid risk model: it implies some portfolio has negative variance, and an optimiser will pile into it. This happens in practice when correlations are estimated pair by pair from different data windows or overridden by hand in a stress test. The check is that the matrix is positive semi-definiteA matrix for which every weighted combination of the assets has a variance of zero or more., and the equal-correlation case is the cleanest example of the rule.
Where candidates lose it
The usual loss is answering -1, because that is the floor for any single pair. The interviewer is testing whether you see that the pairs constrain each other.
The second loss is getting -1/2 by intuition but having no proof. Say the variance of the sum in one line. It takes ten seconds, it generalises to n immediately, and it is the answer the follow-ups build on.
What the interviewer asks next
Two assets have correlation 0.9 with a third. What is the lowest possible correlation between the first two?
A stress test sets every pairwise correlation in a 20-asset book to -0.1. Is that a valid matrix?
How would you repair a correlation matrix that is not positive semi-definite?
036A regime model says a bull month is followed by another bull month 90% of the time, and a bear month by another bear month 80% of the time. In the long run, what share of months are bull months?Random walks and Markov chainsCoreQuant and systematic fundsProp and quant trading firms
Try it first
Long-run share of bull months:
Show the worked solution
Two thirds of months are bull months, whatever the starting state. In the long run the number of months switching from bull to bear must equal the number switching back. 10% of bull months switch out and 20% of bear months switch in, so 0.1 x bull = 0.2 x bear, which makes bull twice as common as bear: 2/3 against 1/3. A second route: bull spells last 10 months on average and bear spells 5.
What has to balance in the long run?
Think of a shop with people walking in and out all day. Once the crowd inside stops growing or shrinking, the number walking in each minute must equal the number walking out. In the long run, the flow from bull to bear must equal the flow from bear to bull, because otherwise one state would keep filling up. The flow out of bull is 10% of bull months; the flow out of bear is 20% of bear months. Setting 0.1 x bull equal to 0.2 x bear, with bull plus bear equal to 1, gives bull = 2/3.
Bull months turn bear 10% of the time and bear months turn bull 20% of the time, so the long-run share of bull months is two thirds, and a chain started in either state is within a few points of two thirds after about a year.
How do you check two thirds another way?
Use the length of each spell. A state you leave with probability p each month lasts 1/p months on average, so bull spells last 10 months and bear spells 5. Spells alternate, so over a long stretch the market spends 10 months bull for every 5 bear: 10 out of 15 is two thirds. Two methods that agree is what the interviewer is listening for.
The relationship
$\pi_{\text{bull}} = \frac{0.2}{0.1 + 0.2} = \frac{2}{3} \qquad \text{gap after } t \text{ months} \propto (0.9 + 0.8 - 1)^t = 0.7^t$
pi_bull the long-run share of bull months
0.1, 0.2 the chances of leaving bull and leaving bear each month
0.7 how much of any starting gap survives each month
What it says in wordsThe long-run share of a state is the chance of entering it divided by the total chance of switching, and the start is forgotten at a rate of 0.7 a month.
Answer the part of the question people skip: why the starting state does not matter. The gap between today's odds and two thirds shrinks by a factor of 0.7 every month, so after 12 months only 1.4% of it is left. A model this sticky still forgets its starting point within about a year, which is why regime forecasts beyond a few months mostly return the long-run average.
Where candidates lose it
The first loss is answering 90%, the one-step persistence, as if it were the long-run share. The second is saying it depends on today's state, which is true for next month and false for the long run.
Set up the flow balance in one line, give two thirds, then check it with spell lengths of 10 and 5 months. If you have time, say how fast the start is forgotten: 0.7 a month.
What the interviewer asks next
Today is a bear month. What is the chance that the month after next is a bull month?
How long does the average bear spell last, and what is the chance one lasts more than a year?
How would you estimate the two transition probabilities from 20 years of monthly data, and how wide would the error be?
037A strategy has an average annual return of 10% and annual volatility of 20%. Roughly what compound annual growth rate should an investor expect over many years?Returns, compounding and feesWarm upFund of funds and allocatorsMulti-manager platforms
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Your estimate of the compound growth rate:
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About 8% a year, two points below the 10% average. Compound growth is roughly the average return minus half the variance: 10% minus 0.5 x 0.20 squared, which is 10% minus 2%. Check with two years of +30% and -10%: the average is 10% and the volatility 20%, but 1.30 x 0.90 is 1.17, which compounds at 8.17% a year. The drag grows with the square of volatility.
Why does the average return overstate what you end up with?
A shopkeeper whose sales rise 50% one month and fall 50% the next has not broken even: 100 becomes 150 and then 75. A loss is applied to a bigger base after a gain, and a gain to a smaller base after a loss, so swings always drag compound growth below the simple average. The bigger the swings, the bigger the drag. The simple average of yearly returns is called the {term('arithmetic mean', 'The plain average of the yearly returns, adding them up and dividing by the number of years.')}; the rate your money actually grows at is the geometric mean, and it is always the lower of the two when returns vary.
How big is the drag, and where does half the variance come from?
For returns that are not too large, the geometric mean is close to the arithmetic mean minus half the variance. With 20% volatility the variance is 0.04, half of that is 0.02, so a 10% average compounds at about 8%. The two-year example makes it concrete: +30% and -10% average 10% with a standard deviation of 20%, and 1.30 x 0.90 = 1.17, a compound rate of 8.17% a year. The rule of thumb says 8.00%, close enough to trust in an interview.
Holding the average return at 10%, compound growth falls to 8% at 20% volatility and to 2% at 40%, because the drag is about half the variance; two years of +30% and -10% average 10% but compound at 8.17% a year.
The relationship
$g \approx \mu - \tfrac{1}{2}\sigma^2 = 0.10 - \tfrac{1}{2}(0.20)^2 = 0.08$
g the compound annual growth rate
mu the average annual return, 10%
sigma annual volatility, 20%
What it says in wordsCompound growth equals the average return less half the variance.
Add why an allocator asks this. Two funds with the same average return and different volatility do not leave investors with the same money. Cutting volatility from 20% to 10% raises compound growth by 1.5 points with no change in the average, which is part of why lower-volatility strategies can be worth more than their averages suggest. The limitation: the half-variance rule is an approximation that weakens for very volatile or fat-tailed returns.
Where candidates lose it
The common loss is answering 10%, treating the average return as the growth rate. The interviewer wants to hear the word compounding and a number for the drag.
The second loss is subtracting the full variance or the volatility itself, giving 6% or -10%. It is half the variance, and 20% squared is 4%, not 40%. Say the rule, give 8%, and check it with a two-year example.
What the interviewer asks next
At what volatility does a 10% average return compound to zero?
Fund A averages 12% with 30% volatility; fund B averages 10% with 15%. Which grows money faster?
How does leverage change the answer, and what leverage maximises compound growth here?
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?
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?
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?

sigma	implied volatility, 30% a year
T	time to expiry in years, 1/12
S	the stock price, Rs 1,000

n	the number of assets, each with variance 1
rho	the common pairwise correlation
n(n-1)	the number of ordered pairs, each contributing one covariance of rho

pi_bull	the long-run share of bull months
0.1, 0.2	the chances of leaving bull and leaving bear each month
0.7	how much of any starting gap survives each month

g	the compound annual growth rate
mu	the average annual return, 10%
sigma	annual volatility, 20%

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