Hedge Funds puzzles, solved step by step

About 3.99, so roughly four. Give each face its own yes-or-no: did it appear at least once? A face is missed in all six rolls with chance (5/6)^6, about 33.5%, so it appears with chance about 66.5%. The expected count is the sum of the six chances: 6 x (1 minus (5/6)^6), which is 3.9906.

Why not work out the chance of seeing exactly 1, 2, up to 6 faces?

You could, but each of those needs a fiddly count. There is a shortcut that works whenever the question asks for an expected count. Think of a teacher checking attendance: the expected number present is simply the sum, over pupils, of each pupil's chance of turning up. Write the total as a sum of yes-or-no indicators, one per face, and the expected total is the sum of their probabilities, whether or not they are independent. That rule is {term('linearity of expectation', 'The expected value of a sum equals the sum of the expected values, even when the parts depend on each other.')}, and it turns a hard count into one line.

What is the chance that a given face shows up?

It is easier to find the chance it does not. Each roll misses the face with chance 5/6, and the six rolls are independent, so the face is missed every time with chance (5/6)^6, which is 15,625 over 46,656, about 33.5%. So each face appears at least once with chance about 66.5%, and six faces give 3.99. The indicators are not independent, since seeing face 1 a lot leaves less room for face 2, but linearity does not care.

Give the general form and a sense check. With n faces and n rolls the answer is n(1 minus (1 minus 1/n)^n), which tends to n(1 minus 1/e), about 63% of n. For a die that predicts 3.79, and the exact 3.99 is a little higher because six is small. The same logic tells a desk how many distinct names a random sample of trades touches.

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.

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.

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.

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.

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.

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).

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

The smaller draw averages 1/3 and the larger averages 2/3. Two random points cut the stretch from 0 to 1 into three pieces, and nothing distinguishes one piece from another, so each has expected length 1/3. The minimum ends the first piece and the maximum ends the second. The two averages add to 1, as they must, because the minimum plus the maximum is just the sum of the two draws.

What is the one-line argument?

Imagine two friends who each arrive at a random minute within the same hour. The wait until the first arrival, the gap between the two arrivals and the time from the second arrival to the end of the hour are three stretches of time. No stretch is special, so each averages a third of the hour, and the first friend arrives, on average, 20 minutes in. The same symmetry puts the minimum of two uniform draws at 1/3 and the maximum at 2/3.

How do you check it with a calculation?

Work from the chance that the minimum sits above some level x. Both draws must be above x, which happens with probability (1 - x) squared. For a quantity that cannot be negative, the average equals the area under its chance of exceeding each level, and the area under (1 - x) squared from 0 to 1 is 1/3. The maximum then follows without a second integral: minimum plus maximum equals X plus Y, which averages 1, so the maximum averages 2/3.

Where does the pattern lead next?

With n draws, the points cut the interval into n + 1 pieces, so the smallest averages 1/(n + 1) and the largest n/(n + 1). The ordered values of uniform draws are evenly spaced on average, and that single fact answers a whole family of follow-ups. With ten bids drawn at random across a range, the highest averages 10/11 of the way up. The quick route here is also the one to say first: the order statisticsThe draws sorted from smallest to largest: the minimum is the first order statistic, the maximum the last. of uniform draws split the interval evenly.

The chance is 8 in 32, exactly 1/4. Sort the qualifying sequences by where their first run of three heads starts. Starting at flip 1: HHH then anything, 4 sequences. Starting at flip 2: flip 1 must be tails, then HHH, then anything, 2 sequences. Starting at flip 3: flip 2 must be tails and flip 1 is free, 2 sequences. Four plus two plus two is eight.

Why does the obvious count go wrong?

Suppose you count guests at a party by asking each room how many people it holds, while some guests stand in doorways. They get counted twice. A run of three can start in three places, but one sequence can contain a run starting at more than one of them, so three places times four fillings, 12, counts HHHHH three times and two other sequences twice. The fix is to give each sequence exactly one home: the position where its first run starts.

How do you make every sequence count once?

Force the flip just before the run to be tails. If the first run starts at flip 2, flip 1 must be tails; if it starts at flip 3, flip 2 must be tails, and flip 1 can be anything. That gives 4, then 2, then 2. Then check a second way by counting the complement: sequences with no run of three follow a Tribonacci-style rule, each term the sum of the three before, giving 1, 2, 4, 7, 13, 24 for 0 to 5 flips. And 32 minus 24 is 8.

Why would a quant fund ask this?

Streaks are what people notice in trading records and backtests. A run of three heads turns up in a quarter of all five-flip sequences of a fair coin, so a short winning streak is weak evidence of skill. The counting habit matters as much as the number: giving each outcome one home is the same discipline that stops you double counting overlapping signals when you tally how often a strategy fired.