Hedge Funds puzzles, solved step by step

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.

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%.

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.

1,225 correlations and 1,275 parameters in all. A covariance matrix is symmetric, so only the cells above the diagonal carry new information: one per pair of stocks, 50 x 49 / 2 = 1,225. The diagonal holds the 50 variances. The count grows with the square of the number of names, which is why large books estimate risk through a handful of factors instead.

Why do you count pairs rather than cells?

In a class of 50, how many handshakes happen if everyone shakes everyone else's hand once? Each person shakes 49 hands, but every handshake has been counted twice, once from each side, so it is 50 x 49 / 2 = 1,225. A correlation is a handshake: it belongs to a pair, and the pair A and B is the same pair as B and A. The diagonal is each stock paired with itself. Its correlation is 1 and needs no estimating, but the diagonal of the covariance matrix holds each stock's variance, which does.

Why does the count become a problem for a big book?

Because it grows with the square of the names. Ten times as many stocks needs about a hundred times as many correlations: 500 names need 124,750 of them plus 500 variances, 125,250 parameters. A year of daily returns on 500 names is 250 x 500 = 125,000 numbers, fewer than the parameters being estimated. With fewer days than stocks the sample matrix is singular: some combinations of positions appear to carry zero risk, and an optimiser will pile into exactly those.

What does a factor model buy you?

A factor model says each stock's return is driven by a few shared drivers, such as the market, its sector and its size, plus noise of its own. With 5 factors, 50 stocks need 250 loadings, 50 specific variances and 15 factor covariances: 315 numbers instead of 1,275. At 500 names it is 3,015 instead of 125,250. The limitation is worth saying: any risk the factors do not name is assumed independent across stocks, and in a crowded unwind that assumption is the first to break.

Light rope one at both ends and rope two at one end at the same moment; when rope one burns out, light rope two's other end, and it burns out at 45 minutes. Two flames always meet after burning 60 minutes of rope between them, so rope one takes 30 minutes however uneven it is. Rope two then has 30 minutes left, which two flames finish in 15.

Why does lighting both ends halve the time on an uneven rope?

Picture two people eating a long, uneven sandwich from opposite ends, each chewing through whatever is at their end. However the filling is spread, they meet once the whole sandwich has been eaten between them, and together they finish in half the time one would take. Two flames consume the rope's total burn time twice as fast, so a 60 minute rope lit at both ends is gone in 30 minutes, wherever the flames happen to meet. Length tells you nothing here; burn time is the only quantity you can trust.

Why is this really a question about information?

The rope hides where its time is stored, much as an order book hides how much size is waiting behind a price. The solution uses only what is known, the total burn time, and never what is not, how it is spread along the rope. Anyone who cuts a rope in half is assuming evenness that the first sentence ruled out. Say that out loud before you give the method: naming what you may not assume is half of a good answer.

Expect the follow-up. The same trick measures 15 minutes as an interval, the gap between rope one going out and rope two going out. Each rope lit from its second end at a known moment halves whatever burn time it has left, and chaining those halvings is how you reach times such as 52.5 minutes with a third rope. Walk through the chain in order, one lighting at a time.

Replicate it with a tight call spread: buy 5 calls at 10,990 and sell 5 at 11,010. The position pays 0 below 10,990 and 100 above 11,010, a steep ramp standing in for the step. So the digital costs about 5 x (C at 10,990 minus C at 11,010). With calls at 212.40 and 203.60 that is 5 x 8.80 = Rs 44. In the limit, the price is minus 100 times the slope of call prices against strike.

Why does a call spread look like a step?

A steep enough ramp can stand in for a stair. A call spread's payoff is a ramp: nothing below the lower strike, rising point for point between the strikes, flat above the upper strike. Narrow the strikes and scale up the size, and the ramp tightens into the step a digital pays. Here the strikes are 20 points apart, so each spread pays at most 20, and five spreads pay at most 100, the digital's payout.

What does the price of the spread tell you?

The spread costs the difference in call prices, so the digital costs five times that. As the strikes close in, the price becomes minus 100 times the slope of the call price against strike, and that slope is the discounted market-implied chance of finishing above the strike. With the illustrative quotes, 8.80 across 20 points is a slope of 0.44, a digital worth Rs 44 and an implied chance of about 44% before discounting.

Which spread does a desk that sold the digital actually buy?

A desk that has sold the digital wants a hedge that pays at least 100 wherever the digital does. The centred spread overpays just below the strike and underpays just above it, so a seller hedges with five 10,980 / 11,000 spreads, which pay the full 100 by the strike and cost a little more. That difference is what the desk charges for an index that settles right at the strike. One more point marks a strong answer: the slope of call prices includes the change in implied volatility across strikes, so with the usual equity skew, where lower strikes carry higher volatility, the digital is worth more than a flat-volatility model says.

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.

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.

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.

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.

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.

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.

About 64.2%. A useless signal clears a 5% bar by luck one time in twenty. The chance that all 20 stay insignificant is 0.95 to the twentieth, about 35.8%, so the chance that at least one looks like a discovery is 64.2%. On average the search turns up one false signal, as 20 x 5% suggests, but at least one appears in roughly two searches out of three.

Why does testing more ideas manufacture a winner?

Ask twenty friends to flip a coin five times each. Any one of them flips five heads only once in 32 tries, yet the chance that at least one of the twenty does is 47.0%, and that friend will look gifted. Each test is a lottery ticket for a false discovery, and buying twenty tickets makes a win likely even when nothing works. A signal chosen because it looked best among twenty has not passed a 5% test; it has passed a 64.2% one.

How do you correct for it?

Tighten the bar to match the number of tries. The Bonferroni correction tests each signal at 5% divided by 20, which is 0.25%, and that brings the chance of any false discovery back to 4.9%. The cost is power: a real but modest signal now struggles to get through. The other defence is data the search never touched: choose the best signal on one period, then test it once on another.

What does a quant fund take from this?

Research teams run thousands of tests, and the ones that get presented are the survivors. Count every test, including the ones you ran and forgot, because the significance of the survivor depends on how many were tried. That is why systematic funds keep research logs and hold data back, and why a backtest with a t-statistic of 2 means much less after a large search than after one planned test. Say the limitation: the 64% assumes independent tests; correlated signals give a lower figure, but rarely a comfortable one.

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.

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.

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.

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.