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.

Do not spin: you survive 75% of the time, against 66.7% if you spin. The empty click puts the cylinder on one of the four empty chambers, each equally likely. Because the two bullets sit together, three of those four empties are followed by another empty and only one is followed by a bullet. A fresh spin throws that information away and gives four chances in six.

What does the empty click tell you?

Picture six people in a queue where two friends always stand together. Pick someone at random who is not one of the friends and ask whether the person behind them is a friend. Only one of the four has a friend behind them: the one standing just in front of the pair. The empty click is information: it tells you which chambers you could be on, and using it is the whole puzzle. The cylinder has no memory, but you do.

Why does the answer flip if the bullets are not adjacent?

Separate the bullets, say into chambers 1 and 4. The four empties are now 2, 3, 5 and 6, and the chambers after them are 3, 4, 6 and 1: two empties and two bullets. Not spinning now survives only 2 times in 4, 50%, so spinning, at 66.7%, becomes the better choice. Adjacency is what bunches both bullets behind a single empty chamber. Ask where the bullets sit before you answer, and say that the answer depends on it.

Where does this reasoning show up on a desk?

The same move, updating on what you have just observed instead of resetting to the base rate, is how a trader reads a fill. Getting filled on your bid tells you something about who was selling, just as the empty click tells you which chamber you are on. Ignoring it is the equivalent of spinning the cylinder: it feels neutral, but it throws away an edge you were handed for free.

The probability is 7/8, and the density of X + Y is a triangle, a tent peaking at 1. Because X and Y are independent and uniform, every point of the unit square is equally likely, so probability is area. The line x + y = 1.5 slices off a corner triangle with legs of 0.5, area 1/8. The sum's density rises in a straight line from 0 to 1 and falls back to 0 at 2.

Why does probability become area here?

Throw a dart at a square board so that every point is equally likely to be hit. The chance it lands in a region is that region's share of the board. Two independent uniforms are exactly that dart: the pair (X, Y) lands evenly on the unit square, so any question about X and Y becomes a question about an area. The condition X + Y below 1.5 is everything under the line x + y = 1.5, which is the whole square except one corner.

How do you get the shape of the sum's density?

Slide the line x + y = s across the square and watch how long it is inside. Near s = 0 it barely clips the corner; at s = 1 it runs corner to corner, the longest it gets; past 1 it shortens again. The density of the sum at s is proportional to the length of that line inside the square, which gives a triangle rising from 0 to a peak at 1 and falling to 2. This is the convolutionThe density of a sum of independent variables, found by adding up every way the two parts can combine to the same total. of two flat densities, and the same reason two dice most often total 7.

Check the first answer with the tent. The area beyond 1.5 is a triangle with base 0.5 and height 0.5, which is 1/8 again. Two routes that agree is the check an interviewer wants to hear before you commit. Add a third uniform and the density becomes three joined curved pieces; add many and the sum looks normal, which is the central limit theorem arriving in slow motion.

Randomise: pay Rs 3.15 with probability 0.159 and Rs 3.14 otherwise. Pi is 3.14159..., which sits 0.159 of the way from 3.14 to 3.15. Paying the higher amount with exactly that probability makes the expected payment 3.14 + 0.01 x 0.1593..., which is pi. So the chance you pay Rs 3.15 is about 15.9%, and over many meals nobody is short-changed.

Why can no fixed amount be fair?

Always round to Rs 3.14 and the restaurant loses 0.159 paise every time; always pay 3.15 and you overpay 0.841 paise. Any fixed amount is unfair to one side, so the only way to be exactly fair is to be fair on average. Two friends who split a Rs 101 bill by taking turns to pay the odd rupee are doing the same thing: neither is exact on any one night, both are exact over time.

How do you actually draw a probability of 0.159?

Use any randomness you can split finely. Draw a uniform number between 0 and 1 and pay Rs 3.15 if it falls below 0.1593. With only a die, paying 3.15 on a six gives 1/6, an expected payment of Rs 3.141667: close, not exact. With only a fair coin you can be exact: toss it to generate the binary digits of a uniform number one at a time and stop as soon as the digits so far settle which side of 0.1593 it falls. Each toss settles it with probability one half, so on average it takes two tosses.

Where does randomised rounding show up in a fund?

Whenever a quantity has to be split in whole units. A fund allocating 1,003 shares across three accounts cannot give each 334.33; handing the odd share out by lottery, or in rotation, keeps each account fair on average. The principle is the same: when the exact amount is impossible, make the expected amount exact and keep the error unbiased. The limitation is that fair on average is not fair every time, which is why allocation policies also cap how far any account can drift.

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.

The price falls by about 2.94%. Spread duration of 6 says a 0.50 percentage point widening costs 6 x 0.50% = 3.00%. Convexity of 50 adds back one half x 50 x 0.005 squared, about 0.06%, because the price curve bends upwards. On a bond priced at 100 that is a move to about 97.06. At 50 basis points the convexity term is small; at 300 or 500 it is not.

What do duration and convexity each measure?

Picture a playground slide that curves and flattens towards the bottom. Judge the drop from the steepness at the top and you overstate it, because the slide levels off as you go. Spread durationThe percentage change in a bond price for a one percentage point change in its credit spread, holding the risk-free rate fixed. is the steepness at today's spread; convexity is the flattening, so the straight-line estimate always overstates the loss when spreads widen. Duration gives the first-order move, 6 x 0.50% = 3.00% down; convexity corrects it by a term that depends on the square of the move.

When does the convexity term start to matter?

It grows with the square of the move. At 50 basis points convexity is worth 0.06% against a 3.00% duration loss; at 300 basis points duration says -18% and convexity adds back 2.25%, which is no longer small. That is why a credit desk can run duration-only risk for everyday moves but needs convexity for stress scenarios. One more distinction marks a strong answer: for a fixed-coupon bond spread duration and rate duration are close, but a floating-rate note has almost no rate duration and still carries several years of spread duration.

Say the limitation plainly. Both numbers are local, measured at today's spread, and a distressed bond stops behaving like this long before default, when its price starts tracking the expected recovery instead. For a bond trading near par, as here, the two-term estimate is good to a few hundredths of a per cent for moves of this size.

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.