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.

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.

997 x 1,003 = 999,991 and 67 squared = 4,489. The first is a difference of squares: the numbers sit 3 either side of 1,000, so the product is 1,000,000 minus 9. The second uses the same identity the other way round: 67 squared is 64 x 70 plus 3 squared, which is 4,480 plus 9. Both take one line once you spot the round number nearby.

Why does multiplying around a round number work?

Take a square garden 10 metres a side and reshape it to 13 by 7. The fence is the same length, but the plot shrinks from 100 square metres to 91. It always shrinks by the square of how far you moved each side, here 3 squared, 9. Two numbers spaced equally around a midpoint multiply to the midpoint squared minus the gap squared. For 997 x 1,003 the midpoint is 1,000 and the gap is 3, so the answer is 1,000,000 minus 9.

How do you square a number like 67 in your head?

Push it to a round neighbour and repair the difference. Move 3 down to 64 and 3 up to 70, multiply those, then add back the 3 squared you took away: 64 x 70 = 4,480, plus 9 is 4,489. You choose the gap so one factor is round. The expansion route agrees: 67 is 70 minus 3, so 67 squared is 4,900 - 420 + 9, the same 4,489. Two methods landing on one number is your check.

Why would a fund ask arithmetic at all?

A trader checks prices, spreads and position sizes in their head all day, and a slip costs money before any spreadsheet catches it. Interviewers use speed on arithmetic like this as a proxy for how quickly you would catch a quote that does not add up. Say the identity as you use it, so that if you slip, the interviewer can see where and you can recover out loud.

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.

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.

One round trip makes at most 16; unlimited round trips make 26. For one trade, walk the prices once, carrying the lowest price so far and the best sale against it: buy at 96, sell at 112. For many trades, add every day-on-day rise and skip every fall: 8 + 11 + 7 = 26. Without short selling the falls are simply sat out, never profited from.

How do you find the best single trade without checking every pair?

Imagine walking down a street of shops that all sell the same phone, planning to buy once and sell once further along. You do not need to compare every pair of shops: carry the cheapest price seen so far in your head, and at each shop ask what selling here would make against it. One pass, keeping the running minimum and the best gap found so far, gives the best single trade. Here the running minimum drops to 96 on day 2 and the best gap appears on day 7: 112 - 96 = 16.

Why is the many-trade answer just the sum of the rises?

Any rise from a low to a later high is the sum of the daily steps inside it, and some of those steps may be falls. With no short selling and no costs, the most you can make is the total of every positive day-on-day move, 26 here, because trading only the up steps collects everything a longer trade would and skips its falls. In practice that is three round trips: buy 96, sell 104; buy 99, sell 110; buy 105, sell 112.

What does the interviewer add next?

Costs. Once each round trip costs something, the sum of rises overstates the profit, because small moves stop being worth trading. With a cost of 6 per round trip, the three separate trades net 2 + 5 + 1 = 8, the best two-trade split nets 9, and the single trade from 96 to 112 nets 10, so the single trade now wins. The general version is a short dynamic programme that tracks the best profit on each day while holding and while flat.

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.