Hedge Funds puzzles, solved step by step

About 31% to finish above Rs 110, and about 62% to touch it during the year. A 10% move is half of one year's 20% standard deviation, and a normal variable ends more than half a standard deviation up 30.9% of the time. By the reflection principle, every path that touches 110 and ends below has a mirror twin that ends above, so touching is twice as likely as finishing above: 61.7%.

Why is finishing above 110 about a one in three chance?

Scale the move by the volatility. Over one year the price spreads out with a standard deviation of about Rs 20, so Rs 110 is half a standard deviation above the start, and a normal variable finishes more than half a standard deviation up 30.9% of the time. Treating the price as an arithmetic random walk is close enough for a 10% move; a lognormal model, in which prices cannot go negative, gives a slightly lower figure, about 28%. Say you are approximating, and say which way the error runs.

Why is touching twice as likely as finishing above?

Think of a walker on a foggy path who is equally likely to step forward or back. Once she reaches a marker post, her remaining steps are a fair coin again: from the post she is as likely to end past it as short of it. So for every path that touches 110 and ends below, reflecting the part after the touch gives an equally likely path that ends above; touching paths split evenly between the two. Every path that ends above must have touched on the way, so the chance of touching is twice the chance of ending above.

Where does the factor of two matter on a desk?

Anything that triggers on a touch rather than on the finish. A stop-loss set 10% away is hit about twice as often as the price ends beyond it, and an option that pays on a touch is worth roughly twice one that pays only if the price finishes past the same level. The limitation: the factor of two holds for a driftless, continuously watched walk. Drift, jumps and checking the price only at the daily close all move it, and a checked-daily barrier is touched a little less often than a continuous one.

Weight them 9/13 and 4/13, about 69% and 31%, and the blend's error falls to about 1.66%. For independent unbiased forecasts the best weights are proportional to one over each error variance: 1/4 for the 2% forecast and 1/9 for the 3% one. The blended error variance is 1 over (1/4 + 1/9), which is 36/13, so its standard deviation is 1.66%, better than either forecast alone.

Why does blending two forecasts beat the better one?

Ask two people to guess the weight of a pumpkin at a village fair. One is usually closer, but their mistakes are unrelated, so averaging tends to cancel part of each. Independent errors partly cancel when you average, so even a weaker forecast adds information, provided it gets a smaller weight. Throwing the 3% forecast away leaves you at 2%; blending it in well gets you to 1.66%.

Why inverse variance and not inverse error?

The blend's error variance is w squared times 4 plus (1 minus w) squared times 9. Setting its slope to zero gives w = 9/(4 + 9), so the weights follow one over the variance, which penalises the noisier forecast harder than one over the standard deviation would. Check the alternatives: equal weights give an error of 1.80%, weights of 3/5 and 2/5 give 1.70%, and the inverse-variance weights give the minimum, 1.66%.

What would you check before trusting the blend?

Two assumptions carry the answer. The forecasts must be unbiased and their errors independent; if both forecasters lean on the same survey, their errors are correlated and the gain from blending shrinks. With an error correlation of 0.5, the best blend gives the sharper forecast 6/7 of the weight and improves the error only from 2.00% to 1.96%. Ask where each forecast comes from before you average them.

About 10.75% a year, for ever. In a constant-growth model the forward P/E equals the payout ratio divided by (required return minus growth). With a P/E of 40 and a payout of 0.5, r minus g must be 0.5/40 = 1.25%, so g = 12% - 1.25% = 10.75%. Retaining half its earnings, the company would need a return on equity of 21.5% for ever to fund that growth.

How does a P/E hide a growth assumption?

A flat that rents for Rs 30,000 a month and sells for Rs 1.2 crore is priced at 400 months of rent; a buyer paying that is quietly assuming the rent will grow. A price multiple is a compressed forecast: fix the return investors want and the share of earnings paid out, and the multiple pins down the growth the price needs. The constant-growth model, price equals next year's dividend over (r minus g), divided through by earnings, gives P/E = payout/(r - g).

Is 10.75% for ever plausible?

Test it against the business. Growth funded by retained earnings is return on equity times the share retained, so 10.75% growth with half the earnings kept needs a return on equity of 21.5%, held for ever. Few businesses hold returns like that for decades, and no company can outgrow the economy it sells into indefinitely, so compare the figure with the nominal growth you expect for that economy and say it as your assumption. The price is not wrong by arithmetic; it is demanding by assumption.

How sensitive is the answer?

Very. Because r - g is only 1.25%, every point on the required return moves the implied growth by a full point: at 11% the price implies 9.75%, at 13% it implies 11.75%. Using trailing rather than forward earnings shifts it too: 40 = 0.5(1 + g)/(0.12 - g) gives 10.62%. A high multiple rests on a thin gap between two large numbers, so small changes in either swing the value.

A proposes 98 for himself, 0 for B, 1 for C, 0 for D and 1 for E. Work backwards. With two pirates, D's own vote is half, so he keeps all 100. With three, C buys E with 1 coin. With four, B buys D with 1 coin. With five, A needs two votes beyond his own and buys the two pirates who get nothing in the four-pirate split, C and E, for one coin each.

Why start from the end?

Planning a train journey, you work back from when you must arrive, not forward from when you wake up. Each pirate votes by comparing the offer with what he would get if the proposal failed, so you can only price a vote once you know the next round's outcome, which means solving the smallest game first and working upwards. That method, backward induction, is the whole puzzle.

How does each round play out?

Two pirates, D and E: D proposes 100 for himself, and his own vote is half, so it passes. Three pirates: C needs one more vote and buys E, who gets nothing in the two-pirate game, for 1 coin: 99, 0, 1. Four pirates: B needs one more vote and buys D, who gets nothing in the three-pirate game: 99, 0, 1, 0. Five pirates: A needs two more votes and buys C and E, both empty-handed in the four-pirate game: 98, 0, 1, 0, 1.

What is the interviewer really testing?

Whether you reason about the alternative each party faces rather than about fairness. A vote costs exactly one coin more than what the voter gets if the deal fails, so the cheapest supporters are the ones with the worst fallback. The same logic runs through any negotiation: a creditor backs a restructuring plan when it beats their recovery in liquidation, and support is cheapest from those whose alternative is worst. State the assumptions: perfect rationality, and a pirate who is indifferent votes against, which is why one coin, not zero, is needed.

The MLE is 9.0, the largest observation; it is biased low, and the unbiased correction is 5/4 x 9.0 = 11.25. The likelihood is 1 over theta to the fourth for any theta of at least 9.0 and zero below it, so it peaks at the sample maximum. But the maximum of n draws averages n/(n + 1) of theta, here 4/5, so scaling by (n + 1)/n removes the bias.

Why is the MLE the largest observation?

A friend draws four raffle tickets numbered from 1 up to some unknown top number, and the highest you see is 90. The top number is at least 90; guessing higher only spreads your belief over tickets nobody drew. The likelihood, 1 over theta to the n, is zero for any theta below the largest observation and falls as theta rises above it, so it is maximised exactly at the sample maximum, 9.0. This is a case where you do not differentiate: the maximum sits on a boundary, not where a slope is zero.

Why is it biased, and by how much?

The largest draw can never be above theta and is almost always below it. Four points dropped at random on 0 to theta cut it into five gaps of equal expected length, so the largest point sits on average four fifths of the way up, and the MLE underestimates theta by a fifth on average. Multiplying by 5/4 fixes it: 9.0 becomes 11.25. The bias shrinks as n grows, since n/(n + 1) tends to 1, but with four points it is large.

How does it compare with the obvious alternative?

The method of moments doubles the sample mean, since a uniform on 0 to theta averages theta/2: the mean here is 6.175, giving 12.35. Both 11.25 and 12.35 are unbiased, but the corrected maximum has a much smaller variance, theta squared over n(n + 2) against theta squared over 3n, because the largest draw carries the most information about the top of the range. With four points that is theta squared over 24 against theta squared over 12: half the variance.

Take B, C and D: 10 units for Rs 21 crore. Ranking by profit per unit of risk picks B (2.33), then A (2.25), then D (2.00), which uses 9 units for Rs 20 crore and leaves one unit idle. Swapping A for C costs a little ratio but fills the budget and adds Rs 1 crore. With whole ideas, you check the few sets that fill the budget rather than trust the ratio alone.

Why is profit per unit of risk the right place to start?

Packing a suitcase for a flight with a weight limit, you favour the things that give the most use per kilo. When the budget is the scarce thing, the useful measure is profit per unit of budget, not profit alone. E earns the most in rupees but only 1.83 per unit, the worst of the five. B earns 2.33 per unit, A 2.25, and C and D 2.00 each. If ideas could be split, you would simply fill from the top of that list and the answer would be exact.

Why does the ratio ranking fail here?

Ideas come whole. Take B and A and you have used 7 units; D fits for 9 units, and nothing else fits in the last one. A unit left idle earns nothing, so a set with a slightly lower average ratio that uses the full budget can beat the greedy pick. Replace A (4 units) with C (5 units) and the budget is full: B, C and D earn 7 + 10 + 4 = Rs 21 crore against Rs 20 crore. This is the knapsack problemChoosing whole items, each with a size and a value, to maximise total value without exceeding a capacity., and the fix is to check the handful of sets that nearly fill the budget.

With five ideas there are only 31 possible sets, so enumerate the full ones out loud: B, C, D for 21; A, B, D for 20; A, E for 20; A, C for 19. On a real book with hundreds of positions a desk uses an optimiser, but the logic is the same, and interviewers want to hear that you know the ratio rule is a starting point and not a proof.

Say the limitation too. Expected profit is an estimate, and the gap here is Rs 1 crore on Rs 21 crore. If C's estimate is shakier than A's, a portfolio manager could reasonably prefer the greedy set. The arithmetic tells you the best set on the stated numbers; confidence in each number decides whether that edge is real.

About 15%, not 90%. Picture 10,000 trades. 100 are suspicious and the screen flags 90 of them. 9,900 are clean and it wrongly flags 5%, which is 495. So 585 trades are flagged and only 90 are suspicious: 90 over 585 is 15.4%. The false flags swamp the true ones because clean trades are so common.

Why is 90% the wrong number?

A smoke alarm that goes off whenever there is a fire is good. But if it also goes off every time someone makes toast, most of its alarms are toast. The 90% tells you how often a suspicious trade gets flagged; the question asks how often a flag is suspicious, and those run in opposite directions. Mixing them up is called the {term('base rate', 'How common something is before any test is run. Here, one trade in a hundred is suspicious.')} fallacy, and it is exactly what this question is built to catch.

How do you get the number without the formula?

Use counts, not percentages. Start with 10,000 trades because it makes every number whole. Split by the truth first, then by what the screen says, and then read only the flagged column. 1% of 10,000 is 100 suspicious trades; 90% of those, 90, are flagged. 9,900 are clean; 5% of those, 495, are flagged anyway. The flagged column holds 585 trades, and 90 of them are the real thing.

Add the desk point after the number. A second, independent check changes things fast: run the 585 flagged trades through a second screen with the same error rates and the 15.4% prior becomes about 77%. A weak test is still useful as a first filter; it is only misleading when its hit rate is read as its accuracy.

One in four. Three lengths make a triangle only if no piece is longer than the other two together, which for a stick of length 1 means no piece is longer than one half. Plot the two break points as a dot in the unit square. The dots that keep every piece under half fill two triangles of area 1/8 each, so the chance is 1/4.

What exactly has to be true for a triangle?

Try to make a triangle from three sticks where one is longer than the other two laid end to end: the short ones cannot reach each other across the long one. Three lengths form a triangle only if the longest is shorter than the sum of the other two, and with a total of 1 that means every piece must be under one half. That turns a geometry question into a single condition you can test on any pair of break points.

How do you turn two random breaks into an area?

Call the break points x and y, each uniform between 0 and 1. Every possible outcome is a dot in the unit square, and every dot is equally likely, so probability is area. Take the half where x is below y: the pieces are x, y minus x and 1 minus y, and all three are under half only when x is below 1/2, y is above 1/2 and y minus x is below 1/2. Those three lines cut out a triangle with corners (0, 1/2), (1/2, 1/2) and (1/2, 1), area 1/8. The half where y comes first gives its mirror image, another 1/8.

The red regions are also worth naming, because each one is a different piece being too long. Three failure regions of area 1/4 each are the first piece, the middle piece and the last piece exceeding half. They cannot overlap, since two pieces cannot both be longer than half, so the answer is 1 minus 3/4, which is 1/4, a second route to the same number.

Mention the variant, because interviewers often switch to it: break the stick once, then break the longer piece at a random point. The answer changes to about 0.386, which is 2 ln 2 minus 1, because the second break is no longer uniform over the whole stick. Knowing the setup changes the answer shows you are reading the question.

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.

About 25 trading days, roughly five weeks. The market trades 0.2% of the shares a day and you take at most a fifth of that, so you can sell 0.04% of the company a day. A 1% stake divided by 0.04% a day is 25 days. In the desk's language the position is five days of volume, and exiting it without moving the price takes a month.

What is the one division that answers it?

Emptying a water tank through a tap that you are only allowed to open a fifth of the way: how long it takes is the tank size divided by the flow you actually use. Days to exit equals position size divided by your daily selling capacity, and capacity is market volume times your participation cap. Here that is 1% over (0.2% x 20%), or 1% over 0.04%: 25 days. Keeping everything in percent of shares means you never need the share count or the price.

Why does a portfolio manager care about this number?

Because the price can move a long way in 25 days. A position you cannot exit quickly carries more risk than its daily volatility suggests: if the stock falls 2% a day for a week, you have sold only a fifth of it. That is why many desks cap a position at a set number of days of volume, often a few days, and why liquidity sits beside volatility in position sizing. The measure has a name, days to liquidatePosition size divided by the volume you can realistically trade in a day; a common liquidity limit on hedge fund books., and interviewers like hearing it.

Say the limitations. Volume is not steady; it dries up exactly when you most want to sell, and in a sell-off everyone is trying to do the same thing. A 20% participation rate also moves the price against you, so the real exit costs more than the screen price. A stronger answer adds that you would stress the calculation with half the normal volume, which gives 50 days.