Hedge Funds puzzles, solved step by step

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.

21. The number of rolls until the first 6 averages 6, so there are on average 5 non-six rolls plus the 6. A roll known not to be a 6 is equally likely to be 1 to 5, so it averages 3, not 3.5, giving 5 x 3 + 6 = 21. Wald's identity confirms it: 6 expected rolls times 3.5 a roll is also 21, because a stop that looks only at past rolls does not bias the total.

Why is it tempting to get 23.5?

Suppose you keep buying scratch cards until one wins. Every card before the winner is, by definition, a loser, so those cards are worth less than an average card. The stopping rule changes the rolls before the stop: each one is known not to be a 6, so it averages 3, and valuing them at 3.5 overpays by 0.5 a roll, 2.5 in all. Five non-sixes at 3.5 plus a 6 is 23.5, which counts the high side twice: once in the 3.5 and again in the final 6.

How do the two routes agree?

Route one splits the sum: the expected number of non-six rolls times their average, plus the final 6. The count of rolls is a geometric wait with success chance 1/6, so it averages 6, of which 5 are non-sixes: 5 x 3 + 6 = 21. Route two is Wald's identityFor a stopping rule that uses only rolls already seen, the expected total equals the expected number of rolls times the average roll.: the expected total is the expected number of rolls times the average roll, 6 x 3.5 = 21. The low early rolls and the high final roll balance exactly.

Why would a trading firm ask this?

Stopping rules are everywhere on a desk: exit at the first stop-loss hit, rebalance at the first breach of a band. The question checks whether you can tell when a stopping rule biases what you observe, as it does for the early rolls, and when it does not, as for the total. Say the condition too: Wald's identity needs the decision to stop to use only rolls already seen, and the expected number of rolls to be finite. A rule that could peek at the next roll would break it.

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.

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.

14 drops. With k drops available, the first egg should go from floor k: if it breaks, the second egg checks the k minus 1 floors below one at a time. If it survives, you have k minus 1 drops left, so the next gap is one smaller. k drops therefore cover k + (k minus 1) + ... + 1 = k(k + 1)/2 floors. 13 drops cover 91 floors, 14 cover 105, so 14 is the minimum: drop from 14, 27, 39, 50 and so on.

Why does binary search fail here?

Binary search assumes you can keep testing after a failure. With two eggs, the first break leaves you one egg, and one egg can only be used safely by walking up one floor at a time. Once the first egg breaks, every floor below it that has not been ruled out costs one drop of the second egg, so large jumps with the first egg are expensive. Dropping the first egg at floor 50 and seeing it break could cost 49 more drops. It is like searching for a leak with one spare pipe: once the first one bursts, you test the rest slowly.

How do you balance the worst cases?

Make every worst case take the same number of drops. If you allow k drops in total, the first drop should be from floor k, the next k minus 1 floors higher, the next k minus 2 higher, because each first-egg drop used leaves one fewer drop for the second egg's walk. With k = 14 the first egg goes from 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99 and 100. If it breaks at 27, the second egg tests 15 to 26: 2 plus 12 is 14 drops. The same count holds at every step.

Say what an interviewer wants beyond the number. The move is to fix the budget of drops and ask how many floors it can cover, rather than fixing the building and searching for a strategy. That reversal is what makes the problem easy, and it generalises: with three eggs and k drops, the floors covered are the two-egg coverage for each smaller budget, plus one per drop, added up, which is why three eggs need only 9 drops for 100 floors.

Roughly 1.2, not 4.0. Consecutive 12-month returns share 11 months, so 240 monthly rows over 20 years hold only about 20 independent observations. OLS standard errors assume independence and come out too small by roughly the square root of the overlap, root 12, about 3.5. Dividing 4.0 by 3.46 gives about 1.15: the result is no longer significant. A Newey-West or Hansen-Hodrick standard error does this properly.

What does the overlap do to the regression?

Asking twelve friends for restaurant advice sounds like twelve opinions, but if eleven of them only repeat what the first one said, you have heard about one. Each 12-month return shares 11 months with its neighbour, so the rows are mostly the same data counted again, and the regression thinks it has twelve times more independent evidence than it does. The slope estimate is not biased by the overlap. What breaks is the standard error, because the residuals are strongly correlated from one row to the next, and that breaks one of the {term('OLS assumptions', 'The conditions under which ordinary least squares standard errors are correct, including residuals that are uncorrelated across observations.')}.

Why divide by root 12 and not by 12?

The standard error scales with one over the square root of the number of independent observations. If the effective sample is twelve times smaller, the standard error is root 12, about 3.46, times larger, and the t-statistic is 3.46 times smaller: 4.0 becomes about 1.15. This is a rough correction. The exact factor depends on how persistent the signal is: for a slow-moving signal, such as a valuation ratio, it is close to root 12; for a fast-moving one it can be smaller.

Say how you would fix it properly: use Newey-West standard errors with at least 11 lags, or Hansen-Hodrick errors built for exactly this overlap, or run the regression on non-overlapping annual data and accept the smaller sample. Any of those should give a t-statistic well below 4.0, and a researcher who reports only the OLS number has not yet shown the signal works.

The typical buy-and-hold investor loses about 5.1% a year, yet a 25% stake rebalanced yearly grows about 0.6% a year. The average year returns +5%, but a good year and a bad year multiply wealth by 1.5 x 0.6 = 0.9. With 25% in the stock the two years multiply wealth by 1.125 x 0.9 = 1.0125. That 25% is the Kelly fraction, the stake that maximises the average log return.

How can a positive average return shrink your wealth?

Imagine a shop whose sales rise 50% in a good year and fall 40% in a bad one. After one of each it is at 90% of where it began, whatever the order. Wealth compounds by multiplying, so over many years what matters is the typical growth factor, the square root of 1.5 x 0.6, about 0.949, not the average return of +5%. The average is real, but it is carried by rare paths with long lucky streaks. After 20 years the typical investor holds about 0.35 of the starting money while the average across all paths is 2.65 times it.

Why does holding less of the stock help?

Rebalancing to a fixed mix sells after gains and buys after losses, and a smaller stake shrinks the swings. With a fraction f in the stock, a good year multiplies wealth by 1 + 0.5f and a bad year by 1 - 0.4f; typical growth peaks where 0.5/(1 + 0.5f) equals 0.4/(1 - 0.4f), which gives f = 25%. At 25% a pair of years gives 1.125 x 0.9 = 1.0125, about 0.6% a year. The quick check is return over variance: 0.05 divided by 0.45 squared is 0.247.

What is the limit of this answer?

The 25% rests on knowing both outcomes and their odds exactly, on cash earning nothing and on free rebalancing. Change any of those and the fraction moves, and because a real edge is only an estimate, desks size well below the full Kelly number. The lesson to lead with in the room is the gap itself: a positive average return is not a positive growth rate, and position size decides which one you earn. That gap is called volatility dragThe shortfall of the compound growth rate below the average return, roughly half the variance of returns..

Weigh four against four first, then mix suspects with coins you already know are genuine so every later weighing splits the cases three ways. There are 24 possibilities, 12 coins each heavy or light, and three weighings have 27 outcomes. If 1 to 4 balances 5 to 8, weigh 9, 10, 11 against three good coins; if not, weigh 1, 2, 5 against 3, 6, 9. The third weighing settles what is left.

How do you know three weighings can be enough?

A game of twenty questions works because each yes or no halves what is left. A balance is better than a yes or no: it answers left heavy, right heavy or balanced. Three weighings give 3 x 3 x 3 = 27 outcomes, and there are 24 cases to tell apart, 12 coins each possibly heavy or light, so a procedure can exist only if every weighing splits the remaining cases into three near-equal groups. That counting sets the design: the first weighing must leave at most 9 cases on every branch.

What do you do after the first weighing tips?

Say the left pan was heavy: the odd coin is 1, 2, 3 or 4 and heavy, or 5, 6, 7 or 8 and light. Weigh 1, 2 and 5 against 3, 6 and 9, moving some suspects across and bringing in a known good coin, so each outcome points to a different small group. Left heavy again means 1 heavy, 2 heavy or 6 light: weigh 1 against 2, and a balance means 6. Right heavy means 3 heavy or 5 light: weigh 3 against a good coin. A balance means 4 heavy, 7 light or 8 light: weigh 7 against 8.

What if the first weighing balances?

Then coins 1 to 8 are genuine and the odd coin is among 9 to 12, still heavy or light. Weigh 9, 10 and 11 against three good coins: a tip tells you both that the odd coin is among the three and whether it is heavy or light, and a balance points to coin 12. After a tip, weigh 9 against 10: if the odd coin is heavy the heavier of the two is it, if light the lighter, and a balance means 11. After a balance, weigh 12 against a good coin to learn heavy or light.

An information ratio of about 1.0, and about 900 independent bets a year for 1.5. The fundamental law of active management says the information ratio is roughly the information coefficient times the square root of breadth: 0.05 x the square root of 400 = 0.05 x 20 = 1.0. To reach 1.5 the square root must be 30, so breadth must be 900, more than double, because breadth enters under a square root.

Why do many weak calls add up to a strong result?

Picture a cricket pundit who calls the winner right 52.5% of the time. On one match that is nearly useless; over hundreds of independent matches, the small edge becomes a steady record. With independent bets, the expected gain grows in proportion to the number of bets while the noise grows only with its square root, so the ratio of the two grows with the square root of the number of bets. An information coefficientThe correlation between a signal's forecasts and the returns that follow; for a simple up or down call it equals twice the hit rate minus one. of 0.05 is roughly that pundit's edge: a hit rate of 52.5%.

What does the square root mean for building a strategy?

Skill and breadth are not equal levers. Doubling the information coefficient doubles the information ratio; doubling breadth raises it only by about 41%, so matching a doubling of skill needs four times the bets. Going from 1.0 to 1.5 on breadth alone means 2.25 times as many independent bets, 900 against 400. That is why quant funds chase breadth across many stocks and short horizons, and why a small gain in forecast quality is worth so much.

What does the law leave out?

Two things that usually cut the answer. Independence is the hard part: 400 bets on stocks in one sector, or rebalanced so often that they repeat the same view, are far fewer than 400 independent bets. And constraints on position size, shorting and turnover stop a portfolio from fully expressing the signal; a transfer coefficientA number between 0 and 1 measuring how fully a constrained portfolio reflects the signal; it multiplies the fundamental law. of 0.6 would take the expected information ratio from 1.0 to 0.6. State the law, then say which of these you would check first.