Hedge Funds puzzles, solved step by step

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.

About Rs 12,500 crore. Each stock allows 5% of Rs 50 crore, Rs 2.5 crore a day. Across 200 stocks that is Rs 500 crore a day, and across 250 days Rs 1,25,000 crore a year. A book turned over 10 times a year needs 10 rupees of trading for every rupee of capital, so the book can be Rs 1,25,000 crore divided by 10. Treat it as a ceiling, not a working size.

What does capacity actually mean?

Picture a juice stall that can squeeze 500 glasses a day. How many regular customers can it serve? That depends on how often each one comes. If every regular drinks a glass a day, 500; if each comes once in ten days, 5,000. A strategy's capacity is the capital it can run before its own trading exceeds what the market will absorb, so it is tradable volume divided by how often the book must be traded.

How do you build the number out loud in the room?

Go one multiplication at a time and say each one. 5% of Rs 50 crore is Rs 2.5 crore per stock per day. Two hundred stocks make Rs 500 crore a day. Two hundred and fifty days make Rs 1,25,000 crore a year. Dividing by a turnover of 10 gives Rs 12,500 crore, and the step people drop is the last one: capacity is not a day's or a year's volume.

Why is turnover the lever that matters most?

Every input enters in proportion, but the market and the risk team set the others, while the strategy itself sets its turnover, and turnover varies far more across strategies. A strategy that turns over 50 times a year has a capacity of Rs 2,500 crore on this universe; one that turns over twice a year has Rs 62,500 crore. That is why fast strategies close to new money early and slow ones can run large books.

Why is the real capacity lower than this ceiling?

The estimate assumes the strategy trades evenly every day and in every stock. It does not. Trades cluster when signals fire, which is often when others trade too, and the thinner names in the list hit the 5% limit long before the larger ones. The definition of turnover matters as well: if 10 times means buying 10 times the book and also selling it 10 times, the traded value doubles and capacity halves to Rs 6,250 crore. Say which definition you used, and add that trading costs rise before the hard limit, so returns fade well before the ceiling.

50% after the disagreement, and 70% once the third analyst sides with up. Work in odds. Each analyst's call multiplies the odds by 0.7/0.3 = 7/3 in the direction called. One up and one down multiply by 7/3 and 3/7, which cancel, leaving the prior of 1:1. The third call multiplies by 7/3 again: odds of 7:3, a probability of 70%. A two-to-one split is worth one analyst, not two thirds.

Why switch from probabilities to odds?

Think of two friends who read the weather equally well and disagree about rain: you are back where you started, however good they are. With independent signals, each one multiplies your odds by its likelihood ratioHow much more likely a piece of evidence is if the claim is true than if it is false., so in odds form Bayes' rule is just multiplication. An analyst who is right 70% of the time says up with chance 0.7 if the stock will rise and 0.3 if it will fall, so an up call multiplies the odds by 7/3 and a down call by 3/7.

How do the three calls combine?

Start at 1:1. The first analyst says up: 7:3, or 70%. The second says down: 7:3 times 3:7 is 1:1, back to 50%. The third says up: 7:3 again, 70%. Only the net count of calls matters, not the total. Three for up and none against would give 343:27, about 92.7%, which shows how much the one dissenter costs.

What assumption carries all of this?

Independence. If the analysts read the same research and speak to the same management teams, their errors are correlated, and a second agreeing call adds much less than a factor of 7/3. In the extreme where the second analyst simply copies the first, it adds nothing at all. The rule also assumes each analyst is right 70% of the time whichever way the stock moves; an analyst who calls up too often tells you more when calling down. State independence as the assumption, then say how you would test it: by checking how often the analysts' past calls agreed with each other.

About Rs 24.19. Round n pays Rs 2^n with probability 1/2^n, so each round adds exactly Rs 1 of expected value. That holds up to n = 23, since 2^23 is about Rs 84 lakh and 2^24 is above the Rs 1 crore cap. From round 24 on, the payout is stuck at Rs 1 crore, with total probability 1/2^23, which adds about Rs 1.19. So 23 + 1.19, about Rs 24.

Why is the uncapped game worth an infinite amount?

Each round is a doubling bet: the payout doubles while the chance of reaching it halves. Think of a raffle where a ticket twice as valuable is half as likely to win: every prize tier is worth the same to you. Round n pays 2^n with probability 1/2^n, so every round contributes exactly Rs 1 to the expected value, and there are infinitely many rounds. That is the famous paradox: few people would pay even Rs 100, yet the expected value has no bound. The resolution that matters on a desk is not psychology; it is that nobody can pay out an unlimited amount.

How does the cap change the sum?

Find the round where the cap starts to bind. 2^23 is Rs 83,88,608, under Rs 1 crore; 2^24 is Rs 1,67,77,216, over it. So rounds 1 to 23 each add Rs 1, and every round from 24 on pays the capped Rs 1 crore, which together happen with probability 1/2^23 and add 1,00,00,000 / 83,88,608, about Rs 1.19. The fair price is about Rs 24.19. Nearly all of the textbook infinity lives in outcomes the casino cannot pay.

What does a bigger casino buy you?

Very little. The value grows only with the logarithm of the cap: each doubling of the casino's bankroll adds about Rs 1. A cap of Rs 1,000 crore, a thousand times larger, lifts the fair price only to about Rs 34. That is the lesson a risk manager takes away: a payoff whose expected value rests on rare, enormous outcomes is worth what the other side can actually pay, and any estimate built on the tail should be checked against who stands behind it.

From gamma: the hedged option earns on the size of moves, not their direction, and pays theta for the privilege. Each day the P&L is about half gamma times the squared move, minus theta. Theta is priced for 20% volatility, so a daily move of about 1.26% breaks even. At 30%, typical moves are 1.89%, and the squared gain is 2.25 times the decay. Up or down makes no difference; the path still does.

What is left after the delta hedge?

Think of a street vendor who sells umbrellas and sunglasses from the same cart: she no longer cares whether it rains or shines, only whether the weather changes enough to bring people out. A delta hedge sells enough stock to cancel the option's first-order bet on direction, so what remains is the curvature: the option gains a little more on the way up than the hedge loses, and loses a little less on the way down than the hedge gains. That curvature is gamma, and it pays on moves either way.

How big is the daily P&L in numbers?

Take a stock at Rs 1,000 and a three-month at-the-money call priced at 20% volatility, with zero interest rates. Its gamma is about 0.0040 per rupee and its time decay about Rs 0.32 a day over 252 trading days. A day's move of 1.89%, typical of 30% volatility, is Rs 18.9, and half of 0.0040 times 18.9 squared is a gain of Rs 0.71, against Rs 0.32 of decay: about Rs 0.40 a day per option. The decay is exactly what a move of 1.26%, typical of 20% volatility, would pay back.

Why does direction not matter, but the path does?

The move enters squared, so plus Rs 18.9 and minus Rs 18.9 earn the same. The P&L depends on realised volatility against implied, not on where the stock ends up. But gamma is largest near the strike and shrinks as the stock drifts away, so 30% realised in a trend that carries the stock far from the strike early earns less than 30% realised while the stock chops around the strike. Hedging frequency matters too: hedge rarely and the P&L becomes noisy, even if its average barely changes.

About 10% higher. The variance of a sum of 21 daily returns is 21 daily variances plus twice every covariance between pairs of days. With autocorrelation 0.1 at lag one, 0.01 at lag two and so on, the covariance terms add about 4.42 daily variances, lifting the total from 21 to about 25.4, a 21% rise. Volatility, the square root, rises by about 10%.

Why does a small autocorrelation matter over a month?

Think of a queue where each person who joins makes it slightly more likely that the next person joins too. Each nudge is small, but over a month of days they add up to longer queues than pure chance would give. The variance of a sum counts every pair of days, and there are 20 neighbouring pairs in a 21-day month, each counted twice, so a lag-one correlation of 0.1 alone adds 4 daily variances to the 21.

How do you do the sum?

Write the monthly variance as the daily variance times 21, plus twice the sum over every lag. Lag one appears in 20 pairs with correlation 0.1, lag two in 19 pairs with 0.01, lag three in 18 with 0.001: twice (2.0 + 0.19 + 0.018 and so on) is about 4.42. The total is 25.42 daily variances instead of 21, a ratio of 1.210, and the volatility ratio is its square root, 1.100. Over a long horizon the ratio tends to (1 + 0.1)/(1 - 0.1) = 1.222, so a month is already close to the limit.

What does this mean for risk and performance numbers?

Scaling daily volatility by the square root of time silently assumes zero autocorrelation. Positive autocorrelation, which shows up in trend-following returns and in portfolios of thinly traded assets priced from stale quotes, makes the square-root rule understate longer-horizon risk. Negative autocorrelation, typical of mean-reverting strategies, does the reverse: at minus 0.1 the monthly variance is only 0.83 times the rule's. The same effect is why correlations between two assets measured on daily returns can differ from those measured on monthly returns: prices that react to the same news on different days look less related day by day than month by month.

About 0.289, and it settles at 2/7, about 0.286, for large targets. Let p(n) be the chance the total ever equals n. To hit n, the total must first land on one of n - 1 to n - 6 and then roll exactly the gap, each with chance 1/6, so p(n) is the average of the six values before it, with p(0) = 1. Working up gives p(10) = 0.2893. Totals advance 3.5 a roll on average, so they land on 1 number in 3.5.

How do you set up the recursion?

Think about the last roll before the total reaches n. The total can land exactly on n only by first landing on one of n - 1 down to n - 6 and then rolling exactly the gap, each with chance 1/6, so p(n) = (1/6)[p(n - 1) + ... + p(n - 6)]. Start with p(0) = 1, because you begin at zero, and p of any negative number = 0. Then p(1) = 1/6, p(2) = 7/36, p(3) = 0.227, and so on up to p(10) = 0.2893.

Why does the answer settle at 2/7?

Picture stepping stones across a river, where each stride covers 1 to 6 stones with equal chance. Over a long walk you touch about one stone in every 3.5, because that is your average stride. The running total advances 3.5 per roll on average, so in the long run it lands on a fraction 1/3.5 = 2/7 of all numbers, and each far-off target is hit with probability close to 2/7. The early values wobble: p(6) is the highest, 0.360, because 6 is the last total a single roll from zero can reach directly, and the wobbles die out by about 20.

Why would a quant interviewer ask for a table or code here?

Because the recursion is dynamic programmingSolving a problem by building up answers to smaller versions of it and reusing them, instead of recomputing from scratch.: each value reuses the six before it, so a table of ten numbers is faster and safer than listing every sequence of rolls that sums to 10. Say the recursion, compute a few terms out loud, and give the limit with its reason; that is the complete answer. Do not try to enumerate paths: there are 492 ordered ways to reach 10 with rolls of 1 to 6, each with its own probability.

The time-weighted return is +20%; the money-weighted return is about -10.2% a year, and investors as a group lost Rs 60 crore. Rs 100 crore grows to Rs 150 crore, then Rs 400 crore arrives and the Rs 550 crore falls 20% to Rs 440 crore. Chaining 1.5 x 0.8 gives +20%, the manager's record. But Rs 500 crore went in and Rs 440 crore remains, because most of the money arrived just before the loss.

What does each measure answer?

Think of a restaurant that gets a glowing review after a good year, triples its tables, and then has a poor year. Its record over the two years looks fine, but most of the diners ate in the poor year. The time-weighted return measures the manager, chaining each period's return so the timing of money coming in or out does not count; the money-weighted return measures the investors, weighting each period by the money actually exposed to it.

How do you compute the money-weighted return?

Treat the investors' money as a set of cash flows and find the single annual rate that links them: Rs 100 crore in at the start, Rs 400 crore in after a year, Rs 440 crore out at the end. Solve 100(1 + r) squared + 400(1 + r) = 440; the root is 1 + r = 0.8983, so the money-weighted return is about -10.2% a year. It is negative because most of the money sat through the losing year: Rs 550 crore lost 20%, Rs 110 crore, while the good year earned only Rs 50 crore on Rs 100 crore.

Which number should an allocator look at?

Both, for different questions. To judge the manager's skill, use the time-weighted +20%, because the manager did not choose when investors arrived. To judge whether the fund was good for the people in it, or whether money chased performance, use the money-weighted figure. A wide gap between the two, as here, is a warning about hot money: investors piling in after a strong year and bearing the next year's loss on a much bigger base. Had all Rs 500 crore arrived at the start, both measures would agree and the investors would hold Rs 600 crore.

Ridge gives 0.5 and 0.125; lasso gives 0.4 and exactly 0. With orthonormal regressors each coefficient is shrunk on its own. Ridge divides each by 1 + 0.2 = 1.2, so it scales both down and never reaches zero. Lasso subtracts 0.2 from each size and stops at zero, so the small coefficient, 0.15, is removed. This uses the scaling of half the residual sum of squares plus 0.2 times the penalty.

Why do orthonormal regressors make this a one-line problem?

When the regressors are uncorrelated and scaled to unit length, the fit for each coefficient does not depend on the others, so the penalised problem splits into separate one-variable problems. Think of adjusting the volume on two speakers that are not wired together: turning one down does not change the other. For each coefficient you minimise half of (beta minus b) squared plus the penalty, where b is its least squares value, and the answer depends only on that one number. Ridge's penalty is half of beta squared times 0.2; lasso's is the size of beta times 0.2.

What does each penalty do to a coefficient?

Ridge's squared penalty pulls hard on big coefficients and gently on small ones. Setting the slope to zero gives beta x (1 + 0.2) = b, so ridge divides by 1.2: 0.6 becomes 0.5 and 0.15 becomes 0.125, shrunk but never zero. Lasso's absolute-value penalty pulls with the same force, 0.2, whatever the size. Its answer is the sign of b times the larger of (the size of b minus 0.2) and zero: 0.6 becomes 0.4, and 0.15, weaker than the pull of 0.2, lands exactly on 0.

Why does only lasso select variables?

At zero, the squared penalty is flat: its slope is zero, so any small coefficient still earns its place by improving the fit a little. The absolute-value penalty has a corner at zero with a slope of 0.2 on each side, so a coefficient stays at zero unless the fit improves by more than 0.2 per unit, and 0.15 does not. That is why lasso gives sparseHaving many coefficients exactly equal to zero, so the model uses only a few of the available variables. models and ridge does not. The limitation to state: with correlated regressors lasso tends to keep one of a group arbitrarily and drop the rest, which is why desks often blend the two penalties in an elastic net.

1,000 x (1 + 1/2 + 1/3 + ... + 1/n) miles, which grows without limit but only like 1,000 x ln n. Drive all n cars together for 1,000/n miles: together they have burned one tankful, so one car tops up the others and stops. Then n - 1 cars drive 1,000/(n - 1), and so on, until the last car drives a full 1,000. With 4 cars that is about 2,083 miles; with 100, about 5,187.

Why can the fuel not simply be pooled into one car?

Each tank holds exactly 1,000 miles of fuel, so one car can never carry more than that at once. Think of porters carrying water across a desert: the helpers walk part of the way, hand over what they can spare, and drop back. The helper cars exist to keep the lead car's tank full for as long as possible, and they can only do that by travelling with it and burning fuel themselves.

How long is each leg?

With k cars travelling together on full tanks, drive until the group has burned exactly one tankful, which takes 1,000/k miles. Each tank is then 1/k empty, so the k - 1 cars that continue have (k - 1)/k of a tank of space between them, and the car that stops has exactly (k - 1)/k of a tank left to fill it. The legs are 1,000/n, then 1,000/(n - 1), and so on to 1,000 for the last car alone. With four cars: 250 + 333.3 + 500 + 1,000 = 2,083.3 miles. Dropping each helper the moment its fuel can refill the rest keeps as few cars as possible burning fuel at every mile.

What happens as n grows?

The harmonic series never stops growing, so with enough cars there is no ceiling on the distance. But it grows only like the logarithm of n: 10 cars reach about 2,929 miles, 100 cars about 5,187, and each further tenfold increase in cars adds only about 2,303 miles. Reaching 5,000 miles takes 83 cars. That is the pattern worth naming in the room: unlimited in principle, very expensive in practice, the same diminishing return you meet whenever each extra unit of effort adds less than the one before.