Hedge Funds puzzles, solved step by step

It makes no difference: switching and sticking each win half the time. List the six equally likely cases of car position and the host's random pick. Two of them reveal the car, and you have seen that they did not happen. The four that remain split two and two. The knowing host gives two thirds only because he never risks the car, which pushes those two cases into the switch column.

Where does the usual two-thirds answer come from?

In the standard game the host knows where the car is and always opens a goat door. Your first pick is right one time in three, and nothing the knowing host does can change that, so the other two thirds sit on the remaining closed door. His choice carries information because it is forced: when the car is behind door 2, he must open door 3, and when it is behind door 3, he must open door 2.

What changes when the host picks at random?

Picture a friend who does not know the answer to a quiz question and strikes out one option on a whim. If that option happens to be wrong, you have learned less than if someone who knew had struck it. The random host is that friend. Write out six cases: the car behind door 1, 2 or 3, each with the host's coin choosing door 2 or door 3. In two of the six the random host opens the car door, and the goat you saw rules those two out, which removes switch wins rather than stick wins.

Count what is left. The car behind door 1 survives both host choices: two cases where sticking wins. The car behind door 2 survives only when the host opened door 3, and the car behind door 3 only when he opened door 2: two cases where switching wins. Two against two.

Why would an interviewer want the intuitive answer broken rather than recited?

Because the lesson travels to every desk. The same observation carries different information depending on the process that produced it. A strong track record shown by a manager who launched ten funds and closed the nine that did badly is a host choosing which door to open for you. Before you update on evidence, ask whether the source could have shown you something else, and whether it chose what to show.

It cannot be done at any finite speed. Averaging 60 mph over the 120-mile round trip means finishing in 2 hours. The outward 60 miles at 30 mph already took 2 hours, so the return leg would have to take no time at all. Driving back at 90 mph, the tempting answer, gives an average of only 45 mph.

Why is 90 mph the wrong instinct?

Averaging 30 and 90 to get 60 treats the two speeds as if they counted equally. They do not, because the car spends far longer at the slow speed. Think of a student who scores 30% on a three-hour paper and 90% on a ten-minute quiz: nobody would call that a 60% performance. Average speed is total distance over total time, so the slow leg carries more weight because it takes up more of the clock.

How do you prove it cannot be done?

Work in time, not speed. At 60 mph, 120 miles takes exactly 2 hours, and the first leg has already spent those 2 hours. Any return speed, however fast, adds some time, which pushes the average below 60 mph. At 90 mph the return takes 40 minutes and the average is 45 mph; at 300 mph it takes 12 minutes and the average is 54.5 mph. The average creeps towards 60 but never reaches it.

Where does the same mistake show up on a desk?

It appears whenever numbers are averaged without the right weights. A position that falls 50% and then rises 50% does not break even, because the second move works on a smaller base; the average that matters is the one weighted the way the thing actually compounds. Speeds over equal distances call for the harmonic mean, which sits below the simple average whenever the numbers differ. It is the same reason that putting a fixed rupee amount into a fund each month buys units at an average cost below the average price over those months.

About Rs 31.6 crore: Rs 10 crore times the square root of 10. If daily P&L is independent from day to day, with the same volatility and a mean near zero, variances add, so ten-day volatility is root 10 times daily volatility, and a normal quantile scales the same way. The rule also needs the positions held unchanged for ten days and a distribution that keeps its shape over the horizon.

Why not ten times the one-day number?

Picture ten friends each tossing a coin for Rs 100. The worst case is the group losing Rs 1,000, but the typical spread of the group's total is nowhere near ten times one person's, because some win while others lose. Independent daily moves partly cancel, so their variances add while their volatilities do not, and volatility grows with the square root of the number of days. Ten times would need every bad day to line up in the same direction, which is exactly what independence rules out. Value at riskThe loss a book should not exceed over a set horizon at a set confidence level, for example one day at 99%. inherits that square root when the distribution is normal.

What has to be true for the rule to hold?

List the assumptions, because that is the real question. Returns must be independent across days, volatility constant, the mean close to zero, the positions unchanged, and the distribution one that keeps its shape when summed, as the normal does. Break any one and the rule drifts. Positive autocorrelation, where bad days follow bad days, makes the true ten-day number larger. A book that is cut after losses makes it smaller. Fat tails make the one-day 99% quantile a poor guide to the ten-day one.

Which way does the error usually run?

In a calm market the rule is a fair approximation. In stress it tends to understate, because volatility rises and losses cluster just when the ten-day horizon matters. The square root of time is a scaling convenience, not a law, so a risk team checks it against ten-day P&L measured directly. The same assumption sits behind the desk habit of multiplying daily volatility by 16 to get an annual figure, 16 being roughly the square root of the trading days in a year; stretch it to 250 days here and you get Rs 158 crore, a number few would trust.

About 27.3%. At 2 and 20 the manager earns Rs 20 crore of management fee and 20% of the remaining Rs 100 crore gain, Rs 20 crore: Rs 40 crore in all. At 1%, the management fee is Rs 10 crore and the gain after it is Rs 110 crore. To keep Rs 40 crore the performance fee must bring in Rs 30 crore, which is 30/110, or 27.3%. The two deals match only at a 12% gross return.

What does the manager earn today?

Think of a tailor who charges a fixed stitching fee plus a share of whatever the finished suit sells for above cost. Cut the fixed fee and the share must rise to keep the same income, but the share now applies to a slightly larger base. At 2 and 20 on Rs 1,000 crore earning 12%, the manager takes Rs 20 crore of management fee plus 20% of the Rs 100 crore gain left after it, Rs 40 crore in total. Investors keep Rs 80 crore, a net return of 8%.

How do you find the new performance fee?

Keep the total at Rs 40 crore. The management fee falls to Rs 10 crore, so the performance fee must bring in Rs 30 crore, and it is charged on a gain of Rs 110 crore, not Rs 100 crore. 30 divided by 110 is 27.3%. The base grows because less has been taken off the top before the performance fee is worked out. State the assumptions as you go: no hurdle rate, and no earlier losses to recover below a high-water markThe highest value an investor has paid a performance fee on; no new performance fee is charged until the fund climbs back above it..

Is the new deal really the same for investors?

Only at a 12% gross return. The new deal pays the manager less in poor years and more in good ones, so it moves risk from the investors to the manager. At a 4% gross return the old deal pays Rs 24 crore and the new one Rs 18.2 crore; at 20% the old pays Rs 56 crore and the new one Rs 61.8 crore. That is why allocators push for a lower fixed fee even at the price of a higher share: they would rather pay for performance than for size.

The fair multiple falls from 25x to about 22.2x, a compression of about 11%. The denominator is the real yield plus the premium minus growth: 1.5 + 4.5 - 2.0 = 4.0% before, and 2.5 + 4.5 - 2.5 = 4.5% after. Real yields rose by a full point and growth by only half a point, so the net rate rose by half a point, and the multiple, its inverse, fell.

Why is the multiple one over a spread?

Think of a shop that pays you rent forever, rising a little each year. What you would pay for it depends on the return you demand minus how fast the rent grows. For earnings paid out and growing forever, price over earnings is one divided by the required return minus growth, so the multiple depends only on the gap between the two. In real terms the required return is the real yield plus the equity risk premiumThe extra return investors demand for holding shares rather than government bonds.. The question treats all earnings as paid out, which is the assumption to name.

How do you work it out quickly?

Compute the denominator before and after. Before: 1.5 + 4.5 - 2.0 = 4.0%, a multiple of 25x. After: 2.5 + 4.5 - 2.5 = 4.5%, a multiple of 22.2x. Rates went up by 1.0 point and growth by 0.5, so the spread widened by 0.5 point. A 0.5-point rise on a 4.0% base is a 12.5% rise in the denominator, and the multiple falls by 1 minus 1/1.125, about 11.1%.

What does this teach beyond the arithmetic?

Higher yields do not hurt equities one for one if growth rises with them. What matters is whether real yields rise faster or slower than expected growth: faster compresses multiples, slower expands them. Had growth also risen a full point, to 3.0%, the net rate would be 4.0% again and the multiple 25x. The limitation to state is sensitivity: near a 4% net rate, a half-point move shifts the multiple by about 2.8 turns one way and 3.6 the other, so small errors in the premium or the growth guess swamp the answer.

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.

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.