Hedge Funds puzzles, solved step by step

About 64.2%. A useless signal clears a 5% bar by luck one time in twenty. The chance that all 20 stay insignificant is 0.95 to the twentieth, about 35.8%, so the chance that at least one looks like a discovery is 64.2%. On average the search turns up one false signal, as 20 x 5% suggests, but at least one appears in roughly two searches out of three.

Why does testing more ideas manufacture a winner?

Ask twenty friends to flip a coin five times each. Any one of them flips five heads only once in 32 tries, yet the chance that at least one of the twenty does is 47.0%, and that friend will look gifted. Each test is a lottery ticket for a false discovery, and buying twenty tickets makes a win likely even when nothing works. A signal chosen because it looked best among twenty has not passed a 5% test; it has passed a 64.2% one.

How do you correct for it?

Tighten the bar to match the number of tries. The Bonferroni correction tests each signal at 5% divided by 20, which is 0.25%, and that brings the chance of any false discovery back to 4.9%. The cost is power: a real but modest signal now struggles to get through. The other defence is data the search never touched: choose the best signal on one period, then test it once on another.

What does a quant fund take from this?

Research teams run thousands of tests, and the ones that get presented are the survivors. Count every test, including the ones you ran and forgot, because the significance of the survivor depends on how many were tried. That is why systematic funds keep research logs and hold data back, and why a backtest with a t-statistic of 2 means much less after a large search than after one planned test. Say the limitation: the 64% assumes independent tests; correlated signals give a lower figure, but rarely a comfortable 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.

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.

1 over root 2, about 0.71. The covariance of X with X + Y is Var(X) plus Cov(X, Y), which is sigma squared plus zero. The standard deviation of X + Y is root 2 times sigma because the variances add. So the correlation is sigma squared over (sigma x root 2 sigma), which is 1/root 2. X explains half the variance of the sum, and the correlation is the square root of that half.

What is the fastest way to set it up?

A two-member team's score is the sum of both players' scores. If the players are equally good and play independently, knowing one player's score tells you something about the team total, but only half the story. Split the covariance: Cov(X, X + Y) = Cov(X, X) + Cov(X, Y) = sigma squared + 0. The variance of the sum is sigma squared + sigma squared = 2 sigma squared, because independent variances add. Correlation is covariance over the product of standard deviations: sigma squared over (sigma x root 2 sigma) = 1/root 2.

Why is the answer not 0.5?

Because 0.5 is the R squaredThe share of one variable variance explained by another; for a simple regression it is the correlation squared., not the correlation. X explains exactly half of the variance of X + Y, and correlation is the square root of the share of variance explained, so it is root 0.5, about 0.707. The geometric picture makes it stick: treat independent variables as arrows at right angles, and correlation as the cosine of the angle between arrows. X + Y sits at 45 degrees to X, and cos 45 is 0.707.

Give the general version to show you own it. If Y has variance k times X's, the correlation is 1/root(1 + k): the more noise you add, the lower it falls. That is the logic behind a noisy signal: a forecast that is half signal and half independent noise, by variance, correlates about 0.71 with the signal, not 0.5.

About 4 years for a Sharpe of 1.0 and about 16 years for a Sharpe of 0.5. The t-statistic of a mean return is the mean over its standard error, which works out to the annual Sharpe ratio times the square root of the number of years, whatever the data frequency. Setting Sharpe x root(years) = 2 gives years = (2 / Sharpe) squared: 4 for 1.0, 16 for 0.5 and just 1 for 2.0.

Why does the t-statistic grow with the square root of time?

A coin that lands heads 55% of the time looks fair after 20 tosses; you need hundreds before the bias shows through the noise. The average return grows in proportion to time, but the noise around it grows only with the square root of time, so the signal-to-noise ratio, the t-statistic, grows with root time. With monthly data, the t-statistic is the monthly Sharpe times root(12 x years), and the monthly Sharpe is the annual Sharpe divided by root 12, so the twelves cancel: t = annual Sharpe x root(years).

Why does monthly data not shorten the wait?

More frequent data gives more observations but each is noisier relative to its mean. Sampling the same years more often does not add information about the mean return; only more years do. This is why a {term('t-statistic', 'An estimate divided by its standard error; a value around 2 is the usual threshold for saying an effect is unlikely to be pure noise.')} on the average return depends on the span of the data, not the number of rows. Frequency helps you estimate volatility, not the mean.

Say the practical point. Most real strategies have Sharpe ratios well below 1, so their track records are too short to separate skill from luck with any confidence. An allocator looking at a three-year record with a Sharpe of 0.8 sees a t-statistic of about 1.4. The limitation of the rule: it assumes returns are independent and stable over the whole sample, and fat tails or regime changes make the real uncertainty larger.

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.

No. The square root of the unbiased variance underestimates the standard deviation on average. The square root is concave, so by Jensen's inequality the average of the square roots is below the square root of the average. For normal data with two observations the estimate averages about 0.80 sigma; the bias shrinks as the sample grows, to about 6% at five observations and under 1% at thirty.

Why does taking a square root break unbiasedness?

Two square rooms have floor areas of 4 and 16 square metres, so their sides are 2 and 4 metres. Average the areas, 10, and take the root: 3.16 metres. Average the sides instead: 3 metres. Averaging and then taking a square root gives a bigger answer than taking square roots and then averaging, because the square root bends downwards. The sample variance is right on average, so the average of its square roots must fall short of the true standard deviation.

How big is the bias?

It depends on the sample size and on the distribution. For normal data the expected sample standard deviation is c4 times sigma, with c4 about 0.80 at n = 2, 0.94 at n = 5, 0.97 at n = 10 and 0.99 at n = 30. At n = 2 you can check it directly: the sample standard deviation is the gap between the two draws divided by the square root of 2, and the average gap between two normal draws is 2 sigma over the square root of pi, which leaves the square root of 2/pi, about 0.798.

Does it matter in practice?

Sometimes. With a year of daily returns the bias is a rounding error; with a handful of monthly returns for a new fund it is not. A manager with five monthly returns has a volatility estimate that averages about 6% too low under normality, which flatters a Sharpe ratioAverage excess return divided by the standard deviation of returns, a measure of return per unit of risk. before anyone has looked at fat tails. Dividing by c4 removes the bias for normal data, but the fix depends on the distribution, so name the assumption. And unbiased is not the same as most accurate.

About 31 managers, 1,000 halved five times. Each year roughly half the unbeaten managers beat the market by luck, so 500 survive year one, 250 year two, then 125, 62.5 and 31.25. A perfect five-year record is something luck hands to about 3 managers in every 100, so in a large crowd it cannot on its own separate skill from chance.

Why does a crowd produce streaks even without skill?

Fill a stadium with a thousand people and ask each to toss a coin five times. Someone will throw five heads, and about 31 will. A result that is rare for one person is almost certain somewhere in a large group, so the question is never whether a flawless record exists but how many you would expect by chance. Each manager's chance is 1 in 32; across 1,000 managers that is 31.25 expected perfect records.

If some managers really are skilled, how much does a perfect record tell you?

Suppose 5% of the thousand are skilled and beat the market 60% of the time. They produce about 3.9 perfect records, while the 950 unskilled produce about 29.7, so a manager with five perfect years is skilled only about 12% of the time. A 60% manager has only a 8% chance of five perfect years, so most skilled managers do not have flawless records either. The record is weak evidence in both directions.

What should you look at instead?

Longer records, more decisions per year and a reason. Skill shows up more reliably in many independent decisions than in a handful of annual outcomes, and in a process that explains where the edge comes from. Allocators also check how many managers were in the starting pool, because the funds still reporting are the ones that survived; the ones that were closed after bad years have dropped out of the data. That is survivorship biasThe distortion that comes from studying only the survivors of a process, whose results look better than those of the whole starting group., and it makes every surviving record look stronger than it is.

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 0.9. The market reaches the stock by two paths: directly, with a coefficient of 0.5, and through the sector, which moves 0.4 for each unit of market and passes all of it on with a coefficient of 1.0. A regression on the market alone cannot separate the two and reports the total, 0.5 + 1.0 x 0.4 = 0.9. The extra 0.4 is omitted variable bias.

Why does leaving the sector out change the market slope?

Suppose you measure how much ice cream sales rise on hot days, but hot days also tend to be holidays, and holidays sell ice cream too. Leave holidays out and the heat gets the credit for both. A regression gives a left-out variable's effect to whichever included variable moves with it, in proportion to how strongly the two move together. Here the sector moves with the market, so the market's slope absorbs part of the sector's effect.

How do you compute the bias?

Write the sector as 0.4 x market plus a part unrelated to the market, then substitute. Stock = 0.5 x market + 1.0 x (0.4 x market + other) + noise = 0.9 x market + (1.0 x other + noise). The bracket is unrelated to the market, so a regression on the market alone recovers 0.9. The bias is the omitted coefficient times the slope of the omitted variable on the included one, 1.0 x 0.4. A simulation of 20,000 days with these coefficients gives a slope of 0.897, matching the algebra.

Is 0.9 wrong, or answering a different question?

It depends on what you use it for. If you want to hedge the stock with the market alone, 0.9 is the right hedge ratio, because it captures everything the market drags along with it. If you want the stock's exposure holding the sector fixed, say to build a sector-neutral book, 0.9 overstates it and 0.5 is the number you need. The bias can also run the other way: if the sector moved against the market, or the stock loaded negatively on the sector, the short slope would sit below 0.5. Naming both uses is what the interviewer is listening for.