Hedge Funds puzzles, solved step by step

Puzzles: 100
Traced to a firm: 38
Topics: 14
Hard: 30

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022Two independent, unbiased forecasts of next quarter's GDP growth have error standard deviations of 2% and 3%. How should you combine them, and how accurate is the blend?Statistics and estimationCoreQuant and systematic funds
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What weight should the 2% forecast get?
Show the worked solution
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%.
Blending with inverse-variance weights of 9/13 and 4/13 gives an error of 1.66%, lower than the better forecast's 2.00%, while equal weights give 1.80% and weights of 3/5 and 2/5 give 1.70%.
The relationship
$w_A = \frac{1/\sigma_A^2}{1/\sigma_A^2 + 1/\sigma_B^2} = \frac{1/4}{1/4 + 1/9} = \frac{9}{13} \qquad \sigma_{blend} = \sqrt{\frac{1}{1/4 + 1/9}} = \sqrt{\frac{36}{13}} \approx 1.66\%$
\sigma_A, \sigma_B the two forecasts' error standard deviations, 2% and 3%
w_A the weight on the sharper forecast
What it says in wordsEach forecast is weighted by its precision, one over its variance, and the blend's precision is the sum of the two.
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.
Where candidates lose it
Candidates either average equally, which overweights the noisier forecast, or keep only the better one, which throws information away. Both miss that the right weights come from the variances.
The subtler slip is weighting by one over the standard deviation, 3/5 and 2/5. It is close but not optimal; state the inverse-variance rule and show that the blended error beats 2%.
What the interviewer asks next
What if the two forecast errors have a correlation of 0.5?
How would you estimate each forecaster's error variance in practice?
One forecast is biased upwards by 0.5%. What do you do?
048A strategy's true annual Sharpe ratio is 1.0. How many years of monthly returns do you need before its average return shows a t-statistic of 2? What if the true Sharpe is 0.5?Statistics and estimationCoreViking Global InvestorsNew York · 2014
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Years needed for a Sharpe of 0.5:
Show the worked solution
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).
Because the t-statistic equals the Sharpe ratio times the square root of years, a Sharpe of 2.0 clears t = 2 after 1 year, a Sharpe of 1.0 after 4 years and a Sharpe of 0.5 only after 16 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.
The relationship
$t \approx SR_{\text{annual}}\sqrt{Y} \;\Rightarrow\; Y = \left(\frac{2}{SR}\right)^2: \quad SR = 1 \to 4, \quad SR = 0.5 \to 16$
SR the true annual Sharpe ratio
Y years of data
2 the target t-statistic
What it says in wordsThe years needed to prove a strategy grow with the inverse square of its Sharpe ratio.
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.
Where candidates lose it
The common loss is thinking monthly data gives twelve times the evidence, which leads to answers like four months. The twelve cancels, because the monthly Sharpe is smaller by root 12.
The second loss is saying a Sharpe of 0.5 needs twice as long as 1.0. The dependence is on the square: half the Sharpe, four times the data.
What the interviewer asks next
How many years for a Sharpe of 0.3?
You test 20 strategies and pick the best one with t = 2.2. How much do you trust it?
Would daily data change the answer for estimating the Sharpe ratio itself rather than the mean?
Asked at Viking Global Investors, Quantitative Research, New York, 2014 (Wall Street Oasis): how to reject a hypothesis test, what's your structure of your code, what's the sample size
063The sample variance computed with n minus 1 in the denominator is an unbiased estimator of the population variance. Is its square root an unbiased estimator of the standard deviation?Statistics and estimationCoreSquarepoint CapitalLondon · 2026
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Is the square root of the unbiased sample variance unbiased for the standard deviation?
Show the worked solution
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.
Two equally likely variance estimates of 0.04 and 1.96 average to the true variance of 1.0, but their square roots, 0.2 and 1.4, average only 0.8, below the true standard deviation of 1.0, because the square-root curve bends downwards.
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.
The relationship
$E[s] = c_4(n)\,\sigma < \sqrt{E[s^2]} = \sigma, \qquad c_4(2) = \sqrt{2/\pi} \approx 0.798$
s the square root of the unbiased sample variance
sigma the true standard deviation
c4(n) the correction factor for normal data, below 1 for every n
What it says in wordsThe average sample standard deviation is a fixed fraction of the true one, and that fraction is below one.
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.
Where candidates lose it
The trap is assuming unbiasedness carries through any function of an estimate. It carries through straight-line transformations only; the square root is curved, so the property is lost.
The second loss is saying it is biased without the direction or the size. Say biased low, give the Jensen reason in one sentence, and quote about 0.80 at two observations, shrinking towards 1 as the sample grows.
What the interviewer asks next
Is the square of an unbiased estimator of the standard deviation unbiased for the variance?
Why does the sample variance divide by n minus 1 rather than n?
Which estimator of sigma has the lowest mean squared error for normal data?
Asked at Squarepoint Capital, Quantitative Research, London, 2026 (Wall Street Oasis): Is the square root of the unbiased estimator for sample variance unbiased for standard deviation?
073A thousand fund managers have no skill at all: each has a 50% chance of beating the market in any year, independently. How many will beat it five years running, and what does that say about track records?Statistics and estimationCoreQuant and systematic funds
Try it first
How many of the 1,000 unskilled managers beat the market five years in a row?
Show the worked solution
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.
Starting from 1,000 unskilled managers, half fall away each year, leaving 500, 250, 125, 62.5 and finally about 31 with a flawless five-year record produced by chance alone.
The relationship
$E[\text{perfect records}] = 1{,}000 \times \left(\tfrac12\right)^5 = 31.25$
1,000 the number of managers
1/2 each manager's chance of beating the market in a year
5 the number of years
What it says in wordsMultiply the crowd by the chance that one member gets the streak.
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.
Where candidates lose it
The first loss is saying none, or very few, because five in a row sounds impressive. The interviewer wants the crowd arithmetic: rare for one, expected for many.
The second is stopping at 31 without the conclusion. The number is only half the answer; say what it means for reading a track record, and name survivorship bias.
What the interviewer asks next
How many of the 1,000 beat the market in at least four of the five years?
How many years of beating the market would one unskilled manager in 1,000 be expected to reach?
How would you design a test that separates a 60% manager from a 50% one?
088A stock's true model is: stock return = 0.5 x market return + 1.0 x sector return + noise. Regressing the sector's return on the market gives a slope of 0.4. If you regress the stock on the market alone, what slope do you get?Statistics and estimationCoreQuant and systematic funds
Try it first
What does the market-only regression report?
Show the worked solution
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.
The market reaches the stock directly with a coefficient of 0.5 and through the sector with 0.4 x 1.0 = 0.4, so a regression of the stock on the market alone reports 0.9, of which 0.4 is the sector's effect credited to the market.
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.
The relationship
$\hat\beta_{\text{short}} = \beta_M + \beta_S\,\delta = 0.5 + 1.0 \times 0.4 = 0.9$
beta_M the stock's true direct loading on the market, 0.5
beta_S the stock's loading on the sector that was left out, 1.0
delta the slope of the sector's return on the market's, 0.4
What it says in wordsThe short regression's slope is the true slope plus the left-out variable's effect times how much that variable moves with the one you kept.
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.
Where candidates lose it
The fast wrong answer is 0.5: candidates assume a regression recovers the true coefficient whatever else is left out. It does so only when the omitted variable is unrelated to the included one.
The second loss is getting 0.9 and calling it simply wrong. It is the correct total effect of the market and the right number for a market-only hedge; it is wrong only as an estimate of the direct effect.
What the interviewer asks next
What slope do you get if the sector's slope on the market is minus 0.4?
You add the sector to the regression. What happens to the standard error of the market coefficient if the two are highly correlated?
How does this bias show up when you estimate a stock's factor exposures with too few factors?

\sigma_A, \sigma_B	the two forecasts' error standard deviations, 2% and 3%
w_A	the weight on the sharper forecast

SR	the true annual Sharpe ratio
Y	years of data
2	the target t-statistic

s	the square root of the unbiased sample variance
sigma	the true standard deviation
c4(n)	the correction factor for normal data, below 1 for every n

1,000	the number of managers
1/2	each manager's chance of beating the market in a year
5	the number of years

Try it first

Why does blending two forecasts beat the better one?

Why inverse variance and not inverse error?

What would you check before trusting the blend?

Where candidates lose it

What the interviewer asks next

Try it first

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

Why does monthly data not shorten the wait?

Where candidates lose it

What the interviewer asks next

Try it first

Why does taking a square root break unbiasedness?

How big is the bias?

Does it matter in practice?

Where candidates lose it

What the interviewer asks next

Try it first

Why does a crowd produce streaks even without skill?

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

What should you look at instead?

Where candidates lose it

What the interviewer asks next

Try it first

Why does leaving the sector out change the market slope?

How do you compute the bias?

Is 0.9 wrong, or answering a different question?

Where candidates lose it

What the interviewer asks next