Hedge Funds puzzles, solved step by step

Puzzles
100
Traced to a firm
38
Topics
14
Hard
30
Topic
All topicsBetting and sizing5Conditional probability and Bayes7Continuous probability and distributions7Counting and combinatorics7Estimation and mental maths4Expected value and dice games8Logic and brainteasers10Market making and trading games6Options and payoffs5Portfolio and risk maths8Random walks and Markov chains7Returns, compounding and fees7Statistics and estimation11Valuation, accounting and macro riddles8
Level
AnyWarm upCoreHard
Source
AnyReported at a firmStandard
Showing 1–2 of 2 · filtered from 100Clear filters
  1. 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

    Try it first

    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).

    t = Sharpe x root(years): halving the Sharpe quadruples the wait012345048121620Years of monthly returnst = 21 yr4 yrs16 yrsSharpe 2.0Sharpe 1.0Sharpe 0.5
    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≈SRannualY  ⇒  Y=(2SR)2:SR=1→4,SR=0.5→16t \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
    SRthe true annual Sharpe ratio
    Yyears of data
    2the 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

  2. 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

    Try it first

    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.

    Average first, then take the root: the curve sits above the chord00.511.520.511.5Variance estimateits square root0.04 gives 0.21.96 gives 1.4root of the average: 1.0average of the roots: 0.8Average sample SD / true SDnormal data, n observationsn = 20.79820.2% lown = 50.9406.0% lown = 100.9732.7% lown = 300.9910.9% low
    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]=c4(n) σ<E[s2]=σ,c4(2)=2/π≈0.798E[s] = c_4(n)\,\sigma < \sqrt{E[s^2]} = \sigma, \qquad c_4(2) = \sqrt{2/\pi} \approx 0.798
    sthe square root of the unbiased sample variance
    sigmathe 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?