Hedge Funds puzzles, solved step by step

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Showing 1–2 of 2 · filtered from 100Clear filters
  1. 013A researcher tests 20 unrelated trading signals, each at a 5% significance level, and none of them truly works. What is the chance that at least one of them looks significant?Statistics and estimationWarm upQuant and systematic funds

    Try it first

    Your instinct: the chance of at least one false discovery?

    Show the worked solution

    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.

    Test enough useless signals and a false winner becomes likely25%50%75%5.0%122.6%540.1%1051.2%1564.2%205%What each single test promises: 5%What the search delivers at 20 tests: 64.2%Number of useless signals tested, each at 5%
    The chance of at least one false positive rises from 5% for one useless signal to 40.1% for ten and 64.2% for twenty, passing even odds at 14 tests, although every individual test is run at 5%.
    The relationship
    P(at least one false positive)=1−(1−α)m=1−0.9520≈0.642P(\text{at least one false positive}) = 1 - (1-\alpha)^m = 1 - 0.95^{20} \approx 0.642
    \alphathe significance level of each test, 5%
    mthe number of independent tests, 20
    What it says in wordsThe chance that every test stays quiet shrinks with each test added, so the chance of a false winner grows.

    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.

    Where candidates lose it

    The fast wrong answer adds the probabilities: 20 x 5% = 100%, a certainty. Adding only works for events that cannot happen together; here several false positives can appear at once, so go through the complement.

    The quieter error is answering 5%, treating the batch as one test. The interviewer wants you to see that the error rate of the search is not the error rate of each test inside it.

    What the interviewer asks next

    • How many tests at 5% before a false positive is more likely than not?
    • What significance level per test keeps the family-wide chance at 5% across 100 tests?
    • Why does out-of-sample testing help, and what can still go wrong with it?
  2. 038X and Y are independent random variables with the same variance. What is the correlation between X and X + Y?Statistics and estimationWarm upSquarepoint CapitalMontreal · 2026

    Try it first

    Pick one:

    Show the worked solution

    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.

    X is half of X + Y: it explains half the variance, so rho = root(1/2)X, length sigmaYX + Yroot 2 sigma45 degreescos 45 = 0.707Variance of X + Y = 2 sigma squaredfrom X: sigma squaredfrom Y: sigma squaredCov(X, X + Y) = Var X + Cov(X, Y) = sigma squaredsd(X) x sd(X + Y) = sigma x root 2 sigmarho = sigma squared / (root 2 sigma squared)= 1 / root 20.707R squared = 0.5: X explains half of the sum
    Drawn as arrows, independent X and Y sit at right angles and their sum lies at 45 degrees to X, so the correlation is cos 45, about 0.707; equivalently, X supplies half of the variance of X + Y, and the correlation is the square root of one half.

    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.

    Where candidates lose it

    The common loss is answering 0.5 because X is half of the sum. That is the share of variance, and correlation is its square root.

    The other loss is saying zero because X and Y are independent. The sum contains X, so it cannot be independent of X. Split the covariance in one line and the answer falls out.

    What the interviewer asks next

    • What is the correlation between X + Y and X - Y?
    • Y has four times the variance of X. What is corr(X, X + Y) now?
    • What is the correlation between the sum of the first 10 and the sum of the first 20 of a series of independent returns?

    Asked at Squarepoint Capital, Desk Quant Analyst Interview, Montreal, 2026 (Wall Street Oasis): There were also 3-4 basic math/stats questions about mean, covariance, correlation, etc.