Quant interview preparation

Questions

100

Traced to a firm

53

Firms

15

Updated

September 2026

Asked at

All firmsOld Mission Capital12Tower Research Capital10Jump Trading7Akuna Capital5Citadel4DED.E. Shaw3Jane Street3ACAQR Capital Management2DRW2Millennium Management2Schonfeld2SCSquarepoint Capital2Susquehanna International Group2Belvedere Trading1Optiver1

Topic

All topicsProbability10Coins, cards and games6Expected value8Statistics11Market making15Estimation and mental maths4Stochastic processes4Regression5Machine learning6Time series6Programming10Options and derivatives8Fit and motivation7

Level

AnyCoreIntermediateHard

Type

AnyBrainteaserTechnicalCaseMarket viewFit

1. 011A random variable is uniform on the interval zero to ten. What are its expected value and variance?StatisticsCorephone / first roundOld Mission CapitalFinance · New York · 2018

   Say this

   Mean 5, variance 100 over 12, which is 8.33, so standard deviation about 2.89. For a uniform on a to b the mean is the midpoint and the variance is (b minus a) squared over 12.

   Then walk it

   1. Mean by symmetry: the midpoint of 0 and 10 is 5. No integration needed.
   2. Variance from the formula (b-a) squared over 12: 100 over 12 equals 8.33, standard deviation 2.887.
   3. If you want to derive it, E of X squared is the integral of x squared over 10 from 0 to 10, which is 1000/30 equals 33.33. Subtract 25 and you get 8.33. Good to be able to do it either way.
   4. The 1/12 is worth carrying in your head because it recurs: a fair n-sided die has variance (n squared minus 1)/12, and the rounding error of a value rounded to the nearest tick has variance tick squared over 12. That last one comes up in real microstructure work.
   5. Practical note: the uniform has thin support and no tails, so it is a bad default for anything financial. The moment somebody hands you a uniform in a trading context, ask what it is meant to represent.

   Where candidates lose it

   Reaching for integration under time pressure and fumbling the arithmetic. Know the (b-a) squared over 12 form cold. Also do not quote variance when they asked for standard deviation or the other way round, and say which one you are giving.

   Expect next

   • What is the expected value of the maximum of two independent draws?
   • What is the distribution of the sum of two independent uniforms?
   • What is the variance of the rounding error when you round to the nearest penny?

   Reported by candidates at Old Mission Capital (Finance, New York, 2018). Source: Wall Street Oasis.

2. 030You draw n independent uniforms on zero to one. What are the expected values of the maximum and the minimum, and of the kth smallest?StatisticsIntermediatetechnicalQuant researchQuant trading

   Say this

   The maximum has mean n/(n+1), the minimum 1/(n+1), and the kth smallest k/(n+1). The n points cut the interval into n plus 1 gaps that are exchangeable, so each gap averages 1/(n+1).

   Then walk it

   1. Derive the max directly: P(max at most x) is x to the n, so the density is n x to the n minus 1, and the integral of x times that from 0 to 1 is n/(n+1).
   2. The gap argument is faster and generalises. The n order statistics plus the two endpoints create n plus 1 spacings, which are exchangeable with total length 1, so each has mean 1/(n+1). The kth order statistic is the sum of the first k spacings, hence k/(n+1).
   3. The kth order statistic is Beta(k, n minus k plus 1), which gives you the variance too: k(n-k+1) over ((n+1) squared (n+2)).
   4. Numbers: with 10 draws the max averages 0.909 and the min 0.091. With 100 draws the max averages 0.990. The max creeps to the boundary at rate 1/n, which is why extreme-value estimates converge slowly.
   5. Why a quant desk cares: the max of n draws is your model for the best of n signals, the worst drawdown of n periods, and the winning quote in an auction with n bidders. And it explains selection bias, because the best of a hundred backtests looks good even when none of them has any edge.

   Where candidates lose it

   Answering only for the max with a calculus derivation and then being stuck on the general kth. Learn the spacings argument, it gives all of them at once. And be ready to connect it to selection bias, because the practical follow-up is almost always about why the best of many strategies overstates its own quality.

   Expect next

   • What is the variance of the maximum?
   • What is the expected range, max minus min?
   • How does this explain the selection bias in picking the best of a hundred backtests?
3. 036The sample variance with the n minus one correction is unbiased. Is its square root an unbiased estimator of the standard deviation?StatisticsHardtechnicalSCSquarepoint CapitalQuantitative Research · London · 2026

   Say this

   No. The square root is concave, so by Jensen's inequality the expected square root is strictly less than the square root of the expected value. The sample standard deviation is biased downwards, always, for any distribution with positive variance.

   Then walk it

   1. Jensen: for a strictly concave g, E of g(X) is less than g of E of X unless X is degenerate. With g the square root and X the unbiased sample variance, E of s is less than sigma.
   2. Size the bias for normal data. E of s equals c4(n) times sigma, where c4 is a known constant involving gamma functions. At n equal to 2, c4 is about 0.798, so you understate sigma by 20 percent. At n equal to 10 it is 0.9727, a 2.7 percent understatement. At n equal to 30 it is 0.9914.
   3. So the bias is order 1/(4n) and it vanishes as n grows. It is a real problem for short samples and irrelevant for long ones.
   4. Unbiasedness is also not preserved under any nonlinear transform, which is the general lesson. The unbiased estimator of sigma squared does not give you an unbiased estimator of sigma, or of 1/sigma, or of log sigma.
   5. Where this bites on a desk: annualised volatility estimated from a few weeks of data, and any Sharpe ratio, since the Sharpe divides by s. Understating s inflates the Sharpe, so short-sample Sharpes are biased upwards. That is worth saying because it connects a textbook Jensen question to a live problem in strategy evaluation.

   Where candidates lose it

   Saying yes because the variance estimator is unbiased. Unbiasedness does not survive a nonlinear function. Name Jensen explicitly, give the direction of the bias, and quantify it with c4 for at least one small n. The follow-up about Sharpe ratios is where the real conversation is, so get there yourself.

   Expect next

   • How would you correct it?
   • What does that imply for a Sharpe ratio estimated on a short sample?
   • Is the sample correlation coefficient unbiased?

   Reported by candidates at Squarepoint Capital (Quantitative Research, London, 2026). Source: Wall Street Oasis.

4. 037You stand on a road and watch cars drive past. How would you estimate the parameter of the underlying distribution?StatisticsHardtechnicalJump TradingQuantitative Research · Chicago · 2018

   Say this

   First I would state the model: arrivals as a Poisson process with rate lambda, so inter-arrival times are exponential with mean 1 over lambda. Then the maximum likelihood estimate of lambda is just the count divided by the observation time, and its standard error is lambda over the square root of the count.

   Then walk it

   1. Model choice first, and justify it: independent arrivals at a constant rate with no memory gives a Poisson process. That is reasonable on a quiet road, and clearly wrong near a traffic light where cars arrive in platoons.
   2. MLE: for n arrivals in time T, lambda hat is n over T. It is unbiased, and the variance is lambda over T, so the relative standard error is 1 over the square root of n. Twenty-five cars gives you a 20 percent standard error, a hundred cars gives 10 percent.
   3. That tells you the sample size you need before you open your mouth about precision. If someone wants the rate to five percent, you need 400 cars.
   4. Now the diagnostics, which are what a research interview is actually about. Plot the inter-arrival times and check whether they look exponential. Over-dispersion, meaning variance above the mean of the counts, tells you arrivals are clustered and Poisson is wrong. Then I would go to a Cox process or a Hawkes process with self-excitation.
   5. And I would flag the estimation trap: if instead I sampled by picking a random moment and measuring the gap I happened to land in, I would oversample long gaps. That is the inspection paradox, and it biases the mean gap upward by a factor of one plus the squared coefficient of variation. It is the same bias that makes waiting times feel longer than the timetable says.

   Where candidates lose it

   Jumping to a formula without stating the model or checking it. The interviewer wants model, estimator, standard error, then diagnostics. The specific failure mode they are hunting is the inspection paradox, so mention length-biased sampling unprompted. Hawkes processes are the right answer for clustered arrivals and they are also how trade arrivals actually behave in markets.

   Expect next

   • How would you test whether the Poisson assumption holds?
   • What if the cars arrive in clusters?
   • How long do you need to watch to get the rate within five percent?

   Reported by candidates at Jump Trading (Quantitative Research, Chicago, 2018). Source: Wall Street Oasis.

5. 040Here is a dataset. Analyse it using probability metrics and tell me what you find.StatisticsHardcase studyJane StreetCredit Risk · London · 2025

   Say this

   I would spend the first third of the time on the data itself before any modelling: shape, missingness, duplicates, timestamps, and the univariate distributions. Then state a hypothesis, test it, and report the effect size with an honest uncertainty. Narrate every step, because the interviewer is grading the process, not the punchline.

   Then walk it

   1. Start with the boring checks, out loud. Row count, date range, obvious duplicates, missing values and whether they are missing at random, and whether any column is a leak of the outcome. Most real findings in interviews of this kind are data artefacts.
   2. Then univariates: mean, median, standard deviation, skew, kurtosis, and the tails. Plot histograms and the empirical CDF. If a column is heavy-tailed or bimodal, say so, because it changes every subsequent choice.
   3. Then the relationship you were asked about. Give a point estimate plus a confidence interval, and prefer a plot to a coefficient. If the data are time-ordered, check for autocorrelation and regime change before quoting any p-value, because serial dependence inflates significance badly.
   4. Then the discipline: state your null, say what result would change your mind, and count how many hypotheses you have looked at. If you tested twenty things, say so and adjust.
   5. Close with what the data cannot tell you. A credit dataset with survivors only cannot tell you about defaults. Ending on the limitation is what separates an analyst from someone producing numbers, and in a live exercise it is the cheapest way to sound senior.

   Where candidates lose it

   Going straight to a model. Almost every candidate opens a regression and never looks at a histogram, then reports a spurious result driven by three outliers or a broken timestamp. Talk through the data integrity checks first, and say your uncertainty on every number you quote.

   Expect next

   • What would you check before trusting that correlation?
   • How many hypotheses did you test, and how does that change your p-value?
   • What would you want that is not in this dataset?

   Reported by candidates at Jane Street (Credit Risk, London, 2025). Source: Wall Street Oasis.

6. 043State the central limit theorem and tell me where it fails.StatisticsCoretechnicalQuant researchQuant trading

   Say this

   For independent identically distributed variables with finite mean and finite variance, the standardised sample mean converges in distribution to a standard normal. The key conditions are finite variance and enough independence, and both fail regularly in markets.

   Then walk it

   1. Precisely: root n times (X bar minus mu) over sigma converges in distribution to N(0,1). Note it is the standardised mean that converges, and the rate is 1 over root n.
   2. Failure one, infinite variance. A Cauchy distribution has no variance and the sample mean of Cauchys is Cauchy again, no matter how large n is. Averaging buys you nothing. More generally, stable distributions with tail index alpha below 2 converge to a stable law, not a normal.
   3. Failure two, dependence. With strongly autocorrelated data the effective sample size is far below n, so you converge much more slowly and your standard errors are too small. Long-range dependence can break it entirely.
   4. Failure three, the rate in the tails. Even where the CLT holds, convergence is fastest in the middle and slowest in the tails, which is precisely where a risk manager needs accuracy. Berry-Esseen gives an error bound of order 1 over root n times the third absolute moment, so skewed data converges slowly.
   5. The practical version: daily equity returns have kurtosis of 5 to 10 and volatility clustering, so ten-day sums are much closer to normal than daily returns, but a 99.9 percent quantile computed from a normal assumption will still understate the tail badly. That is why value at risk models use empirical or extreme-value tails rather than leaning on the CLT.

   Where candidates lose it

   Stating the theorem without the finite variance condition, or claiming everything becomes normal for large n. Also do not confuse it with the law of large numbers, which is about convergence of the mean to a constant and needs only finite mean. Be ready to say what happens with infinite variance, because that is the follow-up.

   Expect next

   • What happens with a Cauchy distribution?
   • How is that different from the law of large numbers?
   • How large does n have to be in practice for returns data?
7. 044What is a p-value, and what is it not?StatisticsCoretechnicalQuant researchRisk

   Say this

   It is the probability of seeing data at least as extreme as what you saw, assuming the null hypothesis is true. It is not the probability that the null is true, and it is not the probability you are wrong.

   Then walk it

   1. The conditioning runs the wrong way from what people assume. A p-value is P(data given null), and what you actually want is P(null given data). Those are different objects and Bayes tells you the second depends on your prior.
   2. Concretely: if you test a thousand strategies of which fifty genuinely work, at a five percent significance level you get roughly 47 true discoveries and 47 false ones. A p-value of 0.05 in that setting means a coin flip on whether the finding is real.
   3. It also says nothing about effect size. With a million observations a completely useless one-basis-point edge will have a p-value of 0.0001. Significance is not importance, and in high-frequency data everything is significant.
   4. And it is only valid for a pre-specified test. Choosing the test after looking at the data, or stopping data collection when the p-value crosses 0.05, invalidates it completely.
   5. What I would report instead on a desk: the effect size with a confidence interval, out-of-sample performance, and how many specifications I tried. A p-value on its own is close to useless in a research process where hundreds of hypotheses get screened.

   Where candidates lose it

   Defining it as the probability the null is true. That is the single most common statistical error in finance interviews and it is disqualifying at a research shop. Also be ready with the multiple-testing consequence, because the interviewer's real target is whether you understand why published anomalies do not replicate.

   Expect next

   • So what significance level would you use if you screened a thousand signals?
   • Explain the false discovery rate.
   • What would you report instead of a p-value?
8. 045You test two hundred signals and three come back significant at the five percent level. What do you conclude?StatisticsHardtechnicalQuant researchQuant trading

   Say this

   That you have found nothing. Under a pure null you would expect ten false positives from two hundred tests at five percent, so three is fewer than chance. If anything the result is evidence against there being any signal at all.

   Then walk it

   1. Expected false positives are 200 times 0.05 equals 10. Getting three significant results is below what noise alone produces, so the finding is not just unimpressive, it is worse than random.
   2. The right frame is family-wise error or false discovery rate. Bonferroni sets the threshold at 0.05 over 200, which is 0.00025, brutal but valid. Benjamini-Hochberg controls the expected proportion of false discoveries among the rejections and is much less conservative, which is usually the better choice when you are screening.
   3. The subtlety with financial signals: they are heavily correlated with each other, so the effective number of independent tests is far below 200. Bonferroni is then too harsh. I would estimate the effective number of tests, for example from the eigenvalue spectrum of the signal correlation matrix, or use a permutation or block-bootstrap null that preserves the correlation structure.
   4. The right test of whether anything survived is not a p-value at all. It is out-of-sample: hold back a period, or better a different market, and see whether the three signals still work with the sign you predicted.
   5. And the disclosure discipline, which is the answer a research head wants to hear: I would report the number of specifications tried alongside the result. The deflated Sharpe ratio and Harvey and Liu's work on multiple testing in finance both exist because the profession spent decades not doing this.

   Where candidates lose it

   Getting excited about the three and building a strategy on them. The whole question is whether you compute the expected number of false positives before you get attached. Say ten out of two hundred immediately, then talk about correlated tests, because that is where the technical depth is.

   Expect next

   • How would you estimate the effective number of independent tests?
   • What is the deflated Sharpe ratio?
   • How would you set up the experiment properly from the start?
9. 049You need a covariance matrix for five hundred assets and you have two years of daily data. What is the problem and how do you fix it?StatisticsHardsuperdayQuant researchRisk

   Say this

   You have 500 assets and roughly 500 observations, so the sample covariance matrix is nearly singular and its smallest eigenvalues are garbage. Any optimiser will load up on exactly those directions, so you have to shrink or impose factor structure.

   Then walk it

   1. Count the parameters: 500 times 501 over 2 is about 125,000 numbers estimated from 250,000 data points. The ratio of assets to observations, roughly one here, is what governs the damage, and the sample eigenvalue spectrum is badly biased even at a ratio of a quarter.
   2. Marchenko-Pastur describes exactly how the eigenvalues spread out. The largest are overstated and the smallest understated, and the smallest ones are the low-variance directions a mean-variance optimiser will concentrate in. That is why naive optimisers produce absurd leveraged long-short positions.
   3. Fix one, shrinkage. Ledoit-Wolf shrinks the sample matrix towards a structured target like a constant-correlation matrix, with an optimal intensity derived in closed form. Cheap, well-behaved and hard to beat as a default.
   4. Fix two, factor structure. Model returns as exposures to a few factors plus idiosyncratic noise, so the covariance is B times F times B transpose plus a diagonal. You have gone from 125,000 parameters to a few thousand. This is what every commercial risk model does.
   5. Fix three, random matrix filtering: keep the eigenvalues above the Marchenko-Pastur bulk edge as signal and replace the bulk with its average. Then state the practical check, which is out-of-sample portfolio variance rather than any in-sample fit statistic, because in-sample the sample matrix always wins and is always wrong.

   Where candidates lose it

   Saying you would just use the sample covariance matrix because two years is a lot of data. It is not, relative to 500 assets. The interviewer is testing whether you know that estimation error in the covariance matrix, not in the means, is what breaks portfolio optimisation in practice, and whether you can name shrinkage or factor models as the fix.

   Expect next

   • Why does the optimiser concentrate in the smallest eigenvalue directions?
   • How do you choose the shrinkage intensity?
   • How would you test whether your covariance matrix is any good?
10. 056What is maximum likelihood estimation, and when would you prefer method of moments?StatisticsIntermediatetechnicalQuant researchRisk

    Say this

    MLE picks the parameters that make the observed data most probable under your assumed distribution. It is asymptotically efficient if the model is right, which is exactly the condition that makes method of moments attractive when it is not.

    Then walk it

    1. MLE: maximise the log likelihood, which is the sum of log densities. Under regularity conditions it is consistent, asymptotically normal, and attains the Cramer-Rao bound, with variance given by the inverse Fisher information.
    2. Method of moments: match sample moments to their theoretical expressions and solve. Generalised method of moments extends this to more moment conditions than parameters, weighting them optimally, and it needs no full distributional assumption.
    3. So the tradeoff is efficiency versus robustness. MLE uses the whole density, so it extracts every bit of information and pays for it with sensitivity to misspecification. GMM uses only the moments you trust.
    4. Concrete case: fitting a distribution to daily returns. MLE under a normal assumption gives you the sample mean and variance and will be badly misled by the tails. MLE under a Student t estimates the degrees of freedom and is much better behaved. GMM on a few robust moments avoids committing to either.
    5. Practical points worth raising: MLE can be biased in small samples even when consistent, the classic example being the variance estimator with n rather than n minus 1 in the denominator. And numerically you should always check the Hessian at the optimum, because a flat likelihood means your parameter is not identified, which is common in GARCH and regime models.

    Where candidates lose it

    Describing MLE as the best estimator without the qualifier if the model is correctly specified. That caveat is the entire content of the comparison. Also be ready for the small-sample bias point, since MLE being biased while still consistent catches people who have only memorised the asymptotic properties.

    Expect next

    • Give me an example where MLE is biased.
    • What is the Cramer-Rao bound?
    • How would you fit a Student t to returns, and what does the estimated degrees of freedom tell you?

Firm tags come from public, anonymous candidate reports on Wall Street Oasis: strong signal, not sworn testimony. Firms are named as the places a question was reported, not as partners of Fin Maverick. Answers are written for this page to show how to think out loud; they are not scripts to recite.