Quant Research Statistics Interview: 7 Worked Cases

A quant research statistics interview can start with a small table and become a discussion about what the table actually proves. Calculating a mean is one step. Choosing the right comparison, recognizing dependence and explaining uncertainty are often the more interesting steps.

These seven original practice cases train that conversation. All datasets and research scenarios below are invented for teaching. Try each prompt aloud before opening its explanation, then answer the follow-up with a changed assumption.

The format reflects a real preparation need: Two Sigma's official research interview guide names data analysis, coding and statistics among its discussion areas, and encourages candidates to explain an initial approach and improve it. The precise interview depends on your role and recruiter instructions.

A repeatable way to start an open-ended question

Before choosing a formula, identify the unit of observation, the quantity you want to estimate and how the data were collected. Ask whether observations are independent, whether groups were assigned randomly, and what information was available when a decision was made.

Then give a simple first answer, explain its assumptions and propose a check. You do not need to list every conceivable statistical problem. Choose the issue that could change the conclusion most.

For these cases, spend roughly three minutes on the first answer and another two on the follow-up. That is a practice suggestion, not an employer time limit. If the arithmetic is familiar but your explanation becomes vague, work on naming the population, comparison and uncertainty explicitly.

Case 1: a better overall result can hide worse results in every group

Two fictional models predict whether an observation belongs to a positive class. Their results are:

Which model is better? What would you ask before selecting one?

Show answer and explanation

Overall, A is correct on 83 of 100 observations, or 83%. B is correct on 51 of 100, or 51%. However, B is better within both groups: 95% versus 90% on easy observations, and 40% versus 20% on hard observations.

The overall rates compare different mixes of difficulty. They do not isolate model quality. If a common evaluation population is 50% easy and 50% hard, the weighted accuracies are 55% for A and 67.5% for B. If the common population is 90% easy, they are 83% and 89.5%, respectively.

Ask for both models' predictions on the same held-out observations. Also ask whether accuracy captures the cost of errors. A missed rare positive might matter more than a correctly identified negative.

Follow-up: Does B's within-group advantage prove it will outperform in production? No. These are small observed samples, particularly A's ten hard observations. The population mix could change, the samples could differ in other ways, and estimation uncertainty remains. The comparison reveals a design problem; it does not eliminate uncertainty.

Case 2: a statistically suggestive mean with six observations

A fictional signal produces these six daily returns, in percentage points: 1, 2, −1, 0, 3, 1. Estimate the mean and explain how certain you are that its true mean is positive.

Show answer and explanation

The sample mean is 1 percentage point. The deviations from the mean are 0, 1, −2, −1, 2 and 0; their squared sum is 10. The sample variance is 10/5 = 2, so the sample standard deviation is approximately 1.414 percentage points.

Under independent, identically distributed observations, the estimated standard error of the mean is √(2/6), approximately 0.577 percentage points. With independent normal observations, a two-sided 95% Student-t interval uses five degrees of freedom and a critical value of approximately 2.571. The interval is therefore 1 ± 2.571 × 0.577, or approximately −0.48 to 2.48 percentage points.

That interval includes zero. It is wide because the sample is tiny. It does not say there is a 95% probability that this fixed population mean lies inside this particular frequentist interval.

Follow-up: What if returns persist across adjacent days? Positive serial dependence can make the independent-observation standard error too optimistic. Six observations offer little basis for estimating that dependence reliably. Say what additional history and diagnostics you would request before making a stronger claim. Do not silently annualize six daily observations into a persuasive performance statistic.

Case 3: two features explain the same information

A regression includes an intercept and two predictors. The values of x₁ vary across observations, but x₂ = 2x₁ for every observation. What happens to the fitted coefficients?

Show answer and explanation

The linear prediction is intercept + β₁x₁ + β₂x₂, which equals intercept + (β₁ + 2β₂)x₁. Only the combined coefficient is identified from these data.

If β₁ + 2β₂ = 3, the pairs (3, 0), (1, 1) and (−1, 2) produce identical predictions. The design matrix lacks full column rank, so ordinary least squares has no unique coefficient solution. A software package might remove a column or select a solution using a pseudoinverse; that choice does not give the separate coefficients a causal interpretation.

Remove redundant information or choose a parameterization that matches the question. Ridge regularization with a positive penalty can choose a unique solution for the penalized slopes, but the resulting allocation reflects the penalty and feature scales as well as the data.

Follow-up: What if the equality is only approximate? Coefficients can become unstable even when predictions remain relatively stable. Check correlations, scaling, coefficient sensitivity and held-out performance. Predictive usefulness and interpretable effects are different objectives.

Case 4: the best result among 100 experiments

A researcher tries 100 independent tests. Assume every null hypothesis is true and each valid test rejects with probability 5%. What is the chance of at least one rejection?

Show answer and explanation

The chance that all tests avoid rejection is 0.95¹⁰⁰. Therefore the probability of at least one rejection is 1 − 0.95¹⁰⁰, approximately 99.41%. The expected number of false rejections is 100 × 0.05 = 5, a different quantity.

A researcher who shows only the winning result hides the selection process. Ask how many alternatives were tried, whether the test set influenced selection, and whether the proposed result survives genuinely untouched evaluation.

A Bonferroni threshold of 0.05/100 = 0.0005 per test controls the family-wise error rate at no more than 5% when the individual p-values are valid. That bound does not require independence, but can be conservative. It also cannot repair a test whose assumptions or data handling were already invalid.

Follow-up: Does correlation between tests change 99.41%? Yes; the exact expression assumed independence. Correlated experiments also make the effective amount of new information difficult to judge. Recording the full search history remains useful even when an exact correction is unavailable.

Case 5: a clean train/test split still leaks information

You split observations chronologically into training and test periods. Before fitting a model, you use both periods together to select the 20 features most correlated with the target. Is the evaluation clean?

Show answer and explanation

No. Feature selection uses test targets, so the test period helps choose the model before evaluation. A chronological row split does not undo that information transfer.

Perform feature selection using training data only. For cross-validation, repeat selection within each training fold. Keep the final holdout outside feature decisions and tuning. Any transformation that learns from data needs an explicit fit boundary; whether a particular transformation causes serious optimism depends on what it learns, but the boundary should remain auditable.

Also check the prediction horizon. If a training label contains a return extending into the validation period, a boundary between rows may still allow overlapping information. Leave an appropriate gap or otherwise remove that overlap according to the target definition.

Follow-up: Should you randomly split daily market data? A random split can mix earlier and later observations and obscure time dependence or regime shifts. Choose a validation design that represents the intended future-use setting, and explain its limits. For auditing a full portfolio project, use our separate research-project review checklist.

Case 6: more rows do not always mean more independent evidence

You have 10,000 rows, built from 100 assets on each of 100 days. A model's errors are strongly correlated across assets on the same day. Can you treat the rows as 10,000 independent observations?

Show answer and explanation

No. A shared market shock can affect many rows at once. Independence-based uncertainty calculations can exaggerate precision, and a random row split can place observations from the same day in both training and evaluation.

First specify the estimand: performance on a new asset, a future day, or both. For future-day performance, hold out time blocks rather than arbitrary rows. Depending on the dependence structure and sample size, uncertainty estimation might use day-level aggregation, clustered methods or a block bootstrap.

There is no universal rule that the effective sample size equals 100. Correlation can occur across days and within assets too. The appropriate method depends on the data-generating process, not just the table dimensions.

Follow-up: What would you plot first? Errors or losses by day, by asset and through time can reveal concentration, common shocks and changing performance. A plot diagnoses structure; it does not automatically establish a valid inference procedure.

Case 7: better forecasts can produce a worse trading result

Two hypothetical strategies have these per-trade assumptions, with gross returns and costs measured on the same notional base:

Which has the higher expected net return per trade? What is missing from that answer?

Show answer and explanation

A has an expected net return of 5 basis points; B has 2. Under the stated assumptions, A is better on this single measure despite B's higher gross return.

That does not settle overall performance. You still need exposure, risk, turnover, trade frequency, capital constraints, capacity and uncertainty in both forecasts and costs. Costs might increase with trade size or worsen exactly when the signal is strongest. A point estimate that assumes a fixed cost is only a starting scenario.

Follow-up: Can improved classification accuracy guarantee higher returns? No. The magnitudes of wins and losses, selection of trades and execution conditions matter. Explain the decision rule connecting a prediction to an action rather than treating an ML metric as a trading result.

Turn your answers into a preparation plan

Score each explanation on four checks: did you define the comparison, calculate correctly, state the key assumption, and propose a test that could change your conclusion? Record the weakest check rather than only counting correct numerical answers.

For foundational practice, revisit conditional probability and expected value. For planning across coding, probability and research discussion, use the four-week interview study plan.

If you can calculate the answers but struggle to defend a modeling choice, bring a case and your target role to a free quant consultation. Discuss the preparation and feedback that would help you explain your work more clearly.