11 Quant Statistics Interview Questions and Answers

Short answer: A p-value is the probability of observing a test statistic at least as extreme as the one you got, assuming the null hypothesis is true: P(data ≥ observed | H₀). It is a statement about the data under H₀ — not about the probability of the hypothesis itself.

In depth:

1. Direction of the conditioning — the p-value conditions on H₀. Getting P(H₀ | data) requires a prior and Bayes' theorem; you cannot extract it from a p-value alone.
2. Not the probability of the null — "p = 0.03 means H₀ is true with probability 3%" is the classic trap interviewers specifically listen for.
3. Not an effect size — with huge n, a tiny, economically meaningless effect yields p < 0.001. Statistical significance ≠ practical importance.

p-value  =  P(T ≥ t_obs | H₀ true)        ✓
p-value  ≠  P(H₀ true | data)             ✗
1 − p    ≠  P(the effect is real)         ✗

⚠️ Common mistake: "1 − p = 97% probability the effect is real." A p-value says nothing about the probability of hypotheses — only how surprising the data are if H₀ holds.

Short answer: Bessel's correction: we measure spread around the sample mean x̄ rather than the true μ, and x̄ adapts to the data, eating one degree of freedom. Mathematically E[Σ(Xᵢ−X̄)²] = (n−1)σ², so dividing by n−1 is exactly what makes the estimator unbiased.

In depth:

1. Where the bias comes from — x̄ minimizes Σ(xᵢ−a)² over a, so Σ(xᵢ−x̄)² ≤ Σ(xᵢ−μ)²: deviations from x̄ are systematically smaller than deviations from the true μ.
2. Degrees of freedom — of the n deviations only n−1 are free: they are tied by the identity Σ(xᵢ−x̄) = 0.
3. Size of the effect — dividing by n understates the variance by a factor (n−1)/n; cosmetic for large n, already 20% at n = 5.

E[Σ(Xᵢ−X̄)²] = Σ E[(Xᵢ−μ)²] − n·E[(X̄−μ)²]
             = n·σ² − n·(σ²/n) = (n−1)·σ²
⇒ s² = Σ(Xᵢ−X̄)² / (n−1) is unbiased

⚠️ Common mistake: calling the /n estimator "wrong". It is the MLE in the normal model — biased low but consistent; choosing between /n and /(n−1) is a question of criterion, not arithmetic.

Short answer: Unknown — and by default, probably not. Under 100 tests of pure noise you expect exactly 100 × 0.05 = 5 false positives, so the result is perfectly consistent with zero real signals. This is the multiple-testing problem — the central statistics question in quant research.

In depth:

1. Expectation under the null — the false-positive count is ~ Binomial(100, 0.05), mean 5. Five "discoveries" is exactly what noise delivers.
2. Bonferroni — tighten the threshold to α/m = 0.05/100 = 0.0005; strictly controls FWER but sacrifices a lot of power.
3. FDR (Benjamini–Hochberg) — controls the share of false discoveries among those selected; the standard when screening many signals and factors.
4. Out-of-sample — the final arbiter: a signal must survive on data the selection procedure never saw.

⚠️ Common mistake: reporting the best backtest out of a hundred with no correction. That is p-hacking/data snooping: the Sharpe of the best of 100 noise signals always looks impressive.

Short answer: μ̂ = x̄ and σ̂² = (1/n)Σ(xᵢ−x̄)². Write the log-likelihood, take partial derivatives in μ and σ², set them to zero. Note: the variance MLE divides by n, i.e. it is biased low.

In depth:

1. Log-likelihood — the sum of log normal densities; drop constants, what remains depends on μ and σ².
2. In μ — the derivative vanishes when the sum of deviations is zero ⇒ μ̂ = x̄, regardless of σ².
3. In σ² — plug in μ̂ and solve for σ² (differentiate treating σ² as the variable, not σ).
4. Check — the Hessian at the critical point is negative definite, so it is a maximum, not a saddle.

ℓ(μ, σ²) = −(n/2)·ln(2πσ²) − Σ(xᵢ−μ)² / (2σ²)

∂ℓ/∂μ  = Σ(xᵢ−μ)/σ² = 0            ⇒  μ̂ = x̄
∂ℓ/∂σ² = −n/(2σ²) + Σ(xᵢ−μ̂)²/(2σ⁴) = 0
                                    ⇒  σ̂² = (1/n)·Σ(xᵢ−x̄)²

⚠️ Common mistake: calling σ̂² unbiased. E[σ̂²] = ((n−1)/n)·σ² — biased low; the unbiased version divides by n−1 (Bessel's correction), but the MLE is the /n one.

Short answer: No. The Cauchy distribution has neither a mean nor a variance, and the CLT requires finite variance. Worse: the average of n Cauchy samples is distributed as a single Cauchy with the same scale — averaging accomplishes nothing at all.

In depth:

1. CLT precondition — iid and finite σ². For Cauchy the integral E[X] diverges, so both the law of large numbers and the CLT fail to apply.
2. Cauchy stability — X̄ₙ ~ Cauchy(x₀, γ): the characteristic function exp(itx₀ − γ|t|) is unchanged by averaging n draws. n = 1000 is no better than n = 1.
3. Finance relevance — returns are fat-tailed; on heavy-tailed data the sample mean and Sharpe converge slowly and deceptively. The working tools: medians, rank-based and trimmed statistics.

import numpy as np
x = np.random.standard_cauchy(1_000_000)
np.mean(x[:1_000]), np.mean(x)  # both wander, no convergence
np.median(x)                    # ≈ 0 — the median works

⚠️ Common mistake: "n = 1000 is large, so the mean is approximately normal." The CLT is not large-n magic but a theorem with preconditions; checking that moments are finite is step one.

Short answer: No. Correlation only captures linear dependence. The classic counterexample: X ~ N(0,1), Y = X² — Corr(X, Y) = 0 even though Y is fully determined by X. The converse holds in exactly one special case: for jointly Gaussian variables, uncorrelated ⇔ independent.

In depth:

1. Counterexample — Cov(X, X²) = E[X³] − E[X]·E[X²] = 0 for any symmetric X: the dependence is perfect but nonlinear, and Pearson cannot see it.
2. The exception — if the vector (X, Y) is jointly Gaussian, zero covariance really does give independence — a property of the multivariate normal, not of correlation itself.
3. The trap inside the trap — "both marginally normal" ≠ "jointly Gaussian". Example: Y = X·S where S = ±1 with probability ½, independent of X. Then Y ~ N(0,1) and Corr(X, Y) = 0, yet |Y| = |X| — total dependence.

import numpy as np
x = np.random.randn(100_000)
y = x**2
np.corrcoef(x, y)[0, 1]  # ≈ 0.00, although y = f(x)

⚠️ Common mistake: saying "for normal variables zero correlation = independence" without the word "jointly". Marginal normality is not enough.

Short answer: z/2. Conditional on the sum, X and Y are exchangeable, so E[X | S=z] = E[Y | S=z], and the two conditional expectations must add up to z. The interviewer is listening for the symmetry argument, not for an integration of the joint density.

In depth:

1. Symmetry — the pairs (X, Y) and (Y, X) have the same distribution, and S = X + Y is invariant to the swap ⇒ the conditional expectations of X and Y given S coincide.
2. Linearity — E[X|S] + E[Y|S] = E[X + Y|S] = S, hence each equals S/2.
3. General case — for Gaussians with unequal variances, E[X | S=z] = Cov(X,S)/Var(S)·z = σ_X²/(σ_X² + σ_Y²)·z — the regression of X on S; equal variances give z/2 again. And for z/2 you don't even need normality — iid (exchangeability) suffices.

E[X|S] = E[Y|S]        (X and Y are exchangeable)
E[X|S] + E[Y|S] = S    (linearity of cond. expectation)
⇒ E[X|S] = S/2

⚠️ Common mistake: diving into the joint density and grinding out the integral. The answer comes out right, but the signal is weak: symmetry solves it in two lines and works beyond normality.

Short answer: It is a statement about a procedure, not about the parameter: under repeated sampling, 95% of intervals constructed this way will cover the true value. For one specific computed interval, the parameter is either inside or it isn't — frequentism offers no "95% probability it's inside".

In depth:

1. The interval is random, not the parameter — θ is fixed; the endpoints are random because they are functions of the sample. Coverage is a long-run frequency property.
2. The Bayesian alternative — "θ lies in this interval with probability 95%" describes a credible interval: built from the posterior and requiring a prior.
3. Why the distinction isn't pedantry — with a skewed prior or a small sample the two intervals can differ substantially; in risk-calibration problems conflating them leads to wrong bets.

⚠️ Common mistake: "the true value is in this interval with probability 95%." That is the definition of a credible interval, not a CI — interviewers listen for exactly this slip.

Short answer: About 9%, not 99%. The prior rarity of the disease outweighs the test's accuracy — the base-rate fallacy; it computes directly via Bayes' theorem.

In depth:

1. Intuition with 100,000 people — 100 are sick, of whom the test catches 99. Of the 99,900 healthy, ~999 test falsely positive. Total positives ≈ 1098, only 99 of them actually sick → ≈ 9%.
2. What gets confused — the test's accuracy is P(+ | sick), but the question asks P(sick | +); you cannot flip the conditioning without accounting for the base rate.
3. Quant parallel — "my strategy passed the backtest" works the same way: if a priori most candidate strategies are noise, a positive backtest only moderately raises the odds the strategy is real. Prior matters.

P(sick | +) = P(+ | sick)·P(sick) / P(+)
            = 0.99·0.001 / (0.99·0.001 + 0.01·0.999)
            = 0.00099 / 0.01098 ≈ 0.09

⚠️ Common mistake: answering "99%" and ignoring prevalence. Conflating P(+ | sick) with P(sick | +) is exactly the trap.

Short answer: Regression to the mean. Performance = skill + luck; conditioning on an extreme outcome selects not just for skill but for a large positive draw of noise. The noise doesn't repeat next period — and performance falls back toward the skill level.

In depth:

1. Model — R_t = skill + ε_t. Landing in the top decile means that, with high probability, both skill is above average and ε₁ was strongly positive.
2. Conditioning on the extreme — E[skill | R₁ in the top] is above average, but E[ε₂] = 0, so E[R₂ | R₁ in the top] sits well below R₁. The lower the metric's signal-to-noise, the stronger the pullback.
3. Practical consequences — picking managers on past returns systematically disappoints; the sample Sharpe of the best strategies is inflated (selection bias); live signals additionally suffer genuine alpha decay from crowding.

R_t = skill + ε_t,   ε is noise, E[ε] = 0
E[R₂ | R₁ = top decile] ≈ E[skill | R₁] < R₁
lower signal-to-noise ⇒ stronger pull to the mean

⚠️ Common mistake: explaining the pullback as "lost their touch" or "regime change". It arises purely statistically, even with unchanged skill — a noisy metric plus selection on an extreme is enough.

Short answer: When the relationship is monotone but not necessarily linear, and when the data have outliers or fat tails. Spearman is Pearson computed on ranks: it is invariant to monotone transformations, and a single outlier cannot drag the estimate.

In depth:

1. Pearson — measures strictly linear association; outlier-sensitive: one extreme point can flip both the magnitude and the sign.
2. Spearman — replaces values with ranks, so the scale and shape of a monotone relationship don't matter; it captures any monotone dependence.
3. Quant practice — rank IC (the information coefficient on ranks) is the alpha-research standard precisely because returns are fat-tailed and what you need from a signal is the correct ordering of assets, not the scale of the forecast.

⚠️ Common mistake: treating Spearman as a "weaker Pearson". It answers a different question — monotonicity vs linearity: on Y = X³ Pearson is below 1, Spearman is exactly 1.