Quant Interview Preparation
A substantial, optional set of interview-style problems across mental math, probability, expected value, calculus, linear algebra, stochastic calculus, coding, and market intuition. The goal is to teach the reasoning process - each answer shows the method, not just the number.
How to use this
Cover the answer, solve on paper, then reveal. Verbalize your reasoning aloud as if to an interviewer. Log misses (Phase 0.3) and re-attempt on a spaced schedule.
Mental math & estimation
▶ Estimate \(e^{0.05}\) without a calculator.
Use \(e^x\approx 1+x+x^2/2\): \(1+0.05+0.00125=1.05125\). Actual \(\approx 1.05127\). The linear term alone (1.05) is often enough; interviewers want the method and a fast, well-justified approximation.
▶ What is \(\sqrt{2}\) to 3 decimals, and how would you refine an estimate?
≈ 1.414. Newton on \(f(x)=x^2-2\): \(x_{n+1}=\tfrac12(x_n+2/x_n)\). From \(x_0=1.5\): \(1.4167\), then \(1.41422\) - quadratic convergence, so it doubles correct digits each step.
▶ A stock is $40. It moves ±1% daily. Roughly how many days until a ~10% move is likely?
Daily moves compound like a random walk: cumulative std after \(n\) days \(\approx 1\%\sqrt{n}\). Set \(\sqrt{n}\approx 10\) so \(n\approx 100\) days for a one-sigma 10% move. This is the \(\sqrt{t}\) scaling of volatility (Phase 8/15).
Probability puzzles
▶ Two fair dice. Given the sum is even, what is P(sum = 8)?
Even sums: 2,4,6,8,10,12 with counts 1,3,5,5,3,1 = 18 outcomes. Sum 8 has 5. So \(P=5/18\approx 0.278\). Condition by restricting the sample space to the even-sum outcomes.
▶ You flip a fair coin until the first head. What is the expected number of flips?
Geometric with \(p=1/2\): \(\E[N]=1/p=2\). Derive via \(\E[N]=1+\tfrac12\E[N]\) (if tails, restart having used one flip).
▶ Broken-stick: break a unit stick at two uniform points. P(the three pieces form a triangle)?
The triangle inequality fails unless every piece < 1/2. Standard result: \(P=1/4\). Set up with \(x,y\sim U(0,1)\) and require all three parts below 1/2; the favorable region is 1/4 of the unit square.
▶ Expected number of draws to collect all 6 faces of a die (coupon collector)?
\(6\sum_{k=1}^{6}1/k = 6(1+\tfrac12+\tfrac13+\tfrac14+\tfrac15+\tfrac16)=14.7\). Sum of geometric waiting times for each new coupon.
Expected value & decisions
▶ A game: roll a die; you may keep the value or re-roll once (then must keep). Optimal strategy and EV?
Re-roll if the first roll is below the expected re-roll value \(3.5\), i.e. keep 4,5,6 and re-roll 1,2,3. EV \(=\tfrac12\cdot\tfrac{4+5+6}{3}+\tfrac12\cdot 3.5 = 2.5+1.75=4.25\).
▶ You pay $1 to play: flip until first tails; win \(2^{k}\) if the first tails is on flip \(k+1\) (St. Petersburg-lite). Is $1 fair?
Depends on the exact payout rule; the classic St. Petersburg has infinite expectation yet finite willingness-to-pay due to utility. The teaching point: expectation can diverge, and bounded utility (Phase 14.5) resolves the paradox - never quote EV blindly.
Calculus, linear algebra & stochastic
▶ Derivative of \(x^x\)?
Logarithmic differentiation: \(y=x^x\Rightarrow \ln y=x\ln x\Rightarrow y'/y=\ln x+1\), so \(y'=x^x(\ln x+1)\).
▶ Eigenvalues of \(\begin{pmatrix}2&1\\1&2\end{pmatrix}\)?
Symmetric, so real eigenvalues. \(\det(A-\lambda I)=(2-\lambda)^2-1=0\Rightarrow \lambda=1,3\), eigenvectors \((1,-1),(1,1)\). Trace 4 = sum, det 3 = product - quick checks.
▶ Compute \(d(W_t^2)\) where \(W\) is Brownian motion.
By Itô: \(d(W_t^2)=2W_t\,dW_t+dt\). The extra \(dt\) (from \((dW)^2=dt\)) is the whole point of stochastic calculus (Phase 9.2) - a novice forgets it.
▶ Why is a call option’s value convex in the spot?
Payoff \(\max(S-K,0)\) is convex, and the risk-neutral price is an expectation of a convex function; by Jensen and positive gamma, value is convex in \(S\). Gamma \(\Gamma=\partial^2 V/\partial S^2\gt 0\) (Phase 15.1).
Coding, debugging & markets intuition
▶ How would you test a Black–Scholes implementation?
Put–call parity \(C-P=S-Ke^{-rT}\) as an invariant; known closed-form values; monotonicity in \(\sigma\); boundary \(T\to0\) gives intrinsic value; Monte Carlo agreement within 3 standard errors. Property-based tests beat single-point checks (Phase 17.2).
▶ A backtest shows a Sharpe of 4. What’s your first reaction?
Skepticism. Check for look-ahead bias, survivorship bias, transaction costs/slippage, and multiple-testing (data snooping): with many trials, the best in-sample Sharpe is inflated by \(\approx\sqrt{2\ln N /T}\). Insist on out-of-sample / walk-forward validation (Phases 11.7, 12.6, 16.6).
▶ Why can't you just use tomorrow's return to build today's signal?
Look-ahead bias - the signal uses information not available at decision time, producing a fantastic but unrealizable backtest. All features must be point-in-time (Phase 17.6).