Summary
Voleon data scientist interview with three pretty heavy technical questions back to back. The probability and linear algebra stuff felt like a grad school exam more than a job screen. Not sure how I did.
Questions Asked(3)
I knew the basics: non-negative entries, rows sum to one, eigenvalue of 1 always exists.
Suggested Approach
Begin by precisely defining the structural properties of a Markov transition matrix, then systematically build up to the spectral theory (eigenvalues), connect eigenvalues to stationary distributions, and finally discuss convergence of matrix powers. Ground abstract concepts with brief intuitive examples to demonstrate both mathematical rigor and applied understanding.
Define the Transition Matrix
State that a Markov transition matrix P is a square, non-negative matrix where every row sums to 1 (row-stochastic). Clarify that entry P_ij represents the probability of transitioning from state i to state j.
Characterize the Eigenvalues
Explain that all eigenvalues λ satisfy |λ| ≤ 1, and λ = 1 is always an eigenvalue (since P·1 = 1, the all-ones vector is a right eigenvector). Mention that for irreducible aperiodic chains, 1 is the unique eigenvalue of maximum modulus, while periodic chains have additional eigenvalues on the unit circle.
Connect Eigenvalues to Stationary Distributions
Describe the stationary distribution π as the left eigenvector corresponding to eigenvalue 1, satisfying πP = π with π_i ≥ 0 and Σπ_i = 1. Note that for irreducible chains, this stationary distribution is unique (Perron-Frobenius theorem), while reducible chains may have multiple stationary distributions.
Analyze Convergence of P^k
State that for an irreducible, aperiodic (ergodic) chain, P^k converges as k→∞ to a matrix where every row equals π. Explain that the rate of convergence is governed by the second-largest eigenvalue modulus |λ_2|, with the spectral gap (1 - |λ_2|) determining the mixing time.
Discuss Edge Cases and Practical Implications
Address what breaks convergence: periodicity (eigenvalues like -1 cause oscillation) and reducibility (absorbing states or multiple communicating classes). Briefly connect to applications such as PageRank, MCMC, or hidden Markov models to demonstrate applied relevance.
Key Points to Mention
This one surprised me.
Suggested Approach
Break the question into three distinct sub-problems: (1) how OLS coefficients transform under an orthonormal rotation, (2) whether fitted values (and hence predictions) are preserved, and (3) the distributional invariance condition on the second-moment structure of the features. Ground each answer in the linear algebra of OLS and connect to statistical theory to show depth.
Set Up the Rotation Formally
Let X be the n×2 feature matrix and Q a 2×2 orthonormal rotation matrix (Q'Q = I). Define the rotated design matrix X̃ = XQ. Clarify that the intercept column is kept separate and untouched, so the rotation only applies to the two slope columns.
Derive How Slope Coefficients Transform
Show that the OLS estimator in the rotated space is β̃ = (X̃'X̃)⁻¹X̃'y = Q'β, where β are the original slope coefficients. This means the coefficient vector rotates by Q' — the individual slope values change, but in a fully determined, invertible way.
Demonstrate Fitted Value Invariance
The hat matrix H = X̃(X̃'X̃)⁻¹X̃' = XQQ'(QX'XQ')⁻¹Q'X' = X(X'X)⁻¹X', which is identical to the original hat matrix. Therefore ŷ = Hy is unchanged — fitted values and residuals are fully preserved under orthonormal rotation.
Analyze the Sampling Distribution of β̃
Under standard OLS assumptions, Var(β̃) = σ²(X̃'X̃)⁻¹ = σ²Q'(X'X)⁻¹Q. For the sampling distribution of the slope estimator to be invariant in second moments to the choice of Q, we need Q'(X'X)⁻¹Q = (X'X)⁻¹, which holds when X'X ∝ I — i.e., the feature columns are orthogonal and have equal variance (spherical second moments).
State the Distributional Condition Precisely
The required condition is that the population second-moment matrix Σ_X = E[xx'] is proportional to the identity (isotropic / spherically symmetric features). Under this condition, any orthonormal rotation leaves the covariance structure of β̃ unchanged, making the sampling distribution rotation-invariant in second moments.
Key Points to Mention
I stared at this for a bit.
Suggested Approach
Model the joint stopping times of A and B using Markov chains or state-based probability analysis, carefully tracking A's consecutive-heads progress alongside B's single-flip condition. Compute the probability that A's stopping round is strictly less than B's by enumerating joint state transitions and summing over all valid round counts. Verify with a simulation or recursive formula to confirm the closed-form answer.
Define Stopping Conditions Precisely
Clarify that A stops at the first round where the current flip AND the previous flip are both heads (two consecutive heads), while B stops the moment they flip a tail. Confirm both players flip independently each round.
Model A's Progress as a Markov Chain
Define A's states: S0 (start or last flip was tails), S1 (last flip was heads, no consecutive yet), and S_done (two consecutive heads seen). Write out transition probabilities: from S0, flip H (prob 1/2) → S1, flip T → S0; from S1, flip H → S_done, flip T → S0.
Model B's Stopping Distribution
B stops at round k with probability (1/2)^k, since B needs k-1 heads followed by one tail (geometric distribution with p=1/2). Note B's stopping time T_B ~ Geometric(1/2), so P(T_B = k) = (1/2)^k for k ≥ 1.
Compute Joint Probability via State Enumeration
For each round k, compute P(A finishes at round k) using the Markov chain (track probability mass in S_done at exactly round k), then multiply by P(T_B > k) = P(B has flipped all heads through round k) = (1/2)^k. Sum over all k ≥ 2 to get P(T_A < T_B).
Derive and Verify the Closed-Form Answer
Use the Markov transition equations to find P(A done at round k) in closed form, sum the series, and simplify. Cross-validate with a Monte Carlo simulation or recursive probability equations to confirm the exact probability (approximately 1/8 or derive precisely as 1/8).
Key Points to Mention
Discussion(3)
Sign in to join the discussion.
The convergence part is genuinely the hardest piece to state precisely under pressure, so fumbling it isn't embarrassing. The clean version goes like this: a stochastic matrix always has 1 as its largest eigenvalue (by Perron-Frobenius), all other eigenvalues satisfy |λ| ≤ 1, and the stationary distribution π is the left eigenvector corresponding to λ=1. For P^k to converge to the rank-one matrix you half-remembered (every row being π), you need the chain to be irreducible, meaning every state is reachable from every other state, and aperiodic, meaning the GCD of return times to any state is 1. Irreducibility alone gets you a unique stationary distribution but not convergence of the powers. Aperiodicity kills the oscillatory behavior that can come from periodic chains where eigenvalues like -1 or complex roots of unity stay on the unit circle and prevent P^k from settling. The rate of convergence is governed by the second-largest eigenvalue in absolute value, sometimes called the spectral gap, and this comes up constantly in MCMC analysis. If they pushed back asking for actual conditions, they were almost certainly fishing for those two words: irreducible and aperiodic. Voleon does a lot of work adjacent to probabilistic modeling so I'd guess they care that you can connect the linear algebra framing to what it actually means for a chain's long-run behavior, not just recite definitions.
Your setup is right and that's actually most of the work. The Markov chain for A has three states: start (S0), one head seen (S1), and done. From S0 you go to S1 with prob 1/2 and stay at S0 with prob 1/2. From S1 you finish with prob 1/2 and reset to S0 with prob 1/2. B finishes on round k with probability (1/2)^k since every round is an independent tail with prob 1/2.
The cleaner approach than summing over B's stopping time is to compute P(A finishes on exactly round n) directly from the Markov chain and then compute P(A < B) = sum over n of P(A finishes on round n) * P(B has not yet finished before round n), where the latter is P(B > n-1) = (1/2)^(n-1).
P(A finishes on round n) takes a bit of care because you need to count paths of length n through S0/S1 that hit the absorbing state for the first time at step n. The generating function route is cleaner here: if you write down the transition matrix and track the first passage distribution, you avoid a lot of the combinatorial bookkeeping that probably tangled your arithmetic. The final number I got working through it was 1/5, but I'd double-check that because these geometric series sums have a way of hiding an off-by-one. The key thing Voleon is probably watching for is whether you instinctively reach for the Markov structure rather than trying to brute-force enumerate cases.
You basically had it. The covariance being a scalar multiple of the identity is exactly the right frame, and it's cleaner than listing the conditions piecemeal. If Cov(X) = σ²I, then for any orthonormal R, Cov(RX) = R Cov(X) R^T = σ²RR^T = σ²I, so the second-moment structure is unchanged by the rotation and the sampling distribution of the estimator stays the same shape.