Quant interview preparation

Questions

100

Traced to a firm

53

Firms

15

Updated

September 2026

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

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

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1. 027What is a martingale, and how would you use optional stopping to solve a problem?Stochastic processesHardtechnicalQuant researchQuant trading

   Say this

   A martingale is a process whose expected next value, given everything you know now, equals its current value. Optional stopping says that for a suitably bounded stopping time, the expected value at the stopping time equals the starting value, which is what turns a hard path-dependent question into one line of algebra.

   Then walk it

   1. Formally: E of X_{n+1} given the filtration F_n equals X_n. No drift, conditional on history. It is not the same as independence, and increments need not be identically distributed.
   2. Optional stopping needs a condition, and you should name one: bounded stopping time, or bounded increments plus finite expected stopping time, or uniform integrability. Without it the theorem fails, and the classic failure is the doubling strategy, where a stopping time that is finite with probability one still produces E of X_tau equal to 1 rather than 0.
   3. How I use it: find a quantity that is conserved in expectation, then evaluate it at the stopping time. Gambler's ruin falls out immediately from wealth being a martingale.
   4. A second example, expected time in a symmetric random walk: W_n squared minus n is a martingale, so E of tau equals E of W_tau squared. With barriers at 0 and b starting from a, that gives E of tau equal to a(b minus a) in a line.
   5. And the reason it matters beyond puzzles: risk-neutral pricing is exactly the statement that the discounted price is a martingale under the pricing measure. Delta hedging is the construction of that martingale. If you can say that connection, the puzzle answer becomes a conversation about derivatives.

   Where candidates lose it

   Defining a martingale as a fair game and stopping there, or applying optional stopping without checking the integrability condition. Interviewers at the good shops will hand you the doubling strategy specifically to see whether you know why the theorem does not apply. Name the condition before you use the theorem.

   Expect next

   • Why does optional stopping fail for the doubling strategy?
   • Is the square of a martingale a martingale?
   • Connect this to risk-neutral pricing.
2. 028An ant walks randomly along the edges of a cube starting at one corner. What is the expected number of steps to reach the opposite corner?Stochastic processesHardsuperdayQuant tradingQuant research

   Say this

   Ten steps. Collapse the eight vertices into four states by distance from the start, then solve three linear equations. The symmetry reduction is the whole trick.

   Then walk it

   1. By symmetry, all that matters is your graph distance from the target: state 3 is the start, then 2, then 1, then 0 which is the target. Each vertex has three neighbours.
   2. From state 3 all three neighbours are at distance 2, so E3 equals 1 plus E2.
   3. From state 2, one neighbour is at distance 3 and two are at distance 1. So E2 equals 1 plus (1/3)E3 plus (2/3)E1.
   4. From state 1, one neighbour is the target and two are at distance 2. So E1 equals 1 plus (2/3)E2.
   5. Solve: substitute E3 equals 1 plus E2 into the second equation to get E2 equals 1 plus (1 plus E2)/3 plus (2/3)(1 plus (2/3)E2). That yields E2 equal to 9, so E1 equals 7 and E3 equals 10. Sanity check with the general theorem: for a random walk on a regular graph the expected return time to a vertex is the number of vertices, 8, which is the right order of magnitude for a 10-step commute across the diagonal.

   Where candidates lose it

   Trying to track all eight vertices individually and drowning in eight equations. Say the word symmetry, lump the states by distance, and you have three unknowns. The other error is miscounting neighbours in state 2, where it is one back and two forward, not two back and one forward.

   Expect next

   • What is the expected time to return to the starting corner?
   • Do it for a tetrahedron.
   • What if the walk is on a hypercube in n dimensions?

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