Quant interview preparation

Questions

100

Traced to a firm

53

Firms

15

Updated

September 2026

Asked at

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

Topic

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

Level

AnyCoreIntermediateHard

Type

AnyBrainteaserTechnicalCaseMarket viewFit

1. 056What is maximum likelihood estimation, and when would you prefer method of moments?StatisticsIntermediatetechnicalQuant researchRisk

   Say this

   MLE picks the parameters that make the observed data most probable under your assumed distribution. It is asymptotically efficient if the model is right, which is exactly the condition that makes method of moments attractive when it is not.

   Then walk it

   1. MLE: maximise the log likelihood, which is the sum of log densities. Under regularity conditions it is consistent, asymptotically normal, and attains the Cramer-Rao bound, with variance given by the inverse Fisher information.
   2. Method of moments: match sample moments to their theoretical expressions and solve. Generalised method of moments extends this to more moment conditions than parameters, weighting them optimally, and it needs no full distributional assumption.
   3. So the tradeoff is efficiency versus robustness. MLE uses the whole density, so it extracts every bit of information and pays for it with sensitivity to misspecification. GMM uses only the moments you trust.
   4. Concrete case: fitting a distribution to daily returns. MLE under a normal assumption gives you the sample mean and variance and will be badly misled by the tails. MLE under a Student t estimates the degrees of freedom and is much better behaved. GMM on a few robust moments avoids committing to either.
   5. Practical points worth raising: MLE can be biased in small samples even when consistent, the classic example being the variance estimator with n rather than n minus 1 in the denominator. And numerically you should always check the Hessian at the optimum, because a flat likelihood means your parameter is not identified, which is common in GARCH and regime models.

   Where candidates lose it

   Describing MLE as the best estimator without the qualifier if the model is correctly specified. That caveat is the entire content of the comparison. Also be ready for the small-sample bias point, since MLE being biased while still consistent catches people who have only memorised the asymptotic properties.

   Expect next

   • Give me an example where MLE is biased.
   • What is the Cramer-Rao bound?
   • How would you fit a Student t to returns, and what does the estimated degrees of freedom tell you?
2. 057What does stationarity mean, how do you test for it, and why do you care?Time seriesIntermediatetechnicalQuant researchRisk

   Say this

   Weak stationarity means constant mean, constant variance and an autocovariance that depends only on the lag. You care because the standard inference machinery assumes it, and regressing non-stationary series on each other produces spurious relationships with impressive t statistics.

   Then walk it

   1. Prices are not stationary, they are close to a random walk with a unit root. Returns are much closer to stationary, which is why every model works on returns and not on levels.
   2. Tests: augmented Dickey-Fuller and Phillips-Perron test the null of a unit root, KPSS tests the null of stationarity. Run both, because they have opposite nulls and agreeing tests are more convincing than either alone. And these tests have low power, so failing to reject is weak evidence.
   3. Spurious regression is the cost of getting it wrong. Regress one independent random walk on another and you reject the null of no relationship far more often than five percent of the time, with an R squared that looks respectable. Granger and Newbold showed this in 1974 and people still do it.
   4. The exception that matters for trading: cointegration. Two non-stationary series can have a stationary linear combination, which is precisely the statistical statement of a pair trade. Test it with Engle-Granger or Johansen, and then the correct specification is an error-correction model rather than a regression in levels.
   5. The practical honesty: financial series are not stationary even in returns, because volatility and correlation regimes shift. So I treat stationarity as a working approximation over a limited window, and I check parameter stability across subsamples rather than trusting one test on the full history.

   Where candidates lose it

   Answering just difference it until the test passes. Over-differencing destroys the signal, and a cointegrated pair loses its whole tradeable relationship if you difference both series. Say what stationarity buys you, name the spurious regression result, and bring up cointegration unprompted since it is where the money is.

   Expect next

   • What is cointegration and how does it differ from correlation?
   • How do you test it, and what is an error-correction model?
   • What if a series is stationary in one decade and not the next?
3. 058Daily equity returns are not normal. How are they different, and what do you do about it?StatisticsIntermediatetechnicalQuant researchRisk

   Say this

   They have fat tails, negative skew and volatility clustering. Daily equity index kurtosis is typically 5 to 10 against 3 for a normal, so moves the normal says should happen once a century happen every few years.

   Then walk it

   1. Put a number on it. A normal assigns a five standard deviation daily move a probability of about one in 3.5 million, roughly once in 14,000 years of trading. The S&P has had several since 1950. The tails are not slightly wrong, they are wrong by orders of magnitude.
   2. Negative skew: large down moves are bigger and faster than large up moves. That is why index option skew exists and why puts are persistently richer than calls in implied vol terms.
   3. Volatility clustering means part of the unconditional fat tail is a mixture effect. Returns standardised by a GARCH-type conditional volatility are much closer to normal, though still fat-tailed, which tells you some but not all of the kurtosis is time-varying vol rather than genuinely fat conditional tails.
   4. What I would do depends on the use. For risk: empirical quantiles, a Student t or a generalised Pareto fit to the tail via extreme value theory, and expected shortfall rather than value at risk, because expected shortfall is sensitive to how bad the tail is. For pricing: a model with jumps or stochastic volatility rather than plain Black-Scholes.
   5. And the aggregation point: monthly returns are considerably closer to normal than daily returns because of the CLT, so the right distributional assumption depends on your horizon. That is worth saying because it stops the conversation becoming a generic tails are fat sermon.

   Where candidates lose it

   Saying fat tails and stopping. Quantify it, because the five sigma comparison is what makes the point land. Also do not forget the skew, since symmetric fat tails would not explain the option skew, and be ready to distinguish unconditional fat tails from conditional heteroskedasticity.

   Expect next

   • How much of the kurtosis is explained by volatility clustering?
   • What is expected shortfall and why prefer it to value at risk?
   • How does this show up in the option surface?
4. 065What is adverse selection and why is it a market maker's real cost?Market makingIntermediatetechnicalProp trading firmsQuant trading

   Say this

   Adverse selection is the fact that whoever trades with you chose to, and sometimes they chose because they know something you do not. Your quote gets hit disproportionately when it is wrong, so on average the trades you get are worse than the trades you wanted.

   Then walk it

   1. The mechanism: you post a two-sided quote at your fair value. Uninformed flow hits both sides roughly equally and you earn the spread. Informed flow only takes the side that is mispriced, so those trades lose you money immediately.
   2. The measurement is simple and it is what every market making desk tracks: mark your fills against the mid price a few seconds or minutes later. If your buys are systematically below where the market goes, you are being adversely selected. The industry term is markout.
   3. This is why the spread must be wide enough that the profit from uninformed flow covers the loss to informed flow. Glosten and Milgrom's model makes the spread purely a function of the probability of informed trading, with zero inventory risk at all.
   4. It explains observable behaviour. Spreads widen before earnings and economic releases, when the probability of informed flow spikes. Market makers pay for retail order flow precisely because retail flow is less informed, so it is worth more.
   5. And the extreme version is why quotes get pulled. If adverse selection becomes severe enough that no spread compensates, the correct response is not to widen but to stop quoting. That is what a flash crash looks like from the inside, and saying that shows you understand the business rather than just the term.

   Where candidates lose it

   Confusing adverse selection with inventory risk. Inventory risk is the price moving while you hold a position you did not want. Adverse selection is getting the position in the first place precisely when it is wrong. Interviewers ask candidates to distinguish them, so have both definitions crisp and know that markout is how you measure it.

   Expect next

   • How is that different from inventory risk?
   • How would you measure it on your own fills?
   • Why is retail order flow worth paying for?
5. 076You have K sorted arrays on disk, too large to load at once. How do you merge them into one sorted output?ProgrammingIntermediatetechnicalCitadelEquity Capital Markets · New York · 2026

   Say this

   K-way merge with a min heap of size K. Push the first element of each array into the heap, repeatedly pop the minimum and write it out, then push the next element from whichever array the minimum came from. Time is N log K, memory is O(K) plus your buffers.

   Then walk it

   1. The heap holds one candidate per array, each entry tagged with which array it came from and the index within it. Pop the smallest, emit it, and refill from that same array.
   2. Complexity: N total elements, each pushed and popped once, each operation log K. So N log K, which beats concatenate-and-sort at N log N whenever K is much smaller than N.
   3. The disk part is the real content of the question. You do not read element by element, you read blocks. Keep a buffer per array, say a few megabytes each, refill it when it drains, and write the output through a large buffer too. The heap operations are free compared with I/O, so the design goal is sequential reads and few of them.
   4. If K is very large, K times the buffer size exceeds memory, and then you merge in passes: merge groups of, say, 100 files at a time, then merge the results. That is exactly how external merge sort works, and total I/O is N times the number of passes.
   5. Practical notes I would raise: use a tournament tree or a loser tree instead of a binary heap if you want fewer comparisons per element, handle the tie-breaking rule explicitly if stability matters, and if this is a real system, check whether the operating system's readahead is already doing your buffering for you before you build it yourself.

   Where candidates lose it

   Answering merge them pairwise, which is K times N in the worst case, or ignoring the on-disk part entirely. The interviewer put the data on disk deliberately, so talk about block-sized buffered reads and what happens when K is too large to buffer. State the N log K complexity explicitly.

   Expect next

   • What if K is a million?
   • How large would you make the buffers, and why?
   • How would you parallelise it?

   Reported by candidates at Citadel (Equity Capital Markets, New York, 2026). Source: Wall Street Oasis.

6. 077Given an array and a window of size k, return the maximum in each window as it slides.ProgrammingIntermediatetechnicalAkuna CapitalQuant Development · Chicago · 2025

   Say this

   Monotonic deque, O(n) total. Keep a deque of indices whose values are strictly decreasing. Before pushing a new index, pop from the back everything smaller than the new value, and pop from the front anything that has fallen out of the window. The front is always the maximum.

   Then walk it

   1. Why the deque is monotonic: if a new element is larger than something behind it, that older smaller element can never be the maximum of any future window, because the new one is both larger and more recent. So it is safe to discard permanently.
   2. Each index is pushed once and popped once, so the total work is O(n) even though a single step can pop many elements. That amortised argument is the thing to say out loud, because it is what distinguishes this from the naive O(n k).
   3. Store indices, not values, so you can test whether the front has expired by comparing front index against i minus k plus 1.
   4. Alternatives and why they are worse: a max heap gives O(n log k) and needs lazy deletion of expired entries. A balanced BST or a multiset gives O(n log k) too. Both are fine and both are beaten by the deque.
   5. Where this actually matters on a trading system, which is worth mentioning: rolling extremes over a tick window, running high and low for a breakout signal, and rolling maximum drawdown. The same structure with the comparison reversed gives you the rolling minimum, and the O(1) amortised cost per tick is what makes it usable in a hot path.

   Where candidates lose it

   Reaching for a heap and stopping there. The heap answer is acceptable but it is not the answer to this question, and the interviewer is specifically looking for the monotonic deque and the amortised O(n) argument. Also remember to expire the front by index, which is the bug that shows up most often in live coding.

   Expect next

   • Prove the amortised complexity.
   • Now give me the rolling median instead.
   • How would you handle a window defined by time rather than by count?

   Reported by candidates at Akuna Capital (Quant Development, Chicago, 2025). Source: Wall Street Oasis.

7. 079How would you store key-value pairs, and what are the tradeoffs between the implementations?ProgrammingIntermediatetechnicalJump TradingEngineering · Cambridge · 2019

   Say this

   Hash table for O(1) average lookup with no ordering, balanced tree for O(log n) with ordered iteration and range queries, and a flat sorted array if the data is static and you care about cache behaviour. The choice is driven by whether you need ordering and what your access pattern looks like in memory.

   Then walk it

   1. Hash table: O(1) average, O(n) worst case on collisions, no ordering, and rehashing causes an occasional large latency spike. That spike is a real problem on a trading hot path and it is why people pre-size their maps.
   2. Balanced tree, red-black or B-tree: O(log n) guaranteed, ordered traversal, range queries, and predictable latency. Worse constants and worse cache locality because of pointer chasing.
   3. The tradeoff that matters most in practice is memory layout, not big-O. C++ unordered_map uses separate chaining with nodes scattered across the heap, so every lookup is potentially a cache miss. An open-addressing flat hash map keeps everything in one array and is commonly two to three times faster in real workloads at the same asymptotic complexity.
   4. For a mostly-static table, a sorted array with binary search beats both: contiguous memory, no pointers, and for small n a linear scan beats binary search because it is branch-predictable and prefetchable. Under about 16 to 32 entries, linear wins.
   5. And on disk the answer changes completely: B-trees for read-heavy workloads because of the branching factor against block size, LSM trees for write-heavy because they turn random writes into sequential ones. I would want to know the read-write ratio and whether the working set fits in cache before choosing anything.

   Where candidates lose it

   Answering hash map, O(1), done. The question says tradeoffs, so it is a systems question and the interviewer at a trading firm cares about tail latency and cache behaviour more than asymptotic complexity. Mention rehashing spikes and pointer chasing, and ask what the access pattern is.

   Expect next

   • Why is std::unordered_map often slow in practice?
   • How would you avoid latency spikes from rehashing?
   • What changes if the data lives on disk?

   Reported by candidates at Jump Trading (Engineering, Cambridge, 2019). Source: Wall Street Oasis.

8. 083Write an algorithm to find all the primes from one to n, and then optimise it.ProgrammingIntermediatetechnicalACAQR Capital ManagementResearch · Greenwich · 2015

   Say this

   Sieve of Eratosthenes. Mark every multiple of each prime as composite, and the unmarked survivors are the primes. Time is n log log n, which is essentially linear, and memory is n bits.

   Then walk it

   1. The baseline to reject first: trial division on each number up to its square root is about n times root n over log n, far worse. Say why the sieve wins before you write it.
   2. The sieve itself: start at p equal to 2, mark 4, 6, 8 and so on, then advance to the next unmarked number. Two optimisations that come free: start marking at p squared rather than 2p, because smaller multiples are already marked, and stop the outer loop at root n.
   3. Memory optimisations: store only odd numbers, halving memory, use a bit array rather than bytes for an eightfold saving, and if n is large, sieve in cache-sized blocks. That last one matters more than anything else in practice, because a naive sieve over 10 to the 9 is dominated by cache misses, and segmenting it can be several times faster at identical complexity.
   4. Further refinements if pushed: a wheel sieve skipping multiples of 2, 3 and 5 removes about 77 percent of the candidates, and the sieve of Atkin is asymptotically better at n over log log n but is slower in practice and much harder to get right.
   5. And the answer to a different question they may be asking: if you want to test whether one large number is prime rather than enumerate a range, the sieve is the wrong tool entirely and you want Miller-Rabin, which is probabilistic and fast. Recognising that enumerate and test are different problems is worth saying.

   Where candidates lose it

   Giving trial division and calling it done, or giving the sieve with no optimisation when the question explicitly asks for one. The optimisations they want in order are: start at p squared, skip evens, use a bit array, then segment for cache. Naming cache blocking is what marks you out, because it is the one that matters at scale and it is not in the textbook answer.

   Expect next

   • What is the memory cost for n equal to a billion, and how would you reduce it?
   • How would you parallelise the sieve?
   • Now test whether one very large number is prime.

   Reported by candidates at AQR Capital Management (Research, Greenwich, 2015). Source: Wall Street Oasis.

9. 086Explain how you would price an option.Options and derivativesIntermediatetechnicalDRWQuantitative Trading · Chicago · 2025

   Say this

   The core idea is replication. If I can build a portfolio of the underlying and cash that matches the option's payoff in every state of the world, then no-arbitrage says the option must cost what that portfolio costs. Everything else, Black-Scholes included, is a way of computing that cost.

   Then walk it

   1. Start with one period and two states, because it makes the logic visible. Stock at 100 goes to 110 or 90, a call struck at 100 pays 10 or 0. Hold delta shares plus B in cash and solve two equations: delta is (10 minus 0) over (110 minus 90), which is 0.5, and then B falls out. The option price is 0.5 times 100 plus B. No probabilities were used anywhere.
   2. That is the key insight to state explicitly: the price does not depend on the real-world probability of the up move, only on the size of the moves. Rearranging gives the risk-neutral probability, which is the probability that makes the discounted stock a martingale, and pricing becomes a discounted expectation under that measure.
   3. Extend the tree to many steps and you get the binomial model, which handles American exercise naturally because you compare intrinsic against continuation at each node. Take the limit with the step size going to zero and you get Black-Scholes.
   4. Black-Scholes in words: the price is the discounted risk-neutral expectation of the payoff when the stock follows geometric Brownian motion with constant volatility. The formula's two N terms are the risk-neutral probability of finishing in the money and the delta-weighted version of it.
   5. Then the practical truth, which is the answer a trading firm actually wants: nobody uses Black-Scholes to find the price, because the price is on the screen. You use it as a translator from price to implied volatility, then you trade the volatility surface. Constant vol is false, the smile proves it, so the real work is interpolating and extrapolating the surface consistently and hedging the Greeks it implies.

   Where candidates lose it

   Reciting the Black-Scholes formula. Anyone can memorise it. The interviewer wants replication and no-arbitrage, and specifically wants to hear that the real-world probability drops out. Then close by saying the formula is used backwards, to extract implied vol from a market price. That last move is what marks a trader rather than a student.

   Expect next

   • Why does the real-world probability not appear in the price?
   • What are the assumptions, and which one fails hardest?
   • How would you price an American put?

   Reported by candidates at DRW (Quantitative Trading, Chicago, 2025). Source: Wall Street Oasis.

10. 089What is put-call parity, and what would you do if you saw it violated?Options and derivativesIntermediatetechnicalProp trading firmsDerivatives

    Say this

    For European options on a non-dividend-paying stock, call minus put equals spot minus the discounted strike. It is pure arbitrage, no model, because a long call plus a short put plus the discounted strike in cash replicates the stock exactly. If it breaks, you trade both sides and lock a riskless profit.

    Then walk it

    1. The proof is a payoff table. At expiry, long call plus short put pays S minus K in every state, whether S is above or below K. Adding K in cash held to expiry gives you S. So the cost today of call minus put plus K discounted must equal S.
    2. With dividends, subtract the present value of dividends from the spot. With a cost of carry or borrow cost on the short, use the forward: C minus P equals the discounted difference between the forward and the strike.
    3. If I saw a violation, say the call is too expensive: sell the call, buy the put, buy the stock, and borrow the discounted strike. That is a conversion, and the reverse is a reversal. Lock the difference and hold to expiry.
    4. Then the reasons an apparent violation is usually not one, and this is what the question is really testing. Stale quotes on one leg. You are looking at mid prices but must trade at the bid and offer, and the parity gap is usually smaller than the combined spreads. Hard-to-borrow stock making the short leg expensive. American exercise, where early exercise of the put breaks the equality. Discrete dividends you have modelled wrong.
    5. So my actual answer: I would first check whether the apparent edge survives crossing four spreads and paying the borrow. Ninety-nine times out of a hundred it does not, and that is the point of the question. The hundredth time, borrow cost is usually the explanation, and the implied borrow rate you back out of the parity relationship is itself the useful information.

    Where candidates lose it

    Giving the formula and saying you would arbitrage it, with no mention of transaction costs, borrow or American exercise. A trading interviewer asks this specifically to see whether you treat a screen-level inefficiency as free money. Also know that parity holds for European options only, and be able to say why American puts break it.

    Expect next

    • Why does it not hold exactly for American options?
    • How would you back out the implied borrow rate from the option prices?
    • What does a persistent parity gap tell you about the stock?

Firm tags come from public, anonymous candidate reports on Wall Street Oasis: strong signal, not sworn testimony. Firms are named as the places a question was reported, not as partners of Fin Maverick. Answers are written for this page to show how to think out loud; they are not scripts to recite.