Quant puzzles, solved step by step

7/18, about 0.389, for every n from 3 upwards. A gap is left uncoloured only when it is longer than both gaps beside it, because then neither endpoint has it as its nearest. For three points that gap is simply the longest of three pieces, which averages 11/18, so 7/18 is coloured. For larger n the same 11/18 comes out, so the answer does not depend on n.

When is a gap left uncoloured?

Picture people standing round a circular table, each turning to talk to whoever is closer, left or right. A stretch of table between two people stays silent only if both of them turned away, which means each had a closer person on their other side. A gap is uncoloured exactly when it is longer than both of its neighbouring gaps. A gap coloured from both ends is still coloured once, so the question becomes: what is the expected total length of gaps that are local maxima?

How do you get 11/18 for the uncoloured part?

Start with n = 3, the case you can finish in the room. With three gaps, every gap's two neighbours are the other two gaps, so the only uncoloured gap is the longest one. Three random points cut the circle like a stick broken into three, and the longest of three pieces averages (1/3)(1 + 1/2 + 1/3) = 11/18. So the coloured length is 7/18.

For larger n, use the fact that the n gaps behave like n independent exponentialA random length whose chance of ending is the same at every instant; waiting times between random arrivals follow it. lengths rescaled to add up to 1, and that the rescaling is independent of the shape. For three unit exponentials X, Y and Z, the expected value of X counted only when X is the largest is 1 - 2/4 + 1/9 = 11/18. Each of the n gaps contributes that, divided by the expected total of n, and the n gaps sum to 11/18 again. The uncoloured share is 11/18 whatever n is, so the coloured share is always 7/18.

Say the check: a seeded simulation of 40,000 random circles gives 0.389 for n = 3, 0.388 for n = 5 and 0.389 for n = 10. The limitation is that the exponential step is a known result you should name, not derive, in an interview; the n = 3 case is the part you prove on the spot.

25/48, about 0.521. For a stick broken into n pieces, the expected k-th smallest piece is (1/n)(1/n + 1/(n - 1) + ... ) with k terms. For n = 4 the sorted pieces average 3/48, 7/48, 13/48 and 25/48, which add to 1. The longest is (1/4)(1 + 1/2 + 1/3 + 1/4) = 25/48, more than twice the average piece of 1/4.

Why is the longest piece so much longer than a quarter?

Cut a sheet of dough at three random spots and the pieces are rarely even: one is usually a big slab and one a sliver. Random breaks produce uneven pieces, and the longest piece collects the unevenness, so its average sits far above the average piece. The average piece is always 1/4; the question asks about the largest of four correlated lengths, which is an {term('order statistic', 'The k-th smallest value in a sample, for example the minimum, the median or the maximum.')}.

Where does the harmonic formula come from?

Start with the shortest piece. The chance that all four pieces exceed x is (1 - 4x) cubed: take x off every piece and the three breaks must fit into the remaining length 1 - 4x. Integrating that from 0 to 1/4 gives an expected shortest piece of 1/16. Then the key fact: the step from each sorted piece to the next adds on average (1/n) times 1 over the number of pieces still longer. After the shortest, three pieces remain longer, so the next piece averages 1/16 + (1/4)(1/3); then add (1/4)(1/2), then (1/4)(1). The longest piece is (1/4)(1/4 + 1/3 + 1/2 + 1) = 25/48.

The step rule comes from the fact that the pieces behave like independent exponential lengths scaled to total 1, and the gap between successive minima of exponentials is memoryless. You can name that in the room rather than prove it. The check that the formula is right: the four sorted averages add to exactly 1, and a simulation of 100,000 sticks gives 0.521 for the longest. For n pieces in general, the longest averages (1/n) times the n-th harmonic number, which grows like (ln n)/n.

Bluff with the Q one time in three, and call with the K one time in three. You always bet the A and always check the K; I always call with the A and fold the Q. A Q bluff risks 1 more to win the 2 antes, so I must defend two thirds of hands facing it: the A gives half, the K calling one time in three gives the rest. The game is worth 1/18 per hand to you.

Which hands are easy, and where is the real decision?

Clear away the obvious hands first. With the A you always bet, because you win whether I call or fold. With the K, betting only gets called by the A and folds out the Q, which you beat anyway, so you check. As the caller, I always call with the A and always fold the Q. The whole game comes down to two mixed choices: how often you bluff with the Q, and how often I call with the K.

How do the indifference conditions fix both frequencies?

A teacher who spot-checks homework faces the same logic: check every paper and time is wasted, never check and everyone copies, check at the right rate and copying stops paying. In equilibrium each player mixes at the rate that makes the other indifferent between their two options. Your Q loses 1 by checking. Bluffing loses 2 against my A, and against my K loses 2 if I call and wins 1 if I fold. The two are equal only when I call with the K one time in three. My K loses 1 by folding; calling loses 2 against your A and wins 2 against a bluff, which is worth -1 only when you bluff one time in three.

Check against the pot-odds rule. A bluff risks 1 extra to win the 2 antes, so the caller must defend 2/(2 + 1) = 2/3 of the hands facing it; the A already covers half, and the K calling one time in three covers the other sixth. That two thirds is the number people misremember as the K's calling rate. Put the strategies together and the bettor, who acts first with more information about their own hand, earns 1/18 of a chip per hand. The limitation: with a bigger bet or more cards the ratios change, but the method, indifference on both sides, carries over.

A drift shared by both assets for the whole month can outweigh daily noise that moves them in opposite directions. Within a month the drift is constant, so only the noise shows and the correlation is negative. Summed over 21 days the drift's covariance grows with 21 squared but the noise's only with 21. With noise correlation -0.5 and drift standard deviation 0.3% a day, monthly correlation is +0.48.

What does a three-month example look like?

Picture two shops in the same market street. On any one day, a customer who buys from one did not buy from the other, so their daily takings move against each other. But in festival months the whole street is busy and in the rains the whole street is quiet, so their monthly takings rise and fall together. Correlation at one horizon says nothing on its own about another, because a slow common factor and fast opposing noise can sit in the same data.

Make it numerical with five-day months. In month 1 both assets drift at -0.8% a day, in month 2 at +0.2%, in month 3 at +1.2%. On top, asset A gets daily noise of +0.6, -0.3, 0, +0.3, -0.6 and asset B gets -0.3, +0.3, 0, -0.3, +0.3, which move in opposite directions. Within each month the correlation is -0.95. Summed over each month, the noise nets to zero, so both assets return -4%, +1% and +6%: identical, a monthly correlation of +1. Even all 15 days pooled show +0.71, because the month-to-month swing in drift is larger than the noise.

Why does summing more days push the correlation positive?

Write each daily return as the month's drift m plus noise. Over n days the drift adds up to n times m, while the noise adds up to a sum of n separate shocks. The drift's contribution to covariance scales with n squared, the noise's only with n, so the longer the horizon the more the shared drift wins. Take noise with standard deviation 1% a day and a within-month correlation of -0.5, and a drift whose standard deviation across months is 0.3% a day. For a 21-day month the drift adds 39.69 to the covariance and the noise subtracts 10.5, for a monthly correlation of 29.19/60.69 = 0.48.

The crossover sits where n times 0.09 equals 0.5, about 5.6 days, so even weekly returns of five days would still show a slightly negative correlation, -0.03. A hedge sized on daily correlation can therefore fail at a monthly horizon, which is why a desk measures correlation at the frequency it actually holds risk. The other mechanisms worth naming are mean reversion in the spread between the two assets and stale prices that lag by a day; both also make correlation depend on frequency. The limitation of the example is the assumption that drift is constant inside a month and noise is independent from day to day.

One eigenvalue of 2.6 and four of 0.6, so the first principal component explains 52%. Write the matrix as 0.6 times the identity plus 0.4 times a matrix of ones. The all-ones vector is an eigenvector with eigenvalue 0.6 + 5 x 0.4 = 2.6; any vector whose weights sum to zero is killed by the ones matrix and has eigenvalue 0.6. The trace check: 2.6 + 4 x 0.6 = 5.

What structure should you spot before touching a determinant?

Think of five students whose marks all move together when the paper is hard, plus their own good and bad days. There is one shared shock and five private ones. An equicorrelation matrix is exactly that: R = (1 - rho) I + rho J, where J is the matrix of all ones, so its eigenvectors are those of J and you never need a characteristic polynomial. J sends the all-ones vector to 5 times itself and sends any vector whose entries sum to zero to zero. Those two facts give every eigenvalue.

How do the eigenvalues fall out, and how do you check them?

Apply R to the all-ones vector: each row sums to 1 + 4 x 0.4, so the equal weight portfolio has eigenvalue 1 + (n - 1) rho = 2.6. Apply R to any vector with weights summing to zero, such as long asset 1 and short asset 2: the rho J part vanishes and only (1 - rho) = 0.6 is left, and there are four independent such vectors. The eigenvalues must add to the trace, the sum of the diagonal, which is 5: 2.6 + 2.4 = 5.

What does the answer say about a real portfolio?

The first component is the market: equal weights, and its share rises towards rho as you add assets. With 50 assets at the same correlation the first eigenvalue is 1 + 49 x 0.4 = 20.6, 41.2% of the total, while each of the other 49 stays at 0.6. Diversification removes the private shocks but never the common one. The same formula gives a limit: the smallest eigenvalue 1 - rho is always fine, but 1 + (n - 1) rho must stay positive, so five assets cannot all share a correlation below -0.25.

V = S and V = K e^(-r(T-t)) satisfy it; V = S squared does not. The first is the stock itself and the second is a zero-coupon bond paying K at T, both traded assets that must earn r. S squared has gamma 2 and leaves (r + sigma squared) S squared unbalanced. Multiplying by e^((r + sigma squared)(T-t)) fixes it, which is the price of a claim paying S squared at expiry.

What is the equation actually saying?

Think of a household budget rule that any fair arrangement must obey: over one day, what you hold must earn the same as the same money in a savings account, once the risk has been hedged away. The Black-Scholes equation says that for a delta-hedged position, time decay plus the gamma term plus the financing of the hedge equals r times the value. Written in {term('greeks', 'Theta is the change in value with time, delta with the stock price, and gamma is the change in delta with the stock price.')}, it is theta + half sigma squared S squared gamma + r S delta = r V. A candidate price passes only if its greeks balance that line.

Why do the two that pass make sense as trades?

Anything that is itself a traded, self-financing asset must satisfy the equation, because the equation is only the statement that no hedged position earns more than r. V = S is just holding the stock: delta 1, no gamma, no decay, and the financing term rS matches rV. V = K e^(-r(T-t)) is a zero-coupon bond: it does not depend on S at all, and its value grows at exactly r as it approaches T. The stock and the bond are also the two pieces of the call price formula, which is why the check is worth a minute.

Why does S squared fail, and how do you repair it?

S squared has gamma 2, so the half sigma squared S squared gamma term adds sigma squared S squared, the delta term adds 2rS squared, and subtracting rV leaves (r + sigma squared) S squared with nothing to cancel it. A convex payoff gains from every move, so a fair price for it has to decay over time to pay for that gain, and S squared on its own has no decay. Try V = S squared times f(t): the equation forces f' = -(r + sigma squared) f, so the price of a claim paying S squared at T is S squared e^((r + sigma squared)(T - t)). At S = 100, r = 5%, sigma = 20% and one year, that is about 10,942, not 10,000, and the extra is the value of volatility.

1/6, the same as with no information. An odd first roll says nothing about the order in which 2, 4 and 6 first appear, so it is independent of 2 finishing last. A first roll of 4 or 6 raises the chance 2 is last from 1/3 to 1/2, so conditioning on that ending shifts weight onto 4 and 6, which rise to 1/4 each. The odd faces keep 1/6 each; 1/5 wrongly spreads the weight evenly.

Why does the ending tell you anything about the start?

Suppose you hear that a friend reached a party last. That makes it a little more likely they left home late, because leaving late and arriving last go together. It says nothing about whether they wore a blue shirt, which has no bearing on arrival order. Conditioning on an outcome reweights every starting state by how likely that state makes the outcome, and a state that does not affect the outcome keeps its original probability. Here the outcome is 2 finishing last among the evens; the question is which first rolls make that more or less likely.

How do the numbers work out with Bayes?

Odd rolls never change which new even appears next, so only the order of first appearances matters, and without information it is a random ordering of three: 2 is last with chance 1/3. If the first roll is 4, then 2 and 6 are left to race, and each is equally likely to show first, so 2 ends last with chance 1/2. Now weigh: each face has prior 1/6. The joint chance of first roll 1 and 2 last is 1/6 x 1/3 = 1/18; of first roll 4 and 2 last, 1/6 x 1/2 = 1/12. The total is 1/3, so first roll 1 has posterior (1/18)/(1/3) = 1/6 and first roll 4 has (1/12)/(1/3) = 1/4.

Where does the 1/5 intuition go wrong?

It treats the information as simply ruling out one face and renormalising the rest. Ruling out an outcome and conditioning on an event are the same thing only when every remaining outcome makes the event equally likely, and here they do not. A check: the posteriors 1/6, 1/6, 1/6, 1/4, 1/4 and 0 add to 1, while five faces at 1/5 would give 4 and 6 the same weight as 1. On a desk this is the error of reading a trade's outcome as if it said nothing about which signal triggered it.

Pick THH; you win 7 times in 8. HHH can only win if the first three flips are all heads, which has chance 1/8. In any other run, the first HHH is preceded by a tail, and that tail with the next two heads spells THH, which is completed one flip before HHH. So THH wins every game except the one that opens with three heads.

Why is this not a fair race between two 1/8 sequences?

Think of two runners on the same track where one always starts one step ahead of the other on the only route to the finish. Their speeds are identical but the race is not even. In a race between patterns, what matters is not how often each appears but which one tends to appear first, and that depends on how the patterns overlap. THH is built from HHH's own first two heads with a tail placed in front, so it ambushes HHH whenever HHH has not already won at the start.

How do you prove 7/8 without a Markov chain?

Look at the first time HHH appears. If it does not start at flip 1, the flip immediately before it must be a tail, because otherwise an earlier HHH would already have appeared. That tail plus the first two heads of the HHH is THH, finished one flip before HHH. So HHH wins only if flips 1 to 3 are heads, chance 1/8, and THH wins otherwise: 7/8. The proof is one sentence, and interviewers want to hear it rather than a transition matrix.

What is the general lesson for the second mover?

This is Penney's gameA coin sequence race in which the second player, choosing after seeing the first, can always pick a sequence that wins more than half the time., and the second player always has an edge because the winning relation among three-flip sequences is not transitive: every sequence has another that beats it. The recipe: take the opponent's first two flips, and put in front of them the opposite of the opponent's second flip. Against HHH that gives THH at 7/8; against HTH it gives HHT at 2/3. In trading terms, a strategy that looks as good as any other in isolation can still lose systematically to one designed around it.

Keep a 5 or 6 on the first roll, a 4, 5 or 6 on the second, and take whatever the third gives; the game is worth 14/3, about 4.67. Work backwards. The last roll is worth 3.5. With two rolls left, keep anything above 3.5: (4 + 5 + 6)/6 + (1/2)(3.5) = 4.25. With three, keep anything above 4.25: (5 + 6)/6 + (2/3)(4.25) = 14/3.

Why start from the last roll?

Think of house hunting with three viewings booked: whether to accept the first flat depends on what the remaining viewings are likely to offer, and you only know that once you know how you would behave at the last one. The value of continuing at any point is defined by what you would do later, so the only roll whose value you know outright is the last one, and every earlier decision is built on it. That is backward induction, and the interviewer wants to hear the words before any numbers.

How do the values build up?

On the last roll you take the face, worth 3.5. On the second roll, stop if the face beats 3.5, which means 4, 5 or 6; otherwise you get 3.5 from the last roll. The value is the average of the faces you keep plus the chance you continue times the value of continuing: (4 + 5 + 6)/6 + (3/6)(3.5) = 4.25. On the first roll, the bar to beat is now 4.25, so only 5 and 6 are kept: (5 + 6)/6 + (4/6)(4.25) = 11/6 + 17/6 = 14/3, about 4.667.

What does the common shortcut cost, and where does this lead?

The shortcut is to keep anything above the single-roll average of 3.5 at every stage. On the first roll that keeps a 4, which gives 4.625 instead of 4.667. The threshold rises with the number of rolls left, because each spare roll is an option, and an option is worth more the longer it lives. With six rolls the value is 5.27, and with many rolls it approaches 6, since you can wait for a six. The same structure prices an American option: exercise early only when the payoff beats the value of holding on.

Anywhere from about -0.75 to 0.95. The correlation matrix must be positive semidefinite, which bounds the third correlation at 0.2 x 0.5 plus or minus sqrt((1 - 0.2^2)(1 - 0.5^2)), that is 0.1 plus or minus 0.849. Geometrically, correlations are cosines of angles: X sits 78.5 degrees from Y and Z sits 60 degrees from Y, so X and Z are between 18.5 and 138.5 degrees apart.

Why does knowing two correlations restrict the third at all?

Think of three towns on a map. If A is close to B and B is close to C, then A cannot be far from C; if B is only loosely near both, A and C could be almost anywhere. Correlations behave like distances in disguise: each is the cosine of the angle between two returns viewed as vectors, and angles obey a triangle rule. X makes an angle of {TH_A:.1f} degrees with Y, because its cosine is 0.2, and Z makes 60 degrees with Y. The angle between X and Z is therefore at least the difference and at most the sum.

How do you get the bound algebraically?

Any valid correlation matrix must be positive semidefiniteEvery portfolio built from the variables has a variance of zero or more; for a correlation matrix this means its determinant and all leading minors are non-negative., because a portfolio cannot have negative variance. For three variables with correlations a, b and c, the condition is 1 - a^2 - b^2 - c^2 + 2abc of at least zero. Treating that as a quadratic in c gives c = ab plus or minus sqrt((1 - a^2)(1 - b^2)), so the third correlation lies in an interval centred on the product of the other two. With a = 0.2 and b = 0.5 that is 0.1 plus or minus sqrt(0.72), or -0.7485 to 0.9485.

What does the centre value 0.1 mean, and when is the sign forced?

Split X and Z each into a part explained by Y and a leftover. The explained parts always contribute 0.2 x 0.5 = 0.1; the leftovers can be correlated however you like, and they can move the total by up to 0.849 either way. So 0.1 is the answer only if the leftovers are uncorrelated. The sign of corr(X, Z) is forced positive only when the two given correlations are strong, specifically when their squares add to more than 1; with 0.8 and 0.7 the range is about 0.13 to 0.99. On a desk this is why two hedges that each track an index only loosely say almost nothing about each other.