Quant puzzles, solved step by step

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.

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.

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.

The standard deviation is sqrt(2.5), about 1.58%, and the kurtosis is 12.72, against 3 for a normal. Moments of a mixture are weighted averages of the pieces' moments. The variance is 0.9 x 1 + 0.1 x 16 = 2.5. A normal's fourth moment is 3 sigma^4, so the fourth moment is 3 x (0.9 x 1 + 0.1 x 256) = 79.5. Kurtosis is 79.5 / 6.25 = 12.72.

Why does mixing two normals create fat tails?

Think of a city's daily traffic: most days are ordinary, and a few days a year there is a festival or a strike and everything is wild. Averaged across the year, the typical day looks calmer than the average suggests, and the extreme days are far more extreme than a single bell curve would allow. Mixing a calm regime with a rare wild one concentrates most days near zero and puts the rest far out, which is exactly what kurtosis measures: a high peak with heavy tails. Every piece is normal; the mixture is not.

How do you compute the moments without integrating?

Condition on the regime. Any moment of a mixture is the weighted average of that moment in each regime, because the density itself is the weighted average of the two densities. The second moment is 0.9 x 1^2 + 0.1 x 4^2 = 2.5, so the standard deviation is about 1.58%. For the fourth, use the fact that a normal with standard deviation sigma has fourth moment 3 sigma^4: 3 x (0.9 x 1 + 0.1 x 256) = 79.5. Kurtosis is the fourth moment over the variance squared.

What does the number mean for risk?

Kurtosis is 3 times the ratio of the average of sigma^4 to the square of the average of sigma^2, so any variation in volatility pushes kurtosis above 3, and the more uneven the regimes, the further it goes. The practical cost shows in the tails. A risk model that fits one normal with a 1.58% standard deviation expects a move beyond 6% about once in 6,766 days, roughly once every 27 years of trading; the mixture produces one about once in 75 days, several times a year. This is the simplest model of volatility clustering, and the reason daily returns on real assets show kurtosis well above 3.

Centre on today's level, about 20,000, with a 90% range of roughly 16,800 to 23,600. Four months is a third of a year, so one standard deviation is 18% x sqrt(1/3), about 10.4%. In log terms the 90% band is 1.645 of those either side, which gives 16,767 and 23,601. The median sits a little below 20,000 and the upside tail is longer than the downside.

Why is a single number the wrong answer?

Ask a cab driver how long the airport run takes and a good one says forty minutes, maybe an hour in traffic. The range is the useful part, because you plan your flight around it. A trading interviewer asking where an index closes wants a distribution, because a market maker quotes against the spread of outcomes, not against a guess. Your central value should not be a view on the economy either: with no edge, the best central estimate of a traded index is roughly its forward, which for four months is close to today's 20,000 once financing and dividends roughly offset.

The width comes from the implied volatility the market already quotes. Volatility grows with the square root of time, because independent daily moves add their variances, not their standard deviations. Four months is a third of a year, so one standard deviation is 18% x sqrt(1/3) = 10.4%, about 2,078 index points.

Why is the range lopsided, and where does the median sit?

A fall of 10% and a rise of 10% are not mirror images in log space. Normal log returns make the upside tail longer: the 90% band stretches 3,601 points up but only 3,233 points down. The same convexity pushes the median below the mean: if the mean is 20,000, the median is 20,000 x exp(-sigma squared T / 2), about 19,892. That gap of about 108 points is small here, but it grows with volatility and time, and a candidate who names it shows they know the difference between the most central outcome and the average one.

What would you add before quoting a market on it?

Two honest caveats. Implied volatility is a price, not a forecast: it tends to sit above the volatility that is later realised, because option sellers charge for bearing crash risk, so the band built from it is usually a little wide. Against that, real index returns have fatter tails than the lognormal, so the 5% tails are more likely to hold a larger move than the curve suggests. Say both, then give your market: a tight two-way price around 20,000 if asked for the level, and the 90% band as the range you would sell outside of.

76. Split 100 into 4 x 25. Any power of 4 is 0 mod 4. Mod 25, Euler's theorem applies because 4 and 25 share no factor, and 3000 is a multiple of phi(25) = 20, so 4^3000 is 1 mod 25. The numbers below 100 that are 1 mod 25 are 1, 26, 51 and 76, and only 76 is divisible by 4.

Why does the obvious Euler shortcut fail?

Euler's theorem says a to the power phi(n) is 1 mod n, and phi(100) = 40, so it is tempting to say 4^3000 = (4^40)^75 ends in 01. The theorem needs the base and the modulus to share no factor, and 4 and 100 share a factor of 4, so it does not apply. A quick sense check kills 01 anyway: every power of 4 is divisible by 4, and a number is divisible by 4 exactly when its last two digits are, which 01 is not.

How do you split the problem so the theorem does apply?

Think of a clock with 100 hours as two smaller clocks running together, one with 4 hours and one with 25. Knowing where both small clocks point fixes the big one exactly. Mod 4 the answer is 0, since 4^3000 is a multiple of 4; mod 25 the answer is 1, since 4 and 25 share no factor and 3000 is a multiple of phi(25) = 20. Now list the numbers below 100 that are 1 mod 25: 1, 26, 51, 76. Only 76 is a multiple of 4. This step is the Chinese remainder theorem, and naming it earns credit.

What is the fastest check if you have a pencil?

Just list the endings. Multiply each ending by 4 and keep the last two digits: 04, 16, 64, 56, 24, 96, 84, 36, 44, 76, and then 76 x 4 = 304, which ends in 04, so the cycle has length 10. Because 76 x 76 = 5,776 also ends in 76, every power of 76 ends in 76, and 4^3000 = (4^10)^300 must end in 76. On a multiple-choice test, this listing takes about thirty seconds and needs no theorem at all.

10/12, about 0.833. A uniform prior on the win rate, updated with 9 wins and 1 loss, gives a Beta(10, 2) posterior. The chance of winning tomorrow is the posterior mean, (9 + 1)/(10 + 2). The answer is below 0.9 because ten days cannot rule out a lower true win rate, and averaging over that uncertainty pulls the estimate towards one half.

Why is 0.9 too confident?

A new restaurant with nine five-star reviews out of ten looks excellent, but you would not bet that the next diner rates it five stars with 90% certainty. Ten reviews is a small sample, and a restaurant that truly earns five stars 70% of the time could easily post nine out of ten. The right forecast averages over every win rate the evidence still allows, and with only ten days that includes plenty of rates below 0.9. The average of that spread of possibilities is what you should quote for tomorrow.

How does the Bayesian update give 10/12?

Start with every win rate p between 0 and 1 equally likely. The chance of the observed record is proportional to p^9 (1 - p), so the posterior is proportional to that, which is the Beta(10, 2) distribution. The chance of a win tomorrow is the average of p over the posterior, and the mean of a Beta(a, b) is a/(a + b), here 10/12. The shortcut is Laplace's rule of succession: add one imaginary win and one imaginary loss to the record, then divide.

When does the pull towards one half stop mattering?

When the record is long. At 90 wins out of 100 the rule gives 91/102, about 0.892, almost exactly the raw 0.9, because a hundred days of data swamp the one imaginary win and loss. The limitation is the prior itself. A uniform prior says a 99% win rate was as plausible as a 50% one before you saw any data, which no trader believes about a new strategy. With a sceptical prior centred near one half, ten days would pull the estimate even further below 0.9. The posterior also tells you more than one number: the chance of winning both of the next two days is (10 x 11)/(12 x 13), about 0.705, not 0.833 squared.

pi/2, about 1.571. Rotate the sphere so the first point sits at the north pole; nothing changes, because the second point is uniform. On a unit sphere the surface distance is the polar angle theta of the second point. The northern and southern hemispheres are mirror images, so theta is as likely to be pi/2 - t as pi/2 + t, and its mean is pi/2.

Why can you put the first point at the pole?

Ask how far apart two random towns are on a perfectly round planet, and you can simply stand in one of them: the globe looks the same from every spot on it. Symmetry lets you fix one point anywhere, so the problem shrinks to one random point and its angle from the pole. On a sphere of radius 1, the distance along the surface between the pole and a point at polar angle theta is theta itself, measured in radians, so the question becomes: what is the average polar angle of a uniform point?

Where does sin theta come from? The circle of latitude at angle theta from the pole has circumference 2 pi sin theta, so a thin band there holds surface in proportion to sin theta: almost none near the poles, the most at the equator. Archimedes put it more neatly: the area of a band is proportional to its height along the axis, so cos theta is uniform between -1 and 1. The density (1/2) sin theta is a mirror image about pi/2, so the mean is pi/2 without doing the integral. Integration by parts confirms it, a numerical integral gives 1.5708, and a seeded simulation of 100,000 pairs gives 1.570.

If a uniform angle gives the same mean, why does the shape matter?

Because the mean survives by luck of symmetry and almost nothing else does. Choosing theta uniformly on 0 to pi crowds points near the poles. Ask for the chance the two points are within 60 degrees of each other and the correct answer is (1 - cos 60 degrees)/2 = 0.25, while the uniform angle says 0.33. Ask for the expected straight-line chord, 2 sin(theta/2), and the correct density gives 4/3, about 1.333, while the uniform angle gives 4/pi, about 1.273. The simulation gives 1.333 for the chord. This is the limitation of the shortcut: it answers this one question and must not be reused for the next.

Set the ask at 106, and by the same logic the bid at 104. If the stock is worth 110, a buy arrives with probability 0.2 + 0.8 x 0.5 = 0.6; if it is worth 100, with probability 0.4. Given a buy, Bayes puts the chance of 110 at 0.6, so the stock is worth 106 to the market maker selling it. The spread of 2 is the price of trading against informed flow.

Why can't the market maker just quote the expected value of 105?

A second-hand car dealer who pays the average price for every car will find that the owners of good cars go elsewhere and the owners of bad ones queue up. Who chooses to trade with you is information. A market maker does not care what the stock is worth on average; it cares what the stock is worth given that someone has just chosen to buy from it. At an ask of 105, noise buyers are harmless, a loss of 5 when the stock is worth 110 and a gain of 5 when it is worth 100. Informed buyers only appear in the 110 world, and they cost 0.5 per arriving trader on average, so 105 is a losing quote.

What sets the width of the spread?

The share of informed traders and the size of what they know. With a share alpha informed, a buy is alpha + (1 - alpha)/2 likely in the high world and (1 - alpha)/2 in the low world, and the ask works out to 105 + 5 alpha. The spread is a fee for adverse selection: it is zero when nobody is informed and widens to the full 100 to 110 range when everyone is. The table runs the formula for a few shares. Order processing and inventory costs add to this in real markets, but the information component is what makes spreads jump around earnings and news.

What happens after the first trade?

The market maker updates. After one buy, the chance of 110 is 0.6, and if a second buy arrives the same Bayes step lifts it to 0.692, so the next ask is about 106.92. Each order moves the quotes towards the true value, which is how prices come to reflect what the informed traders know. The limitation is that the model has one share size, no inventory risk and no competition between market makers; real desks also skew quotes to manage position, which this puzzle leaves out.