Quant puzzles, solved step by step

Take X equal to -1, 0 or 1 with probability 1/3 each, and Y = X squared. Y is fixed by X, so they are as dependent as variables can be. But E[X] = 0 and E[XY] = E[X cubed] = (-1 + 0 + 1)/3 = 0, so the covariance E[XY] - E[X]E[Y] is zero. Correlation only measures straight-line association, and this relationship is a V.

What does correlation actually measure?

Think of a thermostat that runs the air conditioner hard on very hot days and the heater hard on very cold days. Energy use is clearly driven by temperature, but a straight line through the data is flat: high use at both ends, low in the middle. Correlation measures only how well a straight line summarises the relationship, so any symmetric U or V shape can have zero correlation while being completely determined. Independence is the stronger claim that knowing X tells you nothing about Y at all.

How do you show the covariance is zero in one line?

Write the definition and let symmetry do the work. Cov(X, Y) = E[XY] - E[X]E[Y], and with Y = X squared the first term is E[X cubed], which is zero for any X symmetric about zero; the second term has E[X] = 0 in it. With the three-point version you can even list the products: -1 x 1, 0 x 0 and 1 x 1 add to zero. Yet P(Y = 0 given X = 0) is 1 while P(Y = 0) is 1/3, which is dependence in plain sight.

Why does a quant interviewer care?

Because models quietly substitute zero correlation for no relationship. A delta-hedged option book gains or loses roughly with the square of the underlying's move, so its daily P and L can show near-zero correlation with the market while being entirely driven by it. The same holds for a volatility strategy or any payoff with a kink. The one case where zero correlation does mean independence is when the pair is jointly normal, which is worth adding before the interviewer asks.

You get 0.75, not 1. Regressing y on x1 alone gives a slope of 1 + 0.5 = 1.5, because x1 soaks up the half of x2 that moves with it. The residual is x2 - 0.5x1 + noise, whose slope on x2 is 1 - 0.5 squared = 0.75. To recover 1, residualise x2 on x1 as well and regress the residual of y on the residual of x2: the Frisch-Waugh-Lovell theorem.

Where does the missing quarter go?

Picture two salespeople who often work the same client. If you credit all joint sales to the first before looking at the second, the second looks worse than they are, because some of their work was already booked to the first. Stage one regresses y on x1 alone, and since x2 is correlated with x1, the coefficient on x1 rises to 1.5: it takes credit for 0.5 of x2. That piece has been removed from the residual, so stage two can only find what is left of x2's effect.

How do you get 0.75 exactly?

Write the residual out. y - 1.5x1 = x2 - 0.5x1 + noise, and the slope of that on x2 is its covariance with x2 over the variance of x2: (1 - 0.5 x 0.5)/1 = 0.75. The formula generalises to 1 - rho squared times the true coefficient, so the bias gets worse as the regressors get more correlated: with rho = 0.9 you would find only 0.19. A simulation of 100,000 observations gives 1.506 for stage one and 0.752 for stage two.

What does Frisch-Waugh-Lovell tell you to do?

To get a variable's coefficient from a multiple regression in stages, partial the other regressors out of both y and that variable, then regress residual on residual. Here that means regressing x2 on x1 as well, keeping the orthogonal part x2 - 0.5x1, and regressing the stage-one residual on it. The slope comes back as exactly 1; the simulation gives 1.003. This is why factor-neutralising a signal before testing it, rather than after, matters in quant research: the order of the stages changes the answer.

R-squared is 0.36 for both regressions and the correlation is 0.6. The slope of y on x is r times sd(y)/sd(x); the slope of x on y is r times sd(x)/sd(y). Multiplying them cancels the standard deviations and leaves r squared: 0.8 x 0.45 = 0.36. The correlation is +0.6, positive because both slopes are positive, and the ratio sd(y)/sd(x) is √(0.8/0.45) = 4/3.

Why are the two slopes not reciprocals of each other?

Tall parents tend to have tall children, but a little less tall; and tall children tend to have tall parents, but a little less tall. Both statements are true at once. Each regression predicts toward the mean, so neither slope is the inverse of the other unless the fit is perfect. If the points lay exactly on a line, the slope of x on y would be 1/0.8 = 1.25. It is 0.45 instead, and the size of that shortfall is what measures how loose the relationship is.

How do the two slopes give R-squared?

Write each slope in terms of the correlation. The least squares slope of y on x is the covariance over the variance of x, which is r times sd(y)/sd(x). Swap the roles and the slope of x on y is r times sd(x)/sd(y). The standard deviation ratios are reciprocals, so the product of the two slopes is r squared, and in a one-variable regression r squared is exactly the R-squared. Here 0.8 x 0.45 = 0.36, so r = 0.6; the sign is positive because both slopes are positive, and the two slopes always share a sign.

The figure uses 40 points built with standard deviations 3 and 4 and a correlation of exactly 0.6, and fitting both regressions returns slopes of 0.80 and 0.45. A quick sanity test comes free: the product of the two slopes can never exceed 1. If an interviewer quotes slopes of 0.8 and 1.5, the product 1.2 is impossible, and saying so is worth more than any calculation.

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.

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.

A^10 has 29,525 on the diagonal and 29,524 off it. A has eigenvalue 3 along (1, 1) and eigenvalue 1 along (1, -1). Writing A = Q D Q^T with D = diag(3, 1), the tenth power is Q D^10 Q^T, and only the numbers 3 and 1 get raised to the tenth. The entries are (3^10 + 1)/2 and (3^10 - 1)/2.

Why look for eigenvectors at all?

Think of a photocopier set to 300% on one axis and 100% on the other. Copy a copy ten times and you do not need to simulate every pass: that axis is 3 to the tenth times longer and the other is unchanged. An eigenvector is a direction the matrix only stretches, so applying the matrix ten times along it is just multiplying by the eigenvalue ten times. Symmetric matrices always have a full set of such directions at right angles, which is what makes this matrix easy.

Find them by inspection. Adding the two rows of A gives 3 in each, so A(1, 1) = (3, 3): eigenvalue 3. Subtracting gives 1, so A(1, -1) = (1, -1): eigenvalue 1. The trace is 4 and the determinant is 3, and 3 + 1 = 4 and 3 x 1 = 3, which confirms both in one line.

Is there an even faster route for this particular matrix?

Yes. Write A = I + J, where J is the all-ones matrix. J squared is 2J, so every power of J is a multiple of J, and (I + J)^n collapses to I + ((3^n - 1)/2) J. For n = 10 that is I + 29,524 J, which gives 29,525 on the diagonal and 29,524 off it: the same answer, and a good cross-check to say aloud. A brute-force multiplication in code agrees exactly.

Say why this matters on a desk. A covariance matrix with equal variances and one common correlation has exactly this shape, and its eigenvectors are the market direction and the spread directions. Powers of transition matrices in Markov chains are computed the same way, and the eigenvalue closest to 1 tells you how fast the chain forgets where it started.

1/sqrt(2), about 0.71. The covariance of X with X + Y is Var X plus Cov(X, Y), and the second term is zero, so it is sigma squared. The variance of X + Y is 2 sigma squared, because independent variances add. Dividing sigma squared by sigma times sqrt(2) sigma leaves 1/sqrt(2). Squared, that is 0.5: X explains half of the sum's variance.

Why isn't the answer one half?

Picture two people each tossing a coin for a rupee, and a pot holding their combined winnings. One player's result explains exactly half of the pot's variability, and the other half comes from the other player. Half is the share of variance explained, R squared, and correlation is its square root, so the correlation is 1/sqrt(2), not 1/2. This is the most common slip on the question, and it comes from mixing up the two measures.

How does it change if the variances differ?

Let Var Y be k times Var X. The covariance is still Var X, and Var(X + Y) becomes (1 + k) Var X. The correlation is 1/sqrt(1 + k): the noisier Y is, the less the sum tracks X. At k = 1 you get 0.707; at k = 4, 0.447; at k = 0.25, 0.894. This is exactly the signal-plus-noise model: if a price move is a true signal plus independent noise of equal size, the best-case correlation between your signal and the move is about 0.71.

Where does this show up on a desk?

Any time one piece is part of a total. A stock's return is market return plus its own specific return; if the two had equal variance, the stock would correlate 0.71 with the market. The same arithmetic tells you the ceiling on a predictor: if half of tomorrow's move is unpredictable noise, no model can correlate more than 0.71 with it. The limitation is independence; if X and Y are correlated, add 2 Cov(X, Y) to the variance of the sum and Cov(X, Y) to the covariance.

OLS gives 0.8 and ridge gives 0.64. OLS minimises squared error and its slope is the sum of xy over the sum of x squared, 80/100. Ridge adds lambda times the slope squared to the loss, which puts lambda into the denominator: 80/(100 + 25) = 0.64. That is the OLS slope times 100/125 = 0.8, so ridge shrinks the slope towards zero but never to zero.

Where does lambda end up in the formula?

Think of a new analyst's forecast that you half trust: you do not discard it, you shade it towards zero, and the less data behind it the more you shade. Ridge does that mechanically. It minimises the squared errors plus lambda times the slope squared; setting the derivative to zero gives b = Sxy/(Sxx + lambda). Penalising the size of the slope acts exactly like adding observations whose x squared totals lambda and whose y is zero, data that say the slope is zero. With 100 of real evidence and 25 of make-believe evidence, the slope is 80/125 = 0.64.

Why would you want a slope that is biased towards zero?

Because a smaller, steadier estimate can be closer to the truth on average. Suppose the true slope is 0.5 and the noise variance is 25. OLS is unbiased but its variance is 25/100 = 0.25. Ridge at lambda 25 has variance 0.16 and a bias of -0.1, so its mean squared error is 0.17. Ridge trades a little bias for a larger cut in variance, and when the signal is weak relative to the noise that trade wins. In this one-feature case the best lambda is noise variance over slope squared, 100, which halves the slope and cuts the error to 0.125. In practice the truth is unknown, so lambda is chosen by cross-validation.

How does lasso differ?

Lasso penalises lambda times the absolute slope instead. In one dimension that subtracts lambda/2 from the cross-product rather than adding to the denominator: (80 - 12.5)/100 = 0.675 at lambda 25, and exactly zero once lambda reaches 160. Ridge scales coefficients down; lasso shifts them down and can set them to exactly zero, which is why lasso selects features and ridge does not. Both penalties depend on the scale of x, which is why the feature must be standardised first; with correlated features, ridge spreads the weight across them while lasso tends to keep one.