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  1. 005Construct two random variables that are uncorrelated but clearly dependent, and show that their covariance is zero.Correlation, regression and linear algebraWarm upTwo SigmaNew York · 2025

    Try it first

    Which pair works?

    Show the worked solution

    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.

    Y is fixed by X, yet the best straight line is flatfit: Y = 2/3(-1, 1)(1, 1)(0, 0)X = -1, 0, 1 equally likely, Y = X squaredE[XY] = (-1 + 0 + 1)/3 = 0 = E[X] E[Y]fit: Y = 1/3Y = X squaredX uniform on -1 to 1, Y = X squaredCov(X, Y) = E[X cubed] = 0 by symmetry
    With X symmetric about zero and Y equal to X squared, Y is fixed exactly by X, yet the best straight line through the points is flat, so the covariance and the correlation are both zero.

    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.

    The relationship
    Cov⁡(X,X2)=E[X3]−E[X] E[X2]=0−0⋅23=0P(Y=0∣X=0)=1≠P(Y=0)=13\operatorname{Cov}(X, X^2) = E[X^3] - E[X]\,E[X^2] = 0 - 0\cdot\tfrac{2}{3} = 0 \qquad P(Y=0\mid X=0) = 1 \ne P(Y=0) = \tfrac{1}{3}
    E[X^3]zero because the values of X are symmetric about zero
    P(Y = 0 | X = 0)knowing X changes the odds on Y, so they are dependent
    What it says in wordsThe covariance cancels by symmetry, while a single conditional probability proves the dependence.

    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.

    Where candidates lose it

    Candidates reach for two independent variables, which are uncorrelated but not dependent, or for X and -X, which are dependent but perfectly correlated. Both show the definitions are fuzzy.

    The quieter trap is choosing X uniform on 0 to 1 and Y = X squared. Without symmetry about zero the covariance is positive, 1/12, and the example fails. Centre X first.

    What the interviewer asks next

    • When does zero correlation imply independence?
    • Give an example with zero correlation where Y is not a function of X.
    • If you regress Y on X in the example, what do the fitted line and R squared look like?

    Asked at Two Sigma, Generalist, New York, 2025 (Wall Street Oasis): Come up with two uncorrelated but dependent variables.

  2. 090X and Y are independent random variables with the same variance. What is the correlation between X and X + Y?Correlation, regression and linear algebraWarm upSquarepoint CapitalMontreal · 2026

    Try it first

    Pick the correlation.

    Show the worked solution

    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.

    X supplies half of the variance of X + Y, so the correlation is 1/sqrt(2)XYXYVar Xsigma^2Cov(X, Y)0Cov(Y, X)0Var Ysigma^2X row:Cov(X, X+Y)= sigma^2all four cells: Var(X + Y) = 2 sigma^2corr = Cov / (sd X x sd(X+Y))= sigma^2 / (sigma x sqrt(2) sigma)= 1/sqrt(2) = 0.707Share of Var(X + Y) explained by XR^2 = 0.5Unequal variances: corr = 1/sqrt(1 + k),k = Var Y / Var X; k = 4 gives 0.447
    In the covariance box, X's own variance fills one of the two non-zero cells, so X accounts for half of Var(X + Y), and the correlation between X and X + Y is sigma squared divided by sigma times sqrt(2) sigma, which is 1/sqrt(2), about 0.71.
    The relationship
    ρ=Cov⁡(X,X+Y)σX σX+Y=σ2+0σ⋅2 σ=12≈0.707\rho = \frac{\operatorname{Cov}(X, X+Y)}{\sigma_X\,\sigma_{X+Y}} = \frac{\sigma^2 + 0}{\sigma\cdot\sqrt{2}\,\sigma} = \frac{1}{\sqrt 2} \approx 0.707
    Cov(X, X + Y)Var X plus Cov(X, Y), which is sigma squared plus zero
    sigma_{X+Y}the standard deviation of the sum, sqrt(2) sigma
    What it says in wordsCovariance is linear, so split it into pieces; the only surviving piece is X's own variance.

    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.

    Where candidates lose it

    The frequent slip is answering one half, confusing the share of variance with the correlation. Correlation is the square root of that share.

    The second loss is writing the standard deviation of X + Y as 2 sigma, adding standard deviations instead of variances, which gives one half again by a different road. Independent variances add; standard deviations do not.

    What the interviewer asks next

    • What is the correlation between X + Y and X - Y?
    • If X and Y have correlation 0.5 and equal variance, what is corr(X, X + Y)?
    • What is the correlation between the first die and the total of two dice?

    Asked at Squarepoint Capital, Desk Quant Analyst Interview, Montreal, 2026 (Wall Street Oasis): There were also 3-4 basic math/stats questions about mean, covariance, correlation, etc.