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  1. 098With orthonormal regressors, ordinary least squares gives coefficients of 0.6 and 0.15. What do ridge and lasso with a penalty of 0.2 give for each, and why does only lasso set a coefficient to zero?Statistics and estimationHardCitadelLondon · 2026

    Try it first

    Using half the residual sum of squares plus the penalty, what do the two methods give?

    Show the worked solution

    Ridge gives 0.5 and 0.125; lasso gives 0.4 and exactly 0. With orthonormal regressors each coefficient is shrunk on its own. Ridge divides each by 1 + 0.2 = 1.2, so it scales both down and never reaches zero. Lasso subtracts 0.2 from each size and stops at zero, so the small coefficient, 0.15, is removed. This uses the scaling of half the residual sum of squares plus 0.2 times the penalty.

    Why do orthonormal regressors make this a one-line problem?

    When the regressors are uncorrelated and scaled to unit length, the fit for each coefficient does not depend on the others, so the penalised problem splits into separate one-variable problems. Think of adjusting the volume on two speakers that are not wired together: turning one down does not change the other. For each coefficient you minimise half of (beta minus b) squared plus the penalty, where b is its least squares value, and the answer depends only on that one number. Ridge's penalty is half of beta squared times 0.2; lasso's is the size of beta times 0.2.

    Ridge scales every coefficient down; lasso subtracts and stops at zeroRidge: divide by 1 + 0.20.6 to 0.50.15 to 0.12500.20.40.60.8least squares coefficientpenalisedLasso: subtract 0.2, floor at 00.6 to 0.40.15 is inside the reddead zone below 0.2,so it goes to exactly 000.20.40.60.8least squares coefficientpenalised
    Ridge scales every least squares coefficient by 1/1.2, so 0.6 becomes 0.5 and 0.15 becomes 0.125, while lasso subtracts 0.2 and stops at zero, so 0.6 becomes 0.4 and 0.15, inside the dead zone below 0.2, becomes exactly 0.

    What does each penalty do to a coefficient?

    Ridge's squared penalty pulls hard on big coefficients and gently on small ones. Setting the slope to zero gives beta x (1 + 0.2) = b, so ridge divides by 1.2: 0.6 becomes 0.5 and 0.15 becomes 0.125, shrunk but never zero. Lasso's absolute-value penalty pulls with the same force, 0.2, whatever the size. Its answer is the sign of b times the larger of (the size of b minus 0.2) and zero: 0.6 becomes 0.4, and 0.15, weaker than the pull of 0.2, lands exactly on 0.

    The relationship
    β^ridge=b1+λβ^lasso=sign⁡(b) max⁡(∣b∣−λ, 0)\hat\beta^{\text{ridge}} = \frac{b}{1 + \lambda} \qquad \hat\beta^{\text{lasso}} = \operatorname{sign}(b)\,\max\big(|b| - \lambda,\ 0\big)
    bthe least squares coefficient, 0.6 or 0.15
    lambdathe penalty weight, 0.2
    What it says in wordsRidge divides every coefficient by the same factor; lasso takes the same amount off every coefficient and never goes past zero.

    Why does only lasso select variables?

    At zero, the squared penalty is flat: its slope is zero, so any small coefficient still earns its place by improving the fit a little. The absolute-value penalty has a corner at zero with a slope of 0.2 on each side, so a coefficient stays at zero unless the fit improves by more than 0.2 per unit, and 0.15 does not. That is why lasso gives sparseHaving many coefficients exactly equal to zero, so the model uses only a few of the available variables. models and ridge does not. The limitation to state: with correlated regressors lasso tends to keep one of a group arbitrarily and drop the rest, which is why desks often blend the two penalties in an elastic net.

    Where candidates lose it

    The common slip is to swap the two, saying ridge sets small coefficients to zero because it penalises harder. Ridge's penalty is heavy on big coefficients and almost nothing on small ones, which is exactly why it never zeroes them.

    The second loss is getting the lasso numbers off by a factor of two. Written as the full residual sum of squares plus 0.2 times the absolute values, without the half, the threshold is 0.1, giving 0.5 and 0.05. State your scaling before you give numbers.

    What the interviewer asks next

    • At what penalty does lasso set the 0.6 coefficient to zero as well?
    • The two regressors now have a correlation of 0.9. How do ridge and lasso behave differently?
    • How would you choose the penalty in practice without fitting it to noise?

    Asked at Citadel, Quantitative Research, London, 2026 (Wall Street Oasis): very detailed and difficult questions about regularisation ridge and lasso