This one tripped me up more than I expected.
Recognize that the correlation matrix must be positive semi-definite (PSD). Construct the 3x3 correlation matrix with all off-diagonal entries equal to ρ, then find the minimum ρ such that all eigenvalues are non-negative. The answer is ρ ≥ -1/2.
Pro tip: Mention that this bound generalizes to n variables with ρ ≥ -1/(n-1), and that it's a common trick in quant interviews to test understanding of covariance matrices.
Write the 3x3 matrix with 1s on the diagonal and ρ on all off-diagonals. This matrix must be positive semi-definite (PSD) for any valid set of random variables.
Compute the eigenvalues of the matrix. For a matrix with constant off-diagonal ρ, the eigenvalues are 1+2ρ (multiplicity 1) and 1-ρ (multiplicity 2).
Require all eigenvalues to be non-negative: 1+2ρ ≥ 0 and 1-ρ ≥ 0. The second inequality gives ρ ≤ 1, which is always true for correlations. The first gives ρ ≥ -1/2.
Show that ρ = -1/2 is achievable by constructing three random variables with the given correlation. For example, let X, Y, Z be such that X+Y+Z=0 and each has variance 1, then pairwise correlation is -1/2.
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I knew the relationship existed but blanked on the exact form mid-answer.
Derive the relationship between the two slopes using the definitions of simple linear regression and the coefficient of determination. Show that β(x|y) = r² / β(y|x), where r² = R², and explain the intuition behind this inverse relationship.
Pro tip: Emphasize that this relationship holds only in simple linear regression with an intercept; in multiple regression, the reverse regression slopes are not simply related to R² and the forward slopes. This shows you understand the limitations and can avoid overgeneralizing.
State the simple linear regression model for Y on X: Y = α + βX + ε, and for X on Y: X = γ + δY + ν. Note that β = Cov(X,Y)/Var(X) and δ = Cov(X,Y)/Var(Y).
Use the fact that R² = [Cov(X,Y)]² / (Var(X)Var(Y)) for simple linear regression with an intercept.
Multiply β and δ: β * δ = [Cov(X,Y)/Var(X)] * [Cov(X,Y)/Var(Y)] = [Cov(X,Y)]² / (Var(X)Var(Y)) = R². Therefore, δ = R² / β.
Explain that this relationship holds only for simple linear regression with an intercept. In multiple regression, the reverse regression coefficients are not simply related to R² and the forward coefficients.
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Probably the question I felt best about, mostly because I remembered that Cov(max, min) can be rewritten using Cov(X+Y, XY) type identities or by just computing E[max*min], E[max], E[min] separately.
Use the identity Cov(max, min) = E[max*min] - E[max]E[min]. Compute E[max] and E[min] using order statistics, and compute E[max*min] by recognizing that max*min = X*Y (since one is max and the other min). Then plug into the covariance formula.
Pro tip: Mention that max*min = X*Y is a key simplification that avoids integration. Also, note that the negative covariance makes intuitive sense: when the minimum is high, the maximum must be even higher, but the spread is constrained.
Write Cov(max, min) = E[max*min] - E[max]E[min]. This decomposes the problem into computing three expectations.
For two independent Uniform(0,1), E[max] = 2/3 and E[min] = 1/3. Derive these using order statistics or integration.
Observe that max(X,Y)*min(X,Y) = X*Y. Since X and Y are independent, E[X*Y] = E[X]E[Y] = (1/2)*(1/2) = 1/4.
Plug values: Cov = 1/4 - (2/3)*(1/3) = 1/4 - 2/9 = (9 - 8)/36 = 1/36. Wait, check: 1/4 = 9/36, 2/9 = 8/36, so difference is 1/36. But is it positive? Actually, let's recompute: E[max]E[min] = (2/3)*(1/3)=2/9≈0.222, E[max*min]=0.25, so Cov=0.25-0.222=0.0278=1/36. That's positive. But intuitively, max and min should be negatively correlated? Let's check: For two uniforms, max and min are actually positively correlated? No, think: if min is high, max must be even higher, so they move together? Actually, if min is high, max is also high, so positive correlation. Yes, positive covariance. So 1/36 is correct.
Confirm the result and provide intuition: positive covariance because when the minimum is large, the maximum is also large, and vice versa.
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Start by clarifying that the function is monotone (either increasing or decreasing) and that the transformation is one-to-one. Then use the change-of-variables formula: f_X(x) = f_Y(g(x)) * |g'(x)|, explaining each component and noting the absolute value for the decreasing case. Finally, mention the importance of checking the support of X and verifying that the resulting pdf integrates to 1.
Pro tip: Emphasize that monotonicity ensures a one-to-one mapping, which simplifies the derivation; if the function were not monotone, you'd need to sum over all preimages. Also, mention that in practice, you often work with the CDF method for safety, especially when the transformation is not strictly monotone or when dealing with piecewise functions.
Confirm that g is monotone (increasing or decreasing) and differentiable, and that Y = g(X) is a one-to-one transformation. Define the pdfs f_X and f_Y and their supports.
For monotone g, the pdf of X is f_X(x) = f_Y(g(x)) * |g'(x)|. Explain that the derivative term accounts for the scaling of probability mass under the transformation.
If g is increasing, g'(x) > 0, so |g'(x)| = g'(x). If g is decreasing, g'(x) < 0, so |g'(x)| = -g'(x). Emphasize that the absolute value ensures a positive pdf.
Find the range of X by inverting the transformation: X = g^{-1}(Y). The support of X is the image of the support of Y under g^{-1}. Ensure the pdf is zero outside this support.
Check that the derived pdf integrates to 1 over the support of X. Optionally, mention the CDF method as an alternative derivation: F_X(x) = P(X ≤ x) = P(g(X) ≤ g(x)) or P(g(X) ≥ g(x)) depending on monotonicity, then differentiate.
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