Questions: Covariance and Correlation Coefficients

Question 1

X is drawn uniformly from [−1, 1] and Y = X². A student claims that since Cov(X, Y) = 0, X and Y must be independent. What is wrong with this reasoning?

• A. Nothing — zero covariance always implies independence for continuous random variables
• B. Zero covariance implies independence only for jointly normal variables, not all distributions
• C. Covariance measures linear association only; a nonlinear relationship can produce zero covariance even with complete dependence
• D. The student computed covariance incorrectly — it should be positive since Y is always non-negative

Answer: (C)

Covariance (and correlation) capture only *linear* association. In this example, Y = X² is a perfect deterministic function of X — knowing X tells you exactly what Y is. Yet by symmetry, E[X] = 0 and E[X³] = 0, so Cov(X,Y) = E[XY] − E[X]E[Y] = E[X³] − 0 = 0. The zero covariance reflects the absence of any *linear* trend between X and Y. Independence requires the joint distribution to factor completely, which is a strictly stronger condition than zero covariance.

Question 2

Which of the following transformations would change the covariance Cov(X, Y) but leave the correlation ρ(X, Y) unchanged?

• A. Replacing X with X − E[X] (centering)
• B. Replacing X with 2X (scaling by a constant)
• C. Replacing X with −X (sign flip)
• D. Replacing X with X² (squaring)

Answer: (B)

Scaling X by a constant c multiplies Cov(X,Y) by c, which changes the covariance. But correlation ρ = Cov(X,Y)/(σ_X σ_Y) is unchanged because σ_{cX} = |c|·σ_X, so the c cancels in the ratio. This is why correlation is scale-invariant and preferred for comparing associations across differently-scaled variables. Note: centering (option A) has no effect on covariance at all since Cov(X − μ, Y) = Cov(X,Y); option C flips the sign of both; option D changes the nature of the relationship entirely.

Question 3

If X and Y are independent random variables, then their covariance must equal zero.

• T. True
• F. False

Answer: True

This direction always holds. Independence means the joint distribution factors as f(x,y) = f_X(x)·f_Y(y), which implies E[XY] = E[X]·E[Y]. Therefore Cov(X,Y) = E[XY] − E[X]E[Y] = 0. The crucial point is that this implication runs only one way: independence implies zero covariance, but zero covariance does not imply independence.

Question 4

A correlation of ρ = 0 between two random variables means there is no relationship between them.

• T. True
• F. False

Answer: False

False — correlation only measures *linear* association. Two variables can have a strong nonlinear relationship (like Y = X²) and still have ρ = 0. The classic counterexample: X ~ Uniform(−1,1) and Y = X² have zero correlation but perfect functional dependence. Saying 'no relationship' conflates 'no linear relationship' with 'no relationship at all,' which is an overstatement that has led to many incorrect conclusions in applied statistics.

Question 5

Why is correlation preferred over raw covariance for measuring association between two random variables?

Model answer: Covariance is scale-dependent: its value changes when the units of measurement change (e.g., switching from meters to centimeters). Correlation normalizes covariance by dividing by both standard deviations, making it unit-free and always bounded between −1 and 1. This allows meaningful comparisons of association strength across different pairs of variables measured in different units.

The normalization ρ = Cov(X,Y)/(σ_X σ_Y) removes units entirely. A covariance of 500 kg·cm doesn't tell you whether the relationship is strong or weak without knowing the scales; a correlation of 0.85 immediately communicates a strong positive linear relationship regardless of units. The extreme values ρ = ±1 correspond to a perfect linear relationship (Y = aX + b almost surely), providing an interpretable upper bound.