Two random variables X₁ and X₂ are each marginally normally distributed, and Cov(X₁, X₂) = 0. Does this guarantee that (X₁, X₂) is jointly multivariate normal?
AYes — normal marginals with zero covariance implies independence, which implies joint normality
BNo — marginal normality and zero covariance are necessary but not sufficient; a joint distribution is MVN only if every linear combination a₁X₁ + a₂X₂ is univariate normal
CYes — the covariance matrix fully determines the joint distribution for any pair of normal variables
DNo — but this only matters in the singular case where the covariance matrix is not invertible
This is the central misconception about the MVN. Marginal normality does not imply joint normality, even with zero covariance. A counterexample: let X₁ ~ N(0,1), and let X₂ = X₁ · S where S is ±1 with equal probability independent of X₁. Then X₂ is marginally normal, Cov(X₁, X₂) = 0, but X₁ + X₂ = X₁(1 + S) is not normally distributed (it's a mixture). The defining property of the MVN is that *every* linear combination is normal — this is strictly stronger than normal marginals.
Question 2 Multiple Choice
If X ~ N(μ, Σ) is a k-dimensional multivariate normal random vector and A is an m×k matrix, what is the distribution of Y = AX?
AY ~ N(Aμ, AΣ) — linear transformations scale the covariance matrix by A on one side
BY ~ N(Aμ, AΣAᵀ) — the MVN is closed under linear transformations, with covariance transformed by the congruence AΣAᵀ
CY is approximately normal for large k but not exactly normal in general
DY is normal only if A is square and invertible
Closure under linear transformations is one of the defining properties of the MVN. If X ~ N(μ, Σ) and A is any matrix, then AX ~ N(Aμ, AΣAᵀ). The mean transforms as Aμ (linearity of expectation), and the covariance transforms as AΣAᵀ (standard covariance propagation for linear functions). This result follows immediately from the characteristic function: φ_{AX}(t) = φ_X(Aᵀt) = exp(itᵀAμ − ½tᵀAΣAᵀt), which is the characteristic function of N(Aμ, AΣAᵀ). A need not be square or invertible.
Question 3 True / False
For the multivariate normal distribution, zero covariance between two components implies statistical independence — a property that does not hold for distributions in general.
TTrue
FFalse
Answer: True
In general distributions, zero covariance (uncorrelatedness) does not imply independence — covariance only captures linear relationships. But for the MVN, the characteristic function factorizes whenever the off-diagonal blocks of Σ are zero: φ_{X₁,X₂}(t₁,t₂) = φ_{X₁}(t₁)φ_{X₂}(t₂), which is the condition for independence. This is a special property of the normal distribution arising from the fact that the MVN is entirely determined by its first two moments (mean and covariance). For MVN variables, correlation = 0 is both necessary and sufficient for independence.
Question 4 True / False
A joint distribution is multivariate normal if and mainly if most of its marginal distributions are univariate normal.
TTrue
FFalse
Answer: False
Marginal normality is necessary but not sufficient for joint MVN. The correct characterization is that a joint distribution is MVN if and only if every linear combination of its components is univariate normal. This condition is strictly stronger: it rules out distributions that have normal margins but non-normal joint structure (such as the counterexample with X₂ = X₁ · S). If you only check marginals, you may classify a non-MVN distribution as MVN and make incorrect inferences about conditional distributions and independence.
Question 5 Short Answer
Why is 'every linear combination a'X is univariate normal' a more useful definition of the multivariate normal than the density formula, especially for proving properties of the distribution?
Think about your answer, then reveal below.
Model answer: The linear combination definition directly implies all the key closure properties. Closure under linear transformations follows immediately: if every linear combination of X is normal, then every linear combination of AX = A(linear combination of X) is also normal, so AX is MVN. Marginality is a special case: a marginal is obtained by setting some coefficients to zero, which is a linear combination. Conditional distributions are obtainable via algebraic argument from the same property. The density formula requires the matrix to be invertible and does not extend to degenerate cases, while the linear combination definition works even when Σ is singular.
Definitions that are close to the property you want to exploit make proofs efficient. The linear combination definition is essentially a statement about the behavior of the distribution under projection, which is exactly what closure under linear maps requires. It also clarifies why the MVN is the natural multivariate extension of the univariate normal: it is normal 'in every direction.'