Statistics Graduate Level Quiz: Multivariate Normal Distribution
10 questions · exam conditions
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Multivariate Normal DistributionQuestion 1 of 10

Suppose X=(X1,X2,X3)TX=(X_1,X_2,X_3)^{\mathsf T} is multivariate normal with mean μ=(1,−1,0)T\mu=(1,-1,0)^{\mathsf T} and covariance matrix Σ=(213123336).\Sigma=\begin{pmatrix}2&1&3\\1&2&3\\3&3&6\end{pmatrix}. Which statement is correct?

P(X3=X1−X2)=1P(X_3=X_1-X_2)=1, and the distribution is supported on that affine plane.
P(X3=X1+X2)=0P(X_3=X_1+X_2)=0, because each component has a continuous marginal distribution.
P(X3=X1+X2)=1P(X_3=X_1+X_2)=1, and the joint distribution has no density with respect to three-dimensional Lebesgue measure.
P(X3=X1+X2)=1/2P(X_3=X_1+X_2)=1/2, because the centered linear constraint is symmetric about zero.
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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Multivariate Normal Distribution

Practice Multivariate Normal Distribution in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Multivariate Normal Distribution, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Suppose X=(X1,X2,X3)TX=(X_1,X_2,X_3)^{\mathsf T} is multivariate normal with mean μ=(1,−1,0)T\mu=(1,-1,0)^{\mathsf T} and covariance matrix Σ=(213123336).\Sigma=\begin{pmatrix}2&1&3\\1&2&3\\3&3&6\end{pmatrix}. Which statement is correct?

  1. P(X3=X1−X2)=1P(X_3=X_1-X_2)=1, and the distribution is supported on that affine plane.
  2. P(X3=X1+X2)=0P(X_3=X_1+X_2)=0, because each component has a continuous marginal distribution.
  3. P(X3=X1+X2)=1P(X_3=X_1+X_2)=1, and the joint distribution has no density with respect to three-dimensional Lebesgue measure. (correct answer)
  4. P(X3=X1+X2)=1/2P(X_3=X_1+X_2)=1/2, because the centered linear constraint is symmetric about zero.
Explanation: When a multivariate normal vector has a singular covariance matrix, its distribution collapses onto a lower-dimensional affine subspace — this is the core concept being tested here. Start by checking whether Σ\Sigma is singular. Notice that column 3 of Σ\Sigma equals column 1 plus column 2: (3,3,6)T=(2,1,3)T+(1,2,3)T(3,3,6)^T = (2,1,3)^T + (1,2,3)^T. This means det⁡(Σ)=0\det(\Sigma) = 0, so Σ\Sigma is singular. Now find what linear constraint this imposes. Since Cov(X3−X1−X2)=Var(X3)+Var(X1)+Var(X2)−2Cov(X1,X3)−2Cov(X2,X3)+2Cov(X1,X2)=6+2+2−6−6+2=0\text{Cov}(X_3 - X_1 - X_2) = \text{Var}(X_3) + \text{Var}(X_1) + \text{Var}(X_2) - 2\text{Cov}(X_1,X_3) - 2\text{Cov}(X_2,X_3) + 2\text{Cov}(X_1,X_2) = 6 + 2 + 2 - 6 - 6 + 2 = 0, the random variable X3−X1−X2X_3 - X_1 - X_2 is degenerate (zero variance). Its mean is E[X3−X1−X2]=0−1−(−1)=0E[X_3 - X_1 - X_2] = 0 - 1 - (-1) = 0. A normal random variable with mean 0 and variance 0 equals 0 almost surely, so P(X3=X1+X2)=1P(X_3 = X_1 + X_2) = 1. Because the distribution is confined to a proper affine subspace of R3\mathbb{R}^3, it has no density with respect to 3D Lebesgue measure. This confirms C is correct. A is wrong because it misidentifies the constraint as X3=X1−X2X_3 = X_1 - X_2; computing E[X1−X2−X3]=1−(−1)−0=2≠0E[X_1 - X_2 - X_3] = 1-(-1)-0 = 2 \neq 0, so that relation fails. B is wrong because continuous marginals don't prevent joint degeneracy — each XiX_i individually can be continuous while the joint distribution is singular. D is a fabricated distractor; probabilities of exact linear constraints for jointly normal variables are either 0 or 1, never 1/2. Study tip: Whenever you see a multivariate normal question, immediately check if Σ\Sigma is singular by looking for linearly dependent rows/columns. If it is, identify the null-space direction — that tells you the exact almost-sure linear constraint the vector satisfies.

Question 2

Let UU and VV be independent Uniform⁡(0,1)\operatorname{Uniform}(0,1) variables, define W=(U+V) mod 1W=(U+V)\bmod 1, and set X1=Φ−1(U)X_1=\Phi^{-1}(U), X2=Φ−1(V)X_2=\Phi^{-1}(V), and X3=Φ−1(W)X_3=\Phi^{-1}(W), where Φ\Phi is the standard normal distribution function. Which conclusion is valid?

  1. Every pair is independent bivariate standard normal, but the full vector is not trivariate normal because X3X_3 is a nonlinear function of (X1,X2)(X_1,X_2). (correct answer)
  2. Every pair is bivariate standard normal, so the full vector must be trivariate normal with identity covariance matrix.
  3. Each component is standard normal, but the pair (X1,X3)(X_1,X_3) is not jointly normal because of the modulo operation.
  4. The vector is a singular trivariate normal because a deterministic relation among the components forces the covariance matrix to have rank less than three.
Explanation: This question tests a subtle but important distinction in multivariate statistics: pairwise normality does not imply joint normality, and joint normality does not follow from marginal normality alone. Start by examining each component. Since U,V∼Uniform(0,1)U, V \sim \text{Uniform}(0,1), applying Φ−1\Phi^{-1} to each yields X1,X2,X3∼N(0,1)X_1, X_2, X_3 \sim N(0,1) — that's guaranteed by the probability integral transform. Now, W=(U+V) mod 1W = (U+V) \bmod 1 is itself Uniform(0,1)\text{Uniform}(0,1) (a classical result: the mod-1 sum of independent uniforms is uniform), so X3X_3 is indeed standard normal. For pairwise joint normality: (X1,X2)(X_1, X_2) are independent standard normals — trivially bivariate normal. For (X1,X3)(X_1, X_3) and (X2,X3)(X_2, X_3), one can verify (via characteristic functions or direct argument) that these pairs are also bivariate standard normal, though with nonzero dependence. So A is correct: every pair is bivariate standard normal, yet the full trivariate vector is not jointly normal because X3=Φ−1((U+V) mod 1)X_3 = \Phi^{-1}((U+V) \bmod 1) is a nonlinear, non-affine function of (X1,X2)(X_1, X_2) — and a trivariate normal would require all linear combinations to be normal, which fails here. B commits the classic error of assuming pairwise normality upgrades to full joint normality — it does not. C is partially right (X3X_3 is a nonlinear function) but wrong to say (X1,X3)(X_1, X_3) is not jointly normal — it actually is bivariate normal, which is the deeper subtlety. D is wrong because there's no deterministic linear relation among the components; the covariance matrix is full rank. Study tip: Pairwise bivariate normality is necessary but not sufficient for joint normality — always a reliable trap on graduate-level exams.

Question 3

A zero-mean trivariate normal vector has precision matrix Ω=Σ−1=(2−10−12−10−12).\Omega=\Sigma^{-1}=\begin{pmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{pmatrix}. Which statement correctly describes the relationship between X1X_1 and X3X_3?

  1. They are marginally independent and conditionally correlated given X2X_2, with conditional correlation 1/31/3.
  2. They are both marginally and conditionally independent because the corresponding precision entry is zero.
  3. They are conditionally independent given X2X_2, and also marginally independent because Ω13=0\Omega_{13}=0 implies Σ13=0\Sigma_{13}=0.
  4. They are conditionally independent given X2X_2 and marginally positively correlated, with correlation 1/31/3. (correct answer)
Explanation: Whenever you see a precision matrix (the inverse of the covariance matrix) in a multivariate normal problem, you need to keep two distinct facts sharp in your mind: zero off-diagonal entries in Ω\Omega signal conditional independence, while marginal independence requires zero entries in Σ\Sigma itself — and these are not the same thing. Here, Ω13=0\Omega_{13} = 0 tells you immediately that X1⊥X3∣X2X_1 \perp X_3 \mid X_2 — they are conditionally independent given X2X_2. That confirms the first half of answer D. For the marginal covariance, you must actually invert Ω\Omega. Computing Σ=Ω−1\Sigma = \Omega^{-1}, the (1,3)(1,3) entry works out to Σ13=1/4\Sigma_{13} = 1/4... wait — let's be precise. For this tridiagonal matrix, the inverse gives Σ13=Σ31=14\Sigma_{13} = \Sigma_{31} = \frac{1}{4}, which is strictly positive. The marginal correlation is Σ13Σ11Σ33\frac{\Sigma_{13}}{\sqrt{\Sigma_{11}\Sigma_{33}}}, and since all diagonal entries equal 3/43/4 by symmetry, this gives correlation 1/43/4=1/3>0\frac{1/4}{3/4} = 1/3 > 0. So X1X_1 and X3X_3 are marginally positively correlated, confirming D is correct. A gets the conditional/marginal relationship backwards — they are conditionally independent (not correlated) and marginally correlated (not independent). B is the most dangerous trap: it conflates Ω13=0\Omega_{13} = 0 with both conditional and marginal independence, ignoring that the inverse matrix generally has no zeros where the original does. C makes the same error as B in the second clause — Ω13=0\Omega_{13} = 0 says nothing directly about Σ13\Sigma_{13}. Study tip: Tattoo this onto your brain — zeros in Ω\Omega mean conditional independence; zeros in Σ\Sigma mean marginal independence. You must invert to check the other.

Question 4

Let X=(X1,X2,X3)TX=(X_1,X_2,X_3)^{\mathsf T} be zero-mean multivariate normal with covariance matrix Σ=(420221012).\Sigma=\begin{pmatrix}4&2&0\\2&2&1\\0&1&2\end{pmatrix}. Which residual is independent of (X2,X3)T(X_2,X_3)^{\mathsf T}, and what is its variance?

  1. R=X1−X2R=X_1-X_2 is independent of the predictors and has variance 22.
  2. R=X1−23X2−13X3R=X_1-\frac{2}{3}X_2-\frac{1}{3}X_3 is independent of the predictors and has variance 8/38/3.
  3. R=X1−43X2+23X3R=X_1-\frac{4}{3}X_2+\frac{2}{3}X_3 is independent of the predictors and has variance 4/34/3. (correct answer)
  4. R=X1−43X2−23X3R=X_1-\frac{4}{3}X_2-\frac{2}{3}X_3 is independent of the predictors and has variance 4/34/3.
Explanation: When a question asks which linear combination of predictors leaves a "residual" independent of (X2,X3)T(X_2, X_3)^{\mathsf T}, you should immediately think of the conditional expectation in multivariate normal distributions. For a jointly normal vector, the residual R=X1−βT(X2,X3)TR = X_1 - \boldsymbol{\beta}^{\mathsf T}(X_2, X_3)^{\mathsf T} is independent of (X2,X3)(X_2, X_3) if and only if β\boldsymbol{\beta} equals the regression coefficient vector from projecting X1X_1 onto (X2,X3)(X_2, X_3). To find it, partition Σ\Sigma: you need β=Σ12Σ22−1\boldsymbol{\beta} = \Sigma_{12}\Sigma_{22}^{-1}, where Σ12=(2,0)\Sigma_{12} = (2, 0) and Σ22=(2112)\Sigma_{22} = \begin{pmatrix}2&1\\1&2\end{pmatrix} . The inverse is $$\Sigma_{22}^{-1} = \frac{1}{3}\begin{pmatrix}2&-1\-1&2\end{pmatrix} $$\boldsymbol{\beta} = (2,\,0)\cdot\frac{1}{3}\begin{pmatrix}2&-1\\-1&2\end{pmatrix} = \frac{1}{3}(4,\,-2) = \left(\tfrac{4}{3},\,-\tfrac{2}{3}\right).$$ So the residual is $$R = X_1 - \tfrac{4}{3}X_2 + \tfrac{2}{3}X_3$$, matching **answer C**. Its variance is the conditional variance: $$\text{Var}(R) = \Sigma_{11} - \Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21} = 4 - (2,0)\cdot\tfrac{1}{3}(4,-2)^{\mathsf T} = 4 - \tfrac{8}{3} = \tfrac{4}{3}$$. **A** uses $$\boldsymbol{\beta} = (1, 0)$$, which doesn't satisfy the projection equations — the residual remains correlated with $$(X_2, X_3)$$. **B** gets the signs right conceptually but uses incorrect coefficient values that don't solve $$\Sigma_{22}\boldsymbol{\beta} = \Sigma_{21}$$. **D** has the correct magnitude but the **wrong sign** on $$X_3$$ — a classic trap since off-diagonal signs in $$\Sigma_{22}^{-1}$$ flip relative to $$\Sigma_{22}$$. Always double-check signs when inverting a $$2\times2$$ matrix — that's exactly where these distractors hide.

Question 5

Let X1,…,XnX_1,\ldots,X_n be independent observations from Np(μ,Σ)N_p(\mu,\Sigma), where Σ\Sigma is unknown and positive definite. Define S=(n−1)−1∑i=1n(Xi−Xˉ)(Xi−Xˉ)TS=(n-1)^{-1}\sum_{i=1}^n(X_i-\bar X)(X_i-\bar X)^{\mathsf T} and T2=n(Xˉ−μ)TS−1(Xˉ−μ)T^2=n(\bar X-\mu)^{\mathsf T}S^{-1}(\bar X-\mu). Assuming n>pn>p, which exact null distribution is correct?

  1. n−pp(n−1)T2∼Fp,n−p\frac{n-p}{p(n-1)}T^2\sim F_{p,n-p}. (correct answer)
  2. n−ppnT2∼Fp,n−p+1\frac{n-p}{pn}T^2\sim F_{p,n-p+1}.
  3. T2∼χp2T^2\sim\chi_p^2 because standardization by SS removes the unknown covariance.
  4. n−1p(n−p)T2∼Fn−p,p\frac{n-1}{p(n-p)}T^2\sim F_{n-p,p}.
Explanation: Whenever you encounter Hotelling's T2T^2 statistic in a multivariate normal setting, your goal is to track how it connects to a known pivot distribution through the relationship between the Wishart and F distributions. Start from first principles: n(Xˉ−μ)∼Np(0,Σ)\sqrt{n}(\bar{X}-\mu)\sim N_p(0,\Sigma), and (n−1)S∼Wp(Σ,n−1)(n-1)S\sim W_p(\Sigma, n-1) independently. Hotelling showed that T2=n(Xˉ−μ)TS−1(Xˉ−μ)T^2 = n(\bar{X}-\mu)^\mathsf{T}S^{-1}(\bar{X}-\mu) follows a Tp,n−12T^2_{p,n-1} distribution. The key bridge to the F distribution is the exact result: n−pp(n−1)T2∼Fp, n−p\frac{n-p}{p(n-1)}T^2 \sim F_{p,\,n-p} This is answer A, and it is correct. The degrees of freedom pp in the numerator reflect the dimension of the mean vector, while n−pn-p in the denominator comes from n−1n-1 (Wishart df) minus (p−1)(p-1), a consequence of the multivariate pivot construction. Answer B is wrong on two counts: the scaling uses pnpn instead of p(n−1)p(n-1), and the denominator degrees of freedom n−p+1n-p+1 has no basis in the correct derivation. Answer C is a common and tempting trap. While n(Xˉ−μ)TΣ−1(Xˉ−μ)∼χp2n(\bar{X}-\mu)^\mathsf{T}\Sigma^{-1}(\bar{X}-\mu)\sim\chi^2_p when Σ\Sigma is known, replacing Σ\Sigma with the estimated SS introduces additional variability that prevents a chi-squared result — you need the F distribution to account for this estimation error. Answer D flips the degrees of freedom, writing Fn−p, pF_{n-p,\,p} instead of Fp, n−pF_{p,\,n-p}, which reverses numerator and denominator and gives an entirely different distribution. Study tip: Memorize the exact scaling constant n−pp(n−1)\frac{n-p}{p(n-1)} for Hotelling's T2→Fp,n−pT^2\to F_{p,n-p}, and always remember that substituting an estimated covariance matrix eliminates the chi-squared pivot — that distinction is frequently tested.

Question 6

Let (X,Y)T(X,Y)^{\mathsf T} be standard bivariate normal with correlation 1/21/2. The distribution is truncated by retaining only observations for which Y>0Y>0. Which statement about the conditional distribution of X∣Y>0X\mid Y>0 is correct?

  1. It is a skew-normal distribution with mean 122π\frac{1}{2}\sqrt{\frac{2}{\pi}} and variance 34\frac{3}{4}, because truncation does not affect the variance of XX.
  2. It is generally nonnormal, with mean 122π\frac{1}{2}\sqrt{\frac{2}{\pi}} and variance 1−12π1-\frac{1}{2\pi}. (correct answer)
  3. It is generally nonnormal, with mean 122π\frac{1}{2}\sqrt{\frac{2}{\pi}} and variance 34\frac{3}{4}.
  4. It remains standard normal because conditioning on the sign of YY does not specify its numerical value.
Explanation: When a bivariate normal vector is truncated (not conditioned on an exact value), the marginal distribution of the non-truncated variable becomes a mixture over a range, generally producing a non-normal shape. That's the key conceptual wedge here. Start with the joint structure. Since (X,Y)T(X,Y)^T is standard bivariate normal with correlation ρ=1/2\rho = 1/2, you can write X=12Y+34 ZX = \frac{1}{2}Y + \sqrt{\frac{3}{4}}\,Z where Z⊥YZ \perp Y. Now condition on Y>0Y > 0. The mean becomes E[X∣Y>0]=12E[Y∣Y>0]=12⋅2πE[X \mid Y > 0] = \frac{1}{2}E[Y \mid Y > 0] = \frac{1}{2}\cdot\sqrt{\frac{2}{\pi}}, since the truncated standard normal has mean 2/π\sqrt{2/\pi}. For variance, use the law of total variance: Var(X∣Y>0)=14Var(Y∣Y>0)+34\text{Var}(X \mid Y>0) = \frac{1}{4}\text{Var}(Y\mid Y>0) + \frac{3}{4}. The truncated standard normal has variance 1−2π1 - \frac{2}{\pi}, so the total is 14 ⁣(1−2π)+34=1−12π\frac{1}{4}\!\left(1-\frac{2}{\pi}\right)+\frac{3}{4} = 1 - \frac{1}{2\pi}. This confirms B is correct — and the distribution is non-normal because it's a mixture of normals with different means (indexed by truncated YY). A is wrong on variance: truncation does affect variance through the correlation, reducing it below 1, not locking it at 3/43/4. C gets the mean right but uses 3/43/4 for the variance, forgetting to add 14Var(Y∣Y>0)\frac{1}{4}\text{Var}(Y\mid Y>0)— it's the conditional variance of ZZ's contribution only, not the full answer. D confuses truncation with independence; correlating XX with YY means restricting YY's range genuinely shifts the distribution of XX. Your study tip: always decompose the truncated variable using the bivariate normal regression form, then apply the law of total variance carefully — the truncated component's variance is strictly less than 1, and that reduction propagates.

Question 7

A parameter vector has prior distribution θ∼N2(0,Σ)\theta\sim N_2(0,\Sigma), where Σ=(11/21/22).\Sigma=\begin{pmatrix}1&1/2\\1/2&2\end{pmatrix}. An observation is generated by Y=θ1+θ2+εY=\theta_1+\theta_2+\varepsilon, where ε∼N(0,1)\varepsilon\sim N(0,1) is independent of θ\theta. After observing Y=5Y=5, which posterior mean and covariance are correct?

  1. Mean (3/2, 5/2)T(3/2,\,5/2)^{\mathsf T} and covariance $$ \begin{pmatrix}11/20&-1/4\-1/4&3/4\end{pmatrix} (correct answer)
  2. Mean (3/2, 5/2)T(3/2,\,5/2)^{\mathsf T} and covariance $$ \begin{pmatrix}1&1/2\1/2&2\end{pmatrix}
  3. Mean (15/8, 25/8)T(15/8,\,25/8)^{\mathsf T} and covariance $$ \begin{pmatrix}7/16&-7/16\-7/16&7/16\end{pmatrix}
  4. Mean (1, 2)T(1,\,2)^{\mathsf T} and covariance $$ \begin{pmatrix}4/5&1/10\1/10&6/5\end{pmatrix}
Explanation: Whenever you see a Bayesian updating problem with a multivariate normal prior and a linear observation model, your instinct should be to apply the normal-normal conjugate update formulas directly. The setup: θ∼N2(0,Σ)\theta \sim N_2(0, \Sigma) with Σ=(11/21/22)\Sigma = \begin{pmatrix}1 & 1/2 \\ 1/2 & 2\end{pmatrix} , and Y=c⊤θ+εY = c^\top\theta + \varepsilon where c=(1,1)⊤c = (1,1)^\top and ε∼N(0,1)\varepsilon \sim N(0,1). The marginal variance of YY is c⊤Σc+1=(1+1/2+1/2+2)+1=5c^\top \Sigma c + 1 = (1+1/2+1/2+2)+1 = 5. The cross-covariance is Σc=(3/2,5/2)⊤\Sigma c = (3/2, 5/2)^\top. The posterior mean formula gives μpost=Σc⋅(c⊤Σc+1)−1⋅Y=(3/2,5/2)⊤⋅(5/5)=(3/2,5/2)⊤\mu_{\text{post}} = \Sigma c \cdot (c^\top \Sigma c + 1)^{-1} \cdot Y = (3/2, 5/2)^\top \cdot (5/5) = (3/2, 5/2)^\top. The posterior covariance is Σpost=Σ−Σc(c⊤Σc+1)−1c⊤Σ=Σ−15(3/2,5/2)⊤(3/2,5/2)\Sigma_{\text{post}} = \Sigma - \Sigma c(c^\top \Sigma c+1)^{-1}c^\top\Sigma = \Sigma - \frac{1}{5}(3/2,5/2)^\top(3/2,5/2), which yields $$ \begin{pmatrix}11/20 & -1/4 \ -1/4 & 3/4\end{pmatrix} Choice **B** has the right posterior mean but leaves the covariance unchanged as the prior — it ignores that observing data *shrinks* uncertainty. Choice **C** uses a wrong scaling: dividing $$\Sigma c$$ by $$c^\top\Sigma c$$ alone (forgetting to add the noise variance 1), inflating the update. Choice **D** applies an incorrect formula entirely, producing a covariance that doesn't match any consistent Bayesian update. **Study tip:** Always remember the posterior covariance formula subtracts a rank-one matrix from the prior covariance — if your posterior covariance equals the prior, you made no update at all.

Question 8

Let (X1,X2)T(X_1,X_2)^{\mathsf T} have a nonsingular bivariate normal distribution with arbitrary means, variances σ12\sigma_1^2 and σ22\sigma_2^2, and covariance τ\tau. Define U=X1+X2U=X_1+X_2 and V=X1−X2V=X_1-X_2. Which condition is necessary and sufficient for UU and VV to be independent?

  1. The original variables must be uncorrelated, so the necessary and sufficient condition is τ=0\tau=0.
  2. The original variables must have equal means, so the necessary and sufficient condition is E(X1)=E(X2)E(X_1)=E(X_2).
  3. The original variables must have equal variances, so the necessary and sufficient condition is σ12=σ22\sigma_1^2=\sigma_2^2. (correct answer)
  4. The transformed variables must have equal variances, so the necessary and sufficient condition is τ=(σ22−σ12)/2\tau=(\sigma_2^2-\sigma_1^2)/2.
Explanation: When dealing with linear combinations of jointly normal random variables, the key tool is the covariance criterion: two jointly normal random variables are independent if and only if they are uncorrelated. So your strategy is to compute Cov(U,V)\text{Cov}(U, V) and set it to zero. Since U=X1+X2U = X_1 + X_2 and V=X1−X2V = X_1 - X_2, expand the covariance directly: Cov(U,V)=Cov(X1+X2, X1−X2)=Var(X1)−Var(X2)=σ12−σ22\text{Cov}(U, V) = \text{Cov}(X_1 + X_2,\, X_1 - X_2) = \text{Var}(X_1) - \text{Var}(X_2) = \sigma_1^2 - \sigma_2^2 (The cross-terms involving τ\tau cancel: +τ−τ=0+\tau - \tau = 0.) Setting this to zero gives σ12=σ22\sigma_1^2 = \sigma_2^2, confirming that C is the necessary and sufficient condition — equal variances. A is wrong because τ\tau drops out entirely from Cov(U,V)\text{Cov}(U,V); whether or not X1X_1 and X2X_2 are uncorrelated has no bearing on the independence of UU and VV. This is a tempting distractor because uncorrelatedness is often a key condition, but here the algebra shows it's irrelevant. B is wrong because means have no effect on covariance whatsoever — equal means is neither necessary nor sufficient for anything related to independence in this context. D is wrong because it sets Cov(U,V)=σ12−σ22+something involving τ\text{Cov}(U,V) = \sigma_1^2 - \sigma_2^2 + \text{something involving}\,\tau to zero incorrectly; the actual covariance contains no τ\tau term, so introducing τ\tau into the condition is a miscalculation. Study tip: When checking independence of linear combinations of normals, always compute the covariance explicitly — don't guess which original parameters matter. Let the algebra decide.

Question 9

Let X=(X1,X2,X3)TX=(X_1,X_2,X_3)^{\mathsf T} have a multivariate normal distribution with mean zero and covariance matrix Σ=(210121012).\Sigma=\begin{pmatrix}2&1&0\\1&2&1\\0&1&2\end{pmatrix}. What is the conditional distribution of X2X_2 given X1+X3=2X_1+X_3=2?

  1. N(1,2)N(1,2), because conditioning changes the mean but leaves the marginal variance unchanged.
  2. N(1,1)N(1,1), because the conditioning variable has variance 44 and covariance 22 with X2X_2. (correct answer)
  3. N(2,1)N(2,1), because the regression coefficient of X2X_2 on the conditioning variable is one.
  4. N(1,0)N(1,0), because the value of the sum determines the middle component exactly.
Explanation: When you condition on a linear combination of jointly normal variables, the key tool is the bivariate normal conditional formula. Reframe the problem: let Y=X1+X3Y = X_1 + X_3. Since (X2,Y)(X_2, Y) is jointly normal (linear combinations preserve normality), you just need Var(Y)\text{Var}(Y), Cov(X2,Y)\text{Cov}(X_2, Y), and the conditional mean/variance formulas. First, compute Var(Y)=Var(X1)+Var(X3)+2Cov(X1,X3)=2+2+2(0)=4\text{Var}(Y) = \text{Var}(X_1) + \text{Var}(X_3) + 2\text{Cov}(X_1,X_3) = 2 + 2 + 2(0) = 4. Next, Cov(X2,Y)=Cov(X2,X1)+Cov(X2,X3)=1+1=2\text{Cov}(X_2, Y) = \text{Cov}(X_2, X_1) + \text{Cov}(X_2, X_3) = 1 + 1 = 2. Both means are zero, so the conditional mean is μ2∣Y=0+24(2−0)=1\mu_{2|Y} = 0 + \frac{2}{4}(2 - 0) = 1. The conditional variance is Var(X2)−[Cov(X2,Y)]2Var(Y)=2−44=1\text{Var}(X_2) - \frac{[\text{Cov}(X_2,Y)]^2}{\text{Var}(Y)} = 2 - \frac{4}{4} = 1. This confirms B: N(1,1)N(1,1). Choice A gets the mean right but claims the marginal variance is unchanged — conditioning always reduces (or preserves) variance, never leaves it at the marginal value unless the covariance is zero. Choice C correctly identifies the regression coefficient as 24=12\frac{2}{4} = \frac{1}{2}, not 1, giving mean 12(2)=1\frac{1}{2}(2)=1 — but then misreports the mean as 2 and the reasoning is garbled. Choice D confuses this with a deterministic relationship; X1+X3=2X_1 + X_3 = 2 does not pin down X2X_2 exactly since X2X_2 is not a function of X1+X3X_1 + X_3. Your study tip: always convert "conditional on a linear combination" problems into a two-variable normal problem — compute variance and covariance of the new variable, then apply the standard conditional formulas mechanically.

Question 10

Let X∼N4(μ,I4)X\sim N_4(\mu,I_4) with μ=(1,1,0,0)T\mu=(1,1,0,0)^{\mathsf T}, and define Q=(X1+X2)22+(X3−X4)22.Q=\frac{(X_1+X_2)^2}{2}+\frac{(X_3-X_4)^2}{2}. What is the distribution of QQ?

  1. A noncentral chi-square distribution with 44 degrees of freedom and noncentrality parameter 22.
  2. A noncentral chi-square distribution with 22 degrees of freedom and noncentrality parameter 22. (correct answer)
  3. A noncentral chi-square distribution with 22 degrees of freedom and noncentrality parameter 44.
  4. A central chi-square distribution with 22 degrees of freedom because the two contrasts are orthogonal.
Explanation: When you encounter a quadratic form built from normal random variables, your first move should be to rewrite it as a sum of squared independent standard normals — that's the blueprint for identifying noncentral chi-square distributions. Define two new variables: Y1=X1+X22Y_1 = \frac{X_1 + X_2}{\sqrt{2}} and Y2=X3−X42Y_2 = \frac{X_3 - X_4}{\sqrt{2}}. Since X∼N4(μ,I4)X \sim N_4(\mu, I_4), these are independent normal random variables (the linear combinations involve disjoint, orthogonal coordinate pairs). Their means are E[Y1]=1+12=2E[Y_1] = \frac{1+1}{\sqrt{2}} = \sqrt{2} and E[Y2]=0−02=0E[Y_2] = \frac{0-0}{\sqrt{2}} = 0, and both have variance 1. So Q=Y12+Y22Q = Y_1^2 + Y_2^2, the sum of squares of two independent normals with unit variance. This is exactly a noncentral chi-square with 2 degrees of freedom. The noncentrality parameter is λ=μY12+μY22=(2)2+02=2\lambda = \mu_{Y_1}^2 + \mu_{Y_2}^2 = (\sqrt{2})^2 + 0^2 = 2. The correct answer is B. A is wrong because it claims 4 degrees of freedom — a tempting trap since XX is 4-dimensional, but QQ is built from only 2 independent squared terms. C gets the degrees of freedom right but miscalculates λ\lambda; doubling it to 4 would require μ=(1,1,1,1)T\mu = (1,1,1,1)^T or similar. D is wrong on two counts: orthogonality of the contrasts doesn't erase noncentrality, and QQ is noncentral because Y1Y_1 has nonzero mean. Study tip: Always compute the noncentrality parameter as λ=∑iμZi2\lambda = \sum_i \mu_{Z_i}^2 after standardizing — don't confuse the original μ\mu components with λ\lambda directly.