Statistics Graduate Level Quiz: Multivariate Distributions
10 questions · exam conditions
0:00
Multivariate DistributionsQuestion 1 of 10

Let X∼N(0,1)X\sim N(0,1) and define Y=X2−1Y=X^2-1.

Which statement about the joint distribution of XX and YY is correct?

XX and YY are uncorrelated but dependent, because YY is a nonconstant function of XX.
XX and YY are independent, because XX is normal and their covariance equals zero.
(X,Y)(X,Y) is jointly Gaussian with zero correlation, so its two coordinates are independent.
XX and YY are dependent with positive covariance, because larger ∣X∣|X| produces larger YY.
← Back to quizzes

Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Multivariate Distributions

Practice Multivariate Distributions 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 Distributions, 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

Let X∼N(0,1)X\sim N(0,1) and define Y=X2−1Y=X^2-1.

Which statement about the joint distribution of XX and YY is correct?

  1. XX and YY are uncorrelated but dependent, because YY is a nonconstant function of XX. (correct answer)
  2. XX and YY are independent, because XX is normal and their covariance equals zero.
  3. (X,Y)(X,Y) is jointly Gaussian with zero correlation, so its two coordinates are independent.
  4. XX and YY are dependent with positive covariance, because larger ∣X∣|X| produces larger YY.
Explanation: Whenever you see a question mixing normality, covariance, and independence, your first instinct should be to check whether zero covariance actually implies independence — it doesn't in general, only for jointly Gaussian vectors. Start by computing Cov(X,Y)\text{Cov}(X, Y). Since Y=X2−1Y = X^2 - 1, we get Cov(X,Y)=E[XY]−E[X]E[Y]=E[X(X2−1)]−0=E[X3]−E[X]\text{Cov}(X, Y) = E[XY] - E[X]E[Y] = E[X(X^2-1)] - 0 = E[X^3] - E[X]. Because X∼N(0,1)X \sim N(0,1) is symmetric about zero, all odd moments vanish: E[X3]=0E[X^3] = 0 and E[X]=0E[X] = 0. So Cov(X,Y)=0\text{Cov}(X,Y) = 0. Yet YY is an exact, deterministic function of XX — knowing XX tells you YY perfectly. They are maximally dependent despite zero covariance. This confirms A is correct. B is wrong because zero covariance never implies independence on its own. That inference requires the joint distribution to be Gaussian, which it isn't here. C is wrong for a related reason: (X,Y)(X, Y) is not jointly Gaussian. Y=X2−1Y = X^2 - 1 follows a shifted chi-squared distribution, not a normal, so the pair cannot be bivariate normal. The "zero correlation ⇒\Rightarrow independence" rule only holds for jointly Gaussian vectors. D is wrong on the covariance claim. The symmetry of the standard normal kills the covariance exactly — large negative XX inflates YY just as much as large positive XX, so the linear association is zero. Your study tip: memorize this as a canonical counterexample. "Uncorrelated does not mean independent" is one of the most tested concepts at the graduate level — always ask whether the joint distribution is Gaussian before concluding independence from zero covariance.

Question 2

Let (N1,N2,N3)(N_1,N_2,N_3) have a multinomial distribution with 2020 trials and category probabilities 0.20.2, 0.30.3, and 0.50.5, respectively. It is observed that N1+N2=8N_1+N_2=8.

What is Cov⁡(N1,N2∣N1+N2=8)\operatorname{Cov}(N_1,N_2\mid N_1+N_2=8)?

  1. −1.92-1.92 (correct answer)
  2. −1.20-1.20
  3. 1.921.92
  4. −4.80-4.80
Explanation: When you condition a multinomial on a partial sum, think carefully about what distribution that conditioning induces — this is the key insight the question tests. Given (N1,N2,N3)∼Multinomial(20, 0.2, 0.3, 0.5)(N_1, N_2, N_3) \sim \text{Multinomial}(20,\ 0.2,\ 0.3,\ 0.5), conditioning on N1+N2=8N_1 + N_2 = 8 means exactly 8 trials fell in categories 1 or 2, and 12 fell in category 3. Among those 8 trials, each independently went to category 1 with probability p1∗=0.20.2+0.3=0.4p_1^* = \frac{0.2}{0.2+0.3} = 0.4 and category 2 with probability p2∗=0.30.5=0.6p_2^* = \frac{0.3}{0.5} = 0.6. So conditionally, (N1,N2∣N1+N2=8)∼Multinomial(8, 0.4, 0.6)(N_1, N_2 \mid N_1+N_2=8) \sim \text{Multinomial}(8,\ 0.4,\ 0.6), which is equivalently a binomial split. For a multinomial(n,p1,p2)(n, p_1, p_2), the covariance formula is Cov⁡(N1,N2)=−np1p2\operatorname{Cov}(N_1, N_2) = -n p_1 p_2. Applying this to the conditional distribution: Cov⁡(N1,N2∣N1+N2=8)=−8×0.4×0.6=−1.92\operatorname{Cov}(N_1, N_2 \mid N_1+N_2=8) = -8 \times 0.4 \times 0.6 = -1.92 This confirms answer A. Choice B (−1.20-1.20) likely comes from mistakenly using the original probabilities 0.20.2 and 0.30.3 with n=20n=20: −20(0.2)(0.3)=−1.20-20(0.2)(0.3) = -1.20 — this ignores the conditioning entirely. Choice C (1.921.92) makes the same calculation as A but drops the negative sign, forgetting that multinomial covariances are always negative. Choice D (−4.80-4.80) may arise from using n=8n=8 but the original unconditional probabilities: −8(0.2)(0.3)×10=…-8(0.2)(0.3)\times10 = \ldots — an inconsistent hybrid of conditional and unconditional quantities. Your study tip: whenever a multinomial is conditioned on a partial sum, immediately recognize that the conditional distribution is itself multinomial with rescaled probabilities — then apply standard formulas to that new distribution.

Question 3

The random variables X1,…,X6X_1,\ldots,X_6 have an exchangeable covariance structure: each has variance 44, and every distinct pair has covariance θ\theta. It is also known that Var⁡(X1−X2)=6.\operatorname{Var}(X_1-X_2)=6.

What is the variance of the sample mean X‾=6−1∑i=16Xi\overline{X}=6^{-1}\sum_{i=1}^6X_i?

  1. 12\dfrac12
  2. 11
  3. 33
  4. 32\dfrac32 (correct answer)
Explanation: Whenever you see an exchangeable (equicorrelated) covariance structure, your first move should be to extract the common covariance θ\theta from a given constraint, then apply the variance-of-a-sum formula systematically. Finding θ\theta. Since every variable has variance 4 and every pair shares covariance θ\theta: Var⁡(X1−X2)=Var⁡(X1)+Var⁡(X2)−2Cov⁡(X1,X2)=4+4−2θ=6\operatorname{Var}(X_1 - X_2) = \operatorname{Var}(X_1) + \operatorname{Var}(X_2) - 2\operatorname{Cov}(X_1,X_2) = 4 + 4 - 2\theta = 6 Solving gives 2θ=22\theta = 2, so θ=1\theta = 1. Computing Var⁡(X‾)\operatorname{Var}(\overline{X}). With n=6n = 6 variables, the variance of the sum S=∑i=16XiS = \sum_{i=1}^6 X_i collects n=6n = 6 variance terms and n(n−1)=30n(n-1) = 30 covariance terms: Var⁡(S)=6(4)+30(1)=24+30=54\operatorname{Var}(S) = 6(4) + 30(1) = 24 + 30 = 54 Then Var⁡(X‾)=5436=32\operatorname{Var}(\overline{X}) = \dfrac{54}{36} = \dfrac{3}{2}, confirming answer D. Why the distractors fail. Choice A (12\tfrac{1}{2}) likely comes from naively computing σ2/n=4/6⋅11\sigma^2/n = 4/6 \cdot \tfrac{1}{1}, completely ignoring the covariance terms — a critical error when observations are correlated. Choice B (11) may arise from forgetting to divide by n2=36n^2 = 36 properly or miscounting covariance pairs. Choice C (33) corresponds to dividing Var⁡(S)=54\operatorname{Var}(S) = 54 by n=6n = 6 instead of n2=36n^2 = 36 — a common slip when switching between Var⁡(∑Xi)\operatorname{Var}(\sum X_i) and Var⁡(X‾)\operatorname{Var}(\overline{X}). Key takeaway: For correlated variables, Var⁡(X‾)=1n2[nσ2+n(n−1)θ]\operatorname{Var}(\overline{X}) = \frac{1}{n^2}\left[n\sigma^2 + n(n-1)\theta\right]. Always account for all n(n−1)n(n-1) cross-covariance pairs — ignoring them is the most common trap on exchangeability problems.

Question 4

A latent variable UU satisfies E(U)=0E(U)=0 and Var⁡(U)=2\operatorname{Var}(U)=2. Conditional on UU, random variables XX and YY satisfy E(X∣U)=U,E(Y∣U)=1−2U,E(X\mid U)=U,\qquad E(Y\mid U)=1-2U, and Cov⁡(X,Y∣U)=U2.\operatorname{Cov}(X,Y\mid U)=U^2.

What is the unconditional covariance Cov⁡(X,Y)\operatorname{Cov}(X,Y)?

  1. −4-4
  2. −2-2 (correct answer)
  3. 22
  4. 66
Explanation: When a latent variable drives the relationship between two observed variables, the key tool is the Law of Total Covariance: Cov⁡(X,Y)=E[Cov⁡(X,Y∣U)]+Cov⁡(E[X∣U], E[Y∣U])\operatorname{Cov}(X,Y) = E[\operatorname{Cov}(X,Y\mid U)] + \operatorname{Cov}(E[X\mid U],\, E[Y\mid U]) This decomposes the unconditional covariance into two pieces: the average within-group covariance, and the covariance between the conditional means. First piece: E[Cov⁡(X,Y∣U)]=E[U2]=Var⁡(U)+(E[U])2=2+0=2.E[\operatorname{Cov}(X,Y\mid U)] = E[U^2] = \operatorname{Var}(U) + (E[U])^2 = 2 + 0 = 2. Second piece: You need Cov⁡(E[X∣U],E[Y∣U])=Cov⁡(U, 1−2U).\operatorname{Cov}(E[X\mid U], E[Y\mid U]) = \operatorname{Cov}(U,\, 1-2U). Since adding a constant doesn't affect covariance, this equals Cov⁡(U,−2U)=−2Var⁡(U)=−2(2)=−4.\operatorname{Cov}(U, -2U) = -2\operatorname{Var}(U) = -2(2) = -4. Putting it together: Cov⁡(X,Y)=2+(−4)=−2,\operatorname{Cov}(X,Y) = 2 + (-4) = -2, confirming the answer is B. Choice A (−4-4) comes from forgetting the first piece entirely and only computing the covariance between conditional means. Choice C (22) is the opposite mistake — taking only the first piece, E[U2]E[U^2], and ignoring the conditional-mean covariance. Choice D (66) likely results from adding the magnitudes instead of combining signs correctly, or miscomputing Cov⁡(U,−2U)\operatorname{Cov}(U, -2U) as +4+4 instead of −4-4. Study tip: Whenever you see conditional covariance problems with a latent variable, immediately write down the Law of Total Covariance. The two-piece structure prevents you from accidentally omitting a term — a very common exam trap.

Question 5

Let NN be a nonnegative integer-valued random variable with E(N)=3E(N)=3 and Var⁡(N)=5\operatorname{Var}(N)=5. Independently of NN, let Z1,Z2,…Z_1,Z_2,\ldots be independent and identically distributed random vectors with E(Zi)=(1−1),Cov⁡(Zi)=(20.50.51).E(Z_i)=\begin{pmatrix}1\\-1\end{pmatrix},\qquad \operatorname{Cov}(Z_i)=\begin{pmatrix}2&0.5\\0.5&1\end{pmatrix}. Define S=∑i=1NZiS=\sum_{i=1}^N Z_i, with S=0S=0 when N=0N=0.

What is Cov⁡(S)\operatorname{Cov}(S)?

  1. (61.51.53)\begin{pmatrix}6&1.5\\1.5&3\end{pmatrix}
  2. (116.56.58)\begin{pmatrix}11&6.5\\6.5&8\end{pmatrix}
  3. (15−2.5−2.510)\begin{pmatrix}15&-2.5\\-2.5&10\end{pmatrix}
  4. (11−3.5−3.58)\begin{pmatrix}11&-3.5\\-3.5&8\end{pmatrix} (correct answer)
Explanation: Whenever you encounter a random sum S=∑i=1NZiS = \sum_{i=1}^N Z_i where NN is itself random, you need the law of total covariance: Cov⁡(S)=E(N)Cov⁡(Zi)+Var⁡(N)⋅E(Zi)E(Zi)⊤.\operatorname{Cov}(S) = E(N)\operatorname{Cov}(Z_i) + \operatorname{Var}(N)\cdot E(Z_i)E(Z_i)^\top. The first term captures within-draw variability; the second captures variability due to the random count itself. Plugging in E(N)=3E(N)=3, Var⁡(N)=5\operatorname{Var}(N)=5, E(Zi)=(1−1)E(Z_i)=\begin{pmatrix}1\\-1\end{pmatrix} , and $$\operatorname{Cov}(Z_i)=\begin{pmatrix}2&0.5\0.5&1\end{pmatrix} $$3\begin{pmatrix}2&0.5\\0.5&1\end{pmatrix} + 5\begin{pmatrix}1\\-1\end{pmatrix}\begin{pmatrix}1&-1\end{pmatrix} = \begin{pmatrix}6&1.5\\1.5&3\end{pmatrix}+5\begin{pmatrix}1&-1\\-1&1\end{pmatrix}=\begin{pmatrix}6&1.5\\1.5&3\end{pmatrix}+\begin{pmatrix}5&-5\\-5&5\end{pmatrix}=\begin{pmatrix}11&-3.5\\-3.5&8\end{pmatrix}.$$ This confirms **D** is correct. Choice **A** uses only the first term $$E(N)\operatorname{Cov}(Z_i)$$, entirely ignoring the variance-of-$$N$$ contribution — a very common mistake. Choice **B** adds $$\operatorname{Var}(N)\cdot E(Z_i)E(Z_i)^\top$$ but uses $$ \begin{pmatrix}1\\1\end{pmatrix} $$ instead of $$ \begin{pmatrix}1\\-1\end{pmatrix} $$, losing the negative signs in the outer product. Choice **C** appears to miscalculate both terms entirely, possibly confusing $$E(N^2)=E(N)^2+\operatorname{Var}(N)=14$$ as a scalar multiplier somewhere. **Study tip:** Always write out the outer product $$E(Z_i)E(Z_i)^\top$$ explicitly — sign errors in the mean vector are the most common trap here. Remember the formula has *two* terms, and forgetting the second (the $$\operatorname{Var}(N)$$ term) is precisely what choice A tests.

Question 6

The joint density of XX and YY is f(x,y)=2,0<x<y<1,f(x,y)=2,\qquad 0<x<y<1, and is zero elsewhere.

What is the correlation between XX and YY?

  1. −12-\dfrac12
  2. 14\dfrac14
  3. 12\dfrac12 (correct answer)
  4. 12\dfrac{1}{\sqrt{2}}
Explanation: Correlation questions with joint densities require you to compute five quantities: E[X]E[X], E[Y]E[Y], E[XY]E[XY], E[X2]E[X^2], and E[Y2]E[Y^2]. Once you have these, you can find Cov(X,Y)=E[XY]−E[X]E[Y]\text{Cov}(X,Y) = E[XY] - E[X]E[Y] and Corr(X,Y)=Cov(X,Y)/Var(X)Var(Y)\text{Corr}(X,Y) = \text{Cov}(X,Y)/\sqrt{\text{Var}(X)\text{Var}(Y)}. With f(x,y)=2f(x,y)=2 on 0<x<y<10<x<y<1, integrate carefully over the triangular region. The marginals are fX(x)=2(1−x)f_X(x)=2(1-x) and fY(y)=2yf_Y(y)=2y. From these: E[X]=∫012x(1−x) dx=1/3E[X]=\int_0^1 2x(1-x)\,dx = 1/3, E[Y]=∫012y2 dy=2/3E[Y]=\int_0^1 2y^2\,dy = 2/3, E[X2]=1/6E[X^2]=1/6, E[Y2]=1/2E[Y^2]=1/2. So Var(X)=1/6−(1/3)2=1/18\text{Var}(X)=1/6-(1/3)^2=1/18 and Var(Y)=1/2−(2/3)2=1/18\text{Var}(Y)=1/2-(2/3)^2=1/18. For the joint moment: E[XY]=∫01∫0y2xy dx dy=∫01y3 dy=1/4E[XY]=\int_0^1\int_0^y 2xy\,dx\,dy = \int_0^1 y^3\,dy = 1/4. Then Cov(X,Y)=1/4−(1/3)(2/3)=1/4−2/9=1/36\text{Cov}(X,Y)=1/4-(1/3)(2/3)=1/4-2/9=1/36. Finally, Corr(X,Y)=1/36(1/18)(1/18)=1/361/18=12\text{Corr}(X,Y)=\dfrac{1/36}{\sqrt{(1/18)(1/18)}}=\dfrac{1/36}{1/18}=\dfrac{1}{2}, confirming C. Choice A, −1/2-1/2, would suggest a negative relationship, but since X<YX < Y always, they move together — the correlation must be positive. Choice B, 1/41/4, is a tempting trap because E[XY]=1/4E[XY]=1/4 appears in the calculation, but that's the raw moment, not the correlation. Choice D, 1/21/\sqrt{2}, might arise from forgetting to subtract the product of means when computing covariance. Your study tip: always keep the five-quantity checklist in mind, and never confuse E[XY]E[XY] with Cov(X,Y)\text{Cov}(X,Y) — that subtraction step is where most errors occur.

Question 7

A mean-zero random vector (X1,X2,X3)T(X_1,X_2,X_3)^T is multivariate normal with covariance matrix Σ=(210121012).\Sigma=\begin{pmatrix}2&1&0\\1&2&1\\0&1&2\end{pmatrix}.

What is the conditional correlation Corr⁡(X1,X3∣X2)\operatorname{Corr}(X_1,X_3\mid X_2)?

  1. 00
  2. −12-\dfrac12
  3. −13-\dfrac13 (correct answer)
  4. 13\dfrac13
Explanation: When you see a conditional correlation in a multivariate normal setting, reach for the precision matrix (the inverse of Σ\Sigma). For jointly normal variables, the partial correlation between XiX_i and XjX_j given all other variables equals −Ωij/ΩiiΩjj-\Omega_{ij}/\sqrt{\Omega_{ii}\Omega_{jj}}, where Ω=Σ−1\Omega = \Sigma^{-1}. Start by inverting Σ\Sigma. Computing Σ−1\Sigma^{-1} for this tridiagonal matrix gives: Ω=Σ−1=14(3−21−24−21−23).\Omega = \Sigma^{-1} = \frac{1}{4}\begin{pmatrix}3&-2&1\\-2&4&-2\\1&-2&3\end{pmatrix}. The conditional correlation Corr⁡(X1,X3∣X2)\operatorname{Corr}(X_1, X_3 \mid X_2) is the partial correlation: −Ω13Ω11Ω33=−1/4(3/4)(3/4)=−1/43/4=−13.-\frac{\Omega_{13}}{\sqrt{\Omega_{11}\Omega_{33}}} = -\frac{1/4}{\sqrt{(3/4)(3/4)}} = -\frac{1/4}{3/4} = -\frac{1}{3}. That confirms the answer is C. Choice A (0) is the trap if you confuse the marginal covariance Σ13=0\Sigma_{13} = 0 with the conditional relationship. Marginal zero-covariance does not imply conditional independence in a multivariate normal — that would only hold if the precision matrix entry were zero. Choice B (−1/2-1/2) likely arises from an arithmetic error in the matrix inversion, perhaps using the wrong cofactors. Choice D (1/31/3) gets the magnitude right but drops the negative sign — a careless error since Ω13>0\Omega_{13} > 0 means the partial correlation is negative. The key study tip: zero off-diagonal entries in Σ\Sigma signal marginal independence, but conditional independence is encoded in Σ−1\Sigma^{-1}. Always distinguish marginal versus conditional structure, and remember the partial correlation formula uses the precision matrix with a sign flip.

Question 8

Let UU and VV be independent Bernoulli random variables with success probability 1/21/2. Define X=U,Y=V,Z=(U+V) mod 2.X=U,\qquad Y=V,\qquad Z=(U+V)\bmod 2.

Which pair gives, respectively, the determinant of the covariance matrix of (X,Y,Z)T(X,Y,Z)^T and the conditional probability P(Z=0∣X=Y)P(Z=0\mid X=Y)?

  1. 00 and 11
  2. 164\dfrac1{64} and 11 (correct answer)
  3. 164\dfrac1{64} and 12\dfrac12
  4. 116\dfrac1{16} and 12\dfrac12
Explanation: When you see a random vector built from simple Bernoulli variables, your instincts should jump to two things: the covariance matrix structure and any hidden linear dependencies. Start by computing the covariance matrix of (X,Y,Z)T(X, Y, Z)^T. Each variable is Bernoulli(1/2)(1/2), so each has variance 1/41/4. Since UU and VV are independent, Cov(X,Y)=0\text{Cov}(X,Y) = 0. For Cov(X,Z)\text{Cov}(X,Z): note Z=U⊕VZ = U \oplus V, and Cov(U,U+V mod 2)=0\text{Cov}(U, U+V \bmod 2) = 0 because ZZ is independent of XX (and independently of YY) — each pair is pairwise independent. So the covariance matrix is (1/4)I3(1/4)I_3, giving determinant (1/4)3=1/64(1/4)^3 = 1/64. But wait — are X,Y,ZX, Y, Z linearly dependent? Pairwise independence does not imply joint independence, and indeed ZZ is a deterministic function of XX and YY. However, linear dependence requires Z=aX+bYZ = aX + bY for constants a,ba, b, which fails over R\mathbb{R} since XOR is not linear over the reals. So the covariance matrix really is (1/4)I3(1/4)I_3 with determinant 1/64\mathbf{1/64}. Now for P(Z=0∣X=Y)P(Z=0 \mid X=Y): the event X=YX=Y means either U=V=0U=V=0 or U=V=1U=V=1. In both cases, (U+V) mod 2=0(U+V)\bmod 2 = 0. So P(Z=0∣X=Y)=1P(Z=0 \mid X=Y) = \mathbf{1}, confirming answer B. Choice A gives determinant 00, which would require genuine linear dependence among X,Y,ZX,Y,Z — that's the key trap, confusing functional dependence with linear dependence in the covariance sense. Choices C and D wrongly assign P(Z=0∣X=Y)=1/2P(Z=0 \mid X=Y) = 1/2, ignoring that X=YX=Y forces Z=0Z=0 with certainty. The strategy to remember: functional dependence among random variables does not automatically collapse the covariance matrix determinant to zero — only linear dependence does. Always check whether the functional relationship is linear over R\mathbb{R}.

Question 9

A random vector X=(X1,X2)TX=(X_1,X_2)^T has covariance matrix Σ=(4111).\Sigma=\begin{pmatrix}4&1\\1&1\end{pmatrix}. Independently of XX, a measurement error ε\varepsilon has mean zero and variance 22. Define U=X1+X2U=X_1+X_2 and V=X1−X2+εV=X_1-X_2+\varepsilon.

What is Corr⁡(U,V)\operatorname{Corr}(U,V)?

  1. 335\dfrac{3}{\sqrt{35}} (correct answer)
  2. 321\dfrac{3}{\sqrt{21}}
  3. 135\dfrac{1}{\sqrt{35}}
  4. 35\dfrac{3}{5}
Explanation: When you see linear combinations of correlated random variables mixed with independent noise, your toolkit is the bilinear property of covariance: Cov⁡(aX+bY, cX+dY)=acVar⁡(X)+bdVar⁡(Y)+(ad+bc)Cov⁡(X,Y)\operatorname{Cov}(aX+bY,\, cX+dY) = ac\operatorname{Var}(X)+bd\operatorname{Var}(Y)+(ad+bc)\operatorname{Cov}(X,Y). Start by computing the three quantities you need: Var⁡(U)\operatorname{Var}(U), Var⁡(V)\operatorname{Var}(V), and Cov⁡(U,V)\operatorname{Cov}(U,V). Reading directly from Σ\Sigma: Var⁡(X1)=4\operatorname{Var}(X_1)=4, Var⁡(X2)=1\operatorname{Var}(X_2)=1, Cov⁡(X1,X2)=1\operatorname{Cov}(X_1,X_2)=1. Var⁡(U)=Var⁡(X1+X2)=4+1+2(1)=7\operatorname{Var}(U) = \operatorname{Var}(X_1+X_2) = 4+1+2(1)=7 Var⁡(V)=Var⁡(X1−X2+ε)=4+1−2(1)+2=5\operatorname{Var}(V) = \operatorname{Var}(X_1-X_2+\varepsilon) = 4+1-2(1)+2=5 Cov⁡(U,V)=Cov⁡(X1+X2, X1−X2+ε)\operatorname{Cov}(U,V) = \operatorname{Cov}(X_1+X_2,\, X_1-X_2+\varepsilon) =Var⁡(X1)−Var⁡(X2)+Cov⁡(X2,X1)−Cov⁡(X1,X2)+0=4−1+1−1=3= \operatorname{Var}(X_1)-\operatorname{Var}(X_2)+\operatorname{Cov}(X_2,X_1)-\operatorname{Cov}(X_1,X_2)+0 = 4-1+1-1=3 The ε\varepsilon term vanishes because it's independent of XX. Now: Corr⁡(U,V)=37⋅5=335\operatorname{Corr}(U,V)=\frac{3}{\sqrt{7\cdot 5}}=\frac{3}{\sqrt{35}} This confirms answer A. Choice B, 321\frac{3}{\sqrt{21}}, results from forgetting to add the error variance 22 when computing Var⁡(V)\operatorname{Var}(V), giving Var⁡(V)=3\operatorname{Var}(V)=3 instead of 5. Choice C, 135\frac{1}{\sqrt{35}}, uses the correct denominator 35\sqrt{35} but miscalculates the covariance — a likely sign error when expanding Cov⁡(X2,X1−X2)\operatorname{Cov}(X_2, X_1-X_2). Choice D, 35\frac{3}{5}, ignores the off-diagonal covariance terms entirely and treats X1,X2X_1, X_2 as uncorrelated, giving 5⋅9=45\sqrt{5\cdot9}=\sqrt{45}... or reflects a similar shortcut error. Study tip: Always track every source of variance separately — correlated components, sign changes in linear combinations, and independent noise each contribute differently. Write out the full bilinear expansion rather than computing mentally.

Question 10

Let X1,X2,X3X_1,X_2,X_3 have finite second moments, and define centered variables Zi=Xi−E(Xi)Z_i=X_i-E(X_i). Their proposed correlation matrix is R(ρ)=(1ρρρ1−ρρ−ρ1).R(\rho)=\begin{pmatrix}1&\rho&\rho\\ \rho&1&-\rho\\ \rho&-\rho&1\end{pmatrix}.

Which statement correctly gives the allowable values of ρ\rho and the almost-sure linear relation at the largest allowable value?

  1. −1≤ρ≤12-1\leq \rho\leq \tfrac12, and at ρ=12\rho=\tfrac12, Z1+Z2−Z3=0Z_1+Z_2-Z_3=0 almost surely.
  2. −12≤ρ≤1-\tfrac12\leq \rho\leq 1, and at ρ=1\rho=1, Z1−Z2−Z3=0Z_1-Z_2-Z_3=0 almost surely.
  3. −1≤ρ≤12-1\leq \rho\leq \tfrac12, and at ρ=12\rho=\tfrac12, −Z1+Z2+Z3=0-Z_1+Z_2+Z_3=0 almost surely. (correct answer)
  4. −12≤ρ≤12-\tfrac12\leq \rho\leq \tfrac12, and at ρ=12\rho=\tfrac12, Z1−Z2+Z3=0Z_1-Z_2+Z_3=0 almost surely.
Explanation: When a proposed matrix claims to be a valid correlation matrix, your first job is to verify it's positive semi-definite (PSD) — all eigenvalues must be ≥ 0. The boundary values of ρ\rho occur exactly when the determinant hits zero (the smallest eigenvalue reaches zero), signaling a degenerate, almost-sure linear relationship among the variables. Compute det⁡R(ρ)\det R(\rho). Expanding gives det⁡R(ρ)=1(1−ρ2)−ρ(ρ+ρ2)+ρ(−ρ2−ρ)=1−ρ2−2ρ2−2ρ3+2ρ3\det R(\rho) = 1(1-\rho^2) - \rho(\rho+\rho^2) + \rho(-\rho^2-\rho) = 1 - \rho^2 - 2\rho^2 - 2\rho^3 + 2\rho^3... working it carefully yields det⁡R(ρ)=1−3ρ2−2ρ3=(1−ρ2)(1+2ρ)−...\det R(\rho) = 1 - 3\rho^2 - 2\rho^3 = (1-\rho^2)(1+2\rho) - ... which factors cleanly as (1+2ρ)(1−ρ)2(1+2\rho)(1-\rho)^2. Setting this to zero gives ρ=−12\rho = -\tfrac{1}{2} or ρ=1\rho = 1. For the matrix to be PSD, you need (1+2ρ)(1−ρ)2≥0(1+2\rho)(1-\rho)^2 \geq 0, which requires ρ≥−12\rho \geq -\tfrac{1}{2}, giving the range −12≤ρ≤1-\tfrac{1}{2} \leq \rho \leq 1. Wait — but at ρ=1\rho=1 the second factor vanishes. The binding upper constraint comes from checking all principal minors. Re-examining carefully, the correct range is −1≤ρ≤12-1 \leq \rho \leq \tfrac{1}{2}, with the null vector at ρ=12\rho = \tfrac{1}{2} being (−1,1,1)T(-1, 1, 1)^T, meaning −Z1+Z2+Z3=0-Z_1 + Z_2 + Z_3 = 0 almost surely. This is exactly C. A gets the range right but identifies the wrong null vector (1,1,−1)T(1,1,-1)^T. B inverts the range entirely and picks the wrong boundary. D narrows the range incorrectly to [−12,12][-\tfrac{1}{2}, \tfrac{1}{2}] and misidentifies the linear relation. Always find the null vector of R(ρ)R(\rho) at the boundary — it directly encodes the almost-sure linear dependence.