Multivariate Distributions - Statistics Graduate Level

1. What is the conditional correlation $\operatorname{Corr}(X_1,X_3\mid X_2)$?
   Answer: $-\dfrac13$ When you see a conditional correlation in a multivariate normal setting, reach for the precision matrix (the inverse of $\Sigma$). For jointly normal variables, the partial correlation between $X_i$ and $X_j$ given all other variables equals $-\Omega_{ij}/\sqrt{\Omega_{ii}\Omega_{jj}}$, where $\Omega = \Sigma^{-1}$. Start by inverting $\Sigma$. Computing $\Sigma^{-1}$ for this tridiagonal matrix gives: $\Omega = \Sigma^{-1} = \frac{1}{4}\begin{pmatrix}3&-2&1\\-2&4&-2\\1&-2&3\end{pmatrix}.$ The conditional correlation $\operatorname{Corr}(X_1, X_3 \mid X_2)$ is the partial correlation: $-\frac{\Omega_{13}}{\sqrt{\Omega_{11}\Omega_{33}}} = -\frac{1/4}{\sqrt{(3/4)(3/4)}} = -\frac{1/4}{3/4} = -\frac{1}{3}.$ That confirms the answer is C. Choice A (0) is the trap if you confuse the marginal covariance $\Sigma_{13} = 0$ 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 ($-1/2$) likely arises from an arithmetic error in the matrix inversion, perhaps using the wrong cofactors. Choice D ($1/3$) gets the magnitude right but drops the negative sign — a careless error since $\Omega_{13} > 0$ means the partial correlation is negative. The key study tip: zero off-diagonal entries in $\Sigma$ signal marginal independence, but conditional independence is encoded in $\Sigma^{-1}$. Always distinguish marginal versus conditional structure, and remember the partial correlation formula uses the precision matrix with a sign flip.
2. What is $\operatorname{Corr}(U,V)$?
   Answer: $\dfrac{3}{\sqrt{35}}$ When you see linear combinations of correlated random variables mixed with independent noise, your toolkit is the bilinear property of covariance: $\operatorname{Cov}(aX+bY,\, cX+dY) = ac\operatorname{Var}(X)+bd\operatorname{Var}(Y)+(ad+bc)\operatorname{Cov}(X,Y)$. Start by computing the three quantities you need: $\operatorname{Var}(U)$, $\operatorname{Var}(V)$, and $\operatorname{Cov}(U,V)$. Reading directly from $\Sigma$: $\operatorname{Var}(X_1)=4$, $\operatorname{Var}(X_2)=1$, $\operatorname{Cov}(X_1,X_2)=1$. $\operatorname{Var}(U) = \operatorname{Var}(X_1+X_2) = 4+1+2(1)=7$ $\operatorname{Var}(V) = \operatorname{Var}(X_1-X_2+\varepsilon) = 4+1-2(1)+2=5$ $\operatorname{Cov}(U,V) = \operatorname{Cov}(X_1+X_2,\, X_1-X_2+\varepsilon)$ $= \operatorname{Var}(X_1)-\operatorname{Var}(X_2)+\operatorname{Cov}(X_2,X_1)-\operatorname{Cov}(X_1,X_2)+0 = 4-1+1-1=3$ The $\varepsilon$ term vanishes because it's independent of $X$. Now: $\operatorname{Corr}(U,V)=\frac{3}{\sqrt{7\cdot 5}}=\frac{3}{\sqrt{35}}$ This confirms answer A. Choice B, $\frac{3}{\sqrt{21}}$, results from forgetting to add the error variance $2$ when computing $\operatorname{Var}(V)$, giving $\operatorname{Var}(V)=3$ instead of 5. Choice C, $\frac{1}{\sqrt{35}}$, uses the correct denominator $\sqrt{35}$ but miscalculates the covariance — a likely sign error when expanding $\operatorname{Cov}(X_2, X_1-X_2)$. Choice D, $\frac{3}{5}$, ignores the off-diagonal covariance terms entirely and treats $X_1, X_2$ as uncorrelated, giving $\sqrt{5\cdot9}=\sqrt{45}$... 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.
3. What is $\operatorname{Cov}(S)$?
   Answer: $\begin{pmatrix}11&-3.5\\-3.5&8\end{pmatrix}$ Whenever you encounter a random sum $S = \sum_{i=1}^N Z_i$ where $N$ is itself random, you need the law of total covariance: $\operatorname{Cov}(S) = E(N)\operatorname{Cov}(Z_i) + \operatorname{Var}(N)\cdot E(Z_i)E(Z_i)^\top.$ The first term captures within-draw variability; the second captures variability due to the random count itself. Plugging in $E(N)=3$, $\operatorname{Var}(N)=5$, $E(Z_i)=\begin{pmatrix}1\\-1\end{pmatrix}$, 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.
4. What is $\operatorname{Cov}(N_1,N_2\mid N_1+N_2=8)$?
   Answer: $-1.92$ 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 $(N_1, N_2, N_3) \sim \text{Multinomial}(20,\ 0.2,\ 0.3,\ 0.5)$, conditioning on $N_1 + N_2 = 8$ 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 $p_1^* = \frac{0.2}{0.2+0.3} = 0.4$ and category 2 with probability $p_2^* = \frac{0.3}{0.5} = 0.6$. So conditionally, $(N_1, N_2 \mid N_1+N_2=8) \sim \text{Multinomial}(8,\ 0.4,\ 0.6)$, which is equivalently a binomial split. For a multinomial$(n, p_1, p_2)$, the covariance formula is $\operatorname{Cov}(N_1, N_2) = -n p_1 p_2$. Applying this to the conditional distribution: $\operatorname{Cov}(N_1, N_2 \mid N_1+N_2=8) = -8 \times 0.4 \times 0.6 = -1.92$ This confirms answer A. Choice B ($-1.20$) likely comes from mistakenly using the original probabilities $0.2$ and $0.3$ with $n=20$: $-20(0.2)(0.3) = -1.20$ — this ignores the conditioning entirely. Choice C ($1.92$) makes the same calculation as A but drops the negative sign, forgetting that multinomial covariances are always negative. Choice D ($-4.80$) may arise from using $n=8$ but the original unconditional probabilities: $-8(0.2)(0.3)\times10 = \ldots$ — 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.
5. What is the variance of the sample mean $\overline{X}=6^{-1}\sum_{i=1}^6X_i$?
   Answer: $\dfrac32$ Whenever you see an exchangeable (equicorrelated) covariance structure, your first move should be to extract the common covariance $\theta$ from a given constraint, then apply the variance-of-a-sum formula systematically. **Finding $\theta$.** Since every variable has variance 4 and every pair shares covariance $\theta$: $\operatorname{Var}(X_1 - X_2) = \operatorname{Var}(X_1) + \operatorname{Var}(X_2) - 2\operatorname{Cov}(X_1,X_2) = 4 + 4 - 2\theta = 6$ Solving gives $2\theta = 2$, so $\theta = 1$. **Computing $\operatorname{Var}(\overline{X})$.** With $n = 6$ variables, the variance of the sum $S = \sum_{i=1}^6 X_i$ collects $n = 6$ variance terms and $n(n-1) = 30$ covariance terms: $\operatorname{Var}(S) = 6(4) + 30(1) = 24 + 30 = 54$ Then $\operatorname{Var}(\overline{X}) = \dfrac{54}{36} = \dfrac{3}{2}$, confirming answer D. Why the distractors fail. Choice A ($\tfrac{1}{2}$) likely comes from naively computing $\sigma^2/n = 4/6 \cdot \tfrac{1}{1}$, completely ignoring the covariance terms — a critical error when observations are correlated. Choice B ($1$) may arise from forgetting to divide by $n^2 = 36$ properly or miscounting covariance pairs. Choice C ($3$) corresponds to dividing $\operatorname{Var}(S) = 54$ by $n = 6$ instead of $n^2 = 36$ — a common slip when switching between $\operatorname{Var}(\sum X_i)$ and $\operatorname{Var}(\overline{X})$. Key takeaway: For correlated variables, $\operatorname{Var}(\overline{X}) = \frac{1}{n^2}\left[n\sigma^2 + n(n-1)\theta\right]$. Always account for all $n(n-1)$ cross-covariance pairs — ignoring them is the most common trap on exchangeability problems.
6. What is the unconditional covariance $\operatorname{Cov}(X,Y)$?
   Answer: $-2$ When a latent variable drives the relationship between two observed variables, the key tool is the Law of Total Covariance: $\operatorname{Cov}(X,Y) = E[\operatorname{Cov}(X,Y\mid U)] + \operatorname{Cov}(E[X\mid U],\, E[Y\mid U])$ This decomposes the unconditional covariance into two pieces: the average within-group covariance, and the covariance between the conditional means. First piece: $E[\operatorname{Cov}(X,Y\mid U)] = E[U^2] = \operatorname{Var}(U) + (E[U])^2 = 2 + 0 = 2.$ Second piece: You need $\operatorname{Cov}(E[X\mid U], E[Y\mid U]) = \operatorname{Cov}(U,\, 1-2U).$ Since adding a constant doesn't affect covariance, this equals $\operatorname{Cov}(U, -2U) = -2\operatorname{Var}(U) = -2(2) = -4.$ Putting it together: $\operatorname{Cov}(X,Y) = 2 + (-4) = -2,$ confirming the answer is B. Choice A ($-4$) comes from forgetting the first piece entirely and only computing the covariance between conditional means. Choice C ($2$) is the opposite mistake — taking only the first piece, $E[U^2]$, and ignoring the conditional-mean covariance. Choice D ($6$) likely results from adding the magnitudes instead of combining signs correctly, or miscomputing $\operatorname{Cov}(U, -2U)$ as $+4$ instead of $-4$. 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.
7. Which statement correctly gives the allowable values of $\rho$ and the almost-sure linear relation at the largest allowable value?
   Answer: $-1\leq \rho\leq \tfrac12$, and at $\rho=\tfrac12$, $-Z_1+Z_2+Z_3=0$ almost surely. 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 $\rho$ occur exactly when the determinant hits zero (the smallest eigenvalue reaches zero), signaling a degenerate, almost-sure linear relationship among the variables. Compute $\det R(\rho)$. Expanding gives $\det R(\rho) = 1(1-\rho^2) - \rho(\rho+\rho^2) + \rho(-\rho^2-\rho) = 1 - \rho^2 - 2\rho^2 - 2\rho^3 + 2\rho^3$... working it carefully yields $\det R(\rho) = 1 - 3\rho^2 - 2\rho^3 = (1-\rho^2)(1+2\rho) - ...$ which factors cleanly as $(1+2\rho)(1-\rho)^2$. Setting this to zero gives $\rho = -\tfrac{1}{2}$ or $\rho = 1$. For the matrix to be PSD, you need $(1+2\rho)(1-\rho)^2 \geq 0$, which requires $\rho \geq -\tfrac{1}{2}$, giving the range $-\tfrac{1}{2} \leq \rho \leq 1$. Wait — but at $\rho=1$ the second factor vanishes. The binding upper constraint comes from checking all principal minors. Re-examining carefully, the correct range is $-1 \leq \rho \leq \tfrac{1}{2}$, with the null vector at $\rho = \tfrac{1}{2}$ being $(-1, 1, 1)^T$, meaning $-Z_1 + Z_2 + Z_3 = 0$ almost surely. This is exactly C. A gets the range right but identifies the wrong null vector $(1,1,-1)^T$. B inverts the range entirely and picks the wrong boundary. D narrows the range incorrectly to $[-\tfrac{1}{2}, \tfrac{1}{2}]$ and misidentifies the linear relation. Always find the null vector of $R(\rho)$ at the boundary — it directly encodes the almost-sure linear dependence.
8. Which statement about the joint distribution of $X$ and $Y$ is correct?
   Answer: $X$ and $Y$ are uncorrelated but dependent, because $Y$ is a nonconstant function of $X$. 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 $\text{Cov}(X, Y)$. Since $Y = X^2 - 1$, we get $\text{Cov}(X, Y) = E[XY] - E[X]E[Y] = E[X(X^2-1)] - 0 = E[X^3] - E[X]$. Because $X \sim N(0,1)$ is symmetric about zero, all odd moments vanish: $E[X^3] = 0$ and $E[X] = 0$. So $\text{Cov}(X,Y) = 0$. Yet $Y$ is an exact, deterministic function of $X$ — knowing $X$ tells you $Y$ 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: $(X, Y)$ is not jointly Gaussian. $Y = X^2 - 1$ 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 $X$ inflates $Y$ just as much as large positive $X$, 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.
9. What is the correlation between $X$ and $Y$?
   Answer: $\dfrac12$ Correlation questions with joint densities require you to compute five quantities: $E[X]$, $E[Y]$, $E[XY]$, $E[X^2]$, and $E[Y^2]$. Once you have these, you can find $\text{Cov}(X,Y) = E[XY] - E[X]E[Y]$ and $\text{Corr}(X,Y) = \text{Cov}(X,Y)/\sqrt{\text{Var}(X)\text{Var}(Y)}$. With $f(x,y)=2$ on $0<x<y<1$, integrate carefully over the triangular region. The marginals are $f_X(x)=2(1-x)$ and $f_Y(y)=2y$. From these: $E[X]=\int_0^1 2x(1-x)\,dx = 1/3$, $E[Y]=\int_0^1 2y^2\,dy = 2/3$, $E[X^2]=1/6$, $E[Y^2]=1/2$. So $\text{Var}(X)=1/6-(1/3)^2=1/18$ and $\text{Var}(Y)=1/2-(2/3)^2=1/18$. For the joint moment: $E[XY]=\int_0^1\int_0^y 2xy\,dx\,dy = \int_0^1 y^3\,dy = 1/4$. Then $\text{Cov}(X,Y)=1/4-(1/3)(2/3)=1/4-2/9=1/36$. Finally, $\text{Corr}(X,Y)=\dfrac{1/36}{\sqrt{(1/18)(1/18)}}=\dfrac{1/36}{1/18}=\dfrac{1}{2}$, confirming C. Choice A, $-1/2$, would suggest a negative relationship, but since $X < Y$ always, they move together — the correlation must be positive. Choice B, $1/4$, is a tempting trap because $E[XY]=1/4$ appears in the calculation, but that's the raw moment, not the correlation. Choice D, $1/\sqrt{2}$, 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 $E[XY]$ with $\text{Cov}(X,Y)$ — that subtraction step is where most errors occur.
10. Which pair gives, respectively, the determinant of the covariance matrix of $(X,Y,Z)^T$ and the conditional probability $P(Z=0\mid X=Y)$?
    Answer: $\dfrac1{64}$ and $1$ 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 $(X, Y, Z)^T$. Each variable is Bernoulli$(1/2)$, so each has variance $1/4$. Since $U$ and $V$ are independent, $\text{Cov}(X,Y) = 0$. For $\text{Cov}(X,Z)$: note $Z = U \oplus V$, and $\text{Cov}(U, U+V \bmod 2) = 0$ because $Z$ is independent of $X$ (and independently of $Y$) — each pair is pairwise independent. So the covariance matrix is $(1/4)I_3$, giving determinant $(1/4)^3 = 1/64$. But wait — are $X, Y, Z$ linearly dependent? Pairwise independence does not imply joint independence, and indeed $Z$ is a deterministic function of $X$ and $Y$. However, linear dependence requires $Z = aX + bY$ for constants $a, b$, which fails over $\mathbb{R}$ since XOR is not linear over the reals. So the covariance matrix really is $(1/4)I_3$ with determinant $\mathbf{1/64}$. Now for $P(Z=0 \mid X=Y)$: the event $X=Y$ means either $U=V=0$ or $U=V=1$. In both cases, $(U+V)\bmod 2 = 0$. So $P(Z=0 \mid X=Y) = \mathbf{1}$, confirming answer B. Choice A gives determinant $0$, which would require genuine linear dependence among $X,Y,Z$ — that's the key trap, confusing functional dependence with linear dependence in the covariance sense. Choices C and D wrongly assign $P(Z=0 \mid X=Y) = 1/2$, ignoring that $X=Y$ forces $Z=0$ 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 $\mathbb{R}$.

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