Multivariate Analysis Questions and Answers

Q: Suppose\[ \underset{\sim}{X}_{3 \times 1} = \begin{pmatrix} X_{1} \\ X_{2} \\ X_{3} \end{pmatrix} \text{ Multivariate} \]\[ \text{Normal}_{3}\left(\begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 & 1 & 0.5 \\ 1 & 4 & 1 \\ 0.5 & 1 & 1 \end{bmatrix}\right). \]Hence the probability distribution of \(\begin{pmatrix} X_{1} \\ X_{2} \end{pmatrix}\) given \(X_{3} = 2\).

The conditional distribution is \(\text{Normal}_{2}\Bigg(\begin{bmatrix} 0.5 \\ -1 \end{bmatrix},\) \(\begin{bmatrix} 1.75 & 0.5 \\ 0.5 & 3 \end{bmatrix}\Bigg)\).Partition with \(\Sigma_{12,3} = (0.5, 1)'\) and \(\sigma_{33} = 1\). The conditional mean is\[ \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0.5 \\ 1 \end{bmatrix}(2 - 3) = \begin{bmatrix} 0.5 \\ -1 \end{bmatrix} \]and the conditional covariance is\[ \begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix} - \begin{bmatrix} 0.25 & 0.5 \\ 0.5 & 1 \end{bmatrix} = \begin{bmatrix} 1.75 & 0.5 \\ 0.5 & 3 \end{bmatrix} \]Option B has the right covariance but adds the correction to the mean instead of subtracting it. Option A has a covariance matrix that was not reduced. The distribution is determined, so D is wrong.

Q: Suppose we have n iid observation from multivariate \(\text{Normal}_{P}(\underline{\mu}, \Sigma)\). The likelihood ratio test to test \(H_{0} : \underline{\mu} = \underline{\mu}_{0}\) against \(H_{1} : \underline{\mu} \neq \underline{\mu}_{0}\) uses :

The likelihood ratio test uses Hotelling's \(T^{2}\) statistic, which follows an F distribution.With \(\Sigma\) unknown, the likelihood ratio for \(H_{0}: \underline{\mu} = \underline{\mu}_{0}\) is a monotone function of\[ T^{2} = n(\bar{x} - \mu_{0})' S^{-1} (\bar{x} - \mu_{0}) \]and \(\frac{n - p}{p(n - 1)}T^{2}\) has an F distribution with p and \(n - p\) degrees of freedom under \(H_{0}\).The t statistic is the one-dimensional special case, \(T^{2} = t^{2}\). A \(\chi^{2}\) statistic arises only when \(\Sigma\) is known, which the question does not give.

Q: Suppose \(\underset{\sim}{X}_{3 \times 1} \sim\) Multivariate \(\text{Normal}_{3}(\underset{\sim}{\mu}\ \Sigma)\) and\[ Y = X_{1}^{2} - X_{2}^{2} + X_{3}^{2} + 0.5X_{1}X_{3}. \]Then the probability distribution of y is :

The distribution of Y is not \(\chi^{2}\).A quadratic form \(X'AX\) of a normal vector has a chi-square distribution only under strict conditions (for \(X \sim N(0, \Sigma)\), \(A\Sigma\) must be idempotent). Here Y has the matrix\[ A = \begin{bmatrix} 1 & 0 & 0.25 \\ 0 & -1 & 0 \\ 0.25 & 0 & 1 \end{bmatrix} \]which has a negative eigenvalue (\(-1\), from the \(-X_{2}^{2}\) term). Y can then take negative values with positive probability, which no chi-square variable can, and nothing about the unspecified \(\mu\) and \(\Sigma\) could rescue that.So none of the chi-square options, with 3, 2 or 1 degrees of freedom, can be right.

Q: For three random variables \(X_{1}\), \(X_{2}\), \(X_{3}\) if \(\rho_{12} = 0.6\) and \(\rho_{13} = 0.8\), then \(\rho_{23}\) :

\(\rho_{23}\) has to be in [0, 0.96].The correlation matrix of \(X_{1}, X_{2}, X_{3}\) must be positive semidefinite. Its determinant condition gives the exact range\[ \begin{aligned} \rho_{23} &\in \rho_{12}\rho_{13} \pm \sqrt{(1 - \rho_{12}^{2})(1 - \rho_{13}^{2})} \\ &= 0.48 \pm \sqrt{0.64 \times 0.36} = 0.48 \pm 0.48 \end{aligned} \]so \(0 \le \rho_{23} \le 0.96\).\(\rho_{23}\) cannot be negative, which rules out option A, and [−0.96, 0.96] describes a wider range that includes impossible negative values, so it is not the range \(\rho_{23}\) is confined to. Values above 0.96 are impossible, so 'any value in [0, 1]' is wrong too.

Q: The Mahalanobis distance between the two Bivariate Normal Population\[ N_{2}\left(\begin{bmatrix} 1 \\ -1 \end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 1 & 4 \end{bmatrix}\right) \text{ and} \]\[ N_{2}\left(\begin{bmatrix} 2 \\ -2 \end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 1 & 4 \end{bmatrix}\right) \]is :

The (squared) Mahalanobis distance is 7/3.The difference of means is \(\delta = (2 - 1, -2 + 1)' = (1, -1)'\), and the common covariance matrix has\[ \Sigma^{-1} = \frac{1}{3}\begin{bmatrix} 4 & -1 \\ -1 & 1 \end{bmatrix} \](its determinant is \(4 - 1 = 3\)). Then\[ \begin{aligned} \delta'\Sigma^{-1}\delta &= \frac{1}{3}\left[4(1)^{2} - 2(1)(-1) + (-1)^{2}\right] \\ &= \frac{7}{3} \end{aligned} \]This \(\Delta^{2}\) is what textbooks call the Mahalanobis distance between the two populations.Getting the sign of the cross term wrong gives \(\frac{4 - 2 + 1}{3} = 1\), and \(\frac{1}{3}\), \(\frac{10}{3}\) and \(\frac{13}{3}\) do not come from \(\delta'\Sigma^{-1}\delta\) with this \(\delta\) and \(\Sigma\).

Q: Given n observations on p variables each, it is decided to carry out the principal component analysis such that the proportion of variation explained by k principal components is at most 0.7.

The usual rule is that k is the smallest value for which k principal components explain at least 70% of total variation.Principal components are ordered by decreasing variance (eigenvalue). The cumulative proportion \(\frac{\lambda_{1} + \ldots + \lambda_{k}}{\lambda_{1} + \ldots + \lambda_{p}}\) increases with k, and the standard cut-off rule retains the fewest components whose cumulative proportion reaches the chosen level, here 0.7.The number of components depends on the eigenvalues, not on a fixed fraction of p or of n, so options A and B are wrong. Option D is not a sensible rule: 'the largest' k reaching 70% would simply be p, which defeats the purpose of reducing dimension.

41.

Suppose

\[ \underset{\sim}{X}_{3 \times 1} = \begin{pmatrix} X_{1} \\ X_{2} \\ X_{3} \end{pmatrix} \text{ Multivariate} \]

\[ \text{Normal}_{3}\left(\begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 & 1 & 0.5 \\ 1 & 4 & 1 \\ 0.5 & 1 & 1 \end{bmatrix}\right). \]

Hence the probability distribution of \(\begin{pmatrix} X_{1} \\ X_{2} \end{pmatrix}\) given \(X_{3} = 2\).

A.is \(\text{Normal}_{2}\Bigg(\begin{bmatrix} 0.5 \\ 1 \end{bmatrix},\) \(\begin{bmatrix} 2 & 0.5 \\ 0.5 & 3 \end{bmatrix}\Bigg)\)
B.is \(\text{Normal}_{2}\Bigg(\begin{bmatrix} 1.5 \\ 1 \end{bmatrix},\) \(\begin{bmatrix} 1.75 & 0.5 \\ 0.5 & 3 \end{bmatrix}\Bigg)\)
C.is \(\text{Normal}_{2}\Bigg(\begin{bmatrix} 0.5 \\ -1 \end{bmatrix},\) \(\begin{bmatrix} 1.75 & 0.5 \\ 0.5 & 3 \end{bmatrix}\Bigg)\)
D.cannot be determined

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The conditional distribution is \(\text{Normal}_{2}\Bigg(\begin{bmatrix} 0.5 \\ -1 \end{bmatrix},\) \(\begin{bmatrix} 1.75 & 0.5 \\ 0.5 & 3 \end{bmatrix}\Bigg)\).

Partition with \(\Sigma_{12,3} = (0.5, 1)'\) and \(\sigma_{33} = 1\). The conditional mean is

\[ \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0.5 \\ 1 \end{bmatrix}(2 - 3) = \begin{bmatrix} 0.5 \\ -1 \end{bmatrix} \]

and the conditional covariance is

\[ \begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix} - \begin{bmatrix} 0.25 & 0.5 \\ 0.5 & 1 \end{bmatrix} = \begin{bmatrix} 1.75 & 0.5 \\ 0.5 & 3 \end{bmatrix} \]

Option B has the right covariance but adds the correction to the mean instead of subtracting it. Option A has a covariance matrix that was not reduced. The distribution is determined, so D is wrong.

Discuss in Forum

42.

Suppose we have n iid observation from multivariate \(\text{Normal}_{P}(\underline{\mu}, \Sigma)\). The likelihood ratio test to test \(H_{0} : \underline{\mu} = \underline{\mu}_{0}\) against \(H_{1} : \underline{\mu} \neq \underline{\mu}_{0}\) uses :

A.The t statistic which follows t distribution
B.The statistic which follows \(\chi^{2}\) distribution
C.Hotelling \(T^{2}\) statistic which follows F distribution
D.None of the above

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The likelihood ratio test uses Hotelling's \(T^{2}\) statistic, which follows an F distribution.

With \(\Sigma\) unknown, the likelihood ratio for \(H_{0}: \underline{\mu} = \underline{\mu}_{0}\) is a monotone function of

\[ T^{2} = n(\bar{x} - \mu_{0})' S^{-1} (\bar{x} - \mu_{0}) \]

and \(\frac{n - p}{p(n - 1)}T^{2}\) has an F distribution with p and \(n - p\) degrees of freedom under \(H_{0}\).

The t statistic is the one-dimensional special case, \(T^{2} = t^{2}\). A \(\chi^{2}\) statistic arises only when \(\Sigma\) is known, which the question does not give.

Discuss in Forum

43.

Suppose \(\underset{\sim}{X}_{3 \times 1} \sim\) Multivariate \(\text{Normal}_{3}(\underset{\sim}{\mu}\ \Sigma)\) and

\[ Y = X_{1}^{2} - X_{2}^{2} + X_{3}^{2} + 0.5X_{1}X_{3}. \]

Then the probability distribution of y is :

A.not \(\chi^{2}\)
B.\(\chi^{2}\) with 3 degrees of freedom
C.\(\chi^{2}\) with 2 degrees of freedom
D.\(\chi^{2}\) with 1 degree of freedom

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The distribution of Y is not \(\chi^{2}\).

A quadratic form \(X'AX\) of a normal vector has a chi-square distribution only under strict conditions (for \(X \sim N(0, \Sigma)\), \(A\Sigma\) must be idempotent). Here Y has the matrix

\[ A = \begin{bmatrix} 1 & 0 & 0.25 \\ 0 & -1 & 0 \\ 0.25 & 0 & 1 \end{bmatrix} \]

which has a negative eigenvalue (\(-1\), from the \(-X_{2}^{2}\) term). Y can then take negative values with positive probability, which no chi-square variable can, and nothing about the unspecified \(\mu\) and \(\Sigma\) could rescue that.

So none of the chi-square options, with 3, 2 or 1 degrees of freedom, can be right.

Discuss in Forum

44.

For three random variables \(X_{1}\), \(X_{2}\), \(X_{3}\) if \(\rho_{12} = 0.6\) and \(\rho_{13} = 0.8\), then \(\rho_{23}\) :

A.can be \(\lt 0\)
B.can be any value in [0, 1]
C.has to be in [−0.96, 0.96]
D.has to be in [0, 0.96]

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

\(\rho_{23}\) has to be in [0, 0.96].

The correlation matrix of \(X_{1}, X_{2}, X_{3}\) must be positive semidefinite. Its determinant condition gives the exact range

\[ \begin{aligned} \rho_{23} &\in \rho_{12}\rho_{13} \pm \sqrt{(1 - \rho_{12}^{2})(1 - \rho_{13}^{2})} \\ &= 0.48 \pm \sqrt{0.64 \times 0.36} = 0.48 \pm 0.48 \end{aligned} \]

so \(0 \le \rho_{23} \le 0.96\).

\(\rho_{23}\) cannot be negative, which rules out option A, and [−0.96, 0.96] describes a wider range that includes impossible negative values, so it is not the range \(\rho_{23}\) is confined to. Values above 0.96 are impossible, so 'any value in [0, 1]' is wrong too.

Discuss in Forum

45.

The Mahalanobis distance between the two Bivariate Normal Population

\[ N_{2}\left(\begin{bmatrix} 1 \\ -1 \end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 1 & 4 \end{bmatrix}\right) \text{ and} \]

\[ N_{2}\left(\begin{bmatrix} 2 \\ -2 \end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 1 & 4 \end{bmatrix}\right) \]

is :

A.1/3
B.7/3
C.10/3
D.13/3

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The (squared) Mahalanobis distance is 7/3.

The difference of means is \(\delta = (2 - 1, -2 + 1)' = (1, -1)'\), and the common covariance matrix has

\[ \Sigma^{-1} = \frac{1}{3}\begin{bmatrix} 4 & -1 \\ -1 & 1 \end{bmatrix} \]

(its determinant is \(4 - 1 = 3\)). Then

\[ \begin{aligned} \delta'\Sigma^{-1}\delta &= \frac{1}{3}\left[4(1)^{2} - 2(1)(-1) + (-1)^{2}\right] \\ &= \frac{7}{3} \end{aligned} \]

This \(\Delta^{2}\) is what textbooks call the Mahalanobis distance between the two populations.

Getting the sign of the cross term wrong gives \(\frac{4 - 2 + 1}{3} = 1\), and \(\frac{1}{3}\), \(\frac{10}{3}\) and \(\frac{13}{3}\) do not come from \(\delta'\Sigma^{-1}\delta\) with this \(\delta\) and \(\Sigma\).

Discuss in Forum

46.

Given n observations on p variables each, it is decided to carry out the principal component analysis such that the proportion of variation explained by k principal components is at most 0.7.

A.k is 70% of p
B.k is 70% of n
C.k is the smallest value for which k principal components explain at least 70% of total variation
D.k is the largest value for which at least k principal components explain at least 70% of total variation

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The usual rule is that k is the smallest value for which k principal components explain at least 70% of total variation.

Principal components are ordered by decreasing variance (eigenvalue). The cumulative proportion \(\frac{\lambda_{1} + \ldots + \lambda_{k}}{\lambda_{1} + \ldots + \lambda_{p}}\) increases with k, and the standard cut-off rule retains the fewest components whose cumulative proportion reaches the chosen level, here 0.7.

The number of components depends on the eigenvalues, not on a fixed fraction of p or of n, so options A and B are wrong. Option D is not a sensible rule: 'the largest' k reaching 70% would simply be p, which defeats the purpose of reducing dimension.

Multivariate Analysis Questions and Answers

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