Multivariate Analysis Questions and Answers

Q: Given three random variables \(X_{1}, X_{2}, X_{3}\) with \(\rho_{12} = 0.6\), \(\rho_{13} = 0.8\) and \(\rho_{23} = 0.24\). Hence the partial correlation coefficient between \(X_{2}\) and \(X_{3}\) after removing the effect of \(X_{1}\) is :

The partial correlation is -0.5.\[ \begin{aligned} r_{23.1} &= \frac{r_{23} - r_{21}r_{31}}{\sqrt{(1 - r_{21}^{2})(1 - r_{31}^{2})}} \\ &= \frac{0.24 - 0.6 \times 0.8}{\sqrt{0.64 \times 0.36}} \\ &= \frac{-0.24}{0.48} = -0.5 \end{aligned} \]Although \(X_{2}\) and \(X_{3}\) are positively correlated overall, most of that comes from their shared dependence on \(X_{1}\). Once \(X_{1}\) is held fixed, the remaining association is negative.\(\pm 0.24\) simply reuse the raw correlation \(r_{23}\) and ignore \(X_{1}\), and +0.5 has the sign of the numerator wrong.

Q: Which of the following techniques is used for the dimension reduction of the data ?

The dimension reduction technique is principal component analysis.PCA replaces \(p\) correlated variables by a few uncorrelated linear combinations, the principal components, which are the eigenvectors of the covariance matrix ordered by eigenvalue (variance explained). Keeping the first \(k\) components summarises the data in \(k \lt p\) dimensions with the least loss of variance. It uses no group labels.Linear and quadratic discriminant analysis are classification methods: they build rules to assign an observation to one of known groups. Fisher's linear discriminant does project data onto a few directions as a by-product, but its purpose is separating labelled groups, and QDA does not reduce dimension at all. PCA is the technique designed for dimension reduction, so 'none of the above' is wrong too.

Q: Suppose\[ \underline{X}_{3 \times 1} \sim \text{MNormal}_{3}\left(\underline{0}, \begin{bmatrix} 2 & 0.5 & 2 \\ 0.5 & 1 & 0 \\ 2 & 0 & 4 \end{bmatrix}\right). \]Then the correlation coefficient between \(X_{1} - X_{3}\) and \(2X_{2}\) is :

The correlation is 1/(2√2).Read the variances and covariances from the matrix: \(Var(X_{1}) = 2\), \(Var(X_{2}) = 1\), \(Var(X_{3}) = 4\), \(Cov(X_{1}, X_{2}) = 0.5\), \(Cov(X_{1}, X_{3}) = 2\), \(Cov(X_{2}, X_{3}) = 0\).\[ \begin{aligned} Cov(X_{1} - X_{3}, 2X_{2}) &= 2(0.5 - 0) = 1 \\ Var(X_{1} - X_{3}) &= 2 + 4 - 2(2) = 2 \\ Var(2X_{2}) &= 4 \end{aligned} \]\[ \rho = \frac{1}{\sqrt{2 \times 4}} = \frac{1}{2\sqrt{2}} \]\(1/\sqrt{2}\) forgets the factor 2 in \(2X_{2}\) when computing the variance, and 0 would require the covariance to vanish.

Q: Suppose \(\begin{bmatrix} X \\ Y \end{bmatrix} \sim\) Bivariate Normal \((\underline{\mu}_{2 \times 1}, \Sigma_{2m})\). Then :

The correct statement is E(Y | X = x) = E(Y) + a linear function of x.For a bivariate normal,\[ \begin{aligned} E(Y \mid X = x) &= \mu_{Y} \\ &\quad + \rho\frac{\sigma_{Y}}{\sigma_{X}}(x - \mu_{X}) \end{aligned} \]which is E(Y) plus a linear function of x. The conditional variance is \(\sigma_{Y}^{2}(1 - \rho^{2})\), a constant that does not depend on x at all.So the options making the conditional variance depend on x (through \(x^{2}\) or a quadratic) are wrong, and \(x\,E(Y)\) is not the form of the regression function.

Q: Suppose \(\underline{X}_{3\times 1}\sim\text{MNormal}_{3}(\underline{\mu},\Sigma_{3+3})\) and let \(Y = \underline{X}'A\underline{X}\) where the matrix A has all eigenvalue equal to 1. Hence the probability distribution of Y

The distribution cannot be determined from the given information.If A is symmetric with all eigenvalues 1, then A = I and \(Y = \underline{X}'\underline{X}\). Its distribution depends on \(\underline{\mu}\) and \(\Sigma\), which are not given. Y is \(\chi^{2}_{3}\) only when \(\underline{\mu} = \underline{0}\) and \(\Sigma = I\); with a non-zero mean it is non-central chi-square, and with a general \(\Sigma\) it is a weighted sum of chi-squares.A quadratic form in normal variables is never normal, which rules out option B, and chi-square with 1 degree of freedom would need A of rank 1. So no specific distribution follows.

Q: Corresponding to the three variables X1, X2, X3 the quadrate form \(X_{1}^{2} + 2X_{2}^{2} + 4X_{3}^{2} - X_{1}X_{2} + X_{1}X_{3}\) is :

The quadratic form is positive definite.Its symmetric matrix is\[ A = \begin{pmatrix} 1 & -\frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & 2 & 0 \\ \frac{1}{2} & 0 & 4 \end{pmatrix} \]The leading principal minors are 1, \(2 - \frac{1}{4} = \frac{7}{4}\), and\[ \det A = 8 - 1 - \tfrac{1}{2} = \tfrac{13}{2} \]All three are positive, so by Sylvester's criterion the form is positive definite. Positive semidefinite would allow a zero minor, and the negative cases need the minors to alternate in sign.

Q: For the three random variables X1, X2, X3 let \(\hat{X}_{1}\) and \(\hat{X}_{2}\) be the fitted values of X1 and X2 respectively after regressing on the linear function of X3. The partial correlation coefficient between X1 and X2 after removing the effect of X3 is the simple correlation coefficient between :

The partial correlation is the correlation between \(X_{1} - \hat{X}_{1}\) and \(X_{2} - \hat{X}_{2}\).Removing the effect of \(X_{3}\) means regressing \(X_{1}\) and \(X_{2}\) on \(X_{3}\) and keeping the residuals \(X_{1} - \hat{X}_{1}\) and \(X_{2} - \hat{X}_{2}\), the parts not explained by \(X_{3}\). The partial correlation \(r_{12.3}\) is the simple correlation between these residuals.Correlating \(X_{1}\) with \(\hat{X}_{2}\), or \(\hat{X}_{1}\) with \(X_{2}\), mixes in the part explained by \(X_{3}\). Subtracting \(X_{3}\) itself, as in \(X_{1} - X_{3}\), removes \(X_{3}\) with coefficient 1 instead of the fitted regression coefficient.

Q: The Euclidean distance between the point \(\underline{X}_{3=1} = \begin{bmatrix} 2 \\ -2 \\ 2 \end{bmatrix}\) and a cluster center \(\bar{X}_{3=1} = \begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix}\) is equal to :

The distance is 5.The difference of the two vectors is \((2 - 2, -2 - 1, 2 - (-2)) = (0, -3, 4)\), so\[ d = \sqrt{0^{2} + (-3)^{2} + 4^{2}} = \sqrt{25} = 5 \]4 is just the largest coordinate difference, 2 adds the differences with their signs, and 0 would mean the point is the cluster centre.

Q: Principal component analysis is used for :

Principal component analysis is used for reducing the number of variables in the data.PCA replaces the original correlated variables by uncorrelated linear combinations, the principal components, ordered by the variance they explain. Keeping the first few components retains most of the total variability with far fewer variables, which is dimension reduction.PCA keeps all the observations. It aims to preserve the variability, not to reduce it. Expressing one variable as a function of others is regression, not PCA.

Q: \[ \underline{X}_{2 \times 1} \sim MN_{2}\left(\begin{bmatrix} \mu_{1} \\ \mu_{2} \end{bmatrix}, \begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix}\right), \]then \(\mathrm{var}(X_{1} + X_{2} \mid X_{1} - X_{2} = Y)\) is equal to :

The conditional variance is \(2(1 + \rho)\).Let \(U = X_{1} + X_{2}\) and \(V = X_{1} - X_{2}\). They are jointly normal with\[ \begin{aligned} \mathrm{Cov}(U, V) &= \mathrm{Var}(X_{1}) - \mathrm{Var}(X_{2}) = 0 \\ \mathrm{Var}(U) &= 1 + 1 + 2\rho = 2(1 + \rho) \end{aligned} \]Uncorrelated jointly normal variables are independent, so conditioning on V does not change the distribution of U, and \(\mathrm{var}(U \mid V) = 2(1 + \rho)\).\(1 + \rho\) is half the answer, 2 ignores the covariance term, and \(\rho(1 + \rho)\) does not come from any variance formula here.

31.

Given three random variables \(X_{1}, X_{2}, X_{3}\) with \(\rho_{12} = 0.6\), \(\rho_{13} = 0.8\) and \(\rho_{23} = 0.24\). Hence the partial correlation coefficient between \(X_{2}\) and \(X_{3}\) after removing the effect of \(X_{1}\) is :

A.-0.24
B.+0.24
C.-0.5
D.+0.5

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The partial correlation is -0.5.

\[ \begin{aligned} r_{23.1} &= \frac{r_{23} - r_{21}r_{31}}{\sqrt{(1 - r_{21}^{2})(1 - r_{31}^{2})}} \\ &= \frac{0.24 - 0.6 \times 0.8}{\sqrt{0.64 \times 0.36}} \\ &= \frac{-0.24}{0.48} = -0.5 \end{aligned} \]

Although \(X_{2}\) and \(X_{3}\) are positively correlated overall, most of that comes from their shared dependence on \(X_{1}\). Once \(X_{1}\) is held fixed, the remaining association is negative.

\(\pm 0.24\) simply reuse the raw correlation \(r_{23}\) and ignore \(X_{1}\), and +0.5 has the sign of the numerator wrong.

Discuss in Forum

32.

Which of the following techniques is used for the dimension reduction of the data ?

A.Linear discriminant analysis
B.Quadratic discriminant analysis
C.Principal component analysis
D.None of the above

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The dimension reduction technique is principal component analysis.

PCA replaces \(p\) correlated variables by a few uncorrelated linear combinations, the principal components, which are the eigenvectors of the covariance matrix ordered by eigenvalue (variance explained). Keeping the first \(k\) components summarises the data in \(k \lt p\) dimensions with the least loss of variance. It uses no group labels.

Linear and quadratic discriminant analysis are classification methods: they build rules to assign an observation to one of known groups. Fisher's linear discriminant does project data onto a few directions as a by-product, but its purpose is separating labelled groups, and QDA does not reduce dimension at all. PCA is the technique designed for dimension reduction, so 'none of the above' is wrong too.

Discuss in Forum

33.

Suppose

\[ \underline{X}_{3 \times 1} \sim \text{MNormal}_{3}\left(\underline{0}, \begin{bmatrix} 2 & 0.5 & 2 \\ 0.5 & 1 & 0 \\ 2 & 0 & 4 \end{bmatrix}\right). \]

Then the correlation coefficient between \(X_{1} - X_{3}\) and \(2X_{2}\) is :

A.\(\dfrac{1}{2\sqrt{2}}\)
B.0
C.\(\dfrac{1}{\sqrt{2}}\)
D.1

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The correlation is 1/(2√2).

Read the variances and covariances from the matrix: \(Var(X_{1}) = 2\), \(Var(X_{2}) = 1\), \(Var(X_{3}) = 4\), \(Cov(X_{1}, X_{2}) = 0.5\), \(Cov(X_{1}, X_{3}) = 2\), \(Cov(X_{2}, X_{3}) = 0\).

\[ \begin{aligned} Cov(X_{1} - X_{3}, 2X_{2}) &= 2(0.5 - 0) = 1 \\ Var(X_{1} - X_{3}) &= 2 + 4 - 2(2) = 2 \\ Var(2X_{2}) &= 4 \end{aligned} \]

\[ \rho = \frac{1}{\sqrt{2 \times 4}} = \frac{1}{2\sqrt{2}} \]

\(1/\sqrt{2}\) forgets the factor 2 in \(2X_{2}\) when computing the variance, and 0 would require the covariance to vanish.

Discuss in Forum

34.

Suppose \(\begin{bmatrix} X \\ Y \end{bmatrix} \sim\) Bivariate Normal \((\underline{\mu}_{2 \times 1}, \Sigma_{2m})\). Then :

A.\(E(Y \mid X = x) = x\, E(Y)\)
B.\(Var(Y \mid X = x) = x^{2}\, Var(Y)\)
C.\(E(Y \mid X = x) = E(Y)\) + a linear function of x
D.\(Var(Y \mid X = x) = Var(Y)\) + a quadrate function of x

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The correct statement is E(Y | X = x) = E(Y) + a linear function of x.

For a bivariate normal,

\[ \begin{aligned} E(Y \mid X = x) &= \mu_{Y} \\ &\quad + \rho\frac{\sigma_{Y}}{\sigma_{X}}(x - \mu_{X}) \end{aligned} \]

which is E(Y) plus a linear function of x. The conditional variance is \(\sigma_{Y}^{2}(1 - \rho^{2})\), a constant that does not depend on x at all.

So the options making the conditional variance depend on x (through \(x^{2}\) or a quadratic) are wrong, and \(x\,E(Y)\) is not the form of the regression function.

Discuss in Forum

35.

Suppose \(\underline{X}_{3\times 1}\sim\text{MNormal}_{3}(\underline{\mu},\Sigma_{3+3})\) and let \(Y = \underline{X}'A\underline{X}\) where the matrix A has all eigenvalue equal to 1. Hence the probability distribution of Y

A.cannot be determined based on the given information
B.is normal \((\underline{\mu}'A\underline{\mu}, \Sigma'A\varepsilon)\)
C.is \(\chi^{2}\) with 1 degree of freedom
D.is \(\chi^{2}\) with 3 degrees of freedom

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The distribution cannot be determined from the given information.

If A is symmetric with all eigenvalues 1, then A = I and \(Y = \underline{X}'\underline{X}\). Its distribution depends on \(\underline{\mu}\) and \(\Sigma\), which are not given. Y is \(\chi^{2}_{3}\) only when \(\underline{\mu} = \underline{0}\) and \(\Sigma = I\); with a non-zero mean it is non-central chi-square, and with a general \(\Sigma\) it is a weighted sum of chi-squares.

A quadratic form in normal variables is never normal, which rules out option B, and chi-square with 1 degree of freedom would need A of rank 1. So no specific distribution follows.

Discuss in Forum

36.

Corresponding to the three variables X₁, X₂, X₃ the quadrate form \(X_{1}^{2} + 2X_{2}^{2} + 4X_{3}^{2} - X_{1}X_{2} + X_{1}X_{3}\) is :

A.Positive definite
B.Positive semidefinite
C.Negative definite
D.Negative semidefinite

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The quadratic form is positive definite.

Its symmetric matrix is

\[ A = \begin{pmatrix} 1 & -\frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & 2 & 0 \\ \frac{1}{2} & 0 & 4 \end{pmatrix} \]

The leading principal minors are 1, \(2 - \frac{1}{4} = \frac{7}{4}\), and

\[ \det A = 8 - 1 - \tfrac{1}{2} = \tfrac{13}{2} \]

All three are positive, so by Sylvester's criterion the form is positive definite. Positive semidefinite would allow a zero minor, and the negative cases need the minors to alternate in sign.

Discuss in Forum

37.

For the three random variables X₁, X₂, X₃ let \(\hat{X}_{1}\) and \(\hat{X}_{2}\) be the fitted values of X₁ and X₂ respectively after regressing on the linear function of X₃. The partial correlation coefficient between X₁ and X₂ after removing the effect of X₃ is the simple correlation coefficient between :

A.\(X_{1}\) and \(\hat{X}_{2}\)
B.\(\hat{X}_{1}\) and \(X_{2}\)
C.\((X_{1} - X_{3})\) and \((X_{2} - X_{3})\)
D.\((X_{1} - \hat{X}_{1})\) and \((X_{2} - \hat{X}_{2})\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The partial correlation is the correlation between \(X_{1} - \hat{X}_{1}\) and \(X_{2} - \hat{X}_{2}\).

Removing the effect of \(X_{3}\) means regressing \(X_{1}\) and \(X_{2}\) on \(X_{3}\) and keeping the residuals \(X_{1} - \hat{X}_{1}\) and \(X_{2} - \hat{X}_{2}\), the parts not explained by \(X_{3}\). The partial correlation \(r_{12.3}\) is the simple correlation between these residuals.

Correlating \(X_{1}\) with \(\hat{X}_{2}\), or \(\hat{X}_{1}\) with \(X_{2}\), mixes in the part explained by \(X_{3}\). Subtracting \(X_{3}\) itself, as in \(X_{1} - X_{3}\), removes \(X_{3}\) with coefficient 1 instead of the fitted regression coefficient.

Discuss in Forum

38.

The Euclidean distance between the point \(\underline{X}_{3=1} = \begin{bmatrix} 2 \\ -2 \\ 2 \end{bmatrix}\) and a cluster center \(\bar{X}_{3=1} = \begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix}\) is equal to :

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The distance is 5.

The difference of the two vectors is \((2 - 2, -2 - 1, 2 - (-2)) = (0, -3, 4)\), so

\[ d = \sqrt{0^{2} + (-3)^{2} + 4^{2}} = \sqrt{25} = 5 \]

4 is just the largest coordinate difference, 2 adds the differences with their signs, and 0 would mean the point is the cluster centre.

Discuss in Forum

39.

Principal component analysis is used for :

A.reducing the number of observations in the data
B.reducing the variability present in the data
C.getting information on variable as a function of other variables
D.reducing the number of variables in the data

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

Principal component analysis is used for reducing the number of variables in the data.

PCA replaces the original correlated variables by uncorrelated linear combinations, the principal components, ordered by the variance they explain. Keeping the first few components retains most of the total variability with far fewer variables, which is dimension reduction.

PCA keeps all the observations. It aims to preserve the variability, not to reduce it. Expressing one variable as a function of others is regression, not PCA.

Discuss in Forum

40.

\[ \underline{X}_{2 \times 1} \sim MN_{2}\left(\begin{bmatrix} \mu_{1} \\ \mu_{2} \end{bmatrix}, \begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix}\right), \]

then \(\mathrm{var}(X_{1} + X_{2} \mid X_{1} - X_{2} = Y)\) is equal to :

A.\(1 + \rho\)
B.\(\rho(1 + \rho)\)
C.2
D.\(2(1 + \rho)\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The conditional variance is \(2(1 + \rho)\).

Let \(U = X_{1} + X_{2}\) and \(V = X_{1} - X_{2}\). They are jointly normal with

\[ \begin{aligned} \mathrm{Cov}(U, V) &= \mathrm{Var}(X_{1}) - \mathrm{Var}(X_{2}) = 0 \\ \mathrm{Var}(U) &= 1 + 1 + 2\rho = 2(1 + \rho) \end{aligned} \]

Uncorrelated jointly normal variables are independent, so conditioning on V does not change the distribution of U, and \(\mathrm{var}(U \mid V) = 2(1 + \rho)\).

\(1 + \rho\) is half the answer, 2 ignores the covariance term, and \(\rho(1 + \rho)\) does not come from any variance formula here.

Multivariate Analysis Questions and Answers

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