Multivariate Analysis Questions and Answers

Q: If \(r_{12.3}\) is the correlation coefficient between the variables \(X_{1}\) and \(X_{2}\) after eliminating the linear effect of \(X_{3}\), then which of the following are correct ?(1) \(-1 \le r_{12.3} \le 1\)(2) \(r_{12}^{2} + r_{13}^{2} + r_{23}^{2} - 2r_{12}r_{13}r_{23} \le 1\)(3) \(r_{12.3}^{2} = b_{12.3}\, b_{21.3}\), where b's are partial regression coefficients

All three statements are correct: (1), (2) and (3).(1) \(r_{12.3}\) is an ordinary correlation, between the residuals of \(X_{1}\) and \(X_{2}\) after regressing each on \(X_{3}\), so it lies in [-1, 1].(2) The correlation matrix of \(X_{1}, X_{2}, X_{3}\) is positive semi-definite, so its determinant is non-negative:\[ \begin{aligned} &1 - r_{12}^{2} - r_{13}^{2} - r_{23}^{2} \\ &\quad + 2r_{12}r_{13}r_{23} \ge 0 \end{aligned} \]which rearranges to statement (2).(3) As with simple regression, where \(r^{2} = b_{yx}b_{xy}\), the partial correlation is the geometric mean of the two partial regression coefficients: \(b_{12.3} = r_{12.3}\sigma_{1.3}/\sigma_{2.3}\) and \(b_{21.3} = r_{12.3}\sigma_{2.3}/\sigma_{1.3}\), whose product is \(r_{12.3}^{2}\).Since all three hold, every option naming only two of them is incomplete.

Q: If X be a \(p\)-component random vector with \(E(X) = 0\) and variance covariance matrix \(\Sigma\), positive definite. If X is partitioned into \(X^{(1)}\) of \(p_{1}\) components and \(X^{(2)}\) of \(p_{2}\) components such that \(p_{1} + p_{2} = p\) and \(p_{1} \le p_{2}\). Then the square of canonical correlation are the characteristic roots of the matrix :

The squared canonical correlations are the roots of \(\Sigma_{11}^{-1}\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}\).Canonical correlation analysis maximises the correlation between \(a'X^{(1)}\) and \(b'X^{(2)}\). The first-order conditions lead to the eigenproblem\[ \begin{aligned} &\Sigma_{11}^{-1}\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}\,a \\ &= \rho^{2}a \end{aligned} \]so the squared canonical correlations are its \(p_{1}\) characteristic roots (the smaller set, since \(p_{1} \le p_{2}\)). \(\Sigma\) itself gives principal components. \(\Sigma_{11}^{-1}\Sigma_{12}\) and \(\Sigma_{22}^{-1}\Sigma_{21}\) are regression coefficient matrices, and are not even square in general.

Q: If \(T^{2}\) is Hotelling \(T^{2}\)-statistic, then the distribution of \(\dfrac{T^{2}}{n - 1}\,\dfrac{n - p}{p}\) would be :

The keyed answer is the non-central F-distribution.For \(T^{2} = n(\bar{x} - \mu_{0})'S^{-1}(\bar{x} - \mu_{0})\) from a sample of size \(n\) from \(N_{p}(\mu, \Sigma)\), Anderson's theorem states that \(\dfrac{T^{2}}{n - 1}\dfrac{n - p}{p}\) has a non-central F-distribution with \(p\) and \(n - p\) degrees of freedom and non-centrality \(n(\mu - \mu_{0})'\Sigma^{-1}(\mu - \mu_{0})\).When the null hypothesis \(\mu = \mu_{0}\) holds the non-centrality is zero and the distribution is the central F, which is the special case used to set critical values; the general statement, with no hypothesis assumed, is the non-central one. A chi-square or t-distribution is not the distribution of this ratio.

Q: If \(x_{1}\), \(x_{2}\), ........, \(x_{n}\) \((n \ge p + 1)\) are distributed independently each according to \(N_{p}(\mu, \Sigma)\), then the distribution of\[ \begin{aligned} S = \frac{1}{n - 1}\sum_{\alpha=1}^{n} &(x_{\alpha} - \bar{x}) \\ &(x_{\alpha} - \bar{x})' \end{aligned} \]is :

S has the distribution \(W_{p}(n - 1, \Sigma/(n - 1))\).The matrix of sums of squares and products \(A = \sum(x_{\alpha} - \bar{x})(x_{\alpha} - \bar{x})'\) has the Wishart distribution \(W_{p}(n - 1, \Sigma)\): one degree of freedom is used in estimating the mean. Scaling a Wishart matrix by a constant scales its parameter matrix, so\[ \begin{aligned} S &= \frac{A}{n - 1} \\ &\sim W_{p}\!\left(n - 1, \frac{\Sigma}{n - 1}\right) \end{aligned} \]The condition \(n \ge p + 1\) makes S non-singular. Option B uses \(n\) degrees of freedom, which would be right only if \(\mu\) were known and used instead of \(\bar{x}\). S is a random matrix, so a normal distribution for it (C and D) makes no sense.

Q: If \(X \sim N_{p}(\mu, \Sigma)\), then the distribution of \(N(\bar{X} - \mu)'\Sigma^{-1}(\bar{X} - \mu)\) would be :

The distribution is \(\chi^{2}\) with \(p\) degrees of freedom.The mean \(\bar{X}\) of N independent \(N_{p}(\mu, \Sigma)\) vectors is \(N_{p}(\mu, \Sigma/N)\), so \(Z = \sqrt{N}\,\Sigma^{-1/2}(\bar{X} - \mu)\) is \(N_{p}(0, I)\). Then\[ N(\bar{X} - \mu)'\Sigma^{-1}(\bar{X} - \mu) = Z'Z \]is a sum of \(p\) independent squared standard normals, a \(\chi^{2}_{p}\). It is a scalar, not a multivariate normal vector. \(N - p\) degrees of freedom would arise only with \(\Sigma\) estimated (and then the result is an F, not a chi-square), and N degrees of freedom confuses the sample size with the dimension.

Q: Let X a \(p\)-component random vector is partitioned into \(X^{(1)}\) and \(X^{(2)}\) where \(X^{(1)}\) has \(q\)-components and \(X^{(2)}\) has \((p - q)\) components and \(X \sim N_{p}(\mu, \Sigma)\). If \(\mu\) and \(\Sigma\) are also partitioned as of X, then the variance covariance matrix of the conditional distribution of \(X^{(1)}\) given \(X^{(2)}\) is :

The conditional covariance matrix is \(\Sigma_{11} - \Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}\).For a partitioned multivariate normal, the conditional distribution of \(X^{(1)}\) given \(X^{(2)} = x^{(2)}\) is normal with mean \(\mu^{(1)} + \Sigma_{12}\Sigma_{22}^{-1}(x^{(2)} - \mu^{(2)})\) and covariance \(\Sigma_{11} - \Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}\), the Schur complement of \(\Sigma_{22}\).Conditioning can only reduce variability, so the correction is subtracted, ruling out the two options with a plus sign. The options built from \(\Sigma_{22}\) give the conditional covariance of \(X^{(2)}\) given \(X^{(1)}\), the other way round.

Q: Given \((X, Y) \sim N_{2}(5, 10, 1, 25, \rho)\) and \(P[4 \lt Y \lt 16 | X = 5] = 0.954\), then \(\rho\) is equal to :(It is given that \(\frac{1}{\sqrt{2\pi}}\int_{0}^{2} e^{-\frac{1}{2}z^{2}} dz = 0.477\))

\(\rho\) is \(\pm 0.8\).Here \(\mu_{X} = 5\), \(\mu_{Y} = 10\), \(\sigma_{X}^{2} = 1\), \(\sigma_{Y}^{2} = 25\). Given \(X = x\), Y is normal with mean \(10 + 5\rho(x - 5)\) and variance \(25(1 - \rho^{2})\). At \(x = 5\) the mean is 10, and the interval (4, 16) is \(10 \pm 6\). Using the given value, \(P(|Z| \lt 2) = 2(0.477) = 0.954\), so\[ \begin{aligned} \frac{6}{5\sqrt{1 - \rho^{2}}} &= 2 \\ \sqrt{1 - \rho^{2}} &= 0.6, \quad \rho^{2} = 0.64 \end{aligned} \]Because \(x = 5\) is the mean of X, the sign of \(\rho\) does not move the conditional mean, so the data fix only \(\rho^{2}\): \(\rho = \pm 0.8\). Choosing +0.8 or -0.8 alone claims more than the information allows, and 0.96 is not a solution.

Q: Suppose a random vector \(\underline{X}\) has normal distribution with mean vector \(\underline{\mu}\) and dispersion matrix V. Then which of the following has a chi-square distribution ?

The chi-square variable is \((X - \mu)'V^{-1}(X - \mu)\).If \(X \sim N_{p}(\mu, V)\) with V non-singular, write \(V^{-1} = (V^{-1/2})^{2}\) and set \(Z = V^{-1/2}(X - \mu)\). Then \(Z \sim N_{p}(0, I)\) and\[ \begin{aligned} (X - \mu)'V^{-1}(X - \mu) &= Z'Z \\ &= \sum_{i=1}^{p} Z_{i}^{2} \sim \chi_{p}^{2} \end{aligned} \]Using V instead of \(V^{-1}\) gives \(Z'V^{2}Z\), a weighted sum of chi-squares, which is chi-square only if \(V^{2} = I\). The exponentials are positive transforms of these quadratic forms and are not chi-square at all.

Q: Suppose \(R_{1.23}\) is a multiple correlation coefficient between \(X_{1}\) and \((X_{2}, X_{3})\). Then which of the following statements cannot be true ?

The statement that cannot be true is \(R_{1.23} = -0.4\).The multiple correlation coefficient is the correlation between \(X_{1}\) and its best linear predictor \(\hat{X}_{1}\) from \(X_{2}, X_{3}\), and equivalently the maximum correlation between \(X_{1}\) and any linear combination of \(X_{2}, X_{3}\) (statements A and B, despite the paper's slips in the coefficient names). A maximum of correlations over combinations and their negatives is never negative, so \(0 \le R_{1.23} \le 1\).So 0.4 is a possible value, and \(-0.4\) is impossible.

Q: Let \(X = (X_{1}, X_{2}, X_{3}, X_{4}) \sim N_{4}(\underline{0}, \Sigma)\) where\[ \Sigma = \begin{bmatrix} 1 & \rho & \rho & \rho \\ \rho & 1 & \rho & \rho \\ \rho & \rho & 1 & \rho \\ \rho & \rho & \rho & 1 \end{bmatrix} \]is positive definite.Then which of the following statements is true ?

The true statement is \(X_{1} + X_{2}, X_{2} + X_{3}, X_{3} + X_{4}\) all have the same distribution.Each sum of two different coordinates is normal with mean 0 and variance \(1 + 1 + 2\rho = 2(1 + \rho)\), because every pair has covariance \(\rho\). So all three are \(N(0, 2 + 2\rho)\).The others fail. \(X_{1} - X_{2}\) and \(X_{1} - X_{3}\) are correlated (covariance \(1 - \rho\), correlation 1/2), so their squared ratio is not \(F_{1,1}\). \(X_{1} - X_{3}\) and \(X_{2} - X_{4}\) are independent with variance \(2(1 - \rho)\) each, so the sum of squares over 2 is \((1 - \rho)\chi_{2}^{2}\), equal to \(\chi_{2}^{2}\) only if \(\rho = 0\). And \(\sum X_{i}\) has variance \(4 + 12\rho\), not 4.

If \(r_{12.3}\) is the correlation coefficient between the variables \(X_{1}\) and \(X_{2}\) after eliminating the linear effect of \(X_{3}\), then which of the following are correct ?

(1) \(-1 \le r_{12.3} \le 1\)
(2) \(r_{12}^{2} + r_{13}^{2} + r_{23}^{2} - 2r_{12}r_{13}r_{23} \le 1\)
(3) \(r_{12.3}^{2} = b_{12.3}\, b_{21.3}\), where b's are partial regression coefficients

A.(1) and (2) only
B.(1) and (3) only
C.(2) and (3) only
D.(1), (2) and (3)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

All three statements are correct: (1), (2) and (3).

(1) \(r_{12.3}\) is an ordinary correlation, between the residuals of \(X_{1}\) and \(X_{2}\) after regressing each on \(X_{3}\), so it lies in [-1, 1].

(2) The correlation matrix of \(X_{1}, X_{2}, X_{3}\) is positive semi-definite, so its determinant is non-negative:

\[ \begin{aligned} &1 - r_{12}^{2} - r_{13}^{2} - r_{23}^{2} \\ &\quad + 2r_{12}r_{13}r_{23} \ge 0 \end{aligned} \]

which rearranges to statement (2).

(3) As with simple regression, where \(r^{2} = b_{yx}b_{xy}\), the partial correlation is the geometric mean of the two partial regression coefficients: \(b_{12.3} = r_{12.3}\sigma_{1.3}/\sigma_{2.3}\) and \(b_{21.3} = r_{12.3}\sigma_{2.3}/\sigma_{1.3}\), whose product is \(r_{12.3}^{2}\).

Since all three hold, every option naming only two of them is incomplete.

Discuss in Forum

If X be a \(p\)-component random vector with \(E(X) = 0\) and variance covariance matrix \(\Sigma\), positive definite. If X is partitioned into \(X^{(1)}\) of \(p_{1}\) components and \(X^{(2)}\) of \(p_{2}\) components such that \(p_{1} + p_{2} = p\) and \(p_{1} \le p_{2}\). Then the square of canonical correlation are the characteristic roots of the matrix :

A.\(\Sigma\)
B.\(\Sigma_{11}^{-1}\Sigma_{12}\)
C.\(\Sigma_{22}^{-1}\Sigma_{21}\)
D.\(\Sigma_{11}^{-1}\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The squared canonical correlations are the roots of \(\Sigma_{11}^{-1}\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}\).

Canonical correlation analysis maximises the correlation between \(a'X^{(1)}\) and \(b'X^{(2)}\). The first-order conditions lead to the eigenproblem

\[ \begin{aligned} &\Sigma_{11}^{-1}\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}\,a \\ &= \rho^{2}a \end{aligned} \]

so the squared canonical correlations are its \(p_{1}\) characteristic roots (the smaller set, since \(p_{1} \le p_{2}\)). \(\Sigma\) itself gives principal components. \(\Sigma_{11}^{-1}\Sigma_{12}\) and \(\Sigma_{22}^{-1}\Sigma_{21}\) are regression coefficient matrices, and are not even square in general.

Discuss in Forum

If \(T^{2}\) is Hotelling \(T^{2}\)-statistic, then the distribution of \(\dfrac{T^{2}}{n - 1}\,\dfrac{n - p}{p}\) would be :

A.Non-central F-distribution
B.Central F-distribution
C.Chi-square distribution
D.Student \(t\)-distribution

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The keyed answer is the non-central F-distribution.

For \(T^{2} = n(\bar{x} - \mu_{0})'S^{-1}(\bar{x} - \mu_{0})\) from a sample of size \(n\) from \(N_{p}(\mu, \Sigma)\), Anderson's theorem states that \(\dfrac{T^{2}}{n - 1}\dfrac{n - p}{p}\) has a non-central F-distribution with \(p\) and \(n - p\) degrees of freedom and non-centrality \(n(\mu - \mu_{0})'\Sigma^{-1}(\mu - \mu_{0})\).

When the null hypothesis \(\mu = \mu_{0}\) holds the non-centrality is zero and the distribution is the central F, which is the special case used to set critical values; the general statement, with no hypothesis assumed, is the non-central one. A chi-square or t-distribution is not the distribution of this ratio.

Discuss in Forum

If \(x_{1}\), \(x_{2}\), ........, \(x_{n}\) \((n \ge p + 1)\) are distributed independently each according to \(N_{p}(\mu, \Sigma)\), then the distribution of

\[ \begin{aligned} S = \frac{1}{n - 1}\sum_{\alpha=1}^{n} &(x_{\alpha} - \bar{x}) \\ &(x_{\alpha} - \bar{x})' \end{aligned} \]

is :

A.\(W_{p}(n - 1, \Sigma/(n - 1))\)
B.\(W_{p}(n, \Sigma/n)\)
C.\(N_{p}(\mu, \Sigma)\)
D.\(N_{p}(\mu, S)\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

S has the distribution \(W_{p}(n - 1, \Sigma/(n - 1))\).

The matrix of sums of squares and products \(A = \sum(x_{\alpha} - \bar{x})(x_{\alpha} - \bar{x})'\) has the Wishart distribution \(W_{p}(n - 1, \Sigma)\): one degree of freedom is used in estimating the mean. Scaling a Wishart matrix by a constant scales its parameter matrix, so

\[ \begin{aligned} S &= \frac{A}{n - 1} \\ &\sim W_{p}\!\left(n - 1, \frac{\Sigma}{n - 1}\right) \end{aligned} \]

The condition \(n \ge p + 1\) makes S non-singular. Option B uses \(n\) degrees of freedom, which would be right only if \(\mu\) were known and used instead of \(\bar{x}\). S is a random matrix, so a normal distribution for it (C and D) makes no sense.

Discuss in Forum

If \(X \sim N_{p}(\mu, \Sigma)\), then the distribution of \(N(\bar{X} - \mu)'\Sigma^{-1}(\bar{X} - \mu)\) would be :

A.Multivariate normal distribution
B.\(\chi^{2}\) with \(N - p\) degree of freedom
C.\(\chi^{2}\) with \(p\) degree of freedom
D.\(\chi^{2}\) with N degree of freedom

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The distribution is \(\chi^{2}\) with \(p\) degrees of freedom.

The mean \(\bar{X}\) of N independent \(N_{p}(\mu, \Sigma)\) vectors is \(N_{p}(\mu, \Sigma/N)\), so \(Z = \sqrt{N}\,\Sigma^{-1/2}(\bar{X} - \mu)\) is \(N_{p}(0, I)\). Then

\[ N(\bar{X} - \mu)'\Sigma^{-1}(\bar{X} - \mu) = Z'Z \]

is a sum of \(p\) independent squared standard normals, a \(\chi^{2}_{p}\). It is a scalar, not a multivariate normal vector. \(N - p\) degrees of freedom would arise only with \(\Sigma\) estimated (and then the result is an F, not a chi-square), and N degrees of freedom confuses the sample size with the dimension.

Discuss in Forum

Let X a \(p\)-component random vector is partitioned into \(X^{(1)}\) and \(X^{(2)}\) where \(X^{(1)}\) has \(q\)-components and \(X^{(2)}\) has \((p - q)\) components and \(X \sim N_{p}(\mu, \Sigma)\). If \(\mu\) and \(\Sigma\) are also partitioned as of X, then the variance covariance matrix of the conditional distribution of \(X^{(1)}\) given \(X^{(2)}\) is :

A.\(\Sigma_{11} + \Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}\)
B.\(\Sigma_{11} - \Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}\)
C.\(\Sigma_{22} + \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12}\)
D.\(\Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The conditional covariance matrix is \(\Sigma_{11} - \Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}\).

For a partitioned multivariate normal, the conditional distribution of \(X^{(1)}\) given \(X^{(2)} = x^{(2)}\) is normal with mean \(\mu^{(1)} + \Sigma_{12}\Sigma_{22}^{-1}(x^{(2)} - \mu^{(2)})\) and covariance \(\Sigma_{11} - \Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}\), the Schur complement of \(\Sigma_{22}\).

Conditioning can only reduce variability, so the correction is subtracted, ruling out the two options with a plus sign. The options built from \(\Sigma_{22}\) give the conditional covariance of \(X^{(2)}\) given \(X^{(1)}\), the other way round.

Discuss in Forum

Given \((X, Y) \sim N_{2}(5, 10, 1, 25, \rho)\) and \(P[4 \lt Y \lt 16 | X = 5] = 0.954\), then \(\rho\) is equal to :

(It is given that \(\frac{1}{\sqrt{2\pi}}\int_{0}^{2} e^{-\frac{1}{2}z^{2}} dz = 0.477\))

A.+0.80
B.-0.80
C.±0.8
D.0.96

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

\(\rho\) is \(\pm 0.8\).

Here \(\mu_{X} = 5\), \(\mu_{Y} = 10\), \(\sigma_{X}^{2} = 1\), \(\sigma_{Y}^{2} = 25\). Given \(X = x\), Y is normal with mean \(10 + 5\rho(x - 5)\) and variance \(25(1 - \rho^{2})\). At \(x = 5\) the mean is 10, and the interval (4, 16) is \(10 \pm 6\). Using the given value, \(P(|Z| \lt 2) = 2(0.477) = 0.954\), so

\[ \begin{aligned} \frac{6}{5\sqrt{1 - \rho^{2}}} &= 2 \\ \sqrt{1 - \rho^{2}} &= 0.6, \quad \rho^{2} = 0.64 \end{aligned} \]

Because \(x = 5\) is the mean of X, the sign of \(\rho\) does not move the conditional mean, so the data fix only \(\rho^{2}\): \(\rho = \pm 0.8\). Choosing +0.8 or -0.8 alone claims more than the information allows, and 0.96 is not a solution.

Discuss in Forum

Suppose a random vector \(\underline{X}\) has normal distribution with mean vector \(\underline{\mu}\) and dispersion matrix V. Then which of the following has a chi-square distribution ?

A.\((X - \underline{\mu})'\,V(X - \underline{\mu})\)
B.\((X - \underline{\mu})'\,V^{-1}(X - \underline{\mu})\)
C.\(\exp\{(X - \underline{\mu})'\,V(X - \underline{\mu})\}\)
D.\(\exp\{(X - \underline{\mu})'\,V^{-1}(X - \underline{\mu})\}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The chi-square variable is \((X - \mu)'V^{-1}(X - \mu)\).

If \(X \sim N_{p}(\mu, V)\) with V non-singular, write \(V^{-1} = (V^{-1/2})^{2}\) and set \(Z = V^{-1/2}(X - \mu)\). Then \(Z \sim N_{p}(0, I)\) and

\[ \begin{aligned} (X - \mu)'V^{-1}(X - \mu) &= Z'Z \\ &= \sum_{i=1}^{p} Z_{i}^{2} \sim \chi_{p}^{2} \end{aligned} \]

Using V instead of \(V^{-1}\) gives \(Z'V^{2}Z\), a weighted sum of chi-squares, which is chi-square only if \(V^{2} = I\). The exponentials are positive transforms of these quadratic forms and are not chi-square at all.

Discuss in Forum

Suppose \(R_{1.23}\) is a multiple correlation coefficient between \(X_{1}\) and \((X_{2}, X_{3})\). Then which of the following statements cannot be true ?

A.\(R_{1.23}\) is a maximum correlation between \(X_{1}\) and \((aX_{2} + a_{3}X_{3})\) where \(a_{1}\) and \(a_{2}\) are any real numbers.
B.\(R_{1.23}\) is a simple correlation between \(X_{1}\) and \(\hat{X}_{1}\), where \(\hat{X}_{1}\) is a line of best fit based on \(X_{2}\) and \(X_{3}\).
C.\(R_{1.23} = -0.4\)
D.\(R_{1.23} = 0.4\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The statement that cannot be true is \(R_{1.23} = -0.4\).

The multiple correlation coefficient is the correlation between \(X_{1}\) and its best linear predictor \(\hat{X}_{1}\) from \(X_{2}, X_{3}\), and equivalently the maximum correlation between \(X_{1}\) and any linear combination of \(X_{2}, X_{3}\) (statements A and B, despite the paper's slips in the coefficient names). A maximum of correlations over combinations and their negatives is never negative, so \(0 \le R_{1.23} \le 1\).

So 0.4 is a possible value, and \(-0.4\) is impossible.

Discuss in Forum

10.

Let \(X = (X_{1}, X_{2}, X_{3}, X_{4}) \sim N_{4}(\underline{0}, \Sigma)\) where

\[ \Sigma = \begin{bmatrix} 1 & \rho & \rho & \rho \\ \rho & 1 & \rho & \rho \\ \rho & \rho & 1 & \rho \\ \rho & \rho & \rho & 1 \end{bmatrix} \]

is positive definite.

Then which of the following statements is true ?

A.\(X_{1} + X_{2}, X_{2} + X_{3}, X_{3} + X_{4}\) have all same distributions
B.\(\dfrac{(X_{1} - X_{2})^{2}}{(X_{1} - X_{3})^{2}} \sim F_{11}\)
C.\(\dfrac{(X_{1} - X_{3})^{2} + (X_{2} - X_{4})^{2}}{2} \sim \chi_{2}^{2}\)
D.\(\Sigma X_{i} \sim N(0, 4)\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The true statement is \(X_{1} + X_{2}, X_{2} + X_{3}, X_{3} + X_{4}\) all have the same distribution.

Each sum of two different coordinates is normal with mean 0 and variance \(1 + 1 + 2\rho = 2(1 + \rho)\), because every pair has covariance \(\rho\). So all three are \(N(0, 2 + 2\rho)\).

The others fail. \(X_{1} - X_{2}\) and \(X_{1} - X_{3}\) are correlated (covariance \(1 - \rho\), correlation 1/2), so their squared ratio is not \(F_{1,1}\). \(X_{1} - X_{3}\) and \(X_{2} - X_{4}\) are independent with variance \(2(1 - \rho)\) each, so the sum of squares over 2 is \((1 - \rho)\chi_{2}^{2}\), equal to \(\chi_{2}^{2}\) only if \(\rho = 0\). And \(\sum X_{i}\) has variance \(4 + 12\rho\), not 4.

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