Multivariate Analysis Questions and Answers

Q: Let \((X, Y) \sim N(\underline{0}, \Sigma)\) where \(\Sigma = \begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix}\), \(\rho \neq 0\) and\[ W = \frac{X - Y}{X + Y}\sqrt{\frac{1 + \rho}{1 - \rho}}. \]Then which of the following statements is not correct ?

The statement that is not correct is \(W \sim \chi_{1}^{2}\).\(X - Y \sim N(0, 2(1 - \rho))\) and \(X + Y \sim N(0, 2(1 + \rho))\), and \(\mathrm{Cov}(X - Y, X + Y) = \mathrm{Var}X - \mathrm{Var}Y = 0\), so being jointly normal they are independent (C). Standardising,\[ \begin{aligned} W &= \frac{(X - Y)/\sqrt{2(1 - \rho)}}{(X + Y)/\sqrt{2(1 + \rho)}} \\ &= \frac{Z_{1}}{Z_{2}} \end{aligned} \]a ratio of independent standard normals, so W is standard Cauchy. Then \(W^{2} = Z_{1}^{2}/Z_{2}^{2} \sim F_{1,1}\) (A), and so is \(1/W^{2}\) (B). W takes negative values, so it cannot be chi-square.

Q: If \(S_{1}, S_{2}, \ldots, S_{5}\) are the corrected sum of squares and sum of products matrices based on five independent samples of sizes 100 each coming from a four variate normal population. The distribution of \(\sum_{i=1}^{5} Si\).

\(\sum S_{i}\) is Wishart with \(k = 495\) degrees of freedom.For a sample of size 100 from a \(p\)-variate normal, the corrected sum of squares and products matrix is \(S_{i} \sim W_{4}(99, \Sigma)\), losing one degree of freedom for the estimated mean. The five samples are independent, and sums of independent Wisharts with the same \(\Sigma\) are Wishart with added degrees of freedom:\[ k = 5 \times 99 = 495 \]496 and 499 subtract too few degrees of freedom; 499 treats the five samples as one sample of 500 with a single mean, which they are not.

Q: Let \(M \sim W_{p}(n, \Sigma)\) and \(CC' = \Sigma\) where C is a non-singular matrix. Then \(C^{-1}MC^{-1}\) is distributed as :

The distribution is \(W_{p}(n, I_{p})\).If \(M \sim W_{p}(n, \Sigma)\), then \(BMB' \sim W_{p}(n, B\Sigma B')\) for any non-singular \(B\). Taking \(B = C^{-1}\):\[ C^{-1}\Sigma(C^{-1})' = C^{-1}CC'(C')^{-1} = I_{p} \]The paper prints \(C^{-1}MC^{-1}\); the result as keyed holds for \(C^{-1}M(C^{-1})'\), which is the same thing when C is symmetric, for instance \(C = \Sigma^{1/2}\). The scale matrix becomes the identity, not C, \(C^{-1}\) or \(\Sigma^{-1}\).

Q: The characteristic function of a vector \(\underline{X} \sim N_{p}(\underline{\mu}, \Sigma)\) is given by :

The characteristic function is \(e^{i\mu' t - \frac{1}{2}t'\Sigma t}\).For \(X \sim N_{p}(\mu, \Sigma)\), \(t'X\) is univariate normal with mean \(t'\mu\) and variance \(t'\Sigma t\). The characteristic function of \(N(m, v)\) at 1 is \(e^{im - v/2}\), so\[ E[e^{it'X}] = e^{i\mu' t - \frac{1}{2}t'\Sigma t} \]Option A is the moment generating function evaluated wrongly (the m.g.f. is \(e^{\mu' t + \frac{1}{2}t'\Sigma t}\)). Options C and D put \(i\) on the quadratic term, which would make the characteristic function have modulus 1 everywhere, impossible for a non-degenerate normal.

Q: Let \(\phi_{X}(t)\) be the characteristic function of \(X \sim N_{3}(\mu, \Sigma)\). Then \(EX_{1}X_{2}^{2}X_{3}\) is given by :

The moment is \((-1)^{4}\partial^{4}\phi_{X}(t)/\partial t_{1}\partial t_{2}^{2}\partial t_{3}\) at \(t = 0\).For \(\phi_{X}(t) = E[e^{i t' X}]\), each differentiation with respect to \(t_{j}\) brings down a factor \(iX_{j}\). Differentiating once in \(t_{1}\), twice in \(t_{2}\) and once in \(t_{3}\) and setting \(t = 0\) gives\[ \begin{aligned} &\frac{\partial^{4}\phi_{X}}{\partial t_{1}\partial t_{2}^{2}\partial t_{3}}\Big|_{0} \\ &= i^{4}E[X_{1}X_{2}^{2}X_{3}] \end{aligned} \]Since \(i^{4} = 1 = (-1)^{4}\), the moment equals the fourth derivative at the origin, as option A states. Moments come from derivatives at \(t = 0\), not at \(t = (1, 2, 1)\), and the order of the derivative must match the total power 4, which rules out the third derivative in option B.

Q: If \((X, Y)'\) follows bivariate normal distribution with mean vector \((0, 0)'\) and variance-covariance matrix \(\Sigma = \begin{pmatrix} 1 & 0.5 \\ 0.5 & 1 \end{pmatrix}\), which of the following does not have a \(\chi^{2}\)-distribution ?

\(X^{2} + Y^{2}\) does not have a chi-square distribution.X and Y are each standard normal, so \(X^{2}\) and \(Y^{2}\) are \(\chi^{2}_{1}\). But they are correlated (\(\rho = 0.5\)), so \(X^{2} + Y^{2}\) is not a sum of independent squares. Its distribution is that of \(\lambda_{1}Z_{1}^{2} + \lambda_{2}Z_{2}^{2}\), where \(\lambda_{1} = 1.5\) and \(\lambda_{2} = 0.5\) are the eigenvalues of \(\Sigma\), a weighted sum that is not chi-square.The quadratic form \(\mathbf{x}'\Sigma^{-1}\mathbf{x}\) is always \(\chi^{2}_{2}\) for a bivariate normal with mean 0. Here\[ \Sigma^{-1} = \frac{1}{0.75}\begin{pmatrix} 1 & -0.5 \\ -0.5 & 1 \end{pmatrix},\quad \mathbf{x}'\Sigma^{-1}\mathbf{x} = \frac{4}{3}(X^{2} - XY + Y^{2}) \]so the last option is \(\chi^{2}_{2}\).

Q: Suppose \(\underline{X}_{3 \times 1} \sim \text{MNormal}_{3}\)\[ \left(\underline{0} \begin{bmatrix} 1 & 0.5 & 0 \\ 0.5 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix}\right). \]Hence \(\text{var}(2X_{2} \mid X_{1} = x)\) is equal to :

\(\text{var}(2X_{2} \mid X_{1} = x) = 15\).For a multivariate normal, the conditional variance of \(X_{2}\) given \(X_{1}\) does not depend on \(x\):\[ \begin{aligned} \text{var}(X_{2} \mid X_{1}) &= \sigma_{22} - \frac{\sigma_{21}^{2}}{\sigma_{11}} \\ &= 4 - \frac{0.5^{2}}{1} = 3.75 \\ \text{var}(2X_{2} \mid X_{1}) &= 2^{2} \times 3.75 = 15 \end{aligned} \]\(X_{3}\) is uncorrelated with both, so it plays no part. The option 3.75 forgets the factor \(2^{2}\) from scaling, 16 is the unconditional variance \(4 \times 4\), and 8 doubles instead of squaring the coefficient.

Q: Suppose \(\underline{X}_{2 \times 1} \sim\) Bivariate Normal\[ \left(\begin{bmatrix} 4 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 & -1 \\ -1 & 5 \end{bmatrix}\right). \]Then the distribution of \(3Y = 3X_{1} - 4X_{2}\) is :

\(3X_{1} - 4X_{2}\) is Normal (8, 122).A linear combination \(a'X\) of a bivariate normal vector is normal with mean \(a'\mu\) and variance \(a'\Sigma a\). With \(a = (3, -4)'\):\[ \begin{aligned} \text{mean} &= 3(4) - 4(1) = 8 \\ \text{variance} &= 9(2) + 16(5) - 2(3)(4)(-1) \\ &= 18 + 80 + 24 = 122 \end{aligned} \]The covariance is negative and the coefficients have opposite signs, so the cross term adds to the variance. Ignoring the cross term gives 98, the third option. Normal(8, 62) has the right mean but a variance that does not equal \(a'\Sigma a\), and Normal(-1, 26) has the wrong mean as well.

Q: The quadrate form\[ 9X_{1}^{2}+4X_{2}^{2}+X_{3}^{2}-4X_{1}X_{3}-2X_{2}X_{3} \]is :

The quadratic form is positive definite.Its symmetric matrix and leading principal minors are\[ A = \begin{pmatrix} 9 & 0 & -2 \\ 0 & 4 & -1 \\ -2 & -1 & 1 \end{pmatrix} \]\[ \begin{aligned} D_{1} &= 9 \gt 0,\quad D_{2} = 36 \gt 0 \\ D_{3} &= 9(4 - 1) - 2(0 + 8) = 27 - 16 = 11 \gt 0 \end{aligned} \]All leading principal minors are positive, so by Sylvester's criterion the form is positive definite: it is positive for every nonzero \((X_{1}, X_{2}, X_{3})\).It is not merely semidefinite, because the determinant is nonzero, so no nonzero vector gives the value 0. The negative options are ruled out at once by the positive diagonal entries. (The paper's 'quadrate form' means quadratic form.)

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

Y is not \(\chi^{2}\).A chi-square variable is a sum of squares of independent standard normals and is therefore never negative. \(Y = 2X_{1}^{2} - X_{2}^{2} + X_{3}^{2}\) has a negative coefficient on \(X_{2}^{2}\), so Y takes negative values with positive probability (whenever \(X_{2}^{2}\) is large compared with the other terms), whatever \(\mu\) and \(\Sigma\) are. A variable that can be negative cannot be chi-square.Even with the sign fixed, the coefficient 2 on \(X_{1}^{2}\) would spoil the chi-square form unless the quadratic form matrix were idempotent with respect to \(\Sigma\). So none of the degrees of freedom offered (2, 3 or 6) applies.

21.

Let \((X, Y) \sim N(\underline{0}, \Sigma)\) where \(\Sigma = \begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix}\), \(\rho \neq 0\) and

\[ W = \frac{X - Y}{X + Y}\sqrt{\frac{1 + \rho}{1 - \rho}}. \]

Then which of the following statements is not correct ?

A.\(W^{2} \sim F_{1,1}\)
B.\(1/W^{2} \sim F_{1,1}\)
C.(X - Y) and (X + Y) are independently distributed
D.\(W \sim \chi_{1}^{2}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The statement that is not correct is \(W \sim \chi_{1}^{2}\).

\(X - Y \sim N(0, 2(1 - \rho))\) and \(X + Y \sim N(0, 2(1 + \rho))\), and \(\mathrm{Cov}(X - Y, X + Y) = \mathrm{Var}X - \mathrm{Var}Y = 0\), so being jointly normal they are independent (C). Standardising,

\[ \begin{aligned} W &= \frac{(X - Y)/\sqrt{2(1 - \rho)}}{(X + Y)/\sqrt{2(1 + \rho)}} \\ &= \frac{Z_{1}}{Z_{2}} \end{aligned} \]

a ratio of independent standard normals, so W is standard Cauchy. Then \(W^{2} = Z_{1}^{2}/Z_{2}^{2} \sim F_{1,1}\) (A), and so is \(1/W^{2}\) (B). W takes negative values, so it cannot be chi-square.

Discuss in Forum

22.

If \(S_{1}, S_{2}, \ldots, S_{5}\) are the corrected sum of squares and sum of products matrices based on five independent samples of sizes 100 each coming from a four variate normal population. The distribution of \(\sum_{i=1}^{5} Si\).

A.is Wishart with \(k = 496\)
B.is Wishart with \(k = 495\)
C.is Wishart with \(k = 499\)
D.cannot be decided on the basis of given information

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

\(\sum S_{i}\) is Wishart with \(k = 495\) degrees of freedom.

For a sample of size 100 from a \(p\)-variate normal, the corrected sum of squares and products matrix is \(S_{i} \sim W_{4}(99, \Sigma)\), losing one degree of freedom for the estimated mean. The five samples are independent, and sums of independent Wisharts with the same \(\Sigma\) are Wishart with added degrees of freedom:

\[ k = 5 \times 99 = 495 \]

496 and 499 subtract too few degrees of freedom; 499 treats the five samples as one sample of 500 with a single mean, which they are not.

Discuss in Forum

23.

Let \(M \sim W_{p}(n, \Sigma)\) and \(CC' = \Sigma\) where C is a non-singular matrix. Then \(C^{-1}MC^{-1}\) is distributed as :

A.\(W_{p}(n, C)\)
B.\(W_{p}(n, \Sigma^{-1})\)
C.\(W_{p}(n, I_{p})\)
D.\(W_{p}(n, C^{-1})\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The distribution is \(W_{p}(n, I_{p})\).

If \(M \sim W_{p}(n, \Sigma)\), then \(BMB' \sim W_{p}(n, B\Sigma B')\) for any non-singular \(B\). Taking \(B = C^{-1}\):

\[ C^{-1}\Sigma(C^{-1})' = C^{-1}CC'(C')^{-1} = I_{p} \]

The paper prints \(C^{-1}MC^{-1}\); the result as keyed holds for \(C^{-1}M(C^{-1})'\), which is the same thing when C is symmetric, for instance \(C = \Sigma^{1/2}\). The scale matrix becomes the identity, not C, \(C^{-1}\) or \(\Sigma^{-1}\).

Discuss in Forum

24.

The characteristic function of a vector \(\underline{X} \sim N_{p}(\underline{\mu}, \Sigma)\) is given by :

A.\(e^{\mu't - \frac{1}{2}t'\Sigma t}\)
B.\(e^{i\mu't - \frac{1}{2}t'\Sigma t}\)
C.\(e^{\mu't - \frac{i}{2}t'\Sigma t}\)
D.\(e^{i\mu't - \frac{i}{2}t'\Sigma t}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The characteristic function is \(e^{i\mu' t - \frac{1}{2}t'\Sigma t}\).

For \(X \sim N_{p}(\mu, \Sigma)\), \(t'X\) is univariate normal with mean \(t'\mu\) and variance \(t'\Sigma t\). The characteristic function of \(N(m, v)\) at 1 is \(e^{im - v/2}\), so

\[ E[e^{it'X}] = e^{i\mu' t - \frac{1}{2}t'\Sigma t} \]

Option A is the moment generating function evaluated wrongly (the m.g.f. is \(e^{\mu' t + \frac{1}{2}t'\Sigma t}\)). Options C and D put \(i\) on the quadratic term, which would make the characteristic function have modulus 1 everywhere, impossible for a non-degenerate normal.

Discuss in Forum

25.

Let \(\phi_{X}(t)\) be the characteristic function of \(X \sim N_{3}(\mu, \Sigma)\). Then \(EX_{1}X_{2}^{2}X_{3}\) is given by :

A.\((-1)^{4}\partial^{4}\phi_{X}(t)\) / \((\partial t_{1}\partial t_{2}^{2}\partial t_{3})\) at \(t = 0\)
B.\((-1)^{3}\partial^{3}\phi_{X}(t)\) / \((\partial t_{1}\partial t_{2}\partial t_{3})\) at \(t_{1} = 1, t_{2} = 2, t_{3} = 1\)
C.\((-1)^{3}\partial^{4}\phi_{X}(t)\) / \((\partial t_{1}\partial t_{2}^{2}\partial t_{3})\) at \(t_{1} = 1, t_{2} = 2, t_{3} = 1\)
D.\(\partial^{4}\phi_{X}(t)\) / \((\partial t_{1}\partial t_{2}^{2}\partial t_{3})\) at \(t_{1} = 1, t_{2} = 2, t_{3} = 1\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The moment is \((-1)^{4}\partial^{4}\phi_{X}(t)/\partial t_{1}\partial t_{2}^{2}\partial t_{3}\) at \(t = 0\).

For \(\phi_{X}(t) = E[e^{i t' X}]\), each differentiation with respect to \(t_{j}\) brings down a factor \(iX_{j}\). Differentiating once in \(t_{1}\), twice in \(t_{2}\) and once in \(t_{3}\) and setting \(t = 0\) gives

\[ \begin{aligned} &\frac{\partial^{4}\phi_{X}}{\partial t_{1}\partial t_{2}^{2}\partial t_{3}}\Big|_{0} \\ &= i^{4}E[X_{1}X_{2}^{2}X_{3}] \end{aligned} \]

Since \(i^{4} = 1 = (-1)^{4}\), the moment equals the fourth derivative at the origin, as option A states. Moments come from derivatives at \(t = 0\), not at \(t = (1, 2, 1)\), and the order of the derivative must match the total power 4, which rules out the third derivative in option B.

Discuss in Forum

26.

If \((X, Y)'\) follows bivariate normal distribution with mean vector \((0, 0)'\) and variance-covariance matrix \(\Sigma = \begin{pmatrix} 1 & 0.5 \\ 0.5 & 1 \end{pmatrix}\), which of the following does not have a \(\chi^{2}\)-distribution ?

A.\(X^{2} + Y^{2}\)
B.\(X^{2}\)
C.\(Y^{2}\)
D.\(4/3\,(X^{2} - XY + Y^{2})\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

\(X^{2} + Y^{2}\) does not have a chi-square distribution.

X and Y are each standard normal, so \(X^{2}\) and \(Y^{2}\) are \(\chi^{2}_{1}\). But they are correlated (\(\rho = 0.5\)), so \(X^{2} + Y^{2}\) is not a sum of independent squares. Its distribution is that of \(\lambda_{1}Z_{1}^{2} + \lambda_{2}Z_{2}^{2}\), where \(\lambda_{1} = 1.5\) and \(\lambda_{2} = 0.5\) are the eigenvalues of \(\Sigma\), a weighted sum that is not chi-square.

The quadratic form \(\mathbf{x}'\Sigma^{-1}\mathbf{x}\) is always \(\chi^{2}_{2}\) for a bivariate normal with mean 0. Here

\[ \Sigma^{-1} = \frac{1}{0.75}\begin{pmatrix} 1 & -0.5 \\ -0.5 & 1 \end{pmatrix},\quad \mathbf{x}'\Sigma^{-1}\mathbf{x} = \frac{4}{3}(X^{2} - XY + Y^{2}) \]

so the last option is \(\chi^{2}_{2}\).

Discuss in Forum

27.

Suppose \(\underline{X}_{3 \times 1} \sim \text{MNormal}_{3}\)

\[ \left(\underline{0} \begin{bmatrix} 1 & 0.5 & 0 \\ 0.5 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix}\right). \]

Hence \(\text{var}(2X_{2} \mid X_{1} = x)\) is equal to :

A.16
B.3.75
C.8
D.15

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

\(\text{var}(2X_{2} \mid X_{1} = x) = 15\).

For a multivariate normal, the conditional variance of \(X_{2}\) given \(X_{1}\) does not depend on \(x\):

\[ \begin{aligned} \text{var}(X_{2} \mid X_{1}) &= \sigma_{22} - \frac{\sigma_{21}^{2}}{\sigma_{11}} \\ &= 4 - \frac{0.5^{2}}{1} = 3.75 \\ \text{var}(2X_{2} \mid X_{1}) &= 2^{2} \times 3.75 = 15 \end{aligned} \]

\(X_{3}\) is uncorrelated with both, so it plays no part. The option 3.75 forgets the factor \(2^{2}\) from scaling, 16 is the unconditional variance \(4 \times 4\), and 8 doubles instead of squaring the coefficient.

Discuss in Forum

28.

Suppose \(\underline{X}_{2 \times 1} \sim\) Bivariate Normal

\[ \left(\begin{bmatrix} 4 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 & -1 \\ -1 & 5 \end{bmatrix}\right). \]

Then the distribution of \(3Y = 3X_{1} - 4X_{2}\) is :

A.Normal (-1, 26)
B.Normal (8, 62)
C.Normal (8, 98)
D.Normal (8, 122)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

\(3X_{1} - 4X_{2}\) is Normal (8, 122).

A linear combination \(a'X\) of a bivariate normal vector is normal with mean \(a'\mu\) and variance \(a'\Sigma a\). With \(a = (3, -4)'\):

\[ \begin{aligned} \text{mean} &= 3(4) - 4(1) = 8 \\ \text{variance} &= 9(2) + 16(5) - 2(3)(4)(-1) \\ &= 18 + 80 + 24 = 122 \end{aligned} \]

The covariance is negative and the coefficients have opposite signs, so the cross term adds to the variance. Ignoring the cross term gives 98, the third option. Normal(8, 62) has the right mean but a variance that does not equal \(a'\Sigma a\), and Normal(-1, 26) has the wrong mean as well.

Discuss in Forum

29.

The quadrate form

\[ 9X_{1}^{2}+4X_{2}^{2}+X_{3}^{2}-4X_{1}X_{3}-2X_{2}X_{3} \]

is :

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

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The quadratic form is positive definite.

Its symmetric matrix and leading principal minors are

\[ A = \begin{pmatrix} 9 & 0 & -2 \\ 0 & 4 & -1 \\ -2 & -1 & 1 \end{pmatrix} \]

\[ \begin{aligned} D_{1} &= 9 \gt 0,\quad D_{2} = 36 \gt 0 \\ D_{3} &= 9(4 - 1) - 2(0 + 8) = 27 - 16 = 11 \gt 0 \end{aligned} \]

All leading principal minors are positive, so by Sylvester's criterion the form is positive definite: it is positive for every nonzero \((X_{1}, X_{2}, X_{3})\).

It is not merely semidefinite, because the determinant is nonzero, so no nonzero vector gives the value 0. The negative options are ruled out at once by the positive diagonal entries. (The paper's 'quadrate form' means quadratic form.)

Discuss in Forum

30.

Suppose \(\underline{X}_{3 \times 1} \sim\) multivariable \(\text{Normal}_{3}(\mu, \Sigma)\) and let

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

Then the probability distribution of Y is :

A.\(\chi^{2}\) with 2 degrees of freedom
B.\(\chi^{2}\) with 3 degrees of freedom
C.\(\chi^{2}\) with 6 degrees of freedom
D.not \(\chi^{2}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

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

A chi-square variable is a sum of squares of independent standard normals and is therefore never negative. \(Y = 2X_{1}^{2} - X_{2}^{2} + X_{3}^{2}\) has a negative coefficient on \(X_{2}^{2}\), so Y takes negative values with positive probability (whenever \(X_{2}^{2}\) is large compared with the other terms), whatever \(\mu\) and \(\Sigma\) are. A variable that can be negative cannot be chi-square.

Even with the sign fixed, the coefficient 2 on \(X_{1}^{2}\) would spoil the chi-square form unless the quadratic form matrix were idempotent with respect to \(\Sigma\). So none of the degrees of freedom offered (2, 3 or 6) applies.

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