Section 2.6

The Multiparameter Case and Other Extensions

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Problem 2.6.1

For any random variables (ψ1,…,ψs)\left(\psi_{1}, \ldots , \psi_{s}\right), show that the matrices ∥Eψiψj∥\left\| E \psi_{i} \psi_{j}\right\| and C=∥cov⁡(ψi,ψj)∥C= \left\| \operatorname {cov}\left(\psi_{i}, \psi_{j}\right)\right\| are positive semidefinite.

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Problem 2.6.2

In this problem, we establish some facts about eigenvalues and eigenvectors of square matrices. (For a more general treatment, see, for example, Marshall and Olkin 1979, Chapter 20.) We use the facts that a scalar λ>0\lambda >0 is an eigenvalue of the n×nn \times n symmetric matrix AA if there exists an n×1n \times 1 vector pp, the corresponding eigenvector, satisfying Ap=λpA p=\lambda p. If A is nonsingular, there are nn eigenvalues with corresponding linearly independent eigenvectors.

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(a)

Show that A=P′DλPA=P^{\prime } D_{\lambda } P, where DλD_{\lambda } is a diagonal matrix of eigenvalues of AA and PP is and n×nn \times n matrix whose rows are the corresponding eigenvalues that satisfies P′P=PP′=IP^{\prime } P=P P^{\prime }=I, the identity matrix.

(b)

Show that max⁡xx′Axx′x=\max_{x} \frac{x^{\prime } A x}{x^{\prime } x}= largest eigenvalue of AA.

(c)

If BB is a nonsingular symmetric matrix with eigenvector-eigenvalue representation B=Q′DβQB=Q^{\prime } D_{\beta } Q, then max⁡xx′Axx′Bx=\max_{x} \frac{x^{\prime } A x}{x^{\prime } B x}= largest eigenvalue of A∗A^{*}, where A∗=Dβ−1/2QAQ′Dβ−1/2A^{*}= D_{\beta }^{-1 / 2} Q A Q^{\prime } D_{\beta }^{-1 / 2} and Dβ−1/2D_{\beta }^{-1 / 2} is a diagonal matrix whose elements are the reciprocals of the square roots of the eigenvalues of BB.

(d)

For any square matrices CC and DD, show that the eigenvalues of the matrix CDC D are the same as the eigenvalues of the matrix DCD C, and hence that max⁡xx′Axx′Bx=\max_{x} \frac{x^{\prime } A x}{x^{\prime } B x}= largest eigenvalue of AB−1A B^{-1}.

(e)

If A=aa′A=a a^{\prime }, where aa is a n×1n \times 1 vector ( AA is thus a rank-one matrix), then max⁡xx′aa′xx′Bx=a′B−1a\max_{x} \frac{x^{\prime } a a^{\prime } x}{x^{\prime } B x}= a^{\prime } B^{-1} a.

Problem 2.6.3

An alternate proof of Theorem 6.1 uses the method of Lagrange (or undetermined) multipliers. Show that, for fixed γ\gamma, the maximum value of a′γa^{\prime } \gamma, subject to the constraint that a′Ca=1a^{\prime } C a=1, is obtained by the solutions to

∂∂ai{a′γ−12λ[a′Ca−1]}=0, \frac{\partial }{\partial a_{i}}\left\{ a^{\prime } \gamma -\frac{1}{2} \lambda \left[a^{\prime } C a-1\right]\right\} =0,

where λ\lambda is the undetermined multiplier. (The solution is a=±C−1γ/γ′C−1γa= \pm C^{-1} \gamma / \sqrt{\gamma^{\prime } C^{-1} \gamma }.)

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Problem 2.6.4

Prove (6.11) under the assumptions of the text.

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Problem 2.6.5

Verify

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(a)

the information matrices of Table 6.1 and

(b)

Equations (6.15) and (6.16).

Problem 2.6.6

If p(x)=(1−ε)ϕ(x−ξ)+(ε/τ)ϕ[(x−ξ)/τ]p(x)=(1-\varepsilon ) \phi (x-\xi )+(\varepsilon / \tau ) \phi [(x-\xi ) / \tau ] where ϕ\phi is the standard normal density, find I(ε,ξ,τ)I(\varepsilon , \xi , \tau ).

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Problem 2.6.7

Verify the expressions (6.20) and (6.21).

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Problem 2.6.8

Let A=(A11A12A21A22)A=\left(\begin{array}{ll}A_{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right) be a partitioned matrix with A22A_{22} square and nonsingular, and let

B=(I−A12A22−10I). B=\left(\begin{array}{cc} I-A_{12} A_{22}^{-1} 0 & I \end{array}\right).

Show that ∣A∣=∣A11−A12A22−1A21∣⋅∣A22∣\left|A\right|=\left|A_{11}-A_{12} A_{22}^{-1} A_{21}\right| \cdot \left|A_{22}\right|.

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Problem 2.6.9
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(a)

Let

A=(ab′bC) A=\left(\begin{array}{ll} a & b^{\prime } \\ b & C \\ \end{array}\right)

where aa is a scalar and bb a column matrix, and suppose that AA is positive definite. Show that ∣A∣≤a∣C∣\left|A\right| \leq a\left|C\right| with equality holding if and only if b=0b=0.

(b)

More generally, if the matrix AA of Problem 6.8 is positive definite, show that ∣A∣≤∣A11∣⋅∣A22∣\left|A\right| \leq \left|A_{11}\right| \cdot \left|A_{22}\right| with equality holding if and only if A12=0A_{12}=0.

Problem 2.6.10
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(a)

Show that if the matrix AA is nonsingular, then for any vector x,(x′Ax)(x′A−1x)>(x′x)2x,\left(x^{\prime } A x\right)\left(x^{\prime } A^{-1} x\right)> \left(x^{\prime } x\right)^{2}.

(b)

Show that, in the notation of Theorem 6.6 and the following discussion,

[∂∂θiEθ[δ]]2Iii(θ)=(εi′α)2εi′I(θ)εi, \frac{\left[\frac{\partial }{\partial \theta _{i}} \mathbb {E}_{\theta }\left[\delta \right]\right]^{2}}{I_{i i}(\theta )}=\frac{\left(\varepsilon _{i}^{\prime } \alpha \right)^{2}}{\varepsilon _{i}^{\prime } I(\theta ) \varepsilon _{i}},

and if α=(0,…,0,αi,0,…0),α′I(θ)−1α=(εi′α)2εi′I(θ)−1εi\alpha =\left(0, \ldots , 0, \alpha_{i}, 0, \ldots 0\right), \alpha^{\prime } I(\theta )^{-1} \alpha =\left(\varepsilon_{i}^{\prime } \alpha \right)^{2} \varepsilon_{i}^{\prime } I(\theta )^{-1} \varepsilon_{i}, and hence establish (6.25).

Problem 2.6.11

Prove that (6.26) is necessary for equality in (6.25).

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Problem 2.6.12

Prove the Bhattacharyya inequality (6.29) and show that the condition of equality is as stated.

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