Instructor’s Solutions Manual to accompany Elementary Linear by Grossman, Ragozin

By Grossman, Ragozin

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Example text

Let ak (x) = σk (x)σk (x)∗ , k = 1, 2. Take c(x, y) = σ1 (x)σ2 (y)∗ . The two choices given in the next example are due to T. G. F. Li (1989), respectively. 17 (Coupling by reflection). Let L1= L2 and a(x) = σ(x)σ(x)∗. We have two choices: c(x, y) = σ(x) σ(y)∗ − 2 σ(y)−1 u ¯u ¯∗ , |σ(y)−1 u ¯|2 c(x, y) = σ(x) I − 2¯ uu¯∗ σ(y)∗ , det σ(y) = 0, x = y, x = y, where u ¯ = (x − y)/|x − y|. S. Kendall (1986) and M. Cranston (1991). In the case that x = y, the first and the third couplings here are defined to be the same as the second one.

We are now going to prove the variational formula for the lower bounds (cf. 3) 44 3 New Variational Formulas for the First Eigenvalue where W = {w : w0 = 0, wi ↑↑}, i 0, w ¯i = wi − π(w), Ii (w) = W = {w : wi ↑↑, π(w) 0}, ∞ 1 µi bi (wi+1 − wi ) j=i+1 µj wj , i w∈W, 0, and “↑↑” means strictly increasing. 3), since {w ¯:w∈W}⊂W. (a) First, we prove that Ii (w) > 0 for each w ∈ W and all i 1. ∞ µ w > 0 for all i 0. Otherwise, let i satisfy Equivalently, j j 0 j=i+1 ∞ 0. Then, since wj is strictly increasing, it follows that j=i0 +1 µj wj wi0 < 0, and furthermore, ∞ µj wj = 0 ∞ i0 j=0 i0 µj wj + µj wj µj wj j=i0 +1 j=0 i0 wi0 j=0 µj < 0.

24) for details. This example illustrates the flexibility in the application of couplings. The details of this chapter, except for diffusions, are included in Chapter 5 of the second edition of Chen (1992a). Finally, we mention that the coupling methods are also powerful for timeinhomogeneous Markov processes, not touched on in this book. 14 is valid for Markov jump processes valued in Polish spaces [cf. L. Zheng (1993)]. I. L. I. Zeifman (1997). 2. 2. Then, three sections are used to explain the ideas in detail for the proof in the geometric case.

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