solutions-3 - Solution to Homework#3 36-754 25 February...

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Unformatted text preview: Solution to Homework #3, 36-754 25 February 2006 Exercise 10.1 I need one last revision of the definition of a Markov operator: a linear operator on L 1 satisfying the following conditions. 1. If f ≥ 0 ( μ-a.e.), then Kf ≥ 0 ( μ-a.e.). 2. If f ≤ M ( μ-a.e.), then Kf ≤ M ( μ-a.e.). 3. K 1 Ξ = 1 Ξ . 4. If f n ↓ 0, then Kf n ↓ 0. First, we show that kernels induce operators, then that operators induce kernels. Given a kernel κ , we define the operator Kf ( x ) ≡ R κ ( x,dy ) f ( y ). 1. Clearly, if f ≥ 0 a.e., then R κ ( x,dy ) f ( y ) ≥ 0. 2. If f ≤ M a.e., then M- f ( x ) ≥ 0 a.e., and R κ ( x,dy )( M- f ( y )) = R κ ( x,dy ) M- R κ ( x,dy ) f ( y ) = M- Kf ( x ) ≥ 0 a.e., so M ≥ Kf ( x ) a.e. 3. R κ ( x,dy ) 1 Ξ ( y ) = 1 for all x , so K 1 Ξ ( x ) = 1 Ξ ( x ). 4. If f n ( x ) ↓ 0 pointwise, then, for each x , R κ ( x,dy ) f n ( x ) → 0 by monotone convergence, so Kf n ( x ) → 0. Now for the converse: given an operator K , define κ ( x,B ) = K 1 B ( x ). For fixed x , we need this to be a probability measure. For every B ∈ X , 1 B ∈ L 1 , so K 1 B ( x ) is, for fixed x , a set-function defined over the whole of the σ-algebra in question. Since 1 B ( x ) ≥ 0, we know that K 1 B ( x ) ≥ 0, and this is a non- negative set-function. Next, we check that it’s finitely additive: If A and B are disjoint sets, 1 A ∪ B ( x ) = 1 A ( x ) + 1 B ( x ). Hence, by the linearity of K , K 1 A ∪ B = K 1 A + K 1 B , which by induction extends to any finite collection of disjoint sets. We notice that if A n ↓ ∅ , 1 A n ( x ) ↓ 0 pointwise, so K 1 A n (...
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This note was uploaded on 12/20/2011 for the course STAT 36-754 taught by Professor Schalizi during the Spring '06 term at University of Michigan.

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solutions-3 - Solution to Homework#3 36-754 25 February...

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