J 1 d f j p v e f j f 2 g 2 d 2 p y 2 p 2 g 2 g 2 d 2

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j ± 1 D F j   p v E ¡ F j   ¢ ± ; F 2   g 2   d 2 ± ; p Y | 2   p 2   g 2   g 2   d 2 ± ; p Y | 2   p 2   d 2 ± p Y   Conclusion: ! j ± 1 D 2 j   F j   ! j ± 1 D F j   p v ; 2 p Y | 2   p 2   d 2 p Y   ± ; 2 p 2 | Y   d 2 What’s required of g .   ? F j   ± p ¡ Y | 2 j   ¢ p ¡ 2 j   ¢ g ¡ 2 j   ¢ should satisfy law of large numbers. F j   needs finite variance– g .   must have fatter tails than p Y | 2   p 2   .
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5 Caveat: Always produces an answer, needs to be checked. (1) Try special cases where result is known analytically. (2) Try different g .   to see if get the same result. (3) Use analytic results for components of 2 in order to keep dimension that must be importance- sampled small. I. Bayesian econometrics F. Numerical Bayesian methods 1. Importance sampling 2. The Gibbs sampler
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6 Suppose the parameter vector 2 can be partitioned as 2 U ± 2 1 U , 2 2 U , 2 3 U   with the property that p 2 | Y   is of unknown form but p 2 1 | Y , 2 2 , 2 3   p 2 2 | Y , 2 1 , 2 3   p 2 3 | Y , 2 1 , 2 2   are of known form (same idea works for 2 , 4 , or n blocks) (1) Start with arbitrary initial guesses 2 1 j   , 2 2 j   , 2 3 j   for j ± 1 . (2) Generate: 2 1 j ³ 1   from p 2 1 | Y , 2 2 j   , 2 3 j     2 2 j ³ 1   from p 2 2 | Y , 2 1 j ³ 1   , 2 3 j     2 3 j ³ 1   from p 2 3 | Y , 2 1 j ³ 1   , 2 2 j ³ 1     (3) Repeat step (2) for j ± 1,2,..., D Notice the sequence 2 j   j ± 1 D is a Markov chain with transition kernel = 2 j ³ 1   | 2 j     ± p 2 3 j ³ 1   | Y , 2 1 j ³ 1   , 2 2 j ³ 1     p 2 2 j ³ 1   | Y , 2 1 j ³ 1   , 2 3 j     p 2 1 j
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  • Winter '09
  • JamesHamilton
  • Econometrics, Markov chain, Markov chain Monte Carlo

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