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Unformatted text preview: March 18, 2007 HW4 Problem Solutions #3.4.1. We are given the statespace S = { , . . . , N } . (I am using j to denote what the book writes as E j .) Clearly the transitions are given by: P ij = p I [ j =0] + (1 p ) I [ j = i +1 ≤ N ] if i < N P ij = (1 rp )) I [ j = N ] + rp I [ j =0] if i = N Irreducibility is obvious because 0 7→ 1 7→ ··· 7→ N 1 7→ N 7→ 0. For aperiodicity of an irreducible chain, it is always sufficient to check that some state (here, N or 0) has positive probability of succeeding itself. The stationary equations are: π j = π j 1 (1 p ) = π (1 p ) j , j = 1 , . . . , N 1 π N = π N 1 (1 p ) + π N (1 rp ) = (1 p ) N rp π and (using the fact that π must be a probability vector) π = p ( π + ··· π N 1 ) + rp π N = p (1 (1 r ) π N ) = p (1 p ) N 1 r r π which implies π = p n 1 + (1 p ) N 1 r r o 1 Then, since items are always inspected in states 0 , . . . , N 1 and inspected only with probability r in state N , the longrun proportion of items inspected is 1 π N + rπ N = r/ (1 + (1 r )(1 p ) N ). Finally, since we know a proportion p of all ( iid ) items is defective, the longterm or stationary proportion of defective items inspected is just this same overall proportion of inspected items.items inspected is just this same overall proportion of inspected items....
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This note was uploaded on 09/09/2009 for the course STAT 650 taught by Professor Slud during the Spring '09 term at Maryland.
 Spring '09
 SLUD

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