Handout5-W09

Handout5-W09 - ij = 2 z i z j D 1-z i-z j (1-2 z i )(1-2 z...

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STAT 454/854 Winter 2009 Handout #5 Brewer’s PPS Sampling Method for n = 2 Suppose that 0 < z i < 1 / 2 for all i and N i =1 z i = 1. 1. Draw the first element with probability equal to p i = z i (1 - z i ) / { D (1 - 2 z i ) } , i = 1 , ··· ,N , where D is a constant satisfying the constraint N i =1 p i = 1 and is given by D = N X i =1 z i (1 - z i ) 1 - 2 z i = 1 2 N X i =1 z i { 1 + (1 - 2 z i ) } 1 - 2 z i = 1 2 1 + N X i =1 z i 1 - 2 z i ! . 2. Draw the second element with probability equal to q i | j = z i / (1 - z j ), i = 1 , ··· ,N and i 6 = j , assuming j is selected on the first draw. Properties of Brewer’s (1963) PPS sampling method: (1) First order inclusion probabilities: π i = z i (1 - z i ) D (1 - 2 z i ) + N X j =1 ,j 6 = i z j (1 - z j ) D (1 - 2 z j ) · z i 1 - z j = 2 z i . (2) Second order inclusion probabilities:
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Unformatted text preview: ij = 2 z i z j D 1-z i-z j (1-2 z i )(1-2 z j ) . (3) i j- ij &gt; 0 for all i 6 = j (Rao, 1965): i j- ij = 2 z i z j D (1-2 z i )(1-2 z j ) { 2 D (1-2 z i )(1-2 z j )-(1-z i-z j ) } and 2 D (1-2 z i )(1-2 z j )-(1-z i-z j ) = 1 + N X k =1 z k 1-2 z k ! (1-2 z i )(1-2 z j )-(1-z i-z j ) = N X k =1 ,k 6 = i,j z k 1-2 z k (1-2 z i )(1-2 z j ) &gt; ....
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This note was uploaded on 11/23/2010 for the course STAT 454 taught by Professor Da during the Fall '09 term at Waterloo.

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