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Unformatted text preview: Statistics 610 Homework 1 Solutions (1) Without loss of generality, let the Democrats be labeled 1 through m and the Republicans be labeled m + 1 through N . The number of permutations ( s) for which X 1 = 1 is m ( N 1)!, since the first position can be filled in one of m ways and the remaining n 1 positions can be filled in any order. The total number of permutations is, of course, N !. Hence, P ( X 1 = 1) = m ( N 1)! /N ! = m/N = p . The same argument works for any X i . For X i to equal 1, the i th position must be filled by one of the m Democrats and the remaining n 1 positions can be filled in any order. To work out the distribution of ( X 1 ,X 2 ), we proceed thus. We just work out the case P ( X 1 = 1 ,X 2 = 1). For X 1 = X 2 = 1 to be satisfied, there are m choices for the first position and m 1 choices for the second and then the remaining N 2 slots can be filled up in any order. Thus we have m ( m 1) ( N 2)! permutations satisfying this. The probability is then m ( m 1) ( N 2)! /N ! = m ( m 1) /N ( N 1) = p ( Np 1) / ( N 1). The other cases may be similarly worked out. Notice that the same argument goes through for any ( X i ,X j ) for i < j . The general fact about k tuples follows by obvious extensions of the above arguments. (2) (i) Taking A i = for all i , we get a sequence of mutually disjoint sets with A i = . The additivity property of then implies that ( ) = ( ) + ( ) + ... showing that ( ) = 0. (ii) Let A 1 = B 1 ,A 2 = B 2 B 1 ,A 3 = B 3 B 2 ,... and so on. The A i s are all mutually disjoint, and A i = B i = B . Then, ( B ) = i =1 ( A i ) = lim n n i =1 ( A i ) = lim n n i =1 ( A i ) = lim n ( B n ) where we have used the (easily verified) facts that countable additivity implies...
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This note was uploaded on 04/14/2010 for the course STATS 610 taught by Professor Moulib during the Fall '09 term at University of Michigan.
 Fall '09
 moulib
 Statistics

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