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stat610hw1-solutions

# stat610hw1-solutions - Statistics 610 Homework 1...

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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 ( ω ’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 = φ for all i , we get a sequence of mutually disjoint sets with A i = φ . The additivity property of μ then implies that μ ( φ ) = μ ( φ ) + μ ( φ ) + . . . showing that μ ( φ ) = 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, μ ( B ) = i =1 μ ( A i ) = lim n n i =1 μ ( A i ) = lim n n i =1 μ ( A i ) = lim n μ ( B n ) where we have used the (easily verified) facts that countable additivity implies finite additivity and that B n = n i =1 A i .

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