pset8soln - EECS 310, Fall 2011 Instructor: Nicole...

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Unformatted text preview: EECS 310, Fall 2011 Instructor: Nicole Immorlica Problem Set #8 Due: November 22, 2011 1. (30 points) Let S = { 1 ,...,n } and let π be a random permutation on S (i.e., π : S → S is a random bijection). Let π ( k ) represent the k th element in the permutation π . For a subset A of S let f ( A ) be the minimum of π ( a ) over all elements of a ∈ A . For example, if n = 10, A = { 2 , 5 , 7 } , and π (2) = 3 ,π (5) = 10 ,π (7) = 2, then f ( A ) = 2. Let A and B be two arbitrary subsets of S . Express the following in terms of A , B , and S . Be sure to explain your calculations. (a) (6 points) What is the probability that π (1) = 1? Solution: Using the above notation, this equates to setting v 1 = 1. By the product rule, the number of ways to choose v 2 ,...,v n is ( n- 1)! (there are ( n- 1) choices for v 2 since 1 is already gone, and then ( n- 2) choices for v 3 and so on). The size of the sample space is n !, and so the total probability is ( n- 1)! n ! = 1 n . (b) (12 points) For an arbitrary element a ∈ A , what’s the probability that f ( A ) = π ( a )? Solution: Let k = | A | . We first choose the co-domain of the elements in A , i.e., we choose the numbers v a for a ∈ A . There are k elements in a and so ( n k ) ways to pick the image of the set A . The minimum element of this set must be assigned to a , so we have ( k- 1)! ways to assign the rest of the image. The remaining ( n- k ) elements of the co-domain can be arranged in any order, so this gives ( n- k )! arrangements. By the product rule, the number of ways for f ( A ) to equal π ( a ) is thus ( n k ) ( n- k )!( k- 1)! = n ! /k and the total probability is 1 /k . An alternative solution is to let E be the event that f ( A ) = π ( a ) and F T be the event that the image of A is set T ⊂ S . Then F T ∩ F T = ∅ if T 6 = T so the F T are disjoint. Hence Pr[ E ] = X T ⊂ S, | T | = k Pr[ E | F T ] Pr[ F T ] . Conditioning on the image of A reduces the problem to that in the previous part (but for a problem of size k instead of n ) from whence Pr[ E | F T ] = 1 /k . As the distribution is uniform, Pr[ F T ] = 1 / ( n k ) and hence cancels with the number of terms in the summation giving the answer of Pr[ E ] = 1 /k as above. (c) (12 points) What is the probability f ( A ) = f ( B )? Solution: The key observation is that this happens if and only if the minimum element of A ∪ B is in A ∩ B . Now we can employ either of the above approaches from the previous part to solve the problem. Let E be the event that f ( A ) = f ( B ) and F T be the event that the image of A ∪ B is T . Then as above Pr[ E ] = Pr[ E | F T ] for an arbitrary T . Let k = | A ∩ B | and m = | A ∪ B | . There are k places where we can place the minimum element of T , and it lands in each place with equal probability since π is uniform. Thus the probability is k/m ....
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This note was uploaded on 01/15/2012 for the course ECON 101 taught by Professor N/a during the Fall '11 term at Middlesex CC.

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pset8soln - EECS 310, Fall 2011 Instructor: Nicole...

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