pset8soln

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

## 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.

### Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online