pset 8 solve - Ryan Davis EECS 310 Problem Set 8 1.a) The...

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

View Full Document Right Arrow Icon
Ryan Davis EECS 310 Problem Set 8 1.a) The probability that π(1) = 1 is 1/ |S|. Since π is a random bijection, π(1) is uniformly distributed over the interval [1, n] with possible discrete outcomes {1,…, n}. There are |S| possible outcomes, each that occurs with equal probability, so the probability of a single given outcome is 1/ |S|. b) The probability that f(A) = π(a) is |S| - π(a) C |A|-1 / |S| C |A| . The denominator represents the total number of ways to select a subset A of size |A| from S. We choose |A| elements from a set of |S| elements, the number of ways which to do is simply |S| C |A| . The numerator is significantly less intuitive. It gives the number of ways to choose a subset A such that π(a) is the minimum element in A. Since we know that π(a) is the minimum element of A, A must include |A| - 1 other elements that are greater than π(a). The number of elements that are greater than π(a) is |S| - π(a), which follows simply from the fact that the original set S has elements {1,…., n}. Thus we must choose |A| - 1elements from a set of |S| - π(a), given simply by |S| - π(a) C |A|-1. We give an example to illustrate the correctness of the formula. Consider S = {1, 2, 3, 4, 5}, |A| = 3, and π(a) = 2. The 5 C 3 possibilities for A are {1, 2, 3} {1, 2, 4} {1, 2, 5} {1, 3, 4} {1, 3, 5} {1, 4, 5} {2, 3, 4} {2, 3, 5} {2, 4, 5} {3, 4, 5} There are |S| - π(a) C |A|-1 = 5 – 2 C 2 = 3 C 2 = 3 sets in which π(a) = 2 is the lowest number. c) We can build on the approach described in b to show that the probability that f(A) = f(B) is J=1,…, |S| - 1 |S| - J C |A| - 1 * |S| - J C |B| - 1 ) / ( |S| C |A| * |S| C |B| ). The denominator gives the total number of ways to choose a subset of cardinality |A| and a subset of cardinality |B|, each from a set of cardinality |S|. Assume that A and B both have a smallest element J, so f(A) = f(B). A must have A – 1 other distinct elements, each of which is greater than J. The total number of elements greater than J is |S| - J. Thus the number of ways to choose a subset A in which J is the smallest element is |S| - J C |A| - 1 . We can perform the same analysis to show that for the same J, the number of ways to choose a subset A in which J is the smallest element is |S| - J C |B| - 1 . In the numerator, we sum over all possible smallest elements J to find the total number of ways that we can choose subsets A and B such that J is the smallest element in both A and B.
Background image of page 1

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

View Full DocumentRight Arrow Icon
We illustrate the formula with an example in which |S| = 5, |A| = 3, and |B| = 2. The possibilities for A and B are given
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the 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 / 5

pset 8 solve - Ryan Davis EECS 310 Problem Set 8 1.a) The...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online