Chap07 - Chapter 7 Discrete Probability Distributions 1 1...

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1 Chapter 7 Discrete Probability Distributions
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2 1. Counting 1. Permutations 2. The binomial coefficient 2. Binomial Distribution 3. Expectations 1. Linearity 2. Additivity of means 3. Additivity of independent variances 4. Cumulative Distribution Function (cdf)
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3 Permutations and Combinations • For any set {x 1 , …, x n } of n objects, there are exactly n! = n(n – 1)(n – 2)…(3)(2)(1) (linear) orderings of the objects. Each such ordering is a permutation . Incidentally, 0! = 1 (check your calculator) You can check your calculations using the Eviews command: show @fact(n) e.g. show @fact(6) returns 720
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4 Notice that permutations do not intrinsically involve “orderings”. n! is the number of ways that n objects can paired up in a one-to-one fashion with n categories, slots, other objects, etc. E.g., In a ballroom dance class with 5 men and 5 women, there are exactly 5! (= 120) ways to pair up the dancers. But talk of “orderings” and “lists” is very intuitively appealing, so we’ll often do it.
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5 How many ways are there to create an ordered list of length k from our set of n objects (assume k < n)? n(n – 1)(n – 2)…(n – k + 1)
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6 Notice also that a little algebra shows that: So there are n!/(n – k)! ways to form an k- length sequence using a set of n objects. You can check your work on Eviews using the command: show @fact(n)[email protected](n-k) e.g., show @fact(7)[email protected](7-5) returns 2520. )! ( ! ) 1 ( ) 2 )( 1 ( k n n k n n n n - = + - - -
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7 The binomial coefficient The term is known as the binomial coefficient , aka a combination . It is often written as , and pronounced “n choose k” It expresses: the number of ways that you can select k distinct unordered objects from a set of n objects . Let’s see how this is so in two ways. )! ( ! ! k n k n - k n
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8 First way: Select k objects . The number of ways of selecting k objects is: But since we don’t care about the order of these k objects, any list containing the same k objects should be “identified” For any k objects, we can form k! lists, so we have: ways of choosing k out of n objects ) 1 ( ) 2 )( 1 ( + - - - k n n n n = + - - - ! ) 1 ( ) 2 )( 1 ( k k n n n n = - - + - - - )! ( )! ( ! ) 1 ( ) 2 )( 1 ( k n k n k k n n n n = - )! ( ! ! k n k n k n
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9 Second way: select them all . Start by selecting all n objects in order (you will only keep the first k). There are n! ways of doing this. We don’t care about the ordering of the first k objects, or about the ordering of the last (n – k) objects. We only care about which objects are among the first k (and which are in the remaining (n – k)). So again, we have ways of choosing k out of n objects without = - )! ( ! ! k n k n k n
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Notice that we can cancel out many elements. E.g., calculating is not hard, and you can always eliminate the denominator: 5 15 5 15 ! 10 ! 5 ! 15 = ) 11 )( 13 )( = ) 11 )( 13 )( 7 ( 3 = 003 , 3 =
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11 Notice also that: - = k n n k n ))! ( ( )! ( !
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This note was uploaded on 01/20/2011 for the course ECON 15A taught by Professor Shirey during the Winter '08 term at UC Irvine.

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Chap07 - Chapter 7 Discrete Probability Distributions 1 1...

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