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Stat150_Spring08_Poisson_processes

Stat150_Spring08_Poisson_processes - Statistics 150 Spring...

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Statistics 150: Spring 2007 January 22, 2008 0-1
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NOTE: These slides are not meant to be complete “lecture notes” that replace attending class. Rather, they are intended to act as a basis for the discussion in class and you will need to attend class to flesh out the details. 1
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1 Examples of discrete variables Bernoulli trials. A Bernoulli random variable X takes values 1 and 0 with probability p and q (= 1 - p ) , respectively. Sometimes we think of these values as representing the ’success’ or the ’failure’ of a trial. The mass function is f (0) = 1 - p, f (1) = p, and it follows that E [ X ] = p and var [ X ] = p (1 - p ) . 2
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Binomial distribution. Perform n independent, identically distributed Bernoulli trials X 1 , X 2 , . . . , X n and count the total number of successes Y = X 1 + X 2 + · · · + X n . The mass function of Y is f ( k ) = n k p k (1 - p ) n - k , k = 0 , 1 , . . . , n. and we can compute that E [ Y ] = np, var [ Y ] = np (1 - p ) . 3
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Poisson distribution. A Poisson variable is a random variable with the Poisson mass function f ( k ) = λ k k ! e - λ , k = 0 , 1 , 2 , . . . for some λ 0 . Suppose Y be a bin( n, p ) variable and let n → ∞ and p 0 together in such a way that E [ Y ] = np approaches a non-zero constant λ . Then, for k = 0 , 1 , 2 , . . . , P { Y = k } = n k p k (1 - p ) n - k 1 k ! np 1 - p k (1 - p ) n λ k k ! e - λ . * Check that both the mean and the variance of this distribution are equal to λ . 4
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2 Total variation distance Let F and G be the distribution functions of discrete distributions which place masses f n and g n at the points x n , for n 1 , and define d T V ( F, G ) = k 1 | f k - g k | . The quantity d T V ( F, G ) is called the total variation distance between F and G . 5
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For random variables X and Y , define d T V ( X, Y ) = d T V ( F X , F Y ) . Note that d T V ( X, Y ) = 2 sup A R | P { X A } - P { Y A } | for discrete random variables X, Y . 6
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3 Poisson convergence Theorem 3.1. Let { X r : 1 r n } be independent Bernoulli random variables with respective parameters { p r : 1 r n } , and let S = n r =1 X r . Then d T V ( S, P ) 2 n r =1 p 2 r where P is a random variable having the Poisson distribution with parameter λ = n r =1 p r . 7
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Proof. Let ( X r , Y r ) , 1 r n , be a sequence of independent pairs, where the pair ( X r , Y r ) takes values in the set { 0 , 1 } × { 0 , 1 , 2 , . . . } with mass function P { X r = x, Y r = y } = 1 - p r , if x = y = 0; e - p r - 1 + p r , if x = 1 , y = 0; p y r y ! e - p r , if x = 1 , y 1 . It is easy to check that X r is Bernoulli with parameter p r , and Y r has the Poisson distribution with parameter p r . 8
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Set S = n r =1 X r , P = n r =1 Y r . Note that since the sum of Poissons is also Poisson, P has the Poisson distribution with parameter λ = n r =1 p r . 9
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Now, since | a - b | ≤ | a | + | b | , | P { S = k } - P { P = k } | = | P { S = k, P = k } - P { S = k, P = k } | P { S = k, S = P } + P { P = k, S = P } . Hence d T V ( S, P ) = k | P { S = k } - P { P = k } | ≤ 2 P { S = P } . 10
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Observe that P { S = P } P { X r = Y r for some r } ≤ n r =1 P { X r = Y r } = n r =1 { e - p r - 1 + p r + P { Y r 2 }} = n r =1 p r (1 - e - p r ) n r =1 p 2 r .
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