Stat150_Spring08_Poisson_processes

Stat150_Spring08_Poisson_processes - Statistics 150: Spring...

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Unformatted text preview: Statistics 150: Spring 2007 January 22, 2008 0-1 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 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 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 Poisson distribution. A Poisson variable is a random variable with the Poisson mass function f ( k ) = k k ! e- , k = 0 , 1 , 2 ,... for some . Suppose Y be a bin( n,p ) variable and let n and p together in such a way that E [ Y ] = np approaches a non-zero constant . 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 k k ! e- . * Check that both the mean and the variance of this distribution are equal to . 4 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 TV ( F,G ) = X k 1 | f k- g k | . The quantity d TV ( F,G ) is called the total variation distance between F and G . 5 For random variables X and Y , define d TV ( X,Y ) = d TV ( F X ,F Y ) . Note that d TV ( X,Y ) = 2 sup A R | P { X A } - P { Y A }| for discrete random variables X,Y . 6 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 TV ( S,P ) 2 n X r =1 p 2 r where P is a random variable having the Poisson distribution with parameter = n r =1 p r . 7 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 { , 1 } { , 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 ....
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Stat150_Spring08_Poisson_processes - Statistics 150: Spring...

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