summary_of_exp_and_Poisson_process

summary_of_exp_and_Poisson_process - N s 2 and N t 1 N s 1...

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Summer 2007 STAT333 Summary of the exponential distribution and the Poisson process Exponential distribution: (a) pdf function: f ( x )= λe λx ,x> 0and λ> 0; (b) tail probability: P ( X>x )= e λx ; (c) expectation=1 and variance=1 2 ; (d) no-memory property: P ( X>s + t | X>s )= P ( X>t ), t, s > 0; (understanding: no matter how long we already spent, the remaining time has the same distribution as the original waiting time) (e) Alarm-clock lemma: suppose X 1 ,...,X n are independent exponential random variables and X i Exp ( λ i ), i =1 ,...,n , i. min { X 1 ,...,X n } follows exponential distribution with rate n i =1 λ i ; ii. P ( X i =m in { X 1 ,...,X n } )= λ i n k =1 λ k . Counting process: { N ( t ) ,t 0 } is a counting process if (a) N ( t ) 0; (b) N ( t ) can only be integers; (c) N ( t ) is a increasing function of t . Poisson process (a) { N ( t ) ,t 0 } is a Poisson process with rate λ if i. N ( t ) is a counting process and N (0) = 0; ii. “independent increment”: for 0 s 1 <t 1 s 2 <t 2 , N ( t
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Unformatted text preview: ) N ( s 2 ) and N ( t 1 ) N ( s 1 ) are independent; (intuitively understanding: the numbers of the events in two intervals without overlap are independent) iii. N ( t ) N ( s ) P oisson ( ( t s )). (b) Properties: i. Let T 1 be the waiting time for the 1st event, T i be the waiting time for the i th event after the ( i 1)th event, i = 2 , . . . . Then T 1 , T 2 , . . . are iid exponential random variables with rate . ii. Given X ( t ) = n , then X ( s ) follows Binomial ( n, s/t ) when s t . That is X ( s ) | X ( t ) = n Binomial ( n, s/t ) . iii. lim h P ( N ( h ) 2) h = 0, which means in a very small interval, we can only see one or 0 event. (Try to understand this point from the E matrix in the continuous Markov chain.) 1...
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