307_poisson_lecture

307_poisson_lecture - Lecture Sep 29 The random variable X...

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Unformatted text preview: Lecture Sep 29 The random variable X has the Poisson distribution with parameter λ > 0 if P ( X = k ) = exp ( −λ ) λ k k ! for each k = 0,1, 2,…. Please note: we will show that λ > 0 is the mean or average value of X . This is important in using what follows. This distribution is easy to lay out. The question is, when does it apply? To what sorts of counts? Two attacks on this: 1. “ the law of small numbers”: If X has the binomial distribution with n large and p small with np moderate, then the distribution of X is approximately Poisson with λ = np. Here is the theorem to this effect: Theorem: Let { pn } be a sequence of numbers in ( 0,1) such that npn → λ > 0. Let k be a k fixed positive integer. Then ⎜ ⎟ pn (1 − pn ) ⎛n⎞ ⎝k ⎠ n−k → λk k! exp ( −λ ) . Proof: Let ε n = λ − npn ∀n. Note that pn = k is fixed. Then ⎜ ⎟ pn (1 − pn ) λ − εn n , where ε n → 0. Also keep in mind that k ⎛n⎞ ⎝k ⎠ n−k = ⎜ ⎝ n ( n − 1) k! ( n − k + 1) ⎛ λ − ε n ⎞k ⎛1 − λ − ε n ⎞n−k ⎜ ⎝ n ⎟⎜ ⎠⎝ n ⎟ ⎠ ⎟ ⎠ The second and fourth terms (λ − εn ) = k! k n ( n − 1) nk ( n − k + 1) ⎛1 − λ − ε n ⎞n ⎛1 − λ − ε n ⎞− k . n ⎟⎜ ⎠⎝ n here converge to 1 ; the first term obviously converges to λk k! , and the third term has a well known limit of exp ( −λ ) . (To see this take the logarithm and use L’Hopital’s Rule.) This proves the theorem. 2. Poisson processes: A. Time case: Consider random events occurring in time—incoming phone calls, requests for service, auto accidents etc. for each t > 0, let N ( t ) be the number of events occurring in the time interval [ 0, t ] . We assume i) ii) iii) N ( 0 ) = 0. If s, t > 0 then N ( s + t ) − N ( s ) has the same distribution as N ( t ) . If 0 < t1 < t2 < < tn then N ( t1 ) , N ( t2 ) − N ( t1 ) ,… , N ( tn ) − N ( tn −1 ) are independent random variables. Note: (i) to (iii) tell us that our counting process is stationary with independent increments. (iv) There is a constant λ > 0 such that P N ( h ) = 1 = λ h + o ( h ) and ( ) P ( N ( h ) > 1) = o ( h ) , both as h → 0. Under these assumptions it can be shown that N ( t + s ) − N ( t ) has the Poisson distribution with parameter λ s. Such an N is a Poisson process with intensity parameter λ > 0. B. Space case: We can also model events occurring in two, three or higher dimensional space. A Poisson process with intensity parameter λ is a counting process N (a family of random variables giving a count for each region in the relevant space) such that for disjoint regions R1 , R2 ,… , Rn the random variables N ( R1 ) ,… , N ( Rn ) are independent, and for each region R the random variable N ( R ) has the Poisson distribution with parameter λ vol ( R ) . Examples: 1. A coin comes up heads with probability .01. We toss the coin 200 times. What is the probability we get 0 heads or 1 heads? Binomial dist answer: .404646 Poisson approx: .406006. Darts at Cooter Brown’s Raindrops Continuous r.v.s Def P(X=a) = 0 for every a. Means cdf is continuous Continuous cdf does not mean differentiable. But in practical cases derivative exists, and is the density of X. Explain densities. Examples ...
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