The Poisson Distribution

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1 MATH 382 The Poisso Distributio Dr. Neal, WKU Oe of the importat distributios i probabilistic modelig is the Poisso Process X t that couts the umber of occurreces over a period of t uits of time. This process is based upo a Poisso distributio X ~ Poi( λ) that couts the umber of occurreces durig a fixed period of time, where λ > 0 is the average umber of occurreces per uit time. Geeral Assumptios Whe ca we assume that measuremets follow a Poisso distributio? Basically, the followig geeral coditios must be reasoably satisfied: (i) The average umber of occurreces λ per uit time should be costat ad should remai proportioal to the legth of all time itervals of legth t so that λ t is the average umber of occurreces through t uits of time. (ii) Occurreces should happe idepedetly. (iii) Durig ay small icremet of time, there are a large umber of potetial sample poits that may give rise to a occurrece, but the probability of a actual occurrece from ay of these poits is very small. Uder these coditios, the we ca approximate the probability of exactly k occurreces by i uit time by P(X = k ) e λ for, 1, 2, 3,... ad probability of exactly k occurreces i time t is give by P(X t = k ) = k e λ t for, 1, 2, 3,... Usig the fact that rage of X t equals 1: x k = e x, we ca verify the sum of the probabilities over the (λ t ) k e λ t = e λ t k = e λ t e λ t = 1. There is o closed-form formula for the cumulative probability P(X k) or for computig probabilities such as P( j X k). I each case, the idividual probabilities must be summed, or we must use a calculator/computer commad: k k λ i e P(X k) = P( X = i ) = λ ad P( j X k) = i = 0 i = 0 i! k λ i e λ. i! i = j

2 Techically, the pdf formula for X arises from the limit of the b(, p) pdf as the umber of attempts icreases to, while the probability of occurrece p decreases to 0, but while the average umber of occurreces λ = p remais costat: P( X = k ) = lim P(b(, p) = k) = lim p 0 p=λ = lim! λ ( k )! lim lim e λ k p 0 p=λ p k k (1 p) k k ( 1)... ( k +1) k 1... k + 1 ( ) 1 e λ. k lim k lim Mea, Variace, ad Stadard Deviatio For X ~ Poi( λ ), it is specified i advace that the average value is λ. But we shall verify this result usig the defiitio of expected value ad also show that its stadard deviatio is λ. To do so, we shall work with the geeral process X t ~ Poi( λ t ) ad simply let t = 1 to get the result for X ~ Poi( λ). The derivatio is as follows: E[ X t ] = k P( X t = k ) = k k e λ t = k k e λ t k Rage X t = e λ t k = e λ t k +1 (re - idex ) (k 1)! = e λ t = λ t. k = e λ t (λ t )e λ t (first term cacels ) To fid the variace of X t, we first fid the average square of A t by

3 2 E[ X t ] = k 2 k e λ t = k 2 k e λ t (first term cacels ) = k (λ t )k e λ t = (k + 1) k +1 e λ t (re - idex ) (k 1)! = k k e λ t + (λ t ) (λ t ) k e λ t = E[ X t ] + (λ t ) P( X t = k ) = (λ t ) + (λ t )(1) =(λ t ) 2 + λ t. Dr. Neal, WKU So Var( X t ) = E[ X t 2 ] (E[ Xt ]) 2 = (λ t ) 2 + λ t (λ t ) 2 = λ t, ad its stadard deviatio is σ ( X t ) = σ X = λ. λ t. Lettig t = 1 uit of time, we have E[ X ] = λ for X ~ Poi(λ ) with Mode The mode of X t ~ Poi( λ t ) is the most likely umber of occurreces i time t. This value is always give by the largest iteger k such that k λ t, deoted by k = λ t. However, whe λ t is actually a iteger, the the Poi( λ t ) will be bi-modal. I this case, this value of k = λ t ad the previous iteger k 1 will be the two modes. To derive this expressio for the mode, we let a k = P( X t = k ) = (λ t )k e λ t. We the wish to fid the largest iteger k such that a k 1 a k. (We the have a k > a k +1 > a k+2 >...> a makig k have the highest probability value a k.) We ow cosider the ratio a k / a k 1. To have a k 1 a k, we must have 1 a k / a k 1, which holds if ad oly if 1 a k a k 1 = k e λ t k 1 e λ t (k 1)! = λ t k. This result holds if ad oly if k λ t. Note that we have a k 1 = a k if ad oly if k = λ t, which ca hold if ad oly if λ t is a iteger. Example. Durig eveig hours, a website averages 2.4 ew hits every 30 secods. Let X cout the umber of ew hits i oe miute durig eveig hours. Let X t cout the umber of ew hits i t miutes durig eveig hours.

4 (a) Explai why the coditios of a Poisso distributio are satisfied. (b) For a give miute, compute the probability of there beig (i) exactly 5 ew hits (ii) at least 4 ew hits (iii) from 3 to 5 ew hits (c) What is the most likely umber of ew hits durig a give miute? (d) What is the probability of o ew hits i t miutes time? (e) Derive the maximum amout of time durig which the probability of o ew hits is at least Solutio. (a) We may assume that hits to the website geerally occur idepedetly of each other from differet people all over cyberspace who have o coectio to each other. Durig ay istat of time, there are umerous people olie who may potetially visit the page, but the actual probability of doig so is very small. The average umber of hits is kow ad is assumed to stay costat durig eveig hours. Now we ca say that, o average, there are λ = = 4.8 hits every miute. Thus, X ~ Poi(4.8) ad X t ~ Poi (4.8t ) so that durig a give miute P(X = k ) = 4.8k e 4.8 for, 1, 2, 3,..., ad P(X t = k ) = (4.8t )k e 4.8t durig t miutes time. (b) Usig the pdf of X, we obtai (i) P(X = 5) = 4.85 e 4.8 5! , (ii) P( X 4) = 1 P(0 X 3) = 1 e ! ! = 1 e ! ! ad (iii) P( 3 X 5) = e ! ! ! (c) The most likely umber of ew hits durig a give miute is k = 4.8 = 4 hits. (d) The probability of o ew hits i t miutes time is P(X t = 0) = e 4.8t. (e) We simply solve for t is the iequality P(X t = 0) Usig the result form (d), we have e 4.8t 0.90 which gives 4.8 t l(0. 90) ad the t l(0.90) miutes, or about secods.

5 Calculator Commads Poisso probability values ca be computed with the built-i poissopdf( ad poissocdf( TI commads from the DISTR meu: P( X = k ) = poissopdf(λ,k) P( X k) = poissocdf(λ,k) P( j X k ) = P( X k) P( X j 1) = poissocdf(λ,k) poissocdf(λ,j 1) P( X k) = 1 P( X k 1) = 1 poissocdf(λ,k 1) For X ~ Poi(4.8), we have (i) P(X = 5) = poissopdf(4.8,5) (ii) P( X 4) = 1 P(X 3) = 1 poissocdf(4.8,3) (iii) P( 3 X 5) = poissocdf(4.8,5) poissocdf(4.8,2) Mathematica Commads We also ca use Mathematica commads to compute Poisso probability values: X=PoissoDistributio[4.8]; PDF[X,5] CDF[X,4] Mea[X] StadardDeviatio[X] Exercise Traffic egieers ofte use a Poisso distributio to model the flow of cars i light traffic. For 15 miute itervals i the late morig hours, a average of 2.5 cars pass through a particular stop sig without makig a complete stop. Let X cout the umber of such cars durig ay give hour i the late morig. (a) What is the distributio of X? (b) Durig a give hour, what is the most likely umber of cars to pass through without makig a complete stop? (c) For a give hour, compute the probability that (i) from 8 to 12 cars pass through without makig a complete stop (ii) at least 8 cars pass through without makig a complete stop (d) Derive the miimum amout of time t durig which the probability of at least oe car passig through without makig a complete stop is at least 0.95.

BHW #13 1/ Cooper. ENGR 323 Probabilistic Analysis Beautiful Homework # 13

BHW #13 1/ Cooper. ENGR 323 Probabilistic Analysis Beautiful Homework # 13 BHW # /5 ENGR Probabilistic Aalysis Beautiful Homework # Three differet roads feed ito a particular freeway etrace. Suppose that durig a fixed time period, the umber of cars comig from each road oto the