Derivation of the Poisson distribution

Transcription

1 Gle Cowa RHUL Physics 1 December, 29 Derivatio of the Poisso distributio I this ote we derive the fuctioal form of the Poisso distributio ad ivestigate some of its properties. Cosider a time t i which some umber of evets may occur. Examples are the umber of photos collected by a telescope or the umber of decays of a large sample of radioactive uclei. Suppose that the evets are idepedet, i.e., the occurrece of oe evet has o ifluece o the probability for the occurrece of aother. Furthermore, suppose that the probability of a sigle evet i ay short time iterval δt is P(1;δt) = λδt, (1) where λ is a costat. I Sectio 1 we will show that the probability for evets i the time t is give by where the parameter ν is related to λ by P(;ν) = ν e ν, (2) ν = λt. (3) We will follow the covetio that argumets i a probability distributio to the left of the semi-colo are radom variables, that is, outcomes of a repeatable experimet, such as the umber of evets. Argumets to the right of the semi-colo are parameters, i.e., costats. The Poisso distributio is show i Fig. 1 for several values of the parameter ν. I Sectio 2 we will show that the mea value of the Poisso distributio is give by ad that the stadard deviatio σ is = ν, (4) σ = ν. (5) The mea ν roughly idicates the cetral regio of the distributio, but this is ot the same as the most probable value of. Ideed is a iteger but ν i geeral is ot. The stadard deviatio is a measure of the wih of the distributio. 1 Derivatio of the Poisso distributio Cosider the time iterval t broke ito small subitervals of legth δt. If δt is sufficietly short the we ca eglect the probability that two evets will occur i it. We will fid oe evet with probability 1

2 P (; ν).4.2 ν=2 P (; ν).4.2 ν=5 P (; ν).4.2 ν=1 Figure 1: The Poisso distributio P(;ν) for several values of the mea ν. P(1;δt) = λδt (6) ad o evets with probability P(;δt) = 1 λδt. (7) What we wat to fid is the probability to fid evets i t. We ca start by fidig the probability to fid zero evets i t, P(;t) ad the geeralize this result by iductio. Suppose we kew P(;t). We could the ask what is the probability to fid o evets i the time t + δt. Sice the evets are idepedet, the probability for o evets i both itervals, first oe i t ad the oe i δt, is give by the product of the two idividual probabilities. That is, This ca be rewritte as P(;t+δt) = P(;t)(1 λδt). (8) P(;t+δt) P(;t) δt = λp(;t), (9) which i the limit of small δt becomes a differetial equatio, dp(;t) = λp(;t). (1) Itegratig to fid the solutio gives P(;t) = Ce λt. (11) For a legth of time t = we must have zero evets, i.e., we require the boudary coditio P(;) = 1. The costat C must therefore be 1 ad we obtai 2

3 P(;t) = e λt. (12) Now cosider the case where the umber of evets is ot zero. The probability of fidig evets i a time t+δt is give by the sum of two terms: P(;t+δt) = P(;t)(1 λδt)+p( 1;t)λδt. (13) The first term gives the probability to have all evets i the first subiterval of time t ad the o evets i the fial δt. The secod term correspods to havig 1 evets i t followed by oe evet i the last δt. I the limit of small δt this gives a differetial equatio for P(;t): dp(;t) +λp(;t) = λp( 1;t). (14) We ca solve equatio (14) by fidig a itegratig factor µ(t), i.e., a fuctio which whe multiplied by the left-had side of the equatio results i a total derivative with respect to t. That is, we wat a fuctio µ(t) such that We ca easily show that the fuctio [ ] dp(;t) µ(t) +λp(;t) = d [µ(t)p(;t)]. (15) has the desired property ad therefore we fid µ(t) = e λt (16) d [ ] e λt P(;t) = e λt λp( 1;t). (17) We ca use this result, for example, with = 1 to fid d [ ] e λt P(1;t) = λe λt P(;t) = λe λt e λt = λ, (18) where we substituted our previous result (12) for P(; t). Itegratig equatio (18) gives e λt P(1;t) = λ = λt+c. (19) Now the probability to fid oe evet i zero time must be zero, i.e., P(1;) = ad therefore C =, so we fid P(1;t) = λte λt. (2) We ca geeralize this result to arbitrary by iductio. We assert that the probability to fid evets i a time t is P(;t) = (λt) e λt. (21) 3

4 We have already show that this is true for = as well as for = 1. Usig the differetial equatio (17) with +1 o the left-had side ad substitutig (21) o the right, we fid d [ e λt P(+1;t) Itegratig equatio (22) gives ] = e λt λp(;t) = e λt λ (λt) e λt = λ (λt). (22) e λt P(+1;t) = λ (λt) = (λt)+1 +C. (23) (+1)! Imposig the boudary coditio P(+1;) = implies C = ad therefore P(+1;t) = (λt)+1 (+1)! e λt. (24) Thus the assertio (21) for also holds for +1 ad the result is proved by iductio. 2 Mea ad stadard deviatio of the Poisso distributio First we ca verify that the sum of the probabilities for all is equal to uity. Usig ow ν = λt, we fid P(;ν) = = e ν = e ν e ν ν e ν ν = 1, (25) where we have idetified the fial sum with the Taylor expasio of e ν. The mea value (or expectatio value) of a discrete radom variable is defied as = P(), (26) where P() is the probability to observe ad the sum exteds over all possible outcomes. I the case of the Poisso distributio this is = P(;ν) = ν e ν. (27) To carry out the sum ote first that the = term is zero ad therefore 4

5 = e ν =1 = νe ν =1 = νe ν = νe ν e ν m= ν ν 1 ( 1)! ν m m! = ν. (28) Here i the third lie we simply relabelled the idex with the replacemet m = 1 ad the we agai idetified the Taylor expasio of e ν. To fid the stadard deviatio σ of we use the defiig relatio σ 2 = 2 2. (29) We already have, ad we ca fid 2 usig the followig trick: 2 = ( 1) +. (3) We ca fid ( 1) i a maer similar that used to fid, amely, ( 1) = ( 1) ν e ν = ν 2 e ν =2 = ν 2 e ν = ν 2 e ν e ν m= ν 2 ( 2)! ν m m! = ν 2, (31) where we used the fact that the = ad = 1 terms are zero. I the third lie we relabelled the idex usig m = 2 ad idetified the resultig series with e ν. Puttig this ito equatio (29) for σ 2 gives σ 2 = ν 2 +ν ν 2 = ν or σ = ν. (32) This is the importat result that the stadard deviatio of a Poisso distributio is equal to the square root of its mea. 5