---------- ---------- ---------- ---------- ---------- ---------- ---------- ----------Other examples of distributions(seehere) # The time between the arrival at a traffic light (or at a highway toll booth) of a car # and the arrival of the next car, the time between the arrival of one phone call and # the other at the switchboard for a television broadcast, … on the one hand, the # number of car or the number of phone calls that arrive in a fixed time interval, … # on the other are example of two other laws of distribution: thenegative exponential#distributionand thePoisson distribution. Let's remember them through an example.# A 1 long wall; a w wide hole in the wall; every second a zombie arrives in a # completely random position of the wall. # Let w be 1/10; let's simulate 8.5 hours. # (1) The distribution (in 1 sec intervals) of the waiting times between a pass through # the hole of a zombie and the next pass. # (2) The distribution of the number of zombies that pass through the hole in 60 sec(1)(2)zombie = function() { w<<-1/10; TZ<<-NULL; h=8.5; sec=60*60*h i=1; n=0; NZ<<-NULL; u=0 while (i <= sec) { t=0; j=0; while(j<1) {j=ifelse(runif(1) < w, 1,0); t=t+1; r=i/60; u=u+j if(r==floor(r)) {NZ[r] <<- u; u=0}; i=i+1} n=n+1; TZ[n] <<- t } }# The assignment<<-also applies outside of the function definition# ifelse(runif(1)<w,1,0) is 1 if the zombie pass through the hole, 0 otherwise; t: time # between 2 passes; TZ: the waiting Times; NZ: the Number of zombies that pass in 60 sBF=3; HF=2.5 zombie(); noClass=1; histogram(TZ) # histogram of a continuous variable c(min(NZ),max(NZ))#1 15 noClass=1; Histogram(NZ, 0.5,15.5, 1) # histogram of a discrete variable# If we want the histogram of the density (ie the histogram of area 1): max( hist(TZ, right=FALSE, probability=TRUE, plot=FALSE)$density )#0.06664432 Plane(0,80, 0,0.07) hist(TZ, right=FALSE, probability=TRUE, col="grey",add=TRUE)(1) (2)statistics(TZ)#Min. 1st Qu. Median Mean 3rd Qu. Max.#1.00 3.00 7.00 10.27 14.00 78.00#The brown dots are 5^ and 95^ percentiles sqm(TZ)#9.86031 max( hist(NZ, seq(0.5,15.5,1), probability=TRUE, plot=FALSE)$density )#0.1666667 Plane(0,16, 0,0.17) hist(NZ, seq(0.5,15.5,1), probability=TRUE, col="grey",add=TRUE) statistics(NZ)#Min. 1st Qu. Median Mean 3rd Qu. Max.#1.000 4.000 6.000 5.847 7.000 15.000#The brown dots are 5^ and 95^ percentiles#The red dot is the mean sqm(NZ)#2.363145# It can be shown that, in the hypotheses made, TZ has the density function: # x ->w·exp(-w·x)[negativeexponentialdistribution with parameterw] # and that NZ has the law of distribution: # Pr(NZ=n) =exp(-a)·a[^{n}/n!Poissondistribution with parametera] # (a= average number of zombies that pass through the hole in 60 s; # usually, the parameter of the Poisson distribution is denoted bylambda) # In our case a=w*60=6. For the density function of the Poisson distribution I can use: k=0:100; P =dpois(k, lambda = 6)EN = function(x) w*exp(-w*x) Plane(0,80, 0,0.07) hist(TZ, right=FALSE, probability=TRUE, col="grey",add=TRUE) graph(EN,0,80, "brown"); integral(EN,0,Inf)#1 OK: the integral must be 1 # the mean: g = function(x) EN(x)*x; m = integral(g,0,Inf); m#10 (but I already knew that m = 10) # the Sd (or sqm): h = function(x) EN(x)*(x-m)^2; S = sqrt(integral(h,0,Inf)); S#10 # Thenegativehas both theexponentialdistributionmeanand theSdequal tow(1) (2)Plane(0,16, 0,0.17) hist(NZ, seq(0.5,15.5,1), probability=TRUE, col="grey",add=TRUE) polyline(k, p, "red"); POINT(k, p, "brown") sum(p[0:40]) # I must calcuate p[0]+p[1]+p[2]+... and the sum must be 1, but#1 OK it is enough p[0]+ … +p[40]: p[40]=2.7e-19, p[41]=4.1e-20, … # the mean: Wmean(0:100,dpois(0:100, lambda = 6))#6 # the variance: WVar(0:100,dpois(0:100, lambda = 6))#6 # ThePoissondistributionhas both themeanand thevarianceequal tolambda# It can be shown that, at the growth of lambda, the graph of Poisson distribution # tendes to confuse with a Gauss curve.