search for: ppoi

Le 04/12/2018 ? 11:27, I?aki Ucar a ?crit?: > On Tue, 4 Dec 2018 at 11:12, <qweytr1 at mail.ustc.edu.cn> wrote: >> function ppois is a function calculate the CDF of Poisson distribution, it should generate a non-decreasing result, but what I got is: >> >>> any(diff(ppois(0:19,lambda=0.9))<0) >> [1] TRUE >> >> Actually, >> >>> ppois(19,lambda=0.9)<ppois(18,lambda=0.9) >...

Full_Name: Murray H Smith Version: 0.63.3 OS: Windows NT Submission from: (NULL) (130.216.5.57) The ppois function is displaced by -0.5. Try: > ppois(-0.5,1) [1] 0.3678794 > ppois(-0.51,1) [1] 0 > ppois(0,1) [1] 0.3678794 and > par(mfrow=c(2,1)) > x<-seq(-1,5,0.01) > plot(x,ppois(x,1),type="s",ylab="F(x)",main="Poisson CDF?") > plot(x,pbinom(x,1...

function ppois is a function calculate the CDF of Poisson distribution, it should generate a non-decreasing result, but what I got is: > any(diff(ppois(0:19,lambda=0.9))<0) [1] TRUE Actually, > ppois(19,lambda=0.9)<ppois(18,lambda=0.9) [1] TRUE Which could not be TRUE. Code is tested in b...

>>>>> Serguei Sokol >>>>> on Tue, 4 Dec 2018 11:46:32 +0100 writes: > Le 04/12/2018 ? 11:27, I?aki Ucar a ?crit?: >> On Tue, 4 Dec 2018 at 11:12, <qweytr1 at mail.ustc.edu.cn> wrote: >>> function ppois is a function calculate the CDF of Poisson distribution, it should generate a non-decreasing result, but what I got is: >>> >>>> any(diff(ppois(0:19,lambda=0.9))<0) >>> [1] TRUE >>> >>> Actually, >>> >&gt...

Could somebody help me to understand the syntax of R's ppois function? I'm looking to calculate the cumulative probability density of an observed value (y) given the expected mean (mu) and the level of significance (alpha). I'm coming from using SAS to do this and don't recognize the descriptions of the arguments for ppois. The definitions of la...

...ing "inexactness": We work with R Version 1.8.1 (2003-11-21) on Windows systems (NT, 2000, XP). Try the code: ## Kolmogorov distance between two methods to ## determine P(Poisson(lambda)<=x) Kolm.dist <- function(lam, eps){ x <- seq(0,qpois(1-eps, lambda=lam), by=1) max(abs(ppois(x, lambda=lam)-cumsum(dpois(x, lambda=lam)))) } erg<-optimize(Kolm.dist, lower=900, upper=1000, maximum=TRUE, eps=1e-15) erg Kolm1.dist <- function(lam, eps){ x <- seq(0,qpois(1-eps, lambda=lam), by=1) which.max(abs(ppois(x, lambda=lam)-cumsum(dpois(x, lambda=lam)))) } Kolm1.dist(lam...

A quick question! The number of episodes per year of otitis media follows a Possion distribution with lambda = 1.6 episodes per year. Wouldn't the probability of getting 3 or more episodes of otitis media in the first 2 years of life be: > ppois(q=3, lambda=1.6*2, lower.tail = TRUE, log.p = FALSE) [1] 0.6025197 I am confused with the lambda and 3 or more.. thx much

...De manera muy comedida les pido ayuda con la función indicada abajo. La usan en un artículo de inventarios. No veo dónde estoy errado. Gracias de antemano. César Escalante C. > lamb<-50 > L<-1 > h.<-10 > p<-25 > K<-5 > integ<-function(y){integrate(function(x)ppois(x,lamb*L),0,y)$value} > G<-function(y){(h.+p)*integ(y) + p*(lamb*L-y)} > G(0:43) [1] 1250 1225 1200 1175 1150 1125 1100 1075 1050 1025 1000 975 950 925 900 875 850 825 800 775 750 725 700 675 650 625 600 575 550 525 [31] 500 475 450 425 400 375 350 325 300 275...

The expected number of bladder cancer over next 20 years a tire industry is 1.8. Poission distribution is assumed to hold and 6 reported deaths are caused by bladder cancer among the employees. Trying to find how unusual this event is. > ppois(q=6, lambda=1.8, lower.tail = TRUE, log.p = FALSE) [1] 0.9974306 not sure if ppois is the right one to use and the parameters... thx much

...std::cout << pf(1.5, 2.0, 3.0, 1, 0) << std::endl; Outputs 1, expect 0.646447 std::cout << pchisq(1.5, 2.0, 1, 0) << std::endl; Outputs 1, expect 0.527633 std::cout << pgamma(1.5, 2.0, 3.0, 1, 0) << std::endl; Outputs 1, expect 0.090204 std::cout << ppois(2.0, 5.0, 1, 0) << std::endl; Outputs 0, expect 0.124652 std::cout << qchisq(0.5, 2.0, 1, 0) << std::endl; Outputs -0.61379, expect 1.38629 std::cout << qgamma(0.5, 2.0, 3.0, 1, 0) << std::endl; Outputs 0.0282703, expect 5.03504 I should stress that within the R...

Hello, I am performing a sensitivity analysis using a Latin Hypercube sampling. However, I have difficulty to draw a Hypercube sample for one variable. I?ve generated this variable from a Poisson distribution as follows: set.seed(5) mortality_probability <- round(ppois(seq(0, 7, by = 1), lambda = 0.9), 2) barplot(mortality_probability, names.arg = seq(0, 7, by = 1), xlab = "Age class", ylab = "Probability") How can I draw a Hypercube sample for the variable ?mortality_probability? so that this variable exhibits the same pattern as the observ...

...dunif.o dweibull.o fmax2.o fmin2.o ftrunc.o gamma.o gamma_cody.o i1mach.o imax2.o imin2.o lbeta.o lgamma.o lgammacor.o mlutils.o pbeta.o pbinom.o pcauchy.o pchisq.o pexp.o pf.o pgamma.o pgeom.o phyper.o plnorm.o plogis.o pnbeta.o pnbinom.o pnchisq.o pnf.o pnmath.o pnorm.o pnt.o polygamma.o ppois.o pt.o ptukey.o punif.o pweibull.o qbeta.o qbinom.o qcauchy.o qchisq.o qexp.o qf.o qgamma.o qgeom.o qhyper.o qlnorm.o qlogis.o qnbeta.o qnbinom.o qnchisq.o qnf.o qnorm.o qnt.o qpois.o qt.o qtukey.o qunif.o qweibull.o sign.o stirlerr.o toms708.o -lgomp -F/Library/Frameworks/R.framework/.....

I have problem evaluating the expression h = exp(x)/(exp(exp(x))-1) for large negative x. This expression is actually the probability that y = 1 when y is a Poisson random variable truncated at 0, hence must satisfy 0 <= h <= 1. However, when x < -18, I may get an h value that is larger than 1 while the true value should be a quantity that is smaller than but very close to 1. For

There is a problem with `rpois'. It does seem to take care about the order of the arguments. This is an example: > rpois(n=1,lambda=2) [1] 3 > rpois(lambda=2,n=1) [1] 2 0 It obviously uses the first argument as the number of samples to be drawn, which is wrong. I used Version 0.49 Beta (April 23, 1997). Fredrik

There is a problem with `rpois'. It does seem to take care about the order of the arguments. This is an example: > rpois(n=1,lambda=2) [1] 3 > rpois(lambda=2,n=1) [1] 2 0 It obviously uses the first argument as the number of samples to be drawn, which is wrong. I used Version 0.49 Beta (April 23, 1997). Fredrik

...ith a given parameter. I would like to know what is the proper way to do a ''goodness of fit'' test using R. I know the steps I''d take if I were to do it ''manually'': grouping the numbers into classes, calculating the expected frequencies using ''ppois'', then calculating Chi_2_obs = Sum (e_i - o_i)^2/e_i) (where e_i represents the expected frequencies and o_i the observeds ones) and then finally calculating the p-value (using pchisq). I''ve read a lot of documentation, also tried googling for ''goodness of fit R'&...

> How can I draw a Hypercube sample for the variable mortality_probability so > that this variable exhibits the same pattern as the observed distribution? One simple way is to use the uniform random output of randomLHS as input to the quantile function for your desired distribution(s). For example: q <- randomLHS(1000, 3) colnames(q) <- c("A", "B",

Thanks for your answer. However, my variable is simulated from the cumulative distribution function of the Poisson distribution. So, the pattern obtained from the function "qpois" is not the same as the observed pattern (i.e., obtained from the function "ppois") set.seed(5) mortality_probability <- round(ppois(seq(0, 7, by = 1), lambda = 0.9), 2) barplot(mortality_probability, names.arg = seq(0, 7, by = 1), xlab = "Age class", ylab = "Probability") library(lhs) set.seed(1) parm <- c("var1", "var2", &q...

...t ######################################### No_of_Frouds<- c(4,1,6,9,9,10,2,4,8,2,3,0,1,2,3,1,3,4,5,4,4,4,9,5,4,3,11,8,12,3,10,0,7) N <- length(No_of_Frouds) # Estimation of Parameter lambda<- sum(No_of_Frouds)/N lambda pmf <- dpois(i, lambda, log = FALSE) step_function <- ppois(i, lambda, lower.tail = TRUE, log.p = FALSE) # Chi-Squared Goodness of Fit Test # Ho: The data follow a Poisson distribution Vs H1: Not Ho Frauds <- c(1:13) counts<- c(2,3,3,5,7,2,1,1,2,3,2,1,1,0) # Observed frequency Expected <-c(0.251005528,1.224602726,2.987288468,4.85811559,5.9...

...pt.o qt.o rt.o dnt.o dchisq.o pchisq.o qchisq.o rchisq.o rnchisq.o dbinom.o pbinom.o qbinom.o rbinom.o rmultinom.o dcauchy.o pcauchy.o qcauchy.o rcauchy.o dexp.o pexp.o qexp.o rexp.o dgeom.o pgeom.o qgeom.o rgeom.o dhyper.o phyper.o qhyper.o rhyper.o dnbinom.o pnbinom.o qnbinom.o rnbinom.o dpois.o ppois.o qpois.o rpois.o dweibull.o pweibull.o qweibull.o rweibull.o dlogis.o plogis.o qlogis.o rlogis.o dnchisq.o pnchisq.o qnchisq.o dnbeta.o pnbeta.o qnbeta.o pnf.o pnt.o qnf.o qnt.o ptukey.o qtukey.o toms708.o wilcox.o signrank.o ranlib libnmath.a config.status: creating src/unix/Makefile make: /u...