search for: probabilites

Dear list member, I struggle with the problem, why the probabilities of choosing one of three mutually exclusive alternatives don?t sum up to 1! Let?s assume we have three alternatives X, Y, and Z. Let?s further assume we know their respective utilities: uX, uY, uZ. I?m interested in calculating the probability of choosing X, Y, and Z. Since I assume that the alternatives are mutually

I would like to compute the MC test (rfm.test) available in the package MarkedPointProcess (for the data BITOEK for example) in order to test the dependence between the marks and their locations. Why the syntax of rfm.test is false here? I have the message : ****************************** ML WARNING! Forbidden values! -- if there are too many warnings try narrower lower and upper bounds for the

...ins). This works fine. T is a factored vector with levels cancer, noncancer. Proteins are numeric. Now, I want to use predict.glm to predict a new data. predict(model, newdata=testsamples, type="response") (testsamples is a small set of new samples). The result is a vector of the probabilites for each sample in testsamples. But probabilty WHAT for? To belong to the first level in T? To belong to second level in T? Is this fallowing expression factor(predict(model, newdata=testsamples, type="response") >= 0.5) TRUE, when the new sample is classified to Cancer or when it...

...an association between events A:{a,not(a)} and B:{b,not(b)} | b | not(b) | --------+-----+--------+ a | 1 | 13714 | --------+-----+--------+ not(a) | 506 | 878702 | --------+-----+--------+ I am worried that prop.test and chisq.test are not valid given the low counts and low probabilites associated with 'sucess' in each category. Is it safe to use them, and what is the alternative? (given that fisher.test can't handle this data... hold the phone... I just found fisher.test can handle this data if the test is one-tailed and not two-tailed. I don't understand the d...

Maybe its slightly off-topic, but can anybody help with computing entropy on matrix of probabilities? Guess we have a matrix of probabilites, A, 2x2, something like this: z x 0 1 2 3 4 0 0.063 0.018 0.019 0.016 0.000 1 0.011 0.162 0.040 0.042 0.003 2 0.015 0.030 0.164 0.033 0.002 3 0.012 0.035 0.036 0.159 0.002 4 0.004 0.021 0.018 0.013 0.082 sum(A)=1 Can i compute entropy like this: entropy<-su...

Hi. I have a vector of quantiles and a vector of probabilites that, when plotted, look very like the gamma cumulative distribution function. I can guess some shape and scale parameters that give a similar result, but I'd rather let the parameters be estimated. Is there a direct way to do this in R? Thanks, Tim. week <- c(0,5,6,7,9,11,14,19,39)...

Hi, I am interested in using a bayesian network as a predictor (machine learning); however, I can't get any of the implementations (deal, nblearn) to learn & predict stuff. Shouldn't there also be probabilites for each node after the learning phase, how can I access these? Cheers, Stephan -- View this message in context: http://www.nabble.com/Prediction-with-Bayesian-Network--tp18220601p18220601.html Sent from the R help mailing list archive at Nabble.com.

Dear experts, I am looking for a package that does logistic regression with corssvalidation and gives me the probabilites of all the corssvalidations so that I can plot them in ROCR. Would also like to know if the corssvalidate model would give me a summary coefficient for the intercept and my to predictors. Thank you. Best regards, Jon Toledo, MD Postdoctoral fellow University of Pennsylvania School of Medicine Cen...

Hi, my question is probably amateurish but I can't seem to find the answer anywhere. In what metric are the MCMCglmm package's posterior means for family = "categorical"? I suppose that they can't be odds ratios and probabilites as my numbers are outside their bounds. So I'm thinking ? are they just basic regression coefficients conceptually equal to those obtained by lme4::glmer? Thank you! [[alternative HTML version deleted]]

I have two problems for the data processing of my large data base (50000 rows). For example, a sample is as follows Num <- c(1,2,3,4,4,4,5,5) Date <- c("1/1/04 0:48","1/1/04 1:52", "1/1/04 1:55", "1/1/04 2:14", "1/1/04 3:09", "1/1/04 8:02", "1/1/04 9:05", "1/1/04 9:06") Place <-

...ed papers on this. One thing I don't know how to do is to standardize the probabilities by race. For example, if I train an LRM on a bunch of variable I get a model. I can then get probability predictions from the model. That works. It seems to me, that for a given race (8-12 horses) the probabilites of my predictions should sum to one. 1) Is there some way to train the LRM to evaluate and then model the subsequent date "per race"?? (Perhaps indicate some kind of grouping variable? 2) Alternately, if I just run my data through a "standard" LRM, is there some way to th...

...le niveau de test ? 10% # Cr?ation d'une liste n11 qui contient les tailles des differents groupes n11=list() for (i in 1:q){ n11[[i]]=rep(as.integer(n/R[i]),R[i]) n11[[i]][R[i]]=n-((R[i]-1)*n11[[i]][1]) } # Cr?ation des listes P11 et P12 qui contient les probabilit?s et # les inverses des probabilites empiriques des differents groupes respectivement P11=list() P12=list() for (i in 1:q){ P11[[i]]=n11[[i]]/n P12[[i]]=n/n11[[i]] } # cr?ation d'une liste contenant les matrices W W=list() for (i in 1:q){ w=matrix(0,n,R[i]) w[1:n11[[i]][1],1]=1 for (j in 2:R[i]){ s1=sum(n11[[i]][1:(j-1)]) w[(1+...

Hi all, I am wondering whether there is a small bug in the binom.test function of the ctest library (I'm using R 1.6.0 on windows 2000, but Splus 2000 seems to have the same behaviour). Or perhaps I've misunderstood something. the command binom.test(11,100,p=0.1) and binom.test(9,100,p=0.1) give different p-values (see below). As 9 and 11 are equidistant from 10, the mean of the

I want to obtain datetime differences in mins in an other column, in front of my datetimes. I have tried this : T1 <- c("12/31/03 23:49","1/1/04 1:14","1/1/04 0:02") T2 <- c("1/1/04 0:58","1/1/04 1:16","") toto <- data.frame(T1,T2) toto y <- strptime(T1,"%m/%d/%y %H:%M") x <- strptime(T2,"%m/%d/%y

Hello Could someone please direct me to the correct commands for adding accents (grave and aigu) to a letter in a plot title, label, or in added text? I'm sure there's a handy list somewhere, but I've failed in coming up with the correct search words to find it. Thank you muchly! Jen

...hey suggested: *"Could you simple take the negative of the logarithm? Since you are dealing with probabilities (i.e. values <= 1), the logarithm is always negative, so negating it will always make it positive."* But this approach wont be a good idea as large values will indicates low probabilites,small values will indicate high probabilities.Hence matcher will tend to skip some good documents from ranked list due to lower weight. Thanks, *-- * with regards Gaurav A. -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://lists.xapian.org/pipermail/xapian...

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, >> >>>

It looks like a homework assignment. It also looks like you didn't read the documentation carefully enough. The 'len.out' argument in seq is solely for specifying the length of a sequence. The 'quantile' function omputes the empirical quantile of raw data in the vector 'x' at cumulative probabilit(y)(ies) given in the weights' argument, with interpolation I'm

Hi all, I was working with some probabilities and wanted to store some small values. For example: > x = 2e-250 > x [1] 2e-250 > y = 2e-300 > y [1] 2e-300 > z = 2e-350 > z [1] 0 Is there any way to store a small value (e.g. 2e-350) in R? thanks! [[alternative HTML version deleted]]

...in a way that return consistent result for ppois()". You have > indicated one way to do so (I have just added exp() in the row): > any(diff(exp(ppois(0:19, lambda=0.9, log.p=TRUE))) < 0) > #[1] FALSE > But may be there is another, more economic way? Well, log probabilites *are* very economic for many such p*() functions. OTOH, I'm a bit surprised that nobody mentioned the 'lower.tail=FALSE' option which here makes so very much sense, and is I think slightly more intuitive than the log-probabilities: It gives much much much more accurate results for suc...