set the lower bound of normal distribution to 0 ?

Dear Tom,

In my opinion you should first transform your data to the log-scale and then calculate the mean and st.dev. of the log-transformed data. Because mean(log(x)) is not equal to log(mean(x)).

HTH,

Thierry


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-----Oorspronkelijk bericht-----
Van: r-help-bounces op r-project.org [mailto:r-help-bounces op r-project.org] Namens Tom Cohen
Verzonden: dinsdag 1 april 2008 14:17
Aan: r-help op stat.math.ethz.ch
Onderwerp: [R] set the lower bound of normal distribution to 0 ?



Tom Cohen <tom.cohen78 op yahoo.se> skrev:    Thanks Prof Brian for your suggestion. 
I should know that for right-skewed data,
one should generate the samples from a lognormal. 
  
My problem is that x and y are two instruments that were thought to 
be measured the same thing but somehow show a wide confidence interval
of the difference between the two intruments.This may be true that
these two measure differently but can also due to the small 
number of observations, so the idea is if I increases the sample size 
then I may get better precision between the two instrument by generating
samples based on the means and standard deviations
from x and y.
  
I am using 'urlnorm' which allows sampling from 
truncated distribution since I want the samples 
to take values from 0 to the max(x) respectively max(y). 
I am unsure how to specify the means and standard deviations
in 'urlnorm'. Based on x- and y-values I have standard deviations
sd_x=0.3372137, sd_y=0.5120841 and the means mean_x=0.3126667 
mean_y=0.4223137 which are not on log scale as required in urlnorm.
  
To covert sd_x, sd_y and mean_x, mean_y on a log-scale I did
sd_logx=sqrt(log(1.3372137))=0.54, sd_logy=sqrt(log(1.5120841))=0.64,
mean_logx=-(0.54^2)/2 and mean_logy=-(0.64^2)/2. Can anyone tell if these 
are correctly calculated? Are these the values to be specified in urlnorm?
Do the lower respectively upper bound have to be on the log-scale as well
or which scale?
   
   set.seed(7)

for(i in 1:len){
s1[[i]]<-cbind.data.frame(x=urlnorm(n*i,meanlog=mean_logx,sdlog=sd_logx, lb=0, ub=max(x)),
y=urlnorm(n*i,meanlog=mean_logy,sdlog=sd_logy, lb=0, ub=max(y)))
}

Thanks again for any suggetions.

Prof Brian Ripley <ripley op stats.ox.ac.uk> skrev:

Dear list,

I have a dataset containing values obtained from two different 
instruments (x and y). I want to generate 5 samples from normal 
distribution for each instrument based on their means and standard 
deviations. The problem is values from both instruments are 
non-negative, so if using rnorm I would get some negative values. Is 
there any options to determine the lower bound of normal distribution to 
be 0 or can I simulate the samples in different ways to avoid the 
negative values?

Well, that would not be a normal distribution.

If you want a _truncated_ normal distribution it is very easy by 
inversion. E.g.

trunc_rnorm <- function(n, mean = 0, sd = 1, lb = 0)
{
lb <- pnorm(lb, mean, sd)
qnorm(runif(n, lb, 1), mean, sd)
}

but I suggest you may rather want samples from a lognormal.

dat

id x y
75 101 0.134 0.1911315
79 102 0.170 0.1610306
76 103 0.134 0.1911315
84 104 0.170 0.1610306
74 105 0.134 0.1911315
80 106 0.170 0.1610306
77 107 0.134 0.1911315
81 108 0.170 0.1610306
82 109 0.170 0.1610306
78 111 0.170 0.1610306
83 112 0.170 0.1610306
85 113 0.097 0.2777778
2 201 1.032 1.5510434
1 202 0.803 1.0631001
5 203 1.032 1.5510434

mu<-apply(dat[,-1],2,mean)
sigma<-apply(dat[,-1],2,sd)
len<-5
n<-20
s1<-vector("list",len)
set.seed(7)
for(i in 1:len){
s1[[i]]<-cbind.data.frame(x=rnorm(n*i,mean=mu[1],sd=sigma[1]),
y=rnorm(n*i,mean=mu[2],sd=sigma[2]))
}

Thanks for any help,
Tom


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Brian D. Ripley, ripley op stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595

    
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