How can I get more significant digits in R? Specifically, I have the following example:
> dpois(50, lambda= 5)
[1] 1.967673e-32
However when I get the p-value:
> 1-ppois(50, lambda= 5)
[1] 0
Obviously, the p-value is not 0. In fact it should greater than 1.967673e-32 since I'm summing a bunch of probabilities. How do I get the extra precision?
Use lower.tail=FALSE:
ppois(50, lambda= 5, lower.tail=FALSE)
## [1] 2.133862e-33
Asking R to compute the upper tail is much more accurate than computing the lower tail and subtracting it from 1: given the inherent limitations of floating point precision, R can't distinguish (1-eps) from 1 for values of eps less than .Machine$double.neg.eps, typically around 10^{-16} (see ?.Machine).
This issue is discussed in ?ppois:
Setting ‘lower.tail = FALSE’ allows to get much more precise
results when the default, ‘lower.tail = TRUE’ would return 1, see
the example below.
Note also that your comment about the value needing to be greater than dpois(50, lambda=5) is not quite right; ppois(x,...,lower.tail=FALSE) gives the probability that the random variable is greater than x, as you can see (for example) by seeing that ppois(0,...,lower.tail=FALSE) is not exactly 1, or:
dpois(50,lambda=5) + ppois(50,lambda=5,lower.tail=FALSE)
## [1] 2.181059e-32
ppois(49,lambda=5,lower.tail=FALSE)
## [1] 2.181059e-32
Related
I want to create a set of 10 logspaced numbers from zero to some big number M, say M=60,000, for example in R.
First, I tried to use lseq() from the package emdbook. The problem with lseq, however, is that it cannot handle 0 as a starting point. (This is due to the fact that it will try to calculate log(0) and then fail).
Next, I tried to use logspace() from the pracma package in the following way:
Numbers <- log(logspace(0,M,10),base=10)
This works fine for values of M up to about 340. From then on the numbers in the set will become infinity because the exponential function becomes too large.
Is there any other way in R to create a set of logspaced numbers from zero to some big number M which will not make most of the numbers in the set infinity and which can actually handle zero as a starting point?
Correct me if I am wrong, but can't you just çalculate the logspaces for lower numbers and then multiply? They should be linearly related right? Just look at this output:
library(pracma)
> log(logspace(0,60, 10), base = 10)[1:5]
[1] 0.000000000000000 6.666666666666667 13.333333333333334 20.000000000000000 26.666666666666668
> log(logspace(0,600, 10), base = 10)[1:5]
[1] 0.000000000000000 66.666666666666671 133.333333333333343 200.000000000000000 266.666666666666686
> x1 <- (log(logspace(0,600, 10), base = 10)*100)[2]
> x1
[1] 6666.666666666667
> x2 <- seq(0 , 9, 1)*x1
> x2
[1] 0.000000000000 6666.666666666667 13333.333333333334 20000.000000000000 26666.666666666668
[6] 33333.333333333336 40000.000000000000 46666.666666666672 53333.333333333336 60000.000000000000
Cans someone explain the results in a typical dt function? The help page says that I should receive the density function. However, in my code below, what does the first value ".2067" represent?The second value?
x<-seq(1,10)
dt(x, df=3)
[1] 0.2067483358 0.0675096607 0.0229720373 0.0091633611 0.0042193538 0.0021748674
[7] 0.0012233629 0.0007369065 0.0004688171 0.0003118082
Two things were confused here:
dt gives you the density, this is why it decreases for large numbers:
x<-seq(1,10)
dt(x, df=3)
[1] 0.2067483358 0.0675096607 0.0229720373 0.0091633611 0.0042193538 0.0021748674
[7] 0.0012233629 0.0007369065 0.0004688171 0.0003118082
pt gives the distribution function. This is the probability of being smaller or equal x.
This is why the values go to 1 as x increases:
pt(x, df=3)
[1] 0.8044989 0.9303370 0.9711656 0.9859958 0.9923038 0.9953636 0.9970069 0.9979617 0.9985521 0.9989358
A "probability density" is not really a true probability, since probabilities are bounded in [0,1] while densities are not. The integral of densities across their domain is normalized to exactly 1. So densities are really the first derivatives of the probability function. This code may help:
plot( x= seq(-10,10,length=100),
y=dt( seq(-10,10,length=100), df=3) )
The value of 0.207 for dt at x=1 says that at x=1 that the probability is increasing at a rate of 0.207 per unit increase in x. (And since the t-distribution is symmetric that is also the value of dt with 3 df at -1.)
A bit of coding to instantiate the dt(x,df=3) function (see ?dt) and then integrate it:
> dt3 <- function(x) { gamma((4)/2)/(sqrt(3*pi)*gamma(3/2))*(1+x^2/3)^-((3+1)/2) }
> dt3(1)
[1] 0.2067483
> integrate(dt3, -Inf, Inf)
1 with absolute error < 7.2e-08
It appears dnorm(x) and dmvnorm(x) (from mvtnorm package v1.0.5) generates the same result when x is of length 1. However, when working with integration function, the result for dmvnorm(x) seems to be quite off. Here is the example, I defined two integrand functions:
integrand1 <- function(x)
{ pnorm((-2-sqrt(0.2)*x)/sqrt(0.8))*dmvnorm(x)}
integrand2 <- function(x)
{ pnorm((-2-sqrt(0.2)*x)/sqrt(0.8))*dnorm(x)}
When evaluating at individual values, integrand1(x) = integrand2(x).
However, integration along these two functions generates totally different results:
integrate(integrand1,lower = -10,upper = 10)$value
[1] 7.567463e-236
integrate(integrand2,lower = -10,upper = 10)$value
[1] 0.02275013
Similar phenomena is also observed for curve function. Just wonder if it is a bug or I use it incorrectly.
It's not a bug. The dmvnorm function expects to evaluate the density distribution function of a multinormal, while dnorm handle single variate distributions. Let's see an example:
dnorm(1:5)
#[1] 2.419707e-01 5.399097e-02 4.431848e-03 1.338302e-04 1.486720e-06
dmvnorm(1:5)
#[1] 1.151999e-14
First difference: dnorm returns a vector of length 5, while dmvnorm a single value. We note also that:
prod(dnorm(1:5))
#[1] 1.151999e-14
basically the same value returned by dmvnorm. In short, these are the difference:
dnorm(1:5) evaluates the density of a normal distribution at 1, then 2, 3, 4 and 5 with mean 0 and standard deviation 1. So we have 5 values.
dmvnorm is made to treat multivariate distributions. We now ask which are the density of a 5-variate multinormal (since we provided 1:5, i.e. a vector of length 5) where the first variable is at 1, the second at 2 and so on. The means are all zeroes and the covariance matrix is diag(5), i.e. a 5 dimensional identity matrix. Both are the default values, since we didn't specify them. See ?dmvnorm.
In the latter case, the five variable are independent and so the density is the product of 5 independent (single-variate) normal distribution at 1, 2, 3, 4 and 5.
Use dnorm when you are dealing with single-variate distributions and dmvnorm otherwise.
I am calculating z-scores to see if a value is far from the mean/median of the distribution.
I had originally done it using the mean, then turned these into 2-side pvalues. But now using the median I noticed that there are some Na's in the pvalues.
I determined this is occuring for values that are very far from the median.
And looks to be related to the pnorm calculation.
"
'qnorm' is based on Wichura's algorithm AS 241 which provides
precise results up to about 16 digits. "
Does anyone know a way around this as I would like the very small pvalues.
Thanks,
> z<- -12.5
> 2-2*pnorm(abs(z))
[1] 0
> z<- -10
> 2-2*pnorm(abs(z))
[1] 0
> z<- -8
> 2-2*pnorm(abs(z))
[1] 1.332268e-15
Intermediately, you are actually calculating very high p-values:
options(digits=22)
z <- c(-12.5,-10,-8)
pnorm(abs(z))
# [1] 1.0000000000000000000000 1.0000000000000000000000 0.9999999999999993338662
2-2*pnorm(abs(z))
# [1] 0.000000000000000000000e+00 0.000000000000000000000e+00 1.332267629550187848508e-15
I think you will be better off using the low p-values (close to zero) but I am not good enough at math to know whether the error at close-to-one p-values is in the AS241 algorithm or the floating point storage. Look how nicely the low values show up:
pnorm(z)
# [1] 3.732564298877713761239e-36 7.619853024160526919908e-24 6.220960574271784860433e-16
Keep in mind 1 - pnorm(x) is equivalent to pnorm(-x). So, 2-2*pnorm(abs(x)) is equivalent to 2*(1 - pnorm(abs(x)) is equivalent to 2*pnorm(-abs(x)), so just go with:
2 * pnorm(-abs(z))
# [1] 7.465128597755427522478e-36 1.523970604832105383982e-23 1.244192114854356972087e-15
which should get more precisely what you are looking for.
One thought, you'll have to use an exp() with larger precision, but you might be able to use log(p) to get slightly more precision in the tails, otherwise you are effectively at 0 for the non-log p values in terms of the range that can be calculated:
> z<- -12.5
> pnorm(abs(z),log.p=T)
[1] -7.619853e-24
Converting back to the p value doesn't work well, but you could compare on log(p)...
> exp(pnorm(abs(z),log.p=T))
[1] 1
pnorm is a function which gives what P value is based on given x. If You do not specify more arguments, then default distribution is Normal with mean 0, and standart deviation 1.
Based on simetrity, pnorm(a) = 1-pnorm(-a).
In R, if you add positive numbers it will round them. But if you add negative no rounding is done. So using this formula and negative numbers you can calculate needed values.
> pnorm(0.25)
[1] 0.5987063
> 1-pnorm(-0.25)
[1] 0.5987063
> pnorm(20)
[1] 1
> pnorm(-20)
[1] 2.753624e-89
As an assignment I had to develop and algorithm and generate a samples for a given geometric distribution with PMF
Using the inverse transform method, I came up with the following expression for generating the values:
Where U represents a value, or n values depending on the size of the sample, drawn from a Unif(0,1) distribution and p is 0.3 as stated in the PMF above.
I have the algorithm, the implementation in R and I already generated QQ Plots to visually assess the adjustment of the empirical values to the theoretical ones (generated with R), i.e., if the generated sample follows indeed the geometric distribution.
Now I wanted to submit the generated sample to a goodness of fit test, namely the Chi-square, yet I'm having trouble doing this in R.
[I think this was moved a little hastily, in spite of your response to whuber's question, since I think before solving the 'how do I write this algorithm in R' problem, it's probably more important to deal with the 'what you're doing is not the best approach to your problem' issue (which certainly belongs where you posted it). Since it's here, I will deal with the 'doing it in R' aspect, but I would urge to you go back an ask about the second question (as a new post).]
Firstly the chi-square test is a little different depending on whether you test
H0: the data come from a geometric distribution with parameter p
or
H0: the data come from a geometric distribution with parameter 0.3
If you want the second, it's quite straightforward. First, with the geometric, if you want to use the chi-square approximation to the distribution of the test statistic, you will need to group adjacent cells in the tail. The 'usual' rule - much too conservative - suggests that you need an expected count in every bin of at least 5.
I'll assume you have a nice large sample size. In that case, you'll have many bins with substantial expected counts and you don't need to worry so much about keeping it so high, but you will still need to choose how you will bin the tail (whether you just choose a single cut-off above which all values are grouped, for example).
I'll proceed as if n were say 1000 (though if you're testing your geometric random number generation, that's pretty low).
First, compute your expected counts:
dgeom(0:20,.3)*1000
[1] 300.0000000 210.0000000 147.0000000 102.9000000 72.0300000 50.4210000
[7] 35.2947000 24.7062900 17.2944030 12.1060821 8.4742575 5.9319802
[13] 4.1523862 2.9066703 2.0346692 1.4242685 0.9969879 0.6978915
[19] 0.4885241 0.3419669 0.2393768
Warning, dgeom and friends goes from x=0, not x=1; while you can shift the inputs and outputs to the R functions, it's much easier if you subtract 1 from all your geometric values and test that. I will proceed as if your sample has had 1 subtracted so that it goes from 0.
I'll cut that off at the 15th term (x=14), and group 15+ into its own group (a single group in this case). If you wanted to follow the 'greater than five' rule of thumb, you'd cut it off after the 12th term (x=11). In some cases (such as smaller p), you might want to split the tail across several bins rather than one.
> expec <- dgeom(0:14,.3)*1000
> expec <- c(expec, 1000-sum(expec))
> expec
[1] 300.000000 210.000000 147.000000 102.900000 72.030000 50.421000
[7] 35.294700 24.706290 17.294403 12.106082 8.474257 5.931980
[13] 4.152386 2.906670 2.034669 4.747562
The last cell is the "15+" category. We also need the probabilities.
Now we don't yet have a sample; I'll just generate one:
y <- rgeom(1000,0.3)
but now we want a table of observed counts:
(x <- table(factor(y,levels=0:14),exclude=NULL))
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 <NA>
292 203 150 96 79 59 47 25 16 10 6 7 0 2 5 3
Now you could compute the chi-square directly and then calculate the p-value:
> (chisqstat <- sum((x-expec)^2/expec))
[1] 17.76835
(pval <- pchisq(chisqstat,15,lower.tail=FALSE))
[1] 0.2750401
but you can also get R to do it:
> chisq.test(x,p=expec/1000)
Chi-squared test for given probabilities
data: x
X-squared = 17.7683, df = 15, p-value = 0.275
Warning message:
In chisq.test(x, p = expec/1000) :
Chi-squared approximation may be incorrect
Now the case for unspecified p is similar, but (to my knowledge) you can no longer get chisq.test to do it directly, you have to do it the first way, but you have to estimate the parameter from the data (by maximum likelihood or minimum chi-square), and then test as above but you have one fewer degree of freedom for estimating the parameter.
See the example of doing a chi-square for a Poisson with estimated parameter here; the geometric follows the much same approach as above, with the adjustments as at the link (dealing with the unknown parameter, including the loss of 1 degree of freedom).
Let us assume you've got your randomly-generated variates in a vector x. You can do the following:
x <- rgeom(1000,0.2)
x_tbl <- table(x)
x_val <- as.numeric(names(x_tbl))
x_df <- data.frame(count=as.numeric(x_tbl), value=x_val)
# Expand to fill in "gaps" in the values caused by 0 counts
all_x_val <- data.frame(value = 0:max(x_val))
x_df <- merge(all_x_val, x_df, by="value", all.x=TRUE)
x_df$count[is.na(x_df$count)] <- 0
# Get theoretical probabilities
x_df$eprob <- dgeom(x_df$val, 0.2)
# Chi-square test: once with asymptotic dist'n,
# once with bootstrap evaluation of chi-sq test statistic
chisq.test(x=x_df$count, p=x_df$eprob, rescale.p=TRUE)
chisq.test(x=x_df$count, p=x_df$eprob, rescale.p=TRUE,
simulate.p.value=TRUE, B=10000)
There's a "goodfit" function described as "Goodness-of-fit Tests for Discrete Data" in package "vcd".
G.fit <- goodfit(x, type = "nbinomial", par = list(size = 1))
I was going to use the code you had posted in an earlier question, but it now appears that you have deleted that code. I find that offensive. Are you using this forum to gather homework answers and then defacing it to remove the evidence? (Deleted questions can still be seen by those of us with sufficient rep, and the interface prevents deletion of question with upvoted answers so you should not be able to delete this one.)
Generate a QQ Plot for testing a geometrically distributed sample
--- question---
I have a sample of n elements generated in R with
sim.geometric <- function(nvals)
{
p <- 0.3
u <- runif(nvals)
ceiling(log(u)/log(1-p))
}
for which i want to test its distribution, specifically if it indeed follows a geometric distribution. I want to generate a QQ PLot but have no idea how to.
--------reposted answer----------
A QQ-plot should be a straight line when compared to a "true" sample drawn from a geometric distribution with the same probability parameter. One gives two vectors to the functions which essentially compares their inverse ECDF's at each quantile. (Your attempt is not particularly successful:)
sim.res <- sim.geometric(100)
sim.rgeom <- rgeom(100, 0.3)
qqplot(sim.res, sim.rgeom)
Here I follow the lead of the authors of qqplot's help page (which results in flipping that upper curve around the line of identity):
png("QQ.png")
qqplot(qgeom(ppoints(100),prob=0.3), sim.res,
main = expression("Q-Q plot for" ~~ {G}[n == 100]))
dev.off()
---image not included---
You can add a "line of good fit" by plotting a line through through the 25th and 75th percentile points for each distribution. (I added a jittering feature to this to get a better idea where the "probability mass" was located:)
sim.res <- sim.geometric(500)
qqplot(jitter(qgeom(ppoints(500),prob=0.3)), jitter(sim.res),
main = expression("Q-Q plot for" ~~ {G}[n == 100]), ylim=c(0,max( qgeom(ppoints(500),prob=0.3),sim.res )),
xlim=c(0,max( qgeom(ppoints(500),prob=0.3),sim.res )))
qqline(sim.res, distribution = function(p) qgeom(p, 0.3),
prob = c(0.25, 0.75), col = "red")