Is it possible to/how can I generate a beta-binomial distribution from an existing vector?
My ultimate goal is to generate a beta-binomial distribution from the below data and then obtain the 95% confidence interval for this distribution.
My data are body condition scores recorded by a veterinarian. The values of body condition range from 0-5 in increments of 0.5. It has been suggested to me here that my data follow a beta-binomial distribution, discrete values with a restricted range.
set1 <- as.data.frame(c(3,3,2.5,2.5,4.5,3,2,4,3,3.5,3.5,2.5,3,3,3.5,3,3,4,3.5,3.5,4,3.5,3.5,4,3.5))
colnames(set1) <- "numbers"
I see that there are multiple functions which appear to be able to do this, betabinomial() in VGAM and rbetabinom() in emdbook, but my stats and coding knowledge is not yet sufficient to be able to understand and implement the instructions provided on the function help pages, at least not in a way that has been helpful for my intended purpose yet.
We can look at the distribution of your variables, y-axis is the probability:
x1 = set1$numbers*2
h = hist(x1,breaks=seq(0,10))
bp = barplot(h$counts/length(x1),names.arg=(h$mids+0.5)/2,ylim=c(0,0.35))
You can try to fit it, but you have too little data points to estimate the 3 parameters need for a beta binomial. Hence I fix the probability so that the mean is the mean of your scores, and looking at the distribution above it seems ok:
library(bbmle)
library(emdbook)
library(MASS)
mtmp <- function(prob,size,theta) {
-sum(dbetabinom(x1,prob,size,theta,log=TRUE))
}
m0 <- mle2(mtmp,start=list(theta=100),
data=list(size=10,prob=mean(x1)/10),control=list(maxit=1000))
THETA=coef(m0)[1]
We can also use a normal distribution:
normal_fit = fitdistr(x1,"normal")
MEAN=normal_fit$estimate[1]
SD=normal_fit$estimate[2]
Plot both of them:
lines(bp[,1],dbetabinom(1:10,size=10,prob=mean(x1)/10,theta=THETA),
col="blue",lwd=2)
lines(bp[,1],dnorm(1:10,MEAN,SD),col="orange",lwd=2)
legend("topleft",c("normal","betabinomial"),fill=c("orange","blue"))
I think you are actually ok with using a normal estimation and in this case it will be:
normal_fit$estimate
mean sd
6.560000 1.134196
Related
I am using the useful gratia package by Gavin Simpson to extract the difference in two smooths for two different levels of a factor variable. The smooths are generated by the wonderful mgcv package. For example
library(mgcv)
library(gratia)
m1 <- gam(outcome ~ s(dep_var, by = fact_var) + fact_var, data = my.data)
diff1 <- difference_smooths(m1, smooth = "s(dep_var)")
draw(diff1)
This give me a graph of the difference between the two smooths for each level of the "by" variable in the gam() call. The graph has a shaded 95% credible interval (CI) for the difference.
Statistical significance, or areas of statistical significance at the 0.05 level, is assessed by whether or where the y = 0 line crosses the CI, where the y axis represents the difference between the smooths.
Here is an example from Gavin's site where the "by" factor variable had 3 levels.
The differences are clearly statistically significant (at 0.05) over nearly all of the graphs.
Here is another example I have generated using a "by" variable with 2 levels.
The difference in my example is clearly not statistically significant anywhere.
In the mgcv package, an approximate p value is outputted for a smooth fit that tests the null hypothesis that the coefficients are all = 0, based on a chi square test.
My question is, can anyone suggest a way of calculating a p value that similarly assesses the difference between the two smooths instead of solely relying on graphical evidence?
The output from difference_smooths() is a data frame with differences between the smooth functions at 100 points in the range of the smoothed variable, the standard error for the difference and the upper and lower limits of the CI.
Here is a link to the release of gratia 0.4 that explains the difference_smooths() function
enter link description here
but gratia is now at version 0.6
enter link description here
Thanks in advance for taking the time to consider this.
Don
One way of getting a p value for the interaction between the by factor variables is to manipulate the difference_smooths() function by activating the ci_level option. Default is 0.95. The ci_level can be manipulated to find a level where the y = 0 is no longer within the CI bands. If for example this occurred when ci_level = my_level, the p value for testing the hypothesis that the difference is zero everywhere would be 1 - my_level.
This is not totally satisfactory. For example, it would take a little manual experimentation and it may be difficult to discern accurately when zero drops out of the CI. Although, a function could be written to search the accompanying data frame that is outputted with difference_smooths() as the ci_level is varied. This is not totally satisfactory either because the detection of a non-zero CI would be dependent on the 100 points chosen by difference_smooths() to assess the difference between the two curves. Then again, the standard errors are approximate for a GAM using mgcv, so that shouldn't be too much of a problem.
Here is a graph where the zero first drops out of the CI.
Zero dropped out at ci_level = 0.88 and was still in the interval at ci_level = 0.89. So an approxiamte p value would be 1 - 0.88 = 0.12.
Can anyone think of a better way?
Reply to Gavin Simpson's comments Feb 19
Thanks very much Gavin for taking the time to make your comments.
I am not sure if using the criterion, >= 0 (for negative diffs), is a good way to go. Because of the draws from the posterior, there is likely to be many diffs that meet this criterion. I am interpreting your criterion as sample the posterior distribution and count how many differences meet the criterion, calculate the percentage and that is the p value. Correct me if I have misunderstood. Using this approach, I consistently got p values at around 0.45 - 0.5 for different gam models, even when it was clear the difference in the smooths should be statistically significant, at least at p = 0.05, because the confidence band around the smooth did not contain zero at a number of points.
Instead, I was thinking perhaps it would be better to compare the means of the posterior distribution of each of the diffs. For example
# get coefficients for the by smooths
coeff.level1 <- coef(gam.model1)[31:38]
coeff.level0 <- coef(gam.model1)[23:30]
# these indices are specific to my multi-variable gam.model1
# in my case 8 coefficients per smooth
# get posterior coefficients variances for the by smooths' coefficients
vp_level1 <- gam.model1$Vp[31:38, 31:38]
vp_level0 <- gam.model1$Vp[23:30, 23:30]
#run the simulation to get the distribution of each
#difference coefficient using the joint variance
library(MASS)
no.draws = 1000
sim <- mvrnorm(n = no.draws, (coeff.level1 - coeff.level0),
(vp_level1 + vp_level0))
# sim is a no.draws X no. of coefficients (8 in my case) matrix
# put the results into a data.frame.
y.group <- data.frame(y = as.vector(sim),
group = c(rep(1,no.draws), rep(2,no.draws),
rep(3,no.draws), rep(4,no.draws),
rep(5,no.draws), rep(6,no.draws),
rep(7,no.draws), rep(8,no.draws)) )
# y has the differences sampled from their posterior distributions.
# group is just a grouping name for the 8 sets of differences,
# (one set for each difference in coefficients)
# compare means with a linear regression
lm.test <- lm(y ~ as.factor(group), data = y.group)
summary(lm.test)
# The p value for the F statistic tells you how
# compatible the data are with the null hypothesis that
# all the group means are equal to each other.
# Same F statistic and p value from
anova(lm.test)
One could argue that if all coefficients are not equal to each other then they all can't be equal to zero but that isn't what we want here.
The basis of the smooth tests of fit given by summary(mgcv::gam.model1)
is a joint test of all coefficients == 0. This would be from a type of likelihood ratio test where model fit with and without a term are compared.
I would appreciate some ideas how to do this with the difference between two smooths.
Now that I got this far, I had a rethink of your original suggestion of using the criterion, >= 0 (for negative diffs). I reinterpreted this as meaning for each simulated coefficient difference distribution (in my case 8), count when this occurs and make a table where each row (my case, 8) is for one of these distributions with two columns holding this count and (number of simulation draws minus count), Then on this table run a chi square test. When I did this, I got a very low p value when I believe I shouldn't have as 0 was well within the smooth difference CI across almost all the levels of the exposure. Maybe I am still misunderstanding your suggestion.
Follow up thought Feb 24
In a follow up thought, we could create a variable that represents the interaction between the by factor and continuous variable
library(dplyr)
my.dat <- my.dat %>% mutate(interact.var =
ifelse(factor.2levels == "yes", 1, 0)*cont.var)
Here I am assuming that factor.2levels has the levels ("no", "yes"), and "no" is the reference level. The ifelse function creates a dummy variable which is multiplied by the continuous variable to generate the interactive variable.
Then we place this interactive variable in the GAM and get the usual statistical test for fit, that is, testing all the coefficients == 0.
#GavinSimpson actually posted a method of how to get the difference between two smooths and assess its statistical significance here in 2017. Thanks to Matteo Fasiolo for pointing me in that direction.
In that approach, the by variable is converted to an ordered categorical variable which causes mgcv::gam to produce difference smooths in comparison to the reference level. Statistical significance for the difference smooths is then tested in the usual way with the summary command for the gam model.
However, and correct me if I have misunderstood, the ordered factor approach causes the smooth for the main effect to now be the smooth for the reference level of the ordered factor.
The approach I suggested, see the main post under the heading, Follow up thought Feb 24, where the interaction variable is created, gives an almost identical result for the p value for the difference smooth but does not change the smooth for the main effect. It also does not change the intercept and the linear term for the by categorical variable which also both changed with the ordered variable approach.
I have an arbitrary CDF that is applied to a point estimate. I have a number of these point estimates with associated CDFs, that I need to simulate random data for a Monte Carlo simulation.
The CDF I'm generating by doing a spline fit to the arbitrary points provided in a table. For example, the quantile 0.1 is a product of 0.13 * point estimate. The quantile 0.9 is a product of 7.57 * point estimate. It is fairly crude and is based on a large study comparing these models to real world system -- ignore that for now please.
I fit the CDF using a spline fit as shown here.
If I take the derivative of this, I get the shape of the pdf (image).
I modified the function "samplepdf" found here, Sampling from an Arbitrary Density, as follows:
samplecdf <- function(n, cdf, spdf.lower = -Inf, spdf.upper=Inf) {
my_fun <- match.fun(cdf)
invcdf <- function(u) {
subcdf <- function(t) my_fun(t) - u
if (spdf.lower == -Inf)
spdf.lower <- endsign(subcdf, -1)
if (spdf.upper == Inf)
spdf.upper <- endsign(subcdf)
return(uniroot(subcdf, c(spdf.lower, spdf.upper))$root)
}
sapply(runif(n), invcdf)
}
This seems to work, OK - when I compare the quantiles I estimate from the randomly generated data they are fairly close to the initial values. However, when I look at the histogram something funny is happening at the tail where it is looks like my function is consistently generating more values than it should according to the pdf. This function consistently does that across all my point-estimates and even though I can look at the individual quantiles and they seem close, I can tell that the overall Monte Carlo simulation is demonstrating higher estimates for the 50% percentile than I expect. Here is a plot of my histogram of the random samples.
Any tips or advice would be very welcome. I think the best route would be to fit an exponential distribution to the CDF, but I'm struggling to do that. All "fitting" assumes that you have data that needs to be fitted -- this is more arbitrary than that.
I am trying to plot the cdf of a uniform distribution in octave but I am not getting the cdf. I am simply getting the original distribution. Also the original distribution, which is meant to be a uniform distribution, is not a uniform distribution at all!
Here is my octave code:
x = unifrnd(0,1,100,1);
hist(x)
cdfPlot = unifcdf(x)
hist(cdfPlot)
The histogram for the 1st one (hist(x)):
and the second one (hist(cdfPlot)) :
I also tried to use cdfplot(x) in octave but it said :
warning: the 'cdfplot' function belongs to the statistics package from
Octave Forge but has not yet been implemented.
Please read http://www.octave.org/missing.html to learn how you can
contribute missing functionality.
please help!
Judging by the submitted code, what you are trying to do is obtain a sample from a uniform distribution and then show a flat (mostly) histogram corresponding to a uniform distribution and a line corresponding to the cumulative distribution of the distribution.
For the first part:
Of course, with 100 samples (and no averaging), you are not going to observe a flat distribution, but if you try:
x=unifrnd(0,1,100000,1);
hist(x);
Then you are more likely to get a flat-looking histogram.
For the second part:
unifcdf(x,A,B) will return the value of a uniform distribution's CDF at some value x, between the interval set by parameters A,B. That is, the value of the CDF model itself, NOT the cumulative sum of the sample's histogram. To obtain that, you need to:
x=unifrnd(0,1,100000,1);
[counts, intervals] = hist(x);
xCDF = cumsum(counts);
bar(xCDF);
Finally, if you are looking for the model values, that is the values that would be returned by a formula describing a distribution, then for the uniform distribution that would be a probability of (1/nBins) between your A, B interval (in this case, 0,1) and a count of (1/nBins)*NSamples, while the CDF would be a line of slope (1/nBins) (i.e. the interval of the density function) and of binNum*((1/nBins)*NSamples). In the example above and using the default nBins for hist which is 10, x is decomposed to 10 intervals each with an approximate number of counts of 10000 items of x and the last value of the cumulative sum is 100000 which is of course the total number of samples in x.
For more information please see this link.
Hope this helps.
So I've got a data set that I want to parameterise but it is not a Gaussian distribution so I can't parameterise it in terms of it's mean and standard deviation. I want to fit a distribution function with a set of parameters and extract the values of the parameters (eg. a and b) that give the best fit. I want to do this exactly the same as the
lm(y~f(x;a,b))
except that I don't have a y, I have a distribution of different x values.
Here's an example. If I assume that the data follows a Gumbel, double exponential, distribution
f(x;u,b) = 1/b exp-(z + exp-(z)) [where z = (x-u)/b]:
#library(QRM)
#library(ggplot2)
rg <- rGumbel(1000) #default parameters are 0 and 1 for u and b
#then plot it's distribution
qplot(rg)
#should give a nice skewed distribution
If I assume that I don't know the distribution parameters and I want to perform a best fit of the probability density function to the observed frequency data, how do I go about showing that the best fit is (in this test case), u = 0 and b = 1?
I don't want code that simply maps the function onto the plot graphically, although that would be a nice aside. I want a method that I can repeatedly use to extract variables from the function to compare to others. GGPlot / qplot was used as it quickly shows the distribution for anyone wanting to test the code. I prefer to use it but I can use other packages if they are easier.
Note: This seems to me like a really obvious thing to have been asked before but I can't find one that relates to histogram data (which again seems strange) so if there's another tutorial I'd really like to see it.
I'm new to R. Having a set of samples along with the target, I want to fit a numeric function to solve the target of new samples. My sample is time in seconds indicating the duration of a user's staying at this place:
>b <- c(101,25711,13451,19442,26,3083,133,184,4403,9713,6918,10056,12201,10624,14984,5241,
+21619,44285,3262,2115,1822,11291,3243,12989,3607,12882,4462,11553,7596,2926,12955,
+1832,3539,6897,13571,16668,813,1824,10304,2508,1493,4407,7820,507,15866,7442,7738,
+5705,2869,10137,11276,12884,11298,...)
Firstly, I convert them to hours dividing by 3600, and I want to fit a function as pdf of the duration:
> b <- b/3600
> hist(c,xlim=c(0,13),prob=T,breaks=seq(0,24,by=0.5))
> lines(density(x), col=red)
I want to fit the red line on the figure, and interpolate new values to find the probability of the specific duration on this place say p(duration = 1.5hours).
Thanks for your attention!
As suggested above, you can fit a distribution with fitdistr in MASS package.
If you use a continuous distribution you will have the probability that the time is within an interval. If you use a discrete distribution, you may compute the probability of a certain time (in hours).
For the continuous case, you can use a Gamma distribution: fitdistr(b, "Gamma") will give you the parameter estimates, and then you can use pgamma with those estimates and an interval.
For the discrete case, you can use a Poisson distribution: fitdistr(b, "Poisson") and then the dpois function with the estimate and the value you want.
To decide which one to use, I'd just plot the pdf with the histogram and take a look.