So I have this discrete set of data my_dat that I am trying to fit a curve over to be able to generate random variables based on my_dat. I had great success using fitdistrplus on continuous data but have many errors when attempting to use it for discrete data.
Table settings:
library(fitdistrplus)
my_dat <- c(2,5,3,3,3,1,1,2,4,6,
3,2,2,8,3,4,3,3,4,4,
2,1,5,3,1,2,2,4,3,4,
2,4,1,6,2,3,2,1,2,4,
5,1,2,3,2)
I take a look at the histogram of the data first:
hist(my_dat)
Since the data's discrete, I decide to try a binomial distribution or the negative binomial distribution to fit and this is where I run into trouble: Here I try to define each:
fitNB3 <- fitdist(my_dat, discrete = T, distr = "nbinom" ) #NaNs Produced
fitB3 <- fitdist(my_dat, discrete = T, distr = "binom")
I receive two errors:
fitNB3 seems to run but notes that "NaNs Produced" - can anyone let me
know why this is the case?
fitB3 doesn't run at all and provides me with the error: "Error in start.arg.default(data10, distr = distname) : Unknown starting values for distribution binom." - can anyone point out why this won't work here? I am unclear about providing a starting number given that the data is discrete (I attempted to use start = 1 in the fitdist function but I received another error: "Error in fitdist(my_dat, discrete = T, distr = "binom", start = 1) : the function mle failed to estimate the parameters, with the error code 100"
I've been spinning my wheels for a while on this but I would be take any feedback regarding these errors.
Don't use hist on discrete data, because it doesn't do what you think it's doing.
Compare plot(table(my_dat)) with hist(my_dat)... and then ponder how many wrong impressions you've gotten doing this before. If you must use hist, make sure you specify the breaks, don't rely on defaults designed for continuous variables.
hist(my_dat)
lines(table(my_dat),col=4,lwd=6,lend=1)
Neither of your models can be suitable as both these distributions start from 0, not 1, and with the size of values you have, p(0) will not be ignorably small.
I don't get any errors fitting the negative binomial when I run your code.
The issue you had with fitting the binomial is you need to supply starting values for the parameters, which are called size (n) and prob (p), so
you'd need to say something like:
fitdist(my_dat, distr = "binom", start=list(size=15, prob=0.2))
However, you will then get a new problem! The optimizer assumes that the parameters are continuous and will fail on size.
On the other hand this is probably a good thing because with unknown n MLE is not well behaved, particularly when p is small.
Typically, with the binomial it would be expected that you know n. In that case, estimation of p could be done as follows:
fitdist(my_dat, distr = "binom", fix.arg=list(size=20), start=list(prob=0.15))
However, with fixed n, maximum likelihood estimation is straightforward in any case -- you don't need an optimizer for that.
If you really don't know n, there are a number of better-behaved estimators than the MLE to be found, but that's outside the scope of this question.
Related
I am fitting a Pareto distribution to some data and have already estimated the maximum likelihood estimates for the data. Now I need to create a fitdist (fitdistrplus library) object from it, but I am not sure how to do this. I need a fitdist object because I would like to create qq, density etc. plots with the function such as denscomp. Could someone help?
The reason I calculated the MLEs first is because fitdist does not do this properly - the estimates always blow up to infinity, even if I give the correct MLEs as the starting values (see below). If the earlier option of manually giving fitdist my parameters is not possible, is there an optimization method in fitdist that would allow the pareto parameters to be properly estimated?
I don't have permission to post the original data, but here's a simulation using MLE estimates of a gamma distribution/pareto distribution on the original.
library(fitdistrplus)
library(actuar)
sim <- rgamma(1000, shape = 4.69, rate = 0.482)
fit.pareto <- fit.dist(sim, distr = "pareto", method = "mle",
start = list(scale = 0.862, shape = 0.00665))
#Estimates blow up to infinity
fit.pareto$estimate
If you look at the ?fitdist help topic, it describes what fitdist objects look like: they are lists with lots of components. If you can compute substitutes for all of those components, you should be able to create a fake fitdist object using code like
fake <- structure(list(estimate = ..., method = ..., ...),
class = "fitdist")
For the second part of your question, you'll need to post some code and data for people to improve.
Edited to add:
I added set.seed(123) before your simulation of random data. Then I get the MLE from fitdist to be
scale shape
87220272 9244012
If I plot the log likelihood function near there, I get this:
loglik <- Vectorize(function(shape, scale) sum(dpareto(sim, shape, scale, log = TRUE)))
shape <- seq(1000000, 10000000, len=30)
scale <- seq(10000000, 100000000, len=30)
surface <- outer(shape, scale, loglik)
contour(shape, scale, surface)
points(9244012, 87220272, pch=16)
That looks as though fitdist has made a somewhat reasonable choice, though there may not actually be a finite MLE. How did you find the MLE to be such small values? Are you sure you're using the same parameters as dpareto uses?
I have some time to event data that I need to generate around 200 shape/scale parameters for subgroups for a simulation model. I have analysed the data, and it best follows a weibull distribution.
Normally, I would use the fitdistrplus package and fitdist(x, "weibull") to do so, however this data has been matched using kernel matching and I have a variable of weighting values called km and so needs to incorporate a weight, which isn't something fitdist can do as far as I can tell.
With my gamma distributed data instead of using fitdist I did the calculation manually using the wtd.mean and wtd.var functions from the hsmisc package, which worked well. However, finding a similar formula for the weibull is eluding me.
I've been testing a few options and comparing them against the fitdist results:
test_data <- rweibull(100, 0.676, 946)
fitweibull <- fitdist(test_data, "weibull", method = "mle", lower = c(0,0))
fitweibull$estimate
shape scale
0.6981165 935.0907482
I first tested this: The Weibull distribution in R (ExtDist)
library(bbmle)
m1 <- mle2(y~dweibull(shape=exp(lshape),scale=exp(lscale)),
data=data.frame(y=test_data),
start=list(lshape=0,lscale=0))
which gave me lshape = -0.3919991 and lscale = 6.852033
The other thing I've tried is eweibull from the EnvStats package.
eweibull <- eweibull(test_data)
eweibull$parameters
shape scale
0.698091 935.239277
However, while these are giving results, I still don't think I can fit my data with the weights into any of these.
Edit: I have also tried the similarly named eWeibull from the ExtDist package (which I'm not 100% sure still works, but does have a weibull function that takes a weight!). I get a lot of error messages about the inputs being non-computable (NA or infinite). If I do it with map, so map(test_data, test_km, eWeibull) I get [[NULL] for all 100 values. If I try it just with test_data, I get a long string of errors associated with optimx.
I have also tried fitDistr from propagate which gives errors that weights should be a specific length. For example, if both are set to be 100, I get an error that weights should be length 94. If I set it to 94, it tells me it has to be length of 132.
I need to be able to pass either a set of pre-weighted mean/var/sd etc data into the calculation, or have a function that can take data and weights and use them both in the calculation.
After much trial and error, I edited the eweibull function from the EnvStats package to instead of using mean(x) and sd(x), to instead use wtd.mean(x,w) and sqrt(wtd.var(x, w)). This now runs and outputs weighted values.
I am trying to fit t-distributions to my data but am unable to do so. My first try was
fitdistr(myData, "t")
There are 41 warnings, all saying that NaNs are produced. I don't know how, logarithms seem to be involved. So I adjusted my data somewhat so that all data is >0, but I still have the same problem (9 fewer warnings though...). Same problem with sstdFit(), produces NaNs.
So instead I try with fitdist which I've seen on stackoverflow and CrossValidated:
fitdist(myData, "t")
I then get
Error in mledist(data, distname, start, fix.arg, ...) :
'start' must be defined as a named list for this distribution
What does this mean? I tried looking into the documentation but that told me nothing. I just want to possibly fit a t-distribution, this is so frustrating :P
Thanks!
Start is the initial guess for the parameters of your distribution. There are logs involved because it is using maximum likelihood and hence log-likelihoods.
library(fitdistrplus)
dat <- rt(100, df=10)
fit <- fitdist(dat, "t", start=list(df=2))
I think it's worth adding that in most cases, using the fitdistrplus package to fit a t-distribution to real data will lead to a very bad fit, which is actually quite misleading. This is because the default t-distribution functions in R are used, and they don't support shifting or scaling. That is, if your data has a mean other than 0, or is scaled in some way, then the fitdist function will simply lead to a bad fit.
In real life, if data fits a t-distribution, it is usually shifted (i.e. has a mean other than 0) and / or scaled. Let's generate some data like that:
data = 1.5*rt(10000,df=5) + 0.5
Given this data has been sampled from the t-distribution with 5 degrees of freedom, you'd think that trying to fit a t-distribution to this should work quite nicely. But actually, here is the result. It estimates a df of 2, and provides a bad fit as shown in the qq plot.
> fit_bad <- fitdist(data,"t",start=list(df=3))
> fit_bad
Fitting of the distribution ' t ' by maximum likelihood
Parameters:
estimate Std. Error
df 2.050967 0.04301357
> qqcomp(list(fit_bad)) # generates plot to show fit
When you fit to a t-distribution you want to not only estimate the degrees of freedom, but also a mean and scaling parameter.
The metRology package provides a version of the t-distribution called t.scaled that has a mean and sd parameter in addition to the df parameter [metRology]. Now let's fit it again:
> library("metRology")
> fit_good <- fitdist(data,"t.scaled",
start=list(df=3,mean=mean(data),sd=sd(data)))
> fit_good
Fitting of the distribution ' t.scaled ' by maximum likelihood
Parameters:
estimate Std. Error
df 4.9732159 0.24849246
mean 0.4945922 0.01716461
sd 1.4860637 0.01828821
> qqcomp(list(fit_good)) # generates plot to show fit
Much better :-) The parameters are very close to how we generated the data in the first place! And the QQ plot shows a much nicer fit.
So I am currently trying to draw the confidence interval for a linear model. I found out I should use predict.lm() for this, but I have a few problems really understanding the function and I do not like using functions without knowing what's happening. I found several how-to's on this subject, but only with the corresponding R-code, no real explanation.
This is the function itself:
## S3 method for class 'lm'
predict(object, newdata, se.fit = FALSE, scale = NULL, df = Inf,
interval = c("none", "confidence", "prediction"),
level = 0.95, type = c("response", "terms"),
terms = NULL, na.action = na.pass,
pred.var = res.var/weights, weights = 1, ...)
Now, what I've trouble understanding:
1) newdata
An optional data frame in which to look for variables
with which to predict. If omitted, the fitted values are used.
Everyone seems to use newdata for this, but I cannot quite understand why. For calculating the confidence interval I obviously need the data which this interval is for (like the # of observations, mean of x etc), so cannot be what is meant by it. But then: What is does it mean?
2) interval
Type of interval calculation.
okay.. but what is "none" for?
3a) type
Type of prediction (response or model term).
3b) terms
If type="terms", which terms (default is all terms)
3a: Can I by that get the confidence interval for one specific variable in my model? And if so, what is 3b for then? If I can specify the term in 3a, it wouldn't make sense to do it in 3b again.. so I guess I'm wrong again, but I cannot figure out why.
I guess some of you might think: Why don't just try this out? And I would (even if it would maybe not solve everything here), but I right now don't know how to. As I do not now what the newdata is for, I don't know how to use it and if I try, I do not get the right confidence interval. Somehow it is very important how you choose that data, but I just don't understand!
EDIT: I want to add that my intention is to understand how predict.lm works. By that I mean I don't understand if it works the way I think it does. That is it calculates y-hat (predicted values) and than uses adds/subtracts for each the upr/lwr-bounds of the interval to calculate several datapoints(looking like a confidence-line then) ?? Then I would undestand why it is necessary to have the same lenght in the newdata as in the linear model.
Make up some data:
d <- data.frame(x=c(1,4,5,7),
y=c(0.8,4.2,4.7,8))
Fit the model:
lm1 <- lm(y~x,data=d)
Confidence and prediction intervals with the original x values:
p_conf1 <- predict(lm1,interval="confidence")
p_pred1 <- predict(lm1,interval="prediction")
Conf. and pred. intervals with new x values (extrapolation and more finely/evenly spaced than original data):
nd <- data.frame(x=seq(0,8,length=51))
p_conf2 <- predict(lm1,interval="confidence",newdata=nd)
p_pred2 <- predict(lm1,interval="prediction",newdata=nd)
Plotting everything together:
par(las=1,bty="l") ## cosmetics
plot(y~x,data=d,ylim=c(-5,12),xlim=c(0,8)) ## data
abline(lm1) ## fit
matlines(d$x,p_conf1[,c("lwr","upr")],col=2,lty=1,type="b",pch="+")
matlines(d$x,p_pred1[,c("lwr","upr")],col=2,lty=2,type="b",pch=1)
matlines(nd$x,p_conf2[,c("lwr","upr")],col=4,lty=1,type="b",pch="+")
matlines(nd$x,p_pred2[,c("lwr","upr")],col=4,lty=2,type="b",pch=1)
Using new data allows for extrapolation beyond the original data; also, if the original data are sparsely or unevenly spaced, the prediction intervals (which are not straight lines) may not be well approximated by linear interpolation between the original x values ...
I'm not quite sure what you mean by the "confidence interval for one specific variable in my model"; if you want confidence intervals on a parameter, then you should use confint. If you want predictions for the changes based only on some of the parameters changing (ignoring the uncertainty due to the other parameters), then you do indeed want to use type="terms".
interval="none" (the default) just tells R not to bother computing any confidence or prediction intervals, and to return just the predicted values.
I am currently trying to fit an adaBoost model in R using the gbm.fit model. I have tried everything I could but in the end my model keeps giving me prediction values outside of [0,1]. I understand that type = "response" only works for bernoulli but I keep getting values just outside of 0,1. Any thoughts? Thanks!
GBMODEL <- gbm.fit(
x=training.set,
y=training.responses,
distribution="adaboost",
n.trees=5000,
interaction.depth=1,
shrinkage=0.005,
train.fraction=1,
)
predictionvalues = predict(GBMODEL,
newdata=test.predictors,
n.trees=5000,
type="response")
it is correct to obtain y range outside [0,1] by gbm package choosing "adaboost" as your loss function.
After training, adaboost predicts category by the sign of output.
For instance, for binary class problem, y{-1,1}, the class lable will be signed to the sign of output y. So if you got y=0.9 or y=1.9 will give you the same result-observation belongs to y=1 class. However, y=1.9 simply suggests a more confident conclusion than y=0.9. (if you want to know why, I would suggest you to read margin-based explanation of adaboost, you will find very similar result with SVM).
Hope this can help you.
This may not be completely accurate mathematically, but I just did pnorm( predicted values) and you get values from 0 to 1, because the adaboost predicted values appear to be scaled on a Normal(0,1).