Estimating PDF with monotonically declining density at tails - r

tldr: I am numerically estimating a PDF from simulated data and I need the density to monotonically decrease outside of the 'main' density region (as x-> infinity). What I have yields a close to zero density, but which does not monotonically decrease.
Detailed Problem
I am estimating a simulated maximum likelihood model, which requires me to numerically evaluate the probability distribution function of some random variable (the probability of which cannot be analytically derived) at some (observed) value x. The goal is to maximize the log-likelihood of these densities, which requires them to not have spurious local maxima.
Since I do not have an analytic likelihood function I numerically simulate the random variable by drawing the random component from some known distribution function, and apply some non-linear transformation to it. I save the results of this simulation in a dataset named simulated_stats.
I then use density() to approximate the PDF and approxfun() to evaluate the PDF at x:
#some example simulation
Simulated_stats_ <- runif(n=500, 10,15)+ rnorm(n=500,mean = 15,sd = 3)
#approximation for x
approxfun(density(simulated_stats))(x)
This works well within the range of simulated simulated_stats, see image:
Example PDF. The problem is I need to be able to evaluate the PDF far from the range of simulated data.
So in the image above, I would need to evaluate the PDF at, say, x=50:
approxfun(density(simulated_stats))(50)
> [1] NA
So instead I use the from and to arguments in the density function, which correctly approximate near 0 tails, such
approxfun(
density(Simulated_stats, from = 0, to = max(Simulated_stats)*10)
)(50)
[1] 1.924343e-18
Which is great, under one condition - I need the density to go to zero the further out from the range x is. That is, if I evaluated at x=51 the result must be strictly smaller. (Otherwise, my estimator may find local maxima far from the 'true' region, since the likelihood function is not monotonic very far from the 'main' density mass, i.e. the extrapolated region).
To test this I evaluated the approximated PDF at fixed intervals, took logs, and plotted. The result is discouraging: far from the main density mass the probability 'jumps' up and down. Always very close to zero, but NOT monotonically decreasing.
a <- sapply(X = seq(from = 0, to = 100, by = 0.5), FUN = function(x){approxfun(
density(Simulated_stats_,from = 0, to = max(Simulated_stats_)*10)
)(x)})
aa <- cbind( seq(from = 0, to = 100, by = 0.5), a)
plot(aa[,1],log(aa[,2]))
Result:
Non-monotonic log density far from density mass
My question
Does this happen because of the kernel estimation in density() or is it inaccuracies in approxfun()? (or something else?)
What alternative methods can I use that will deliver a monotonically declining PDF far from the simulated density mass?
Or - how can I manually change the approximated PDF to monotonically decline the further I am from the density mass? I would happily stick some linear trend that goes to zero...
Thanks!

One possibility is to estimate the CDF using a beta regression model; numerical estimate of the derivative of this model could then be used to estimate the pdf at any point. Here's an example of what I was thinking. I'm not sure if it helps you at all.
Import libraries
library(mgcv)
library(data.table)
library(ggplot2)
Generate your data
set.seed(123)
Simulated_stats_ <- runif(n=5000, 10,15)+ rnorm(n=500,mean = 15,sd = 3)
Function to estimate CDF using gam beta regression model
get_mod <- function(ss,p = seq(0.02, 0.98, 0.02)) {
qp = quantile(ss, probs=p)
betamod = mgcv::gam(p~s(qp, bs="cs"), family=mgcv::betar())
return(betamod)
}
betamod <- get_mod(Simulated_stats_)
Very basic estimate of PDF at val given model that estimates CDF
est_pdf <- function(val, betamod, tol=0.001) {
xvals = c(val,val+tol)
yvals = predict(betamod,newdata=data.frame(qp = xvals), type="response")
as.numeric((yvals[1] - yvals[2])/(xvals[1] - xvals[2]))
}
Lets check if monotonically increasing below min of Simulated_stats
test_x = seq(0,min(Simulated_stats_), length.out=1000)
pdf = sapply(test_x, est_pdf, betamod=betamod)
all(pdf == cummax(pdf))
[1] TRUE
Lets check if monotonically decreasing above max of Simulated_stats
test_x = seq(max(Simulated_stats_), 60, length.out=1000)
pdf = sapply(test_x, est_pdf, betamod=betamod)
all(pdf == cummin(pdf))
[1] TRUE
Additional thoughts 3/5/22
As discussed in comments, using the betamod to predict might slow down the estimator. While this could be resolved to a great extent by writing your own predict function directly, there is another possible shortcut.
Generate estimates from the betamod over the range of X, including the extremes
k <- sapply(seq(0,max(Simulated_stats_)*10, length.out=5000), est_pdf, betamod=betamod)
Use the approach above that you were initially using, i.e. a linear interpolation across the density, but rather than doing this over the density outcome, instead do over k (i.e. over the above estimates from the beta model)
lin_int = approxfun(x=seq(0,max(Simulated_stats_)*10, length.out=5000),y=k)
You can use the lin_int() function for prediction in the estimator, and it will be lighting fast. Note that it produces virtually the same value for a given x
c(est_pdf(38,betamod), lin_int(38))
[1] 0.001245894 0.001245968
and it is very fast
microbenchmark::microbenchmark(
list = alist("betamod" = est_pdf(38, betamod),"lin_int" = lint(38)),times=100
)
Unit: microseconds
expr min lq mean median uq max neval
betamod 1157.0 1170.20 1223.304 1188.25 1211.05 2799.8 100
lin_int 1.7 2.25 3.503 4.35 4.50 10.5 100
Finally, lets check the same plot you did before, but using lin_int() instead of approxfun(density(....))
a <- sapply(X = seq(from = 0, to = 100, by = 0.5), lin_int)
aa <- cbind( seq(from = 0, to = 100, by = 0.5), a)
plot(aa[,1],log(aa[,2]))

Related

Multiple random values between specific ranges in R?

I want to pick up 50 samples from (TRUNCATED) Normal Distribution (Gaussian) in a range 15-85 with mean=35, and sd=30. For reproducibility:
num = 50 # number of samples
rng = c(15, 85) # the range to pick the samples from
mu = 35 # mean
std = 30 # standard deviation
The following code gives 50 samples:
rnorm(n = num, mean = mu, sd = std)
However, I want these numbers to be strictly between the range 15-85. How can I achieve this?
UPDATE: Some people made great points in the comment section that this problem can not be solved as this will no longer be Gaussian Distribution. I added the word TRUNCATED to the original post so it makes more sense (Truncated Normal Distribution).
As Limey said in the comments, by imposing a bounded region the distribution is no longer normal. There are several ways to achieve this.
library("MCMCglmm")
rtnorm(n = 50, mean = mu, sd = std, lower = 15, upper = 85)
is one method. If you want a more manual approach you could simulate using uniform distribution within the range and apply the normal distribution function
bounds <- c(pnorm(15, mu, std), pnorm(50, mu, std))
samples <- qnorm(runif(50, bounds[1], bounds[2]), mu, std)
The idea is very basic: Simulate the quantiles of the outcome, and then estimate the value of the specific quantive given the distribution. The value of this approach rather than the approach linked by GKi is that it ensures a "normal-ish" distribution, where simulating and bounding the resulting vector will cause the bounds to have additional mass compared to the normal distribution.
Note the outcome is not normal, as it is bounded.

How to compute some basic statistics in R with the density and support only?

I have no sample and I'd like to compute the variance, mean, median, and mode of a distribution which I only have a vector with it's density and a vector with it's support. Is there an easy way to compute this statistics in R with this information?
Suppose that I only have the following information:
Support
Density
sum(Density) == 1 #TRUE
length(Support)==length(Density)# TRUE
You have to do weighted summations
F.e., starting with #Johann example
set.seed(312345)
x = rnorm(1000, mean=10, sd=1)
x_support = density(x)$x
x_density = density(x)$y
plot(x_support, x_density)
mean(x)
prints
[1] 10.00558
and what, I believe, you're looking for
m = weighted.mean(x_support, x_density)
computes mean as weighted mean of values, producing output
10.0055796130192
There are weighted.sd, weighted.sum functions which should help you with other quantities you're looking for.
Plot
If you don't need a mathematical solution, and an empirical one is all right, you can achieve a pretty good approximation by sampling.
Let's generate some data:
set.seed(6854684)
x = rnorm(50,mean=10,sd=1)
x_support = density(x)$x
x_density = density(x)$y
# see our example:
plot(x_support, x_density )
# the real mean of x
mean(x)
Now to 'reverse' the process we generate a large sample from that density distribution:
x_sampled = sample(x = x_support, 1000000, replace = T, prob = x_density)
# get the statistics
mean(x_sampled)
median(x_sampled)
var(x_sampled)
etc...

GAM with "gp" smoother: how to retrieve the variogram parameters?

I am using the following geoadditive model
library(gamair)
library(mgcv)
data(mack)
mack$log.net.area <- log(mack$net.area)
gm2 <- gam(egg.count ~ s(lon,lat,bs="gp",k=100,m=c(2,10,1)) +
s(I(b.depth^.5)) +
s(c.dist) +
s(temp.20m) +
offset(log.net.area),
data = mack, family = tw, method = "REML")
Here I am using an exponential covariance function with range = 10 and power = 1 (m=c(2,10,1)). How can I retrieve from the results the variogram parameters (nugget, sill)? I couldn't find anything in the model output.
In smoothing approach the correlation matrix is specified so you only estimate variance parameter, i.e., the sill. For example, you've set m = c(2, 10, 1) to s(, bs = 'gp'), giving an exponential correlation matrix with range parameter phi = 10. Note that phi is not identical to range, except for spherical correlation. For many correlation models the actual range is a function of phi.
The variance / sill parameter is closely related to the smoothing parameter in penalized regression, and you can obtain it by dividing the scale parameter by smoothing parameter:
with(gm2, scale / sp["s(lon,lat)"])
#s(lon,lat)
# 26.20877
Is this right? No. There is a trap here: smoothing parameters returned in $sp are not real ones, and we need the following:
gm2_sill <- with(gm2, scale / sp["s(lon,lat)"] * smooth[[1]]$S.scale)
#s(lon,lat)
# 7.7772
And we copy in the range parameter you've specified:
gm2_phi <- 10
The nugget must be zero, since a smooth function is continuous. Using lines.variomodel function from geoR package, you can visualize the semivariogram for the latent Gaussian spatial random field modeled by s(lon,lat).
library(geoR)
lines.variomodel(cov.model = "exponential", cov.pars = c(gm2_sill, gm2_phi),
nugget = 0, max.dist = 60)
abline(h = gm2_sill, lty = 2)
However, be skeptical on this variogram. mgcv is not an easy environment to interpret geostatistics. The use of low-rank smoothers suggests that the above variance parameter is for parameters in the new parameter space rather than the original one. For example, there are 630 unique spatial locations in the spatial field for mack dataset, so the correlation matrix should be 630 x 630, and the full random effects should be a vector of length-630. But by setting k = 100 in s(, bs = 'gp') the truncated eigen decomposition and subsequent low-rank approximation reduce the random effects to length-100. The variance parameter is really for this vector not the original one. This might explain why the sill and the actual range do not agree with the data and predicted s(lon,lat).
## unique locations
loc <- unique(mack[, c("lon", "lat")])
max(dist(loc))
#[1] 15.98
The maximum distance between two spatial locations in the dataset is 15.98, but the actual range from the variogram seems to be somewhere between 40 and 60, which is too large.
## predict `s(lon, lat)`, using the method I told you in your last question
## https://stackoverflow.com/q/51634953/4891738
sp <- predict(gm2,
data.frame(loc, b.depth = 0, c.dist = 0, temp.20m = 0,
log.net.area = 0),
type = "terms", terms = "s(lon,lat)")
c(var(sp))
#[1] 1.587126
The predicted s(lon,lat) only has variance 1.587, but the sill at 7.77 is way much higher.

Sample from a custom likelihood function

I have the following likelihood function which I used in a rather complex model (in practice on a log scale):
library(plyr)
dcustom=function(x,sd,L,R){
R. = (log(R) - log(x))/sd
L. = (log(L) - log(x))/sd
ll = pnorm(R.) - pnorm(L.)
return(ll)
}
df=data.frame(Range=seq(100,500),sd=rep(0.1,401),L=200,U=400)
df=mutate(df, Likelihood = dcustom(Range, sd,L,U))
with(df,plot(Range,Likelihood,type='l'))
abline(v=200)
abline(v=400)
In this function, the sd is predetermined and L and R are "observations" (very much like the endpoints of a uniform distribution), so all 3 of them are given. The above function provides a large likelihood (1) if the model estimate x (derived parameter) is in between the L-R range, a smooth likelihood decrease (between 0 and 1) near the bounds (of which the sharpness is dependent on the sd), and 0 if it is too much outside.
This function works very well to obtain estimates of x, but now I would like to do the inverse: draw a random x from the above function. If I would do this many times, I would generate a histogram that follows the shape of the curve plotted above.
The ultimate goal is to do this in C++, but I think it would be easier for me if I could first figure out how to do this in R.
There's some useful information online that helps me start (http://matlabtricks.com/post-44/generate-random-numbers-with-a-given-distribution, https://stats.stackexchange.com/questions/88697/sample-from-a-custom-continuous-distribution-in-r) but I'm still not entirely sure how to do it and how to code it.
I presume (not sure at all!) the steps are:
transform likelihood function into probability distribution
calculate the cumulative distribution function
inverse transform sampling
Is this correct and if so, how do I code this? Thank you.
One idea might be to use the Metropolis Hasting Algorithm to obtain a sample from the distribution given all the other parameters and your likelihood.
# metropolis hasting algorithm
set.seed(2018)
n_sample <- 100000
posterior_sample <- rep(NA, n_sample)
x <- 300 # starting value: I chose 300 based on your likelihood plot
for (i in 1:n_sample){
lik <- dcustom(x = x, sd = 0.1, L = 200, R =400)
# propose a value for x (you can adjust the stepsize with the sd)
x.proposed <- x + rnorm(1, 0, sd = 20)
lik.proposed <- dcustom(x = x.proposed, sd = 0.1, L = 200, R = 400)
r <- lik.proposed/lik # this is the acceptance ratio
# accept new value with probablity of ratio
if (runif(1) < r) {
x <- x.proposed
posterior_sample[i] <- x
}
}
# plotting the density
approximate_distr <- na.omit(posterior_sample)
d <- density(approximate_distr)
plot(d, main = "Sample from distribution")
abline(v=200)
abline(v=400)
# If you now want to sample just a few values (for example, 5) you could use
sample(approximate_distr,5)
#[1] 281.7310 371.2317 378.0504 342.5199 412.3302

confidence interval around predicted value from complex inverse function

I'm trying to get a 95% confidence interval around some predicted values, but am not capable of achieving this.
Basically, I estimated a growth curve like this:
set.seed(123)
dat=data.frame(size=rnorm(50,10,3),age=rnorm(50,5,2))
S <- function(t,ts,C,K) ((C*K)/(2*pi))*sin(2*pi*(t-ts))
sommers <- function(t,Linf,K,t0,ts,C)
Linf*(1-exp(-K*(t-t0)-S(t,ts,C,K)+S(t0,ts,C,K)))
model <- nls(size~sommers(age,Linf,K,t0,ts,C),data=dat,
start=list(Linf=10,K=4.7,t0=2.2,C=0.9,ts=0.1))
I have independent size measurements, for which I would like to predict the age. Therefore, the inverse of the function, which is not very straightforward, I calculated like this:
model.out=coef(model)
S.out <- function(t)
((model.out[[4]]*model.out[[2]])/(2*pi))*sin(2*pi*(t-model.out[[5]]))
sommers.out <- function(t)
model.out[[1]]*(1-exp(-model.out[[2]]*(t-model.out[[3]])-S.out(t)+S.out(model.out[[3]])))
inverse = function (f, lower = -100, upper = 100) {
function (y) uniroot((function (x) f(x) - y), lower = lower, upper = upper)[1]
}
sommers.inverse = inverse(sommers.out, 0, 25)
x= sommers.inverse(10) #this works with my complete dataset, but not with this fake one
Although this works fine, I need to know the confidence interval (95%) around this estimate (x). For linear models there is for example "predict(... confidence=)". I could also bootstrap the function somehow to get the quantiles associated with the parameters (didn't find how), to then use the extremes of those to calculate the maximum and minimum values predictable. But that doesn't really look like the good way of doing this....
Any help would be greatly appreciated.
EDIT after answer:
So this worked (explained in the book of Ben Bolker, see answer):
vmat = mvrnorm(1000, mu = coef(mfit), Sigma = vcov(mfit))
dist = numeric(1000)
for (i in 1:1000) {dist[i] = sommers_inverse(9.938,vmat[i,])}
quantile(dist, c(0.025, 0.975))
On the rather bad fake data I gave, this works of course rather horrible. But on the real data (which I have a problem recreating), this is ok!
Unless I'm mistaken, you're going to have to use either regular (parametric) bootstrapping or a method called either "population predictive intervals" (e.g., see section 5 of chapter 7 of Bolker 2008), which assumes that the sampling distributions of your parameters are multivariate Normal. However, I think you may have bigger problems, unless I've somehow messed up your model in adapting it ...
Generate data (note that random data may actually bad for testing your model - see below ...)
set.seed(123)
dat <- data.frame(size=rnorm(50,10,3),age=rnorm(50,5,2))
S <- function(t,ts,C,K) ((C*K)/(2*pi))*sin(2*pi*(t-ts))
sommers <- function(t,Linf,K,t0,ts,C)
Linf*(1-exp(-K*(t-t0)-S(t,ts,C,K)+S(t0,ts,C,K)))
Plot the data and the initial curve estimate:
plot(size~age,data=dat,ylim=c(0,16))
agevec <- seq(0,10,length=1001)
lines(agevec,sommers(agevec,Linf=10,K=4.7,t0=2.2,ts=0.1,C=0.9))
I had trouble with nls so I used minpack.lm::nls.lm, which is slightly more robust. (There are other options here, e.g. calculating the derivatives and providing the gradient function, or using AD Model Builder or Template Model Builder, or using the nls2 package.)
For nls.lm we need a function that returns the residuals:
sommers_fn <- function(par,dat) {
with(c(as.list(par),dat),size-sommers(age,Linf,K,t0,ts,C))
}
library(minpack.lm)
mfit <- nls.lm(fn=sommers_fn,
par=list(Linf=10,K=4.7,t0=2.2,C=0.9,ts=0.1),
dat=dat)
coef(mfit)
## Linf K t0 C ts
## 10.6540185 0.3466328 2.1675244 136.7164179 0.3627371
Here's our problem:
plot(size~age,data=dat,ylim=c(0,16))
lines(agevec,sommers(agevec,Linf=10,K=4.7,t0=2.2,ts=0.1,C=0.9))
with(as.list(coef(mfit)), {
lines(agevec,sommers(agevec,Linf,K,t0,ts,C),col=2)
abline(v=t0,lty=2)
abline(h=c(0,Linf),lty=2)
})
With this kind of fit, the results of the inverse function are going to be extremely unstable, as the inverse function is many-to-one, with the number of inverse values depending sensitively on the parameter values ...
sommers_pred <- function(x,pars) {
with(as.list(pars),sommers(x,Linf,K,t0,ts,C))
}
sommers_pred(6,coef(mfit)) ## s(6)=9.93
sommers_inverse <- function (y, pars, lower = -100, upper = 100) {
uniroot(function(x) sommers_pred(x,pars) -y, c(lower, upper))$root
}
sommers_inverse(9.938, coef(mfit)) ## 0.28
If I pick my interval very carefully I can get back the correct answer ...
sommers_inverse(9.938, coef(mfit), 5.5, 6.2)
Maybe your model will be better behaved with more realistic data. I hope so ...

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