I have been trying to generate R code for maximum likelihood estimation from a log likelihood function in a paper (equation 9 in page 609). Authors in the paper estimated it using MATLAB, which I am not familiar with. So I tried to generate codes in R.
Here is the snapshot of the log likelihood function in the paper:
, where
r: Binary decision (0 or 1) indicating infested plant(s) detection (1) or not (0).
e: Inspection efficiency. This is known.
n: Sample size
The overall objective is to estimate plant infestation rate (gamma: γ) and epsilon (e) based on binary decision of presence and absence of infested plants instead of using infested plant(s) detected. So, the function has only binary information (r) of infested plant detection and sample size. Since epsilon (e) is known or fixed, the actual goal is to estimate gamma (γ) in a population.
Another objective is to compare estimated infestation rates from above with ones in hypergeometric sampling formula in another paper (in page 6). The formula is:
This formula generates required sample size to detect infested plants with selected probability (e.g., 95) given an infested rate. For example:
# Sample size calculation function
fosgate.sample1 <- function(box, p, ci){ # Note: box represent total plant number
ninf <- p*box
sample.size <- round(((1-(1-ci)^(1/ninf))*(box-(ninf-1)/2)))
#sample.size <- ceiling(((1-(1-ci)^(1/ninf))*(box-(ninf-1)/2)))
sample.size
}
fosgate.sample1(box=100, p = .05, ci = .95) # where box: population or total plants, p: infestation rate, and ci: probability of detection
## 44
The idea is if sample size (e.g., 44) and binary decision data are provided the log-likelihood function can be used to estimate infestation rate and the rate may be close to anticipated rate (e.g., .05). Ultimately, I would like to compare plant infestation rates (gamma: γ) estimated from the log likelihood function above and D/N in the sample size calculation formula (second) or p in the sample size code below.
I generated R code for the log-likelihood described above.
### MLE with stat4
library(stats4)
# Log-likelihood function
plant.inf.lik <- function(inf.rate){
logl <- suppressWarnings(
sum((1-insp.result)*n*log(1-inf.rate) +
insp.result*log(1-(1-inf.rate)^n))
)
return(-logl)
}
Using the sample size function (i.e., fosgate.sample1) I generated sample sizes for various cases of total plant (or box) and anticipated detection rate (p) in the function. Since I am also interested in error/confidence ranges of estimated plant infestation rates, I used bootstrapping to calculate range of estimates (I am not sure if this is appropriate/acceptable). Here is the final code I generated:
### MLE and CI with bootstrapping with multiple scenarios
plant <- c(100, 500, 1000, 5000, 10000, 100000) # Total plant number
ir <- seq(.01, .2, by = .01) # Plant infestation rate
df.result <- data.frame(expand.grid(plant=plant, inf.rate = ir))
df.result$sample.size <- fosgate.sample1(box=df.result$plant, p=df.result$inf.rate, ci=.95) # Sample size
df.result$insp.result <- 1000 # Shipment number (can be replaced with random integers)
df.result <- df.result[order(df.result$plant, df.result$inf.rate, df.result$sample.size), ]
rownames(df.result) <- 1:nrow(df.result)
df.result$est.mean <- 0
#df.result$est.median <- 0
df.result$est.lower.ci <- 0
df.result$est.upper.ci <- 0
df.result$nsim <- 0
str(df.result)
head(df.result)
# Looping
est <- rep(NA, 1000)
for(j in 1:nrow(df.result)){
for(i in 1:1000){
insp.result <- sample(c(rep(1, df.result$insp.result[j]-df.result$insp.result[j]*df.result$inf.rate[j]),
rep(0, df.result$insp.result[j]*df.result$inf.rate[j])))
ir <- df.result$inf.rate[j]
n <- df.result$sample.size[j]
insp.result <- sample(insp.result, replace = TRUE)
est[i] <- mle(plant.inf.lik, start = list(inf.rate = ir*.9), method = "BFGS", nobs = length(insp.result))#coef
df.result$est.mean[j] <- mean(est, na.rm = TRUE)
# df.result$est.median[j] <- median(est, na.rm = TRUE)
df.result$est.lower.ci[j] <- quantile(est, prob = .025, na.rm = TRUE)
df.result$est.upper.ci[j] <- quantile(est, prob = .975, na.rm = TRUE)
df.result$nsim[j] <- length(est)
}
}
# Significance test result
sig <- ifelse(df.result$inf.rate >= df.result$est.lower.ci & df.result$inf.rate <= df.result$est.upper.ci, "no sig", "sig")
table(sig)
# Plot
library(ggplot2)
library(reshape2)
df.result$num <- ave(df.result$inf.rate, df.result$plant, FUN=seq_along)
df.result.m <- melt(df.result, id.vars=c("plant", "sample.size", "insp.result", "est.lower.ci", "est.upper.ci", "nsim", "num"))
df.result.m$est.lower.ci <- ifelse(df.result.m$variable == "inf.rate", NA, df.result.m$est.lower.ci)
df.result.m$est.upper.ci <- ifelse(df.result.m$variable == "inf.rate", NA, df.result.m$est.upper.ci)
str(df.result.m)
ggplot(data = df.result.m, aes(x = num, y = value, group=variable, color=variable, shape=variable))+
geom_point()+
geom_errorbar(aes(ymin = est.lower.ci, ymax = est.upper.ci), width=.5)+
scale_y_continuous(breaks = seq(0, .2, .02))+
xlab("Index")+
ylab("Plant infestation rate")+
facet_wrap(~plant, ncol = 3)
When I ran the code, I was able to obtain results and to compare estimated (est.mean) and anticipated (inf.rate) infestation rates as shown in the plot below.
If results are correct, plot indicates that estimation looks fine but off for greater infestation rates.
Also, I always got warning messages without "suppressWarnings" function and occasionally error messages below. I have no clue how to fix them.
## Warning messages
## 29: In log(1 - (1 - inf.rate)^n) : NaNs produced
## 30: In log(1 - inf.rate) : NaNs produced
## Error message (occasionally)
## Error in solve.default(oout$hessian) :
## Lapack routine dgesv: system is exactly singular: U[1,1] = 0
My questions are:
Is R function (plant.inf.lik) for maximum likelihood estimation of the log-likelihood function appropriate?
Should I take care of warning and error messages? If yes, how? Again, I have no clue how to fix...
Is bootstrapping (resampling?) method appropriate to estimate CI ranges and/or standard error?
I found this link useful for alternative approach. Although I am still working both approaches together, results seem different (maybe following question).
Any suggestion would be greatly appreciated.
Concerning your last question about estimating CI ranges, there are three common methods for ML estimators:
Variance estimation from the inverted Hessian matrix.
Jackknife estimator for the variance (simpler and more stable, if the Hessian is estimated numerically, but computationally more expensive)
Bootstrap CIs (the computatianally most expensive approach).
For bootstrap CIs, you do not need to implement them yourself (bias correction, e.g. can be tricky), but can rely on the R library boot.
Incidentally, I have written a summary with R code for all three approaches two years ago: Construction of Confidence Intervals (see section 5). For the method utilizing the Hessian Matrix, e.g., the outline is as follows:
lnL <- function(theta1, theta2, ...) {
# definition of the negative (!)
# log-likelihood function...
}
# starting values for the optimization
theta0 <- c(start1, start2, ...)
# optimization
p <- optim(theta0, lnL, hessian=TRUE)
if (p$convergence == 0) {
theta <- p$par
covmat <- solve(p$hessian)
sigma <- sqrt(diag(covmat))
}
The function mle from stats4 already wraps the covrainace matrix estimation and retruns it in vcov. In the practical use cases in which I have tried this (paired comparison models), though, this estimation was rather unstable, and I have resorted to the jackknife method instead.
Related
I am trying to set up a simple numerical MLE estimation of a multinomial distribution.
The multinomial has one constraint - all the cell probabilities need to add up to one.
Usually the way to have this constraint is to re-express one of the probabilities as (1 - sum of the others)
When I run this however, I have a problem as during the optimization procedure, I might have logarithm of a negative value.
Any thoughts of how to fix this? I tried using another optimization package (Rsolnp) and it worked, but I am trying to make it work with the simple default R optim in order to avoid constrained/nonlinear optimization.
Here is my code (I know that I can get the result in this particular case analytically, but this is a toy example, my actual problem is bigger than this here).
set.seed(1234)
test_data <- rmultinom(n = 1, size = 1000, prob = rep(1/4, 4))
N <- test_data
loglik_function <- function(theta){
output <- -1*(N[1]*log(theta[1]) + N[2]*log(theta[2]) + N[3]*log(theta[3]) + N[4]*log(1- sum(theta)))
return(output)
}
startval <- rep(0.1, 3)
my_optim <- optim(startval, loglik_function, lower = 0.0001, upper = 0.9999, method = "L-BFGS-B")
Any thoughts or help would be very much appreciated. Thanks
Full heads-up: I know you asked about (constrained) ML estimation, but how about doing this the Bayesian way à la Stan/rstan. I will remove this if it's not useful/missing the point.
The model is only a few lines of code.
library(rstan)
model_code <- "
data {
int<lower=1> K; // number of choices
int<lower=0> y[K]; // observed choices
}
parameters {
simplex[K] theta; // simplex of probabilities, one for every choice
}
model {
// Priors
theta ~ cauchy(0, 2.5); // weakly informative
// Likelihood
y ~ multinomial(theta);
}
generated quantities {
real ratio;
ratio = theta[1] / theta[2];
}
"
You can see how easy it is to implement the simplex constraint on the thetas using the Stan data type simplex. In the Stan language, simplex allows you to easily implement a probability (unit) simplex
where K denotes the number of parameters (here: choices).
Also note how we use the generated quantities code block, to calculate derived quantities (here ratio) based on the parameters (here theta[1] and theta[2]). Since we have access to the posterior distributions of all parameters, calculating the distribution of derived quantities is trivial.
We then fit the model to your test_data
fit <- stan(model_code = model_code, data = list(K = 4, y = test_data[, 1]))
and show a summary of the parameter estimates
summary(fit)$summary
# mean se_mean sd 2.5% 25%
#theta[1] 0.2379866 0.0002066858 0.01352791 0.2116417 0.2288498
#theta[2] 0.26 20013 0.0002208638 0.01365478 0.2358731 0.2526111
#theta[3] 0.2452539 0.0002101333 0.01344665 0.2196868 0.2361817
#theta[4] 0.2547582 0.0002110441 0.01375618 0.2277589 0.2458899
#ratio 0.9116350 0.0012555320 0.08050852 0.7639551 0.8545142
#lp__ -1392.6941655 0.0261794859 1.19050097 -1395.8297494 -1393.2406198
# 50% 75% 97.5% n_eff Rhat
#theta[1] 0.2381541 0.2472830 0.2645305 4283.904 0.9999816
#theta[2] 0.2615782 0.2710044 0.2898404 3822.257 1.0001742
#theta[3] 0.2448304 0.2543389 0.2722152 4094.852 1.0007501
#theta[4] 0.2545946 0.2638733 0.2822803 4248.632 0.9994449
#ratio 0.9078901 0.9648312 1.0764747 4111.764 0.9998184
#lp__ -1392.3914998 -1391.8199477 -1391.3274885 2067.937 1.0013440
as well as a plot showing point estimates and CIs for the theta parameters
plot(fit, pars = "theta")
Update: Constrained ML estimation using maxLik
You can in fact implement constrained ML estimation using methods provided by the maxLik library. I found it a bit "fiddly", because convergence seems to be quite sensitive to changes in the starting values and the optimisation method used.
For what it's worth, here is a reproducible example:
library(maxLik)
x <- test_data[, 1]
Define the log-likelihood function for a multinomial distribution; I've included an if statement here to prevent theta < 0 cases from throwing an error.
loglik <- function(theta, x)
if (all(theta > 0)) sum(dmultinom(x, prob = theta, log = TRUE)) else 0
I use the Nelder-Mead optimisation method here to find the maximum of the log-likelihood function. The important bit here is the constraints argument that implements a constraint in the form of the equality A theta + B = 0, see ?maxNM for details and examples.
res <- maxNM(
loglik,
start = rep(0.25, length(x)),
constraints = list(
eqA = matrix(rep(1, length(x)), ncol = length(x)),
eqB = -1),
x = x)
We can inspect the results
summary(res)
--------------------------------------------
Nelder-Mead maximization
Number of iterations: 111
Return code: 0
successful convergence
Function value: -10.34576
Estimates:
estimate gradient
[1,] 0.2380216 -0.014219040
[2,] 0.2620168 0.012664714
[3,] 0.2450181 0.002736670
[4,] 0.2550201 -0.002369234
Constrained optimization based on SUMT
Return code: 1
penalty close to zero
1 outer iterations, barrier value 5.868967e-09
--------------------------------------------
and confirm that indeed the sum of the estimates equals 1 (within accuracy)
sum(res$estimate)
#[1] 1.000077
Sample data
set.seed(1234)
test_data <- rmultinom(n = 1, size = 1000, prob = rep(1/4, 4))
This is my first post, so if there is any improvement to be made or further information that I can give please let me know.
I am using R 3.4.3 on a mac running 10.11.6.
I am working with count data. I am interested in the theta/ dispersion/ size parameter (k) of the negative binomial distribution (my understanding is that these terms are used interchangeably). I am fitting the NB distribution to these data estimating the parameters via maximum likelihood using the fitdistrplus package, with the fitdist function. I am interested if different groups of my data are better modelled using different distributions. I therefore fit the distribution to all the data. The data is then split based on one 2 level factor and the distribution fit to these two separate groups.
When I fit the distribution to the whole data set, I get an estimate of mu and size with standard errors. I then split the data. This same approach works fine for half of the data (Group A), but not the other half (Group B), which should, in theory, be structurally the same. Instead I get an estimate of mu and size but with NA for the standard errors.
Imposing lower = c(0,0) and upper = c(inf, inf) for the optim function behind fitdist also does not achieve anything, and as the output is NA not NaN I don't think it has anything to do with it trying to estimate negative numbers anyway (error 100 that is often discussed).
And just out of interest I removed all zero's in case it was something to do with that and that did nothing either.
So my question is why are the NA's being produced (and ultimately how do I get standard error's for the estimates)?
Here is my data and my code:
require(fitdistrplus)
data.set <- read.csv("data.set.csv")
count.A <- subset(data.set, category == "A")
count.B <- subset(data.set, category == "B")
# All Data
plotdist(data.set$count, histo = TRUE, demp = TRUE)
count.nb <- fitdist(data.set$count, "nbinom")
plot(count.nb)
LL.nb <- logLik(count.nb)
count.p <- fitdist(data.set$count, "pois")
plot(count.p)
LL.p <- logLik(count.p)
cdfcomp(list(count.p, count.nb),legendtext = c("Poisson", "negative binomial"))
gofstat(list(count.p, count.nb),fitnames = c("Poisson", "negative binomial"))
# Group A
plotdist(count.A$count, histo = TRUE, demp = TRUE)
A.count.nb <- fitdist(count.A$count, "nbinom")
plot(A.count.nb)
A.LL.nb <- logLik(A.count.nb)
A.count.p <- fitdist(count.A$count, "pois")
plot(A.count.p)
A.LL.p <- logLik(A.count.p)
cdfcomp(list(A.count.p, A.count.nb),legendtext = c("Poisson", "negative binomial"))
gofstat(list(A.count.p, A.count.nb),fitnames = c("Poisson", "negative binomial"))
# Group B
plotdist(count.B$count, histo = TRUE, demp = TRUE)
B.count.nb <- fitdist(count.B$count, "nbinom", method = "mle")
plot(B.count.nb)
B.LL.nb <- logLik(B.count.nb)
B.count.p <- fitdist(count.B$count, "pois")
plot(B.count.p)
B.LL.p <- logLik(B.count.p)
cdfcomp(list(B.count.p, B.count.nb),legendtext = c("Poisson", "negative binomial"))
gofstat(list(B.count.p, B.count.nb),fitnames = c("Poisson", "negative binomial"))
here is my code
I am trying to make a similar analysis to McNeil & Frey in their paper 'Estimation of tail-related risk measures for heteroscedastic financial time series: an extreme value approach' but I am stuck with a problem when implementing the models.
The approach is to fit a AR(1)-GARCH(1,1) model in order to estimate the the one-day ahead forecast of the VaR using a window of 1000 observations.
I have simulated data that should work fine with my model, and I assume that if I would be doing this correct, the observed coverage rate should be close to the theoretical one. However it is always below the theoretical coverage rate, and I don´t know why.
I beleive that this is how the calculation of the estimated VaR is done
VaR_hat = mu_hat + sigma_hat * qnorm(alpha)
, but I might be wrong. I have tried to find related questions here at stack but I have not found any.
How I approach this can be summarized in three steps.
Simulate 2000 AR(1)-GARCH(1,1) observations and fit a corresponding model and extract the one day prediction of the conditional mean and standard deviation using a window of 1000 observations.(Thereby making 1000 predictions)
Use the predicted values and the normal quantile to calculate the VaR for the wanted confidence level.
Check if the coverage rate is close to the theoretical one.
If someone could help me I would be extremely thankful, and if I'm unclear in my formalation please just tell me and I'll try to come up with a better explanation to the problem.
The code I'm using is attached below.
Thank you in advance
library(fGarch)
nObs <- 2000 # Number of observations.
quantileLevel <- 0.95 # Since we expect 5% exceedances.
from <- seq(1,1000) # Lower index vector for observations in model.
to <- seq(1001,2000) # Upper index vector for observations in model.
VaR_vec <- rep(0,(nObs-1000)) # Empty vector for storage of 1000 VaR estimates.
# Specs for simulated data (including AR(1) component and all components for GARC(1,1)).
spec = garchSpec(model = list(omega = 1e-6, alpha = 0.08, beta = 0.91, ar = 0.10),
cond.dist = 'norm')
# Simulate 1000 data points.
data_sim <- c(garchSim(spec, n = nObs, n.start = 1000))
for (i in 1:1000){
# The rolling window of 1000 observations.
data_insert <- data_sim[from[i]:to[i]]
# Fitting an AR(1)-GARCH(1,1) model with normal cond.dist.
fitted_model <- garchFit(~ arma(1,0) + garch(1,1), data_insert,
trace = FALSE,
cond.dist = "norm")
# One day ahead forecast of conditional mean and standard deviation.
predict(fitted_model, n.ahead = 1)
prediction_model <- predict(fitted_model, n.ahead = 1)
mu_pred <- prediction_model$meanForecast
sigma_pred <- prediction_model$standardDeviation
# Calculate VaR forecast
VaR_vec[i] <- mu_pred + sigma_pred*qnorm(quantileLevel)
if (length(to)-i != 0){
print(c('Countdown, just',(length(to) - i),'iterations left'))
} else {
print(c('Done!'))
}
}
# Exctract only the estiamtes ralated to the forecasts.
compare_data_sim <- data_sim[1001:length(data_sim)]
hit <- rep(0,length(VaR_vec))
# Count the amount of exceedances.
for (i in 1:length(VaR_vec)){
hit[i] <- sum(VaR_vec[i] <= compare_data_sim[i])
}
plot(data_sim[1001:2000], type = 'l',
ylab = 'Simulated data', main = 'Illustration of one day ahead prediction of 95%-VaR')
lines(VaR_vec, col = 'red')
cover_prop <- sum(hit)/length(hit)
print(sprintf("Diff theoretical level and VaR coverage = %f", (1-quantileLevel) - cover_prop))
I want to estimate the scale, shape and threshold parameters of a 3p Weibull distribution.
What I've done so far is the following:
Refering to this post, Fitting a 3 parameter Weibull distribution in R
I've used the functions
EPS = sqrt(.Machine$double.eps) # "epsilon" for very small numbers
llik.weibull <- function(shape, scale, thres, x)
{
sum(dweibull(x - thres, shape, scale, log=T))
}
thetahat.weibull <- function(x)
{
if(any(x <= 0)) stop("x values must be positive")
toptim <- function(theta) -llik.weibull(theta[1], theta[2], theta[3], x)
mu = mean(log(x))
sigma2 = var(log(x))
shape.guess = 1.2 / sqrt(sigma2)
scale.guess = exp(mu + (0.572 / shape.guess))
thres.guess = 1
res = nlminb(c(shape.guess, scale.guess, thres.guess), toptim, lower=EPS)
c(shape=res$par[1], scale=res$par[2], thres=res$par[3])
}
to "pre-estimate" my Weibull parameters, such that I can use them as initial values for the argument "start" in the "fitdistr" function of the MASS-Package.
You might ask why I want to estimate the parameters twice... reason is that I need the variance-covariance-matrix of the estimates which is also estimated by the fitdistr function.
EXAMPLE:
set.seed(1)
thres <- 450
dat <- rweibull(1000, 2.78, 750) + thres
pre_mle <- thetahat.weibull(dat)
my_wb <- function(x, shape, scale, thres) {
dweibull(x - thres, shape, scale)
}
ml <- fitdistr(dat, densfun = my_wb, start = list(shape = round(pre_mle[1], digits = 0), scale = round(pre_mle[2], digits = 0),
thres = round(pre_mle[3], digits = 0)))
ml
> ml
shape scale thres
2.942548 779.997177 419.996196 ( 0.152129) ( 32.194294) ( 28.729323)
> ml$vcov
shape scale thres
shape 0.02314322 4.335239 -3.836873
scale 4.33523868 1036.472551 -889.497580
thres -3.83687258 -889.497580 825.374029
This works quite well for cases where the shape parameter is above 1. Unfortunately my approach should deal with the cases where the shape parameter could be smaller than 1.
The reason why this is not possible for shape parameters that are smaller than 1 is described here: http://www.weibull.com/hotwire/issue148/hottopics148.htm
in Case 1, All three parameters are unknown the following is said:
"Define the smallest failure time of ti to be tmin. Then when γ → tmin, ln(tmin - γ) → -∞. If β is less than 1, then (β - 1)ln(tmin - γ) goes to +∞ . For a given solution of β, η and γ, we can always find another set of solutions (for example, by making γ closer to tmin) that will give a larger likelihood value. Therefore, there is no MLE solution for β, η and γ."
This makes a lot of sense. For this very reason I want to do it the way they described it on this page.
"In Weibull++, a gradient-based algorithm is used to find the MLE solution for β, η and γ. The upper bound of the range for γ is arbitrarily set to be 0.99 of tmin. Depending on the data set, either a local optimal or 0.99tmin is returned as the MLE solution for γ."
I want to set a feasible interval for gamma (in my code called 'thres') such that the solution is between (0, .99 * tmin).
Does anyone have an idea how to solve this problem?
In the function fitdistr there seems to be no opportunity doing a constrained MLE, constraining one parameter.
Another way to go could be the estimation of the asymptotic variance via the outer product of the score vectors. The score vector could be taken from the above used function thetahat.weibul(x). But calculating the outer product manually (without function) seems to be very time consuming and does not solve the problem of the constrained ML estimation.
Best regards,
Tim
It's not too hard to set up a constrained MLE. I'm going to do this in bbmle::mle2; you could also do it in stats4::mle, but bbmle has some additional features.
The larger issue is that it's theoretically difficult to define the sampling variance of an estimate when it's on the boundary of the allowed space; the theory behind Wald variance estimates breaks down. You can still calculate confidence intervals by likelihood profiling ... or you could bootstrap. I ran into a variety of optimization issues when doing this ... I haven't really thought about wether there are specific reasons
Reformat three-parameter Weibull function for mle2 use (takes x as first argument, takes log as an argument):
dweib3 <- function(x, shape, scale, thres, log=TRUE) {
dweibull(x - thres, shape, scale, log=log)
}
Starting function (slightly reformatted):
weib3_start <- function(x) {
mu <- mean(log(x))
sigma2 <- var(log(x))
logshape <- log(1.2 / sqrt(sigma2))
logscale <- mu + (0.572 / logshape)
logthres <- log(0.5*min(x))
list(logshape = logshape, logsc = logscale, logthres = logthres)
}
Generate data:
set.seed(1)
dat <- data.frame(x=rweibull(1000, 2.78, 750) + 450)
Fit model: I'm fitting the parameters on the log scale for convenience and stability, but you could use boundaries at zero as well.
tmin <- log(0.99*min(dat$x))
library(bbmle)
m1 <- mle2(x~dweib3(exp(logshape),exp(logsc),exp(logthres)),
data=dat,
upper=c(logshape=Inf,logsc=Inf,
logthres=tmin),
start=weib3_start(dat$x),
method="L-BFGS-B")
vcov(m1), which should normally provide a variance-covariance estimate (unless the estimate is on the boundary, which is not the case here) gives NaN values ... not sure why without more digging.
library(emdbook)
tmpf <- function(x,y) m1#minuslogl(logshape=x,
logsc=coef(m1)["logsc"],
logthres=y)
tmpf(1.1,6)
s1 <- curve3d(tmpf,
xlim=c(1,1.2),ylim=c(5.9,tmin),sys3d="image")
with(s1,contour(x,y,z,add=TRUE))
h <- lme4:::hessian(function(x) do.call(m1#minuslogl,as.list(x)),coef(m1))
vv <- solve(h)
diag(vv) ## [1] 0.002672240 0.001703674 0.004674833
(se <- sqrt(diag(vv))) ## standard errors
## [1] 0.05169371 0.04127558 0.06837275
cov2cor(vv)
## [,1] [,2] [,3]
## [1,] 1.0000000 0.8852090 -0.8778424
## [2,] 0.8852090 1.0000000 -0.9616941
## [3,] -0.8778424 -0.9616941 1.0000000
This is the variance-covariance matrix of the log-scaled variables. If you want to convert to the variance-covariance matrix on the original scale, you need to scale by (x_i)*(x_j) (i.e. by the derivatives of the transformation exp(x)).
outer(exp(coef(m1)),exp(coef(m1))) * vv
## logshape logsc logthres
## logshape 0.02312803 4.332993 -3.834145
## logsc 4.33299307 1035.966372 -888.980794
## logthres -3.83414498 -888.980794 824.831463
I don't know why this doesn't work with numDeriv - would be very careful with variance estimates above. (Maybe too close to boundary for Richardson extrapolation to work?)
library(numDeriv)
hessian()
grad(function(x) do.call(m1#minuslogl,as.list(x)),coef(m1)) ## looks OK
vcov(m1)
The profiles look OK ... (we have to supply std.err because the Hessian isn't invertible)
pp <- profile(m1,std.err=c(0.01,0.01,0.01))
par(las=1,bty="l",mfcol=c(1,3))
plot(pp,show.points=TRUE)
confint(pp)
## 2.5 % 97.5 %
## logshape 0.9899645 1.193571
## logsc 6.5933070 6.755399
## logthres 5.8508827 6.134346
Alternately, we can do this on the original scale ... one possibility would be to use the log-scaling to fit, then refit starting from those parameters on the original scale.
wstart <- as.list(exp(unlist(weib3_start(dat$x))))
names(wstart) <- gsub("log","",names(wstart))
m2 <- mle2(x~dweib3(shape,sc,thres),
data=dat,
lower=c(shape=0.001,sc=0.001,thres=0.001),
upper=c(shape=Inf,sc=Inf,
thres=exp(tmin)),
start=wstart,
method="L-BFGS-B")
vcov(m2)
## shape sc thres
## shape 0.02312399 4.332057 -3.833264
## sc 4.33205658 1035.743511 -888.770787
## thres -3.83326390 -888.770787 824.633714
all.equal(unname(coef(m2)),unname(exp(coef(m1))),tol=1e-4)
About the same as the values above.
We can fit with a small shape, if we are a little more careful to bound the paraameters, but now we end up on the boundary for the threshold, which will cause lots of problems for the variance calculations.
set.seed(1)
dat <- data.frame(x = rweibull(1000, .53, 365) + 100)
tmin <- log(0.99 * min(dat$x))
m1 <- mle2(x ~ dweib3(exp(logshape), exp(logsc), exp(logthres)),
lower=c(logshape=-10,logscale=0,logthres=0),
upper = c(logshape = 20, logsc = 20, logthres = tmin),
data = dat,
start = weib3_start(dat$x), method = "L-BFGS-B")
For censored data, you need to replace dweibull with pweibull; see Errors running Maximum Likelihood Estimation on a three parameter Weibull cdf for some hints.
Another possible solution is to do Bayesian inference. Using scale priors on the shape and scale parameters and a uniform prior on the location parameter, you can easily run Metropolis-Hastings as follows. It might be adviceable to reparameterize in terms of log(shape), log(scale) and log(y_min - location) because the posterior for some of the parameters becomes strongly skewed, in particular for the location parameter. Note that the output below shows the posterior for the backtransformed parameters.
library(MCMCpack)
logposterior <- function(par,y) {
gamma <- min(y) - exp(par[3])
sum(dweibull(y-gamma,exp(par[1]),exp(par[2]),log=TRUE)) + par[3]
}
y <- rweibull(100,shape=.8,scale=10) + 1
chain0 <- MCMCmetrop1R(logposterior, rep(0,3), y=y, V=.01*diag(3))
chain <- MCMCmetrop1R(logposterior, rep(0,3), y=y, V=var(chain0))
plot(exp(chain))
summary(exp(chain))
This produces the following output
#########################################################
The Metropolis acceptance rate was 0.43717
#########################################################
Iterations = 501:20500
Thinning interval = 1
Number of chains = 1
Sample size per chain = 20000
1. Empirical mean and standard deviation for each variable,
plus standard error of the mean:
Mean SD Naive SE Time-series SE
[1,] 0.81530 0.06767 0.0004785 0.001668
[2,] 10.59015 1.39636 0.0098738 0.034495
[3,] 0.04236 0.05642 0.0003990 0.001174
2. Quantiles for each variable:
2.5% 25% 50% 75% 97.5%
var1 0.6886083 0.768054 0.81236 0.8608 0.9498
var2 8.0756210 9.637392 10.50210 11.4631 13.5353
var3 0.0003397 0.007525 0.02221 0.0548 0.1939
I am working with the cumulative emergence of flies over time (taken at irregular intervals) over many summers (though first I am just trying to make one year work). The cumulative emergence follows a sigmoid pattern and I want to create a maximum likelihood estimation of a 3-parameter Weibull cumulative distribution function. The three-parameter models I've been trying to use in the fitdistrplus package keep giving me an error. I think this must have something to do with how my data is structured, but I cannot figure it out. Obviously I want it to read each point as an x (degree days) and a y (emergence) value, but it seems to be unable to read two columns. The main error I'm getting says "Non-numeric argument to mathematical function" or (with slightly different code) "data must be a numeric vector of length greater than 1". Below is my code including added columns in the df_dd_em dataframe for cumulative emergence and percent emergence in case that is useful.
degree_days <- c(998.08,1039.66,1111.29,1165.89,1236.53,1293.71,
1347.66,1387.76,1445.47,1493.44,1553.23,1601.97,
1670.28,1737.29,1791.94,1849.20,1920.91,1967.25,
2036.64,2091.85,2152.89,2199.13,2199.13,2263.09,
2297.94,2352.39,2384.03,2442.44,2541.28,2663.90,
2707.36,2773.82,2816.39,2863.94)
emergence <- c(0,0,0,1,1,0,2,3,17,10,0,0,0,2,0,3,0,0,1,5,0,0,0,0,
0,0,0,0,1,0,0,0,0,0)
cum_em <- cumsum(emergence)
df_dd_em <- data.frame (degree_days, emergence, cum_em)
df_dd_em$percent <- ave(df_dd_em$emergence, FUN = function(df_dd_em) 100*(df_dd_em)/46)
df_dd_em$cum_per <- ave(df_dd_em$cum_em, FUN = function(df_dd_em) 100*(df_dd_em)/46)
x <- pweibull(df_dd_em[c(1,3)],shape=5)
dframe2.mle <- fitdist(x, "weibull",method='mle')
Here's my best guess at what you're after:
Set up data:
dd <- data.frame(degree_days=c(998.08,1039.66,1111.29,1165.89,1236.53,1293.71,
1347.66,1387.76,1445.47,1493.44,1553.23,1601.97,
1670.28,1737.29,1791.94,1849.20,1920.91,1967.25,
2036.64,2091.85,2152.89,2199.13,2199.13,2263.09,
2297.94,2352.39,2384.03,2442.44,2541.28,2663.90,
2707.36,2773.82,2816.39,2863.94),
emergence=c(0,0,0,1,1,0,2,3,17,10,0,0,0,2,0,3,0,0,1,5,0,0,0,0,
0,0,0,0,1,0,0,0,0,0))
dd <- transform(dd,cum_em=cumsum(emergence))
We're actually going to fit to an "interval-censored" distribution (i.e. probability of emergence between successive degree day observations: this version assumes that the first observation refers to observations before the first degree-day observation, you could change it to refer to observations after the last observation).
library(bbmle)
## y*log(p) allowing for 0/0 occurrences:
y_log_p <- function(y,p) ifelse(y==0 & p==0,0,y*log(p))
NLLfun <- function(scale,shape,x=dd$degree_days,y=dd$emergence) {
prob <- pmax(diff(pweibull(c(-Inf,x), ## or (c(x,Inf))
shape=shape,scale=scale)),1e-6)
## multinomial probability
-sum(y_log_p(y,prob))
}
library(bbmle)
I should probably have used something more systematic like the method of moments (i.e. matching the mean and variance of a Weibull distribution with the mean and variance of the data), but I just hacked around a bit to find plausible starting values:
## preliminary look (method of moments would be better)
scvec <- 10^(seq(0,4,length=101))
plot(scvec,sapply(scvec,NLLfun,shape=1))
It's important to use parscale to let R know that the parameters are on very different scales:
startvals <- list(scale=1000,shape=1)
m1 <- mle2(NLLfun,start=startvals,
control=list(parscale=unlist(startvals)))
Now try with a three-parameter Weibull (as originally requested) -- requires only a slight modification of what we already have:
library(FAdist)
NLLfun2 <- function(scale,shape,thres,
x=dd$degree_days,y=dd$emergence) {
prob <- pmax(diff(pweibull3(c(-Inf,x),shape=shape,scale=scale,thres)),
1e-6)
## multinomial probability
-sum(y_log_p(y,prob))
}
startvals2 <- list(scale=1000,shape=1,thres=100)
m2 <- mle2(NLLfun2,start=startvals2,
control=list(parscale=unlist(startvals2)))
Looks like the three-parameter fit is much better:
library(emdbook)
AICtab(m1,m2)
## dAIC df
## m2 0.0 3
## m1 21.7 2
And here's the graphical summary:
with(dd,plot(cum_em~degree_days,cex=3))
with(as.list(coef(m1)),curve(sum(dd$emergence)*
pweibull(x,shape=shape,scale=scale),col=2,
add=TRUE))
with(as.list(coef(m2)),curve(sum(dd$emergence)*
pweibull3(x,shape=shape,
scale=scale,thres=thres),col=4,
add=TRUE))
(could also do this more elegantly with ggplot2 ...)
These don't seem like spectacularly good fits, but they're sane. (You could in principle do a chi-squared goodness-of-fit test based on the expected number of emergences per interval, and accounting for the fact that you've fitted a three-parameter model, although the values might be a bit low ...)
Confidence intervals on the fit are a bit of a nuisance; your choices are (1) bootstrapping; (2) parametric bootstrapping (resample parameters assuming a multivariate normal distribution of the data); (3) delta method.
Using bbmle::mle2 makes it easy to do things like get profile confidence intervals:
confint(m1)
## 2.5 % 97.5 %
## scale 1576.685652 1777.437283
## shape 4.223867 6.318481
dd <- data.frame(degree_days=c(998.08,1039.66,1111.29,1165.89,1236.53,1293.71,
1347.66,1387.76,1445.47,1493.44,1553.23,1601.97,
1670.28,1737.29,1791.94,1849.20,1920.91,1967.25,
2036.64,2091.85,2152.89,2199.13,2199.13,2263.09,
2297.94,2352.39,2384.03,2442.44,2541.28,2663.90,
2707.36,2773.82,2816.39,2863.94),
emergence=c(0,0,0,1,1,0,2,3,17,10,0,0,0,2,0,3,0,0,1,5,0,0,0,0,
0,0,0,0,1,0,0,0,0,0))
dd$cum_em <- cumsum(dd$emergence)
dd$percent <- ave(dd$emergence, FUN = function(dd) 100*(dd)/46)
dd$cum_per <- ave(dd$cum_em, FUN = function(dd) 100*(dd)/46)
dd <- transform(dd)
#start 3 parameter model
library(FAdist)
## y*log(p) allowing for 0/0 occurrences:
y_log_p <- function(y,p) ifelse(y==0 & p==0,0,y*log(p))
NLLfun2 <- function(scale,shape,thres,
x=dd$degree_days,y=dd$percent) {
prob <- pmax(diff(pweibull3(c(-Inf,x),shape=shape,scale=scale,thres)),
1e-6)
## multinomial probability
-sum(y_log_p(y,prob))
}
startvals2 <- list(scale=1000,shape=1,thres=100)
m2 <- mle2(NLLfun2,start=startvals2,
control=list(parscale=unlist(startvals2)))
summary(m2)
#graphical summary
windows(5,5)
with(dd,plot(cum_per~degree_days,cex=3))
with(as.list(coef(m2)),curve(sum(dd$percent)*
pweibull3(x,shape=shape,
scale=scale,thres=thres),col=4,
add=TRUE))