I am attempting to run a two state HMM using a lognormal distribution. I have read Michelot and Langrock (2019) regarding choosing starting parameters through inspecting the data in a histogram and then running iterations in parallel, which has worked for my gamma distribution. Identifying the starting parameters for the lognormal distribution is troubling me however. Do I plot the log of my step length distribution then attempt extracting starting parameters or use the same starting parameters as my gamma distribution and rely on stepDist="lnorm"?
My code for the lognormal attempt currently looks like this:
ncores <- detectCores() - 1
cl <- makeCluster(getOption("cl.cores", ncores))
clusterExport(cl, list("data", "fitHMM"))
niter <- 20
allPar0 <- lapply(as.list(1:niter), function(x) {
stepMean0 <- runif(2,
min = c(x,y),
max = c(y,z))
stepSD0 <- runif(2,
min = c(x,y),
max = c(y,z))
angleMean0 <- c(0, 0)
angleCon0 <- runif(2,
min = c(a,b),
max = c(a,b))
stepPar0 <- c(stepMean0, stepSD0)
anglePar0 <- c(angleMean0, angleCon0)
return(list(step = stepPar0, angle = anglePar0))
})
# Fit the niter models in parallel
logP <- parLapply(cl = cl, X = allPar0, fun = function(par0) {
m <- fitHMM(data = data, nbStates = 2, stepDist = "lnorm", stepPar0 = par0$step,
anglePar0 = par0$angle)
return(m)
})
# Extract likelihoods of fitted models
likelihoodL <- unlist(lapply(logP, function(m) m$mod$minimum))
likelihoodL
# Index of best fitting model (smallest negative log-likelihood)
whichbestpL <- which.min(likelihoodL)
bestL <- logP[[whichbestpL]]
bestL
If I use negative values from plotting the log of the step length of the data then I get the error:
Error in checkForRemoteErrors(val) :
7 nodes produced errors; first error: Check the step parameters bounds (the initial parameters should be strictly between the bounds of their parameter space).
If I use the same starting parameter values that I used for my gamma distribution then I get the error
Error in unserialize(node$con) :
embedded nul in string: 'X\n\0\0\0\003\0\004\002\0\0\003\005\0\0\0'
Please could someone shed some light on how I'm failing at this?
Thank you!
Unfortunately, I can't tell for sure what the problem is from the code you included. If you don't get an error when you run fitHMM outside of parLapply, then it suggests that the problem is in how you choose the values of x, y, and z in your code.
The first parameter of the log-normal distribution can be negative or positive, and it is actually the mean of the logarithm of the step length. So, to find good starting values for this, you should look at a histogram of the log step lengths (e.g., following the dedicated moveHMM vignette). The second parameter is the standard deviation of the log step lengths, and this should be strictly positive (but could also be chosen based on the spread of the histogram of log step lengths).
To summarise, you should choose all the initial values based on plots of the log step lengths (rather than the step lengths themselves), and you should not use the same ranges of values for stepMean0 and stepSD0 (because the former can be negative or positive, whereas the latter is positive). Hopefully, this should help you choose x, y, and z.
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.
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 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))
I am trying to determine whether there is a significant difference between two Gamm distributions. One distribution has (shape, scale)=(shapeRef,scaleRef) while the other has (shape, scale)=(shapeTarget,scaleTarget). I try to do analysis of variance with the following code
n=10000
x=rgamma(n, shape=shapeRef, scale=scaleRef)
y=rgamma(n, shape=shapeTarget, scale=scaleTarget)
glmm1 <- gam(y~x,family=Gamma(link=log))
anova(glmm1)
The resulting p values keep changing and can be anywhere from <0.1 to >0.9.
Am I going about this the wrong way?
Edit: I use the following code instead
f <- gl(2, n)
x=rgamma(n, shape=shapeRef, scale=scaleRef)
y=rgamma(n, shape=shapeTarget, scale=scaleTarget)
xy <- c(x, y)
anova(glm(xy ~ f, family = Gamma(link = log)),test="F")
But, every time I run it I get a different p-value.
You will indeed get a different p-value every time you run this, if you pick different realizations every time. Just like your data values are random variables, which you'd expect to vary each time you ran an experiment, so is the p-value. If the null hypothesis is true (which was the case in your initial attempts), then the p-values will be uniformly distributed between 0 and 1.
Function to generate simulated data:
simfun <- function(n=100,shapeRef=2,shapeTarget=2,
scaleRef=1,scaleTarget=2) {
f <- gl(2, n)
x=rgamma(n, shape=shapeRef, scale=scaleRef)
y=rgamma(n, shape=shapeTarget, scale=scaleTarget)
xy <- c(x, y)
data.frame(xy,f)
}
Function to run anova() and extract the p-value:
sumfun <- function(d) {
aa <- anova(glm(xy ~ f, family = Gamma(link = log),data=d),test="F")
aa["f","Pr(>F)"]
}
Try it out, 500 times:
set.seed(101)
r <- replicate(500,sumfun(simfun()))
The p-values are always very small (the difference in scale parameters is easily distinguishable), but they do vary:
par(las=1,bty="l") ## cosmetic
hist(log10(r),col="gray",breaks=50)