when starting a standard example from the stan webpage like the following:
schools_code <- '
data {
int<lower=0> J; // number of schools
real y[J]; // estimated treatment effects
real<lower=0> sigma[J]; // s.e. of effect estimates
}
parameters {
real theta[J];
real mu;
real<lower=0> tau;
}
model {
theta ~ normal(mu, tau);
y ~ normal(theta, sigma);
}
'
schools_dat <- list(J = 8,
y = c(28, 8, -3, 7, -1, 1, 18, 12),
sigma = c(15, 10, 16, 11, 9, 11, 10, 18))
fit <- stan(model_code = schools_code, data = schools_dat,
iter = 1000, n_chains = 4)
(this has been obtained from here)
however this does only provide me with the quantiles of the posterior of the parameters. so my question is: how to obtain other percentiles? i guess it should be similar to bugs(?)
remark: i tried to introduce the tag stan however, i have too little reputation ;) sorry for that
As from rstan v1.0.3 (not released yet), you will be able to utilize the workhorse apply() function directly on an object of stanfit class that is produced by the stan() function. If fit is an object obtained from stan(), then for example,
apply(fit, MARGIN = "parameters", FUN = quantile, probs = (1:100) / 100)
or
apply(as.matrix(fit), MARGIN = 2, FUN = quantile, probs = (1:100) / 100)
The former applies FUN to each parameter in each chain, while the latter combines the chains before applying FUN to each parameter. If you were only interested in one parameter, then something like
beta <- extract(fit, pars = "beta", inc_warmup = FALSE, permuted = TRUE)[[1]]
quantile(beta, probs = (1:100) / 100)
is an option.
here's my attempt hope this is correct:
suppose fit is an object obtained from stan(...). then the posterior for any percentile is obtained from:
quantile(fit#sim$sample[[1]]$beta, probs=c((1:100)/100))
where the number in square brackets is the chain i guess. in case this hasn't been clear: i use rstan
Related
I hope you are well and having a nice day. I am attempting to code an MH algorithm for a multiple linear regression model. I am following a tutorial online for simple linear regression using an MH algorithm and I plan on applying the principles myself to my multiple linear regression model, but I am running into a problem figuring our how to specify prior distributions for the parameters. Below, is the code.
trueA <- 5
trueB <- 0
trueSd <- 10
sampleSize <- 31
# create independent x-values
x <- (-(sampleSize-1)/2):((sampleSize-1)/2)
# create dependent values according to ax + b + N(0,sd)
y <- trueA * x + trueB + rnorm(n=sampleSize,mean=0,sd=trueSd)
likelihood <- function(param){
a = param[1]
b = param[2]
sd = param[3]
pred = a*x + b
singlelikelihoods = dnorm(y, mean = pred, sd = sd, log = T)
sumll = sum(singlelikelihoods)
return(sumll)
}
This is the portion where I am running into problems. In the tutorial, the author now specifies that they define prior distributions for the parameters. In this case, I would like to specify uninformative prior distributions for the slopes, coefficients, and standard deviation. The author says that they used uniform/normal distributions.
I found an example using Poisson regression, which looks like:
LogPriorFunction <- function(param){
beta0 <- param[1]
beta1 <- param[2]
beta0prior <- dnorm(beta0, 0, sqrt(100), log=TRUE)
beta1prior <- dnorm(beta1, 0, sqrt(100), log=TRUE)
return(beta0prior + beta1prior) # Logarithm of prior distributions
}
But I am having trouble adapting it to the simple linear regression. Here is my attempt:
prior <- function(param){
a = param[1]
b = param[2]
sd = param[3]
aprior = dnorm(a, 0, 1000, log=TRUE)
bprior = dnorm(b, 0, 1000, log=TRUE)
sdprior = 1/rgamma(sd, shape = .001, scale = .001)
return(aprior + bprior + sdprior) # Logarithm of prior distributions
}
The rest of the code is:
posterior <- function(param){
return (likelihood(param) + prior(param))
}
proposalfunction <- function(param){
return(rnorm(3,mean = param, sd= c(0.1,0.5,0.3)))
}
run_metropolis_MCMC <- function(startvalue, iterations){
chain = array(dim = c(iterations+1,3))
chain[1,] = startvalue
for (i in 1:iterations){
proposal = proposalfunction(chain[i,])
probab = exp(posterior(proposal) - posterior(chain[i,]))
if (runif(1) < probab){
chain[i+1,] = proposal
}else{
chain[i+1,] = chain[i,]
}
}
return(chain)
}
startvalue = c(4,0,10)
chain = run_metropolis_MCMC(startvalue, 10000)
burnIn = 5000
acceptance = 1-mean(duplicated(chain[-(1:burnIn),]))
Finally, along with specifying the priors, I keep getting the warning message:
Error in if (runif(1) < probab) { : missing value where TRUE/FALSE needed
In addition: Warning message:
In if (runif(1) < probab) { :
the condition has length > 1 and only the first element will be used
I am not sure if this is related to my incorrectly specifying the prior distributions, as the author of the article gets similar results from the MH algorithm as running a simple linear regression model using the lm() function. Any help would with these issues would be greatly appreciated. I look forward to hearing from you. Thank you.
As a reference, here is the link to the article:
https://khayatrayen.github.io/MCMC.html#defining_the_prior
Finally, here is a link to the article from where I tried to emulate the priors, but does Poisson regression and not linear regression:
https://rpubs.com/SaraGarcesCespedes/586440
I´m working on a STM Model (topicmodelling) and i´d like to evaluate and verify the model, but i´m not sure how to do it. My code is:
Corpus.STM <- readCorpus(dtm, type = "slam")
Model choice:
BestM1. <- searchK(Corpus.STM$documents, Corpus.STM$vocab, K=c(10,20, 30, 40, 50, 60), proportion = .4, heldout.seed = 1, prevalence=~ cvJahr+ cvDienstgrad+ cvLand, data=Jahr.Land )
BestM2. <- searchK(Corpus.STM$documents, Corpus.STM$vocab, K=c(85,110), proportion = .4, heldout.seed = 1, prevalence=~ cvJahr+ cvDienstgrad+ cvLand, data=Jahr.Land )
BestM3. <- searchK(Corpus.STM$documents, Corpus.STM$vocab, K=c(20,21,22,23,24,25,26,27,28,29,30), proportion = .4, heldout.seed = 1, prevalence=~ cvJahr+ cvDienstgrad+ cvLand, data=Jahr.Land )
str(BestM1.)
plot.searchK(BestM1.)
plot.STM(BestM2)
plot.searchK(BestM3.)
#27 seems to be a good choice
#Heldout
set.seed(1)
heldout<- make.heldout(Corpus.STM$documents, Corpus.STM$vocab, proportion = .5,seed = 1)
stm.mod1 <- stm(heldout$documents, heldout$vocab, K =27, seed = 1, init.type = "Spectral", max.em.its = 100 )
heldout.evaluation <- eval.heldout(stm.mod1, heldout$missing)
heldout.evaluation
#evaluation heldout
labelTopics(stm.mod1)
plot.STM(stm.mod1, type="labels", n=5, frexweight = 0.25)
cloud(stm.mod1, topic=5)
plot.STM(stm.mod1, type="summary", labeltype="frex", topics=c(1:5), n=8)
I´m not sure how to interpret the output of "eval.heldout". Additional I want to make sure that the model doesn´t overfit, but i´m not sure how it could work.
eval.heldout() calculates the held-out log-likelihood using document completion. The number you want is the heldout.evaluation$expected.heldout which is the average of the held-out log-likelihood values for each document. Unfortunately there is no unambiguous measure of whether or not the model is "overfit." The plot.searchK() call you have will give you a plot of the held-out log-likelihood over different values of K and certainly if that number is decreasing as K goes up one explanation is overfitting.
Sorry to not have a clearer answer but unfortunately there are no hard and fast rules here.
I am using 'KFAS' package from R to estimate a state-space model with the Kalman filter. My measurement and transition equations are:
y_t = Z_t * x_t + \eps_t (measurement)
x_t = T_t * x_{t-1} + R_t * \eta_t (transition),
with \eps_t ~ N(0,H_t) and \eta_t ~ N(0,Q_t).
So, I want to estimate the variances H_t and Q_t, but also T_t, the AR(1) coefficient. My code is as follows:
library(KFAS)
set.seed(100)
eps <- rt(200, 4, 1)
meas <- as.matrix((arima.sim(n=200, list(ar=0.6), innov = rnorm(200)*sqrt(0.5)) + eps),
ncol=1)
Zt <- 1
Ht <- matrix(NA)
Tt <- matrix(NA)
Rt <- 1
Qt <- matrix(NA)
ss_model <- SSModel(meas ~ -1 + SSMcustom(Z = Zt, T = Tt, R = Rt,
Q = Qt), H = Ht)
fit <- fitSSM(ss_model, inits = c(0,0.6,0), method = 'L-BFGS-B')
But it returns: "Error in is.SSModel(do.call(updatefn, args = c(list(inits, model), update_args)),: System matrices (excluding Z) contain NA or infinite values, covariance matrices contain values larger than 1e+07"
The NA definitions for the variances works well, as documented in the package's paper. However, it seems this cannot be done for the AR coefficients. Does anyone know how can I do this?
Note that I am aware of the SSMarima function, which eases the definition of the transition equation as ARIMA models. Although I am able to estimate the AR(1) coef. and Q_t this way, I still cannot estimate the \eps_t variance (H_t). Moreover, I am migrating my Kalman filter codes from EViews to R, so I need to learn SSMcustom for other models that are more complicated.
Thanks!
It seems that you are missing something in your example, as your error message comes from the function fitSSM. If you want to use fitSSM for estimating general state space models, you need to provide your own model updating function. The default behaviour can only handle NA's in covariance matrices H and Q. The main goal of fitSSM is just to get started with simple stuff. For complex models and/or large data, I would recommend using your self-written objective function (with help of logLik method) and your favourite numerical optimization routines manually for maximum performance. Something like this:
library(KFAS)
set.seed(100)
eps <- rt(200, 4, 1)
meas <- as.matrix((arima.sim(n=200, list(ar=0.6), innov = rnorm(200)*sqrt(0.5)) + eps),
ncol=1)
Zt <- 1
Ht <- matrix(NA)
Tt <- matrix(NA)
Rt <- 1
Qt <- matrix(NA)
ss_model <- SSModel(meas ~ -1 + SSMcustom(Z = Zt, T = Tt, R = Rt,
Q = Qt), H = Ht)
objf <- function(pars, model, estimate = TRUE) {
model$H[1] <- pars[1]
model$T[1] <- pars[2]
model$Q[1] <- pars[3]
if (estimate) {
-logLik(model)
} else {
model
}
}
opt <- optim(c(1, 0.5, 1), objf, method = "L-BFGS-B",
lower = c(0, -0.99, 0), upper = c(100, 0.99, 100), model = ss_model)
ss_model_opt <- objf(opt$par, ss_model, estimate = FALSE)
Same with fitSSM:
updatefn <- function(pars, model) {
model$H[1] <- pars[1]
model$T[1] <- pars[2]
model$Q[1] <- pars[3]
model
}
fit <- fitSSM(ss_model, c(1, 0.5, 1), updatefn, method = "L-BFGS-B",
lower = c(0, -0.99, 0), upper = c(100, 0.99, 100))
identical(ss_model_opt, fit$model)
I'm a newbie in R!
I would like to find the best gamma distribution parameters to fit my experimental counts data. The optim function's help file says the first argument of the function should be the parameters to be optimized. So I tried :
x = as.matrix(seq(1,20,0.1))
yexp = dgamma(x,2,1)*100 + rnorm(length(x),0,1)
f = function(p,x,yexp) {sum((p[1]*dgamma(x,p[2],scale=p[3]) - yexp)^2)}
mod = optim(c(50,2,1),f(p,x,yexp))
I get the error message :
Error in f(p, x, yexp) : object 'p' not found
Any hint where I'm wrong?
Supplementary question : Is there any other way to fit counts data with standard distribution (gamma, inverse gaussian, etc?)
optim expects its second argument to be a function. Also, the second and third arguments to f are fixed and need to be specified:
optim(c(50, 1, 2), f, x = x, yexp = yexp)
This would also work:
optim(c(50, 1, 2), function(p) f(p, x, yexp))
You could also use nls with default Nelder-Mead algorithm:
nls(yexp ~ a * dgamma(x, sh, scale=sc), start = list(a = 50, sh = 2, sc = 1))
or with plinear in which case no starting value is needed for the first parameter:
nls(c(yexp) ~ dgamma(x, sh, scale=sc), start = list(sh = 2, sc = 1), alg = "plinear")
I want to find Lethal Dose (LD50) with its confidence interval in R. Other softwares line Minitab, SPSS, SAS provide three different versions of such confidence intervals. I could not find such intervals in any package in R (I also used findFn function from sos package).
How can I find such intervals? I coded for one type of intervals based on Delta method (as not sure about it correctness) but would like to use any established function from R package. Thanks
MWE:
dose <- c(10.2, 7.7, 5.1, 3.8, 2.6, 0)
total <- c(50, 49, 46, 48, 50, 49)
affected <- c(44, 42, 24, 16, 6, 0)
finney71 <- data.frame(dose, total, affected)
fm1 <- glm(cbind(affected, total-affected) ~ log(dose),
family=binomial(link = logit), data=finney71[finney71$dose != 0, ])
summary(fm1)$coef
Estimate Std. Error z value Pr(>|z|)
(Intercept) -4.886912 0.6429272 -7.601035 2.937717e-14
log(dose) 3.103545 0.3877178 8.004650 1.198070e-15
library(MASS)
xp <- dose.p(fm1, p=c(0.50, 0.90, 0.95)) # from MASS
xp.ci <- xp + attr(xp, "SE") %*% matrix(qnorm(1 - 0.05/2)*c(-1,1), nrow=1)
zp.est <- exp(cbind(xp, attr(xp, "SE"), xp.ci[,1], xp.ci[,2]))
dimnames(zp.est)[[2]] <- c("LD", "SE", "LCL","UCL")
zp.est
LD SE LCL UCL
p = 0.50: 4.828918 1.053044 4.363708 5.343724
p = 0.90: 9.802082 1.104050 8.073495 11.900771
p = 0.95: 12.470382 1.133880 9.748334 15.952512
From the package drc, you can get the ED50 (same calculation), along with confidence intervals.
library(drc) # Directly borrowed from the drc manual
mod <- drm(affected/total ~ dose, weights = total,
data = finney71[finney71$dose != 0, ], fct = LL2.2(), type = "binomial")
#intervals on log scale
ED(mod, c(50, 90, 95), interval = "fls", reference = "control")
Estimated effective doses
(Back-transformed from log scale-based confidence interval(s))
Estimate Lower Upper
1:50 4.8289 4.3637 5.3437
1:90 9.8021 8.0735 11.9008
1:95 12.4704 9.7483 15.9525
Which matches the manual output.
The "finney71" data is included in this package, and your calculation of confidence intervals exactly matches the example given by the drc folks, down to the "# from MASS" comment. You should give credit to them, rather than claiming you wrote the code.
There's a few other ways to figure this out. One is using parametric bootstrap, which is conveniently available through the boot package.
First, we'll refit the model.
library(boot)
finney71 <- finney71[finney71$dose != 0,] # pre-clean data
fm1 <- glm(cbind(affected, total-affected) ~ log(dose),
family=binomial(link = logit),
data=finney71)
And for illustration, we can figure out the LD50 and LD75.
statfun <- function(dat, ind) {
mod <- update(fm1, data = dat[ind,])
coefs <- coef(mod)
c(exp(-coefs[1]/coefs[2]),
exp((log(0.75/0.25) - coefs[2])/coefs[1]))
}
boot_out <- boot(data = finney71, statistic = statfun, R = 1000)
The boot.ci function can work out a variety of confidence intervals for us, using this object.
boot.ci(boot_out, index = 1, type = c('basic', 'perc', 'norm'))
##BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
##Based on 999 bootstrap replicates
##
##CALL :
##boot.ci(boot.out = boot_out, type = c("basic", "perc", "norm"),
## index = 1)
##Intervals :
##Level Normal Basic Percentile
##95% ( 3.976, 5.764 ) ( 4.593, 5.051 ) ( 4.607, 5.065 )
The confidence intervals using the normal approximation are thrown off quite a bit by a few extreme values, which the basic and percentile-based intervals are more robust to.
One interesting thing to note: if the sign of the slope is sufficiently unclear, we can get some rather extreme values (simulated as in this answer, and discussed more thoroughly in this blog post by Andrew Gelman).
set.seed(1)
x <- rnorm(100)
z = 0.05 + 0.1*x*rnorm(100, 0, 0.05) # small slope and more noise
pr = 1/(1+exp(-z))
y = rbinom(1000, 1, pr)
sim_dat <- data.frame(x, y)
sim_mod <- glm(y ~ x, data = sim_dat, family = 'binomial')
statfun <- function(dat, ind) {
mod <- update(sim_mod, data = dat[ind,])
-coef(mod)[1]/coef(mod)[2]
}
sim_boot <- boot(data = sim_dat, statistic = statfun, R = 1000)
hist(sim_boot$t[,1], breaks = 100,
main = "Bootstrap of simulated model")
The delta method above gives us mean = 6.448, lower ci = -36.22, and upper ci = 49.12, and all of the bootstrap CIs give us similarly extreme estimates.
##Level Normal Basic Percentile
##95% (-232.19, 247.76 ) ( -20.17, 45.13 ) ( -32.23, 33.06 )