I want to estimate parameters of negative binomial distribution using MCMC Metropolis-Hastings algorithm. In other words, I have sample:
set.seed(42)
y <- rnbinom(20, size = 3, prob = 0.2)
and I want to write algorithm that will estimate parameter of size and parameter of prob.
My work so far
I defined prior distribution of size as Poisson:
prior_r <- function(r) {
return(dpois(r, lambda = 2, log = T))
}
And prior distribution of prob as uniform on [0, 1]:
prior_prob <- function(prob) {
return(dunif(prob, min = 0, max = 1, log = T))
}
Moreover for simplicity I defined loglikelihood and joint probability functions:
loglikelihood <- function(data, r, prob) {
loglikelihoodValue <- sum(dnorm(data, mean = r, sd = prob, log = T))
return(loglikelihoodValue)
}
joint <- function(r, prob) {
data <- y
return(loglikelihood(data, r, prob) + prior_r(r) + prior_prob(prob))
}
Finally, the whole algorithm:
run_mcmc <- function(startvalue, iterations) {
chain <- array(dim = c(iterations + 1, 2))
chain[1, ] <- startvalue
for (i in 1:iterations) {
proposal_r <- rpois(1, lambda = chain[i, 1])
proposal_prob <- chain[i, 2] + runif(1, min = -0.2, max = 0.2)
quotient <- joint(proposal_r, proposal_prob) - joint(chain[i, 1], chain[i, 2])
if (runif(1, 0, 1) < min(1, exp(quotient))) chain[i + 1, ] <- c(proposal_r, proposal_prob)
else chain[i + 1, ] <- chain[i, ]
}
return(chain)
}
The problem
Problem that I'm having is that when I run it with starting values even very close to correct ones:
iterations <- 2000
startvalue <- c(4, 0.25)
res <- run_mcmc(startvalue, iterations)
I'll obtain posterior distribution which is obviously wrong. For example
> colMeans(res)
[1] 11.963018 0.994533
As you can see, size is located very close to point 12, and probability is located in point 1.
Do you know what's the cause of those phenomeons?
Change dnorm in loglikelihood to dnbinom and fix the proposal for prob so it doesn't go outside (0,1):
set.seed(42)
y <- rnbinom(20, size = 3, prob = 0.2)
prior_r <- function(r) {
return(dpois(r, lambda = 2, log = T))
}
prior_prob <- function(prob) {
return(dunif(prob, min = 0, max = 1, log = TRUE))
}
loglikelihood <- function(data, r, prob) {
loglikelihoodValue <- sum(dnbinom(data, size = r, prob = prob, log = TRUE))
return(loglikelihoodValue)
}
joint <- function(r, prob) {
return(loglikelihood(y, r, prob) + prior_r(r) + prior_prob(prob))
}
run_mcmc <- function(startvalue, iterations) {
chain <- array(dim = c(iterations + 1, 2))
chain[1, ] <- startvalue
for (i in 1:iterations) {
proposal_r <- rpois(1, lambda = chain[i, 1])
proposal_prob <- chain[i, 2] + runif(1, min = max(-0.2, -chain[i,2]), max = min(0.2, 1 - chain[i,2]))
quotient <- joint(proposal_r, proposal_prob) - joint(chain[i, 1], chain[i, 2])
if (runif(1, 0, 1) < min(1, exp(quotient))) {
chain[i + 1, ] <- c(proposal_r, proposal_prob)
} else {
chain[i + 1, ] <- chain[i, ]
}
}
return(chain)
}
iterations <- 2000
startvalue <- c(4, 0.25)
res <- run_mcmc(startvalue, iterations)
colMeans(res)
#> [1] 3.1009495 0.1988177
Related
I tried binary logistic regression with BFGS using maxlik, but i have included the feature as per the syntax i attached below, but the result is, but i get output like this
Maximum Likelihood estimation
BFGS maximization, 0 iterations
*Return code 100: Initial value out of range.
https://docs.google.com/spreadsheets/d/1fVLeJznB9k29FQ_BdvdCF8ztkOwbdFpx/edit?usp=sharing&ouid=109040212946671424093&rtpof=true&sd=true (this is my data)*
library(maxLik)
library(optimx)
data=read_excel("Book2.xlsx")
data$JKLaki = ifelse(data$JK==1,1,0)
data$Daerah_Samarinda<- ifelse(data$Daerah==1,1,0)
data$Prodi2 = ifelse(data$Prodi==2,1,0)
data$Prodi3 = ifelse(data$Prodi==3,1,0)
data$Prodi4 = ifelse(data$Prodi==4,1,0)
str(data)
attach(data)
ll<- function(param){
mu <- param[1]
beta <- param[-1]
y<- as.vector(data$Y)
x<- cbind(1, data$JKLaki, data$IPK, data$Daerah_Samarinda, data$Prodi2, data$Prodi3, data$Prodi4)
xb<- x%*%beta
pi<- exp(xb)
val <- -sum(y * log(pi) + (1 - y) * log(1 - pi),log=TRUE)
return(val)
}
gl<- funtion(param){
mu <- param[1]
beta <- param[-1]
y <- as.vector(data$Y)
x <- cbind(0, data$JKLaki,data$IPK,data$Daerah_Samarinda,data$Prodi2,data$Prodi3,data$Prodi4)
sigma <- x*beta
pi<- exp(sigma)/(1+exp(sigma))
v= y-pi
vx=as.matrix(x)%*%as.vector(v)
gg= colSums(vx)
return(-gg)}
mle<-maxLik(logLik=ll, grad=gl,hess=NULL,
start=c(mu=1, beta1=0, beta2=0, beta3=0, beta4=0, beta5=0, beta6=0,beta7=0), method="BFGS")
summary(mle)
can i get some help, i tired get this solution, please.
I have been able to optimize the log-likelihood with the following code :
library(DEoptim)
library(readxl)
data <- read_excel("Book2.xlsx")
data$JKLaki <- ifelse(data$JK == 1, 1, 0)
data$Daerah_Samarinda <- ifelse(data$Daerah == 1, 1, 0)
data$Prodi2 <- ifelse(data$Prodi == 2, 1, 0)
data$Prodi3 <- ifelse(data$Prodi == 3, 1, 0)
data$Prodi4 <- ifelse(data$Prodi == 4, 1, 0)
ll <- function(param, data)
{
mu <- param[1]
beta <- param[-1]
y <- as.vector(data$Y)
x <- cbind(1, data$JKLaki, data$IPK, data$Daerah_Samarinda, data$Prodi2, data$Prodi3, data$Prodi4)
xb <- x %*% beta
pi <- exp(mu + xb)
val <- -sum(y * log(pi) + (1 - y) * log(1 - pi))
if(is.nan(val) == TRUE)
{
return(10 ^ 30)
}else
{
return(val)
}
}
lower <- rep(-500, 8)
upper <- rep(500, 8)
obj_DEoptim_Iter1 <- DEoptim(fn = ll, lower = lower, upper = upper,
control = list(itermax = 5000), data = data)
lower <- obj_DEoptim_Iter1$optim$bestmem - 0.25 * abs(obj_DEoptim_Iter1$optim$bestmem)
upper <- obj_DEoptim_Iter1$optim$bestmem + 0.25 * abs(obj_DEoptim_Iter1$optim$bestmem)
obj_DEoptim_Iter2 <- DEoptim(fn = ll, lower = lower, upper = upper,
control = list(itermax = 5000), data = data)
obj_Optim <- optim(par = obj_DEoptim_Iter2$optim$bestmem, fn = ll, data = data)
$par
par1 par2 par3 par4 par5 par6 par7
-350.91045436 347.79576145 0.05337466 0.69032735 -0.01089112 0.47465162 0.38284804
par8
0.42125664
$value
[1] 95.08457
$counts
function gradient
501 NA
$convergence
[1] 1
$message
NULL
everyone I am trying to execute the code in found in the book "Flexible Imputation of Missing Data 2ed" in 2.5.3 section, that calculates a confidence interval for two imputation methods. The problem is that I cannot reproduce the results as the result is always NaN
Here is the code
require(mice)
# function randomly draws artificial data from the specified linear model
create.data <- function(beta = 1, sigma2 = 1, n = 50, run = 1) {
set.seed(seed = run)
x <- rnorm(n)
y <- beta * x + rnorm(n, sd = sqrt(sigma2))
cbind(x = x, y = y)
}
#Remove some data
make.missing <- function(data, p = 0.5){
rx <- rbinom(nrow(data), 1, p)
data[rx == 0, "x"] <- NA
data
}
# Apply Rubin’s rules to the imputed data
test.impute <- function(data, m = 5, method = "norm", ...) {
imp <- mice(data, method = method, m = m, print = FALSE, ...)
fit <- with(imp, lm(y ~ x))
tab <- summary(pool(fit), "all", conf.int = TRUE)
as.numeric(tab["x", c("estimate", "2.5 %", "97.5 %")])
}
#Bind everything together
simulate <- function(runs = 10) {
res <- array(NA, dim = c(2, runs, 3))
dimnames(res) <- list(c("norm.predict", "norm.nob"),
as.character(1:runs),
c("estimate", "2.5 %","97.5 %"))
for(run in 1:runs) {
data <- create.data(run = run)
data <- make.missing(data)
res[1, run, ] <- test.impute(data, method = "norm.predict",
m = 2)
res[2, run, ] <- test.impute(data, method = "norm.nob")
}
res
}
res <- simulate(1000)
#Estimate the lower and upper bounds of the confidence intervals per method
apply(res, c(1, 3), mean, na.rm = TRUE)
Best Regards
Replace "x" by tab$term == "x" in the last line of test.impute():
as.numeric( tab[ tab$term == "x", c("estimate", "2.5 %", "97.5 %")])
Consider the following model for the evolution of an asset's price:
This what I have done (in R). I could not find a function that randomly outputs +1 or -1, so I decided to adapt the inbuilt rbinom function.
## This code is in R
rm(list = ls())
library(dplyr)
library(dint)
library(magrittr)
library(stats)
path =
function(T, mu, sigma, p, x0) {
x = rep(NA, T)
x[1] = x0
for(i in 2:T){
z = if_else(rbinom(1,1,p) == 0, -1, 1)
x[i] = x[i-1] * exp(mu + sigma*z)
}
return(x)
}
## Just some testing
x_sim = path(T = 4, mu = 0, sigma = 0.01, p = 0.5, x0 = 100)
## Actual answer
Np = 10000
mc = matrix(nrow = 17, ncol = Np)
for(j in 1:Np){
mc[,j] = path(T = 17, mu = 0, sigma = 0.01, p = 0.5, x0 = 100)
}
test = mc[2:nrow(mc), ] >= 100
sum_test = colSums(test)
comp = sum(sum_test >= 1)/length(sum_test)
prob = 1 - comp
Does this make sense? Any help/tips/advice would be much appreciated. Thanks!
Staying close to your code, I came up with this. Intuitively, if you think about it, the probability should be rather low due to the parameters and I get a probability of about 6.7% which is roughly what I get if I run your code with the parameters from the assignment.
simpath <- function(t, mu, sigma, p, x0, seed){
# set seed
if(!missing(seed)){
set.seed(seed)
}
# set up matrix for storing the results
res <- matrix(c(1:t, rep(NA, t*2)), ncol = 3)
colnames(res) <- c('t', 'z_t', 'x_t')
res[, 'z_t'] <- sample(c(1, -1), size = t, prob = c(p, 1-p), replace = TRUE)
res[1, 3] <- x0
for(i in 2:t){
res[i, 3] <- res[i-1, 3] * exp(mu+sigma*res[i, 2])
}
return(res)
}
x_sim <- simpath(t = 4, mu = 0, sigma = 0.01, p = 0.5, x0 = 100, seed = 123)
x_sim2 <- simpath(t = 36, mu = 0, sigma = 0.03, p = 0.5, x0 = 100, seed = 123)
## Actual answer
Np <- 100000
mc <- matrix(nrow = 36, ncol = Np)
for (j in 1:Np){
mc[, j] <- simpath(t = 36, mu = 0, sigma = 0.03, p = 0.5, x0 = 100)[, 3]
}
test <- mc > 100
sum_test <- colSums(test)
comp = sum(sum_test == 0)/length(sum_test)
prob = comp
> prob
[1] 0.06759
I am trying to conduct a two sided sign test from 10,000 random normal samples of size 30. I am trying to extract the p-values given from the binom.test and put them into a vector but can't quite figure out how to execute this.
set.seed(100)
sample <- matrix(rnorm(300000, mean=0.1, sd=1), 10000, 30)
success <- ifelse(sample>=0, 1, 0)
success
#sample[1,]
#success[1,]
#sum(success[1,])
#for loop
for(i in 1:10000){
pvalue<- binom.test(sum(success[i,]), 30, p=0.5,
alternative = c("two.sided"),
conf.level = 0.95)$p.value
p_values_success <- ifelse(pvalue<=0.05, 1, 0)
}
I guess what you are trying to do is
pvalue <- numeric(length = 1000L)
p_values_success <- numeric(length = 1000L)
for(i in 1:10000) {
pvalue[i] <- binom.test(sum(success[i,]), 30, p=0.5,
alternative = c("two.sided"),
conf.level = 0.95)$p.value
p_values_success[i] <- ifelse(pvalue[i]<=0.05, 1, 0)
}
However, if I had to rewrite you code completely from scratch I would do
set.seed(100)
sample <- matrix(rnorm(300000, mean=0.1, sd=1), 10000, 30)
success[] <- as.integer(sample >=0)
t(apply(success, 1, function(x) {
p_val <- binom.test(sum(x), 30, p=0.5,alternative = c("two.sided"),
conf.level = 0.95)$p.value
c(p_val, as.integer(p_val<=0.05))
}))
This will return a 2-column matrix where 1st column is pvalue and the second one is p_values_success.
You could also do:
apply(success, 1,
FUN = function(x)
ifelse(
binom.test(sum(x), 30, p = 0.5,
alternative = "two.sided", conf.level = 0.95)$p.value <= 0.05, 1, 0
)
)
I have an R function that provides the 95% confidence Interval for the ncp (non-centrality parameter) of a t distribution.
Via simulation in R, is it possible to show that in the long-run the CIs from this R function capture a given TRUE ncp (here "2" same as input t) 95% of the time?
(I appreciate any ideas as to how to do this)
CI.ncp <- function(t, N){
f <- function (ncp, alpha, q, df) {
abs(suppressWarnings(pt(q = t, df = N - 1, ncp, lower.tail = FALSE)) - alpha) }
sapply(c(0.025, 0.975),
function(x) optim(1, f, alpha = x, q = t, df = N - 1, control = list(reltol = (.Machine$double.eps)))[[1]]) }
#Example of Use:
CI.ncp(t = 2, N = 20) # gives: -0.08293755 4.03548862
#(in the long-run 95% of the time, "2" is contained within these
# two numbers, how to show this in R?)
Here is what I have tried with no success:
fun <- function(t = 2, N = 20){
ncp = rt(1, N - 1, t)
CI.ncp(t = 2, N = 20)
mean(ncp <= 2 & 2 <= ncp )
}
R <- 1000
sim <- t(replicate(R, fun()))
coverage <- mean(sim[,1] <= 2 & 2 <= sim[,2])
The problem is the that we need to feed the random ncp obtained from the fun in the CI.ncp:
fun <- function(t = 2, N = 20){ ;
ncp = rt(1, N - 1, t);
CI.ncp(t = ncp, N = 20);
}
R <- 1e4 ;
sim <- t(replicate(R, fun()));
coverage <- mean(sim[,1] <= 2 & 2 <= sim[,2])
I would use package MBESS.
#install.packages("MBESS")
library(MBESS)
fun <- function(t = 2, N = 20, alpha = 0.95){
x = rt(1, N - 1, t)
conf.limits.nct(x, df = N, conf.level = alpha)[c(1, 3)]
}
set.seed(5221)
R <- 1000
sim <- t(replicate(R, fun()))
head(sim)
coverage <- mean(sim[,1] <= 2 & 2 <= sim[,2])
coverage
[1] 0.941