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I want to get a 95% confidence interval for the following question.
I have written function f_n in my R code. I first randomly sample 100 with Normal and then I define function h for lambda. Then I can get f_n. My question is that how to define a function of f_n-chi-square and use uniroot` to find Confidence interval.
# I first get 100 samples
set.seed(201111)
x=rlnorm(100,0,2)
Based on the answer by #RuiBarradas, I try the following code.
set.seed(2011111)
# I define function h, and use uniroot function to find lambda
h <- function(lam, n)
{
sum((x - theta)/(1 + lam*(x - theta)))
}
# sample size
n <- 100
# the parameter of interest must be a value in [1, 12],
#true_theta<-1
#true_sd<- exp(2)
#x <- rnorm(n, mean = true_theta, sd = true_sd)
x=rlnorm(100,0,2)
xmax <- max(x)
xmin <- min(x)
theta_seq = seq(from = 1, to = 12, by = 0.01)
f_n <- rep(NA, length(theta_seq))
for (i in seq_along(theta_seq))
{
theta <- theta_seq[i]
lambdamin <- (1/n-1)/(xmax - theta)
lambdamax <- (1/n-1)/(xmin - theta)
lambda = uniroot(h, interval = c(lambdamin, lambdamax), n = n)$root
f_n[i] = -sum(log(1 + lambda*(x - theta)))
}
j <- which.max(f_n)
max_fn <- f_n[j]
mle_theta <- theta_seq[j]
plot(theta_seq, f_n, type = "l",
main = expression(Estimated ~ theta),
xlab = expression(Theta),
ylab = expression(f[n]))
points(mle_theta, f_n[j], pch = 19, col = "red")
segments(
x0 = c(mle_theta, xmin),
y0 = c(min(f_n)*2, max_fn),
x1 = c(mle_theta, mle_theta),
y1 = c(max_fn, max_fn),
col = "red",
lty = "dashed"
)
I got the following plot of f_n.
For 95% CI, I try
LR <- function(theta, lambda)
{
2*sum(log(1 + lambda*(x - theta))) - qchisq(0.95, df = 1)
}
lambdamin <- (1/n-1)/(xmax - mle_theta)
lambdamax <- (1/n-1)/(xmin - mle_theta)
lambda <- uniroot(h, interval = c(lambdamin, lambdamax), n = n)$root
uniroot(LR, c(xmin, mle_theta), lambda = lambda)$root
The result is 0.07198144. Then the logarithm is log(0.07198144)=-2.631347.
But there is NA in the following code.
uniroot(LR, c(mle_theta, xmax), lambda = lambda)$root
So the 95% CI is theta >= -2.631347.
But the question is that the 95% CI should be a closed interval...
Here is a solution.
First of all, the data generation code is wrong, the parameter theta is in the interval [1, 12], and the data is generated with rnorm(., mean = 0, .). I change this to a true_theta = 5.
set.seed(2011111)
# I define function h, and use uniroot function to find lambda
h <- function(lam, n)
{
sum((x - theta)/(1 + lam*(x - theta)))
}
# sample size
n <- 100
# the parameter of interest must be a value in [1, 12],
true_theta <- 5
true_sd <- 2
x <- rnorm(n, mean = true_theta, sd = true_sd)
xmax <- max(x)
xmin <- min(x)
theta_seq <- seq(from = xmin + .Machine$double.eps^0.5,
to = xmax - .Machine$double.eps^0.5, by = 0.01)
f_n <- rep(NA, length(theta_seq))
for (i in seq_along(theta_seq))
{
theta <- theta_seq[i]
lambdamin <- (1/n-1)/(xmax - theta)
lambdamax <- (1/n-1)/(xmin - theta)
lambda = uniroot(h, interval = c(lambdamin, lambdamax), n = n)$root
f_n[i] = -sum(log(1 + lambda*(x - theta)))
}
j <- which.max(f_n)
max_fn <- f_n[j]
mle_theta <- theta_seq[j]
plot(theta_seq, f_n, type = "l",
main = expression(Estimated ~ theta),
xlab = expression(Theta),
ylab = expression(f[n]))
points(mle_theta, f_n[j], pch = 19, col = "red")
segments(
x0 = c(mle_theta, xmin),
y0 = c(min(f_n)*2, max_fn),
x1 = c(mle_theta, mle_theta),
y1 = c(max_fn, max_fn),
col = "red",
lty = "dashed"
)
LR <- function(theta, lambda)
{
2*sum(log(1 + lambda*(x - theta))) - qchisq(0.95, df = 1)
}
lambdamin <- (1/n-1)/(xmax - mle_theta)
lambdamax <- (1/n-1)/(xmin - mle_theta)
lambda <- uniroot(h, interval = c(lambdamin, lambdamax), n = n)$root
uniroot(LR, c(xmin, mle_theta), lambda = lambda)$root
#> [1] 4.774609
Created on 2022-03-25 by the reprex package (v2.0.1)
The one-sided CI95 is theta >= 4.774609.
I am new to R. I want to do some parameters estimation by using Maximum Likelihood Estimation.
Here is my attempt:
The data are
my_data = c(0.1,0.2,1,1,1,1,1,2,3,6,7,11,12,18,18,18,18,18,21,32,36,40,
45,45,47,50,55,60,63,63,67,67,67,67,72,75,79,82,82,83,
84,84,84,85,85,85,85,85,86,86)
and
lx <- function(p,x){
l <- p[1]
b <- p[2]
a <- p[3]
n <- length(x)
lnL <- n*log(l)+n*log(b)+n*log(a)+(b-1)*sum(log(x))+(a-1)*sum(log(1+l*x^b))+n-sum(1+l*x^b)
return(-lnL)
}
Note: l is λ, b is β, and a is α.
And here is the optim function
optim(p=c(1,1,1),fn = lx, method = "L-BFGS-B",
lower = c(0.0001, 0.0001, 0.0001),
control = list(), hessian = FALSE, x = my_data)
After I run this code, I get an error message:
Error in optim(p = c(1, 1, 1), fn = lx, method = "L-BFGS-B", lower = c(1e-04, :
objective function in optim evaluates to length 50 not 1
What's wrong with my code? Can you help me to fix it? Thanks in advance!
Instead of a log-likelihood, use MASS::fitdistr.
#
# Power Generalized Weibull distribution
#
# x > 0, alpha, beta, lambda > 0
#
dpowergweibull <- function(x, alpha, beta, lambda){
f1 <- lambda * beta * alpha
f2 <- x^(beta - 1)
f3 <- (1 + lambda * x^beta)^(alpha - 1)
f4 <- exp(1 - (1 + lambda * x^beta)^alpha)
f1 * f2 * f3 * f4
}
ppowergweibull <- function(q, alpha, beta, lambda){
1 - exp(1 - (1 + lambda * q^beta)^alpha)
}
my_data <- c(0.1,0.2,1,1,1,1,1,2,3,6,7,11,12,18,18,18,18,18,21,32,36,40,
45,45,47,50,55,60,63,63,67,67,67,67,72,75,79,82,82,83,
84,84,84,85,85,85,85,85,86,86)
start_par <- list(alpha = 0.1, beta = 0.1, lambda = 0.1)
y1 <- MASS::fitdistr(my_data, dpowergweibull, start = start_par),
start_par2 <- list(shape = 1, rate = 1)
y2 <- MASS::fitdistr(my_data, "gamma", start = start_par2)
hist(my_data, freq = FALSE)
curve(dpowergweibull(x, y1$estimate[1], y1$estimate[2], y1$estimate[3]),
from = 0.1, to = 90, col = "red", add = TRUE)
curve(dgamma(x, y2$estimate[1], y2$estimate[2]),
from = 0.1, to = 90, col = "blue", add = TRUE)
I have written the following code.
library(quantreg)
# return the g function:
G = function(m, N, gamma) {
Tm = m * N
k = 1:Tm
Gvalue = sqrt(m) * (1 + k/m) * (k/(m + k))^gamma
return(Gvalue)
}
sqroot <- function(A) {
e = eigen(A)
v = e$vectors
val = e$values
sq = v %*% diag(sqrt(val)) %*% solve(v)
return(t(sq))
}
fa = function(m, N, a) {
Tm = m * N
k = 1:Tm
t = (m + k)/m
f_value = (t - 1) * t * (a^2 + log(t/(t - 1)))
return(sqrt(f_value))
}
m = 50
N = 2
n= 50*3
x1 = matrix(runif(n, 0, 1), ncol = 1)
x = cbind(1, x1)
beta = c(1, 1)
xb = x %*% beta
pr = 1/(1+exp(-xb))
y = rbinom(n,1,pr)
# calculate statistic:
stat = function(y, x, m, N, a) {
y_train = y[1:m]
x_train = x[(1:m),]
y_test = y[-(1:m)]
x_test = x[-(1:m),]
fit = glm(y ~ 0 + x, family="binomial")
coef = coef(fit)
log_predict = predict(fit, type="response")
sigma = sqrt(1/(m-1)* sum((y_train - log_predict)^2))
Jvalue = t(x_train) %*% x_train/m * sigma^2
Jsroot = sqroot(Jvalue)
fvalue = fa(m, N, a)
score1 = apply((x_test * as.vector((y_test - x_test %*% coef))), 2, cumsum)
statvalue1 = t(solve(Jsroot) %*% t(score1))/fvalue/sqrt(m)
statmax1 = pmax(abs(statvalue1[, 1]), abs(statvalue1[, 2]))
result = list(stat = statmax1)
return(result)
}
m =50
N = 2
a = 2.795
value = stat(y, x, m, N, a)
value
I want to perform bootstrap to obtain B = 999 number of statistics. I use the following r code. But it produces an error saying "Error in statistic(data, original, ...) :
argument "m" is missing, with no default"
library(boot)
data1 = data.frame(y = y, x = x1, m = m , N = N, a = a)
head(data1)
boot_value = boot(data1, statistic = stat, R = 999)
Can anyone give me a hint? Also, am I able to get the bootstrap results in a matrix format? Since the stat function gives 100 values.
There are different kinds of bootstrapping. If you want to draw from your data 999 samples with replications of same size of your data you may just use replicate, no need for packages.
We put the data to be resampled into a data frame. It looks to me like m, N, a remain constant, so we just provide it as vectors.
data2 <- data.frame(y=y, x=x)
stat function needs to be adapted to unpack y and x-matrix. At the bottom we remove the list call to get just a vector back. unnameing will just give us the numbers.
stat2 <- function(data, m, N, a) {
y_train <- data[1:m, 1]
x_train <- as.matrix(data[1:m, 2:3])
y_test <- data[-(1:m), 1]
x_test <- as.matrix(data[-(1:m), 2:3])
y <- data[, "y"]
x <- as.matrix(data[, 2:3])
fit <- glm(y ~ 0 + x, family="binomial")
coef <- coef(fit)
log_predict <- predict(fit, type="response")
sigma <- sqrt(1/(m-1) * sum((y_train - log_predict)^2))
Jvalue <- t(x_train) %*% x_train/m * sigma^2
Jsroot <- sqroot(Jvalue)
fvalue <- fa(m, N, a)
score1 <- apply((x_test * as.vector((y_test - x_test %*% coef))), 2, cumsum)
statvalue1 <- t(solve(Jsroot) %*% t(score1))/fvalue/sqrt(m)
statmax1 <- pmax(abs(statvalue1[, 1]), abs(statvalue1[, 2]))
result <- unname(statmax1)
return(result)
}
replicate is a cousin of sapply, designed for repeated evaluation. In the call we just sample the rows 999 times and already get a matrix back. As in sapply we need to transform our result.
res <- t(replicate(999, stat2(data2[sample(1:nrow(data2), nrow(data2), replace=TRUE), ], m, N, a)))
Result
As result we get 999 bootstrap replications in the rows with 100 attributes in the columns.
str(res)
# num [1:999, 1:100] 0.00205 0.38486 0.10146 0.12726 0.47056 ...
The code also runs quite fast.
user system elapsed
3.46 0.01 3.49
Note, that there are different kinds of bootstrapping. E.g. sometimes just a part of the sample is resampled, weights are used, clustering is applied etc. Since you attempted to use boot the method shown should be the default, though.
I followed the mvrnorm function "If n = 1 a vector of the same length as mu, otherwise an n by length(mu) matrix with one sample in each row." and set n=1. length(X[i,])=3. How to fix this error?
n = 100
p = 3
X = matrix(ncol = p, nrow = n)
rho1 = 0.25 #correlation coefficient
rho2 = 0.5
rho3 = 0.75
o1 = 0.5 # sigma
o2 = 0.5
o3 = 0.5
C1 = o1^2 * matrix(c(1, rho1, rho1, 1), nrow = 2)
C2 = o2^2 * matrix(c(1, rho2, rho2, 1), nrow = 2)
C3 = o3^2 * matrix(c(1, rho3, rho3, 1), nrow = 2)
mu1 = c(0,1)
mu2 = c(1,0)
mu3 = c(-1,1)
o = 0.5
y = as.factor(rmultinom(100, size = 1, prob = c(1/3,1/3,1/3)))
for (i in 1:n){
mu = (y[i]==1)*mu1 + (y[i]==2)*mu2+(y[i]==3)*mu3
C = (y[i]==1)*C1 + (y[i]==2)*C2 + (y[i]==3)*C3
X[i,] = mvrnorm(1, mu, C, tol = 1e-6, empirical = FALSE, EISPACK = FALSE)
}
Error in X[i, ] <- mvrnorm(1, mu, C, tol = 1e-06, empirical = FALSE,
EISPACK = FALSE) : number of items to replace is not a multiple of
replacement length
I am using R package costrOptim.nl.
I need to minimize a function with the following constraints:
Alpha < sqrt(2*omega) and omega > 0
In my code expressed as:
theta[3] < sqrt(2*theta[1]) and theta[1] > 0
I write these conditions as:
Image
But when I call optimizer and run it.
I'm getting the following problem:
1: In sqrt(2 * theta[1]) : NaNs produced
Why? Did I set the proper conditions?
This is my whole code.
data <- read.delim(file = file, header = FALSE)
ind <- seq(from = 1, to = NROW(data), by = 1)
data <- data.frame(ind = ind, Ret = data$V1, Ret2 = data$V1^2)
colnames(data)[1] <- "Ind"
colnames(data)[2] <- "Ret"
colnames(data)[3] <- "Ret2"
T <- length(data$Ret)
m <- arima(x = data$Ret2, order = c(3,0,0), include.mean = TRUE, method = c("ML"))
b_not <- m$coef
omega <- 0.1
alpha <- 0.005
beta <- 0.9
theta <- c(omega,beta,alpha) # "some" value of theta
s0 <- theta[1]/(1-theta[2])
theta[3] < sqrt(2*theta[1]) # check whether the Feller condition is verified
N <- 30000
reps <- 1
rho <- -0.8
n <- 100
heston.II <- function(theta){
set.seed(5)
u <- rnorm(n = N*reps,mean = 0, sd = 1)
u1 <- rnorm(n = N*reps,mean = 0, sd = 1)
u2 <- rho*u + sqrt((1-rho^2))*u1
sigma <- matrix(0, nrow = N*reps, ncol = 1)
ret.int <- matrix(0, nrow = N*reps, ncol = 1)
sigma[1,1] <- s0
for (i in 2:(N*reps)) {
sigma[i,1] <- theta[1] + theta[2]*sigma[i-1,1] + theta[3]*sqrt(sigma[i-1,1])*u1[i]
# if(sigma[i,1] < 0.00000001){ sigma[i,1] = s0}
}
for (i in 1:(N*reps)) {
ret.int[i,1] <- sqrt(sigma[i,1])*u2[i]
}
ret <- matrix(0, nrow = N*reps/n, ncol = 1)
ret[1,1] <- sum(ret.int[1:n],1)
for (i in 2:((N*reps)/n)) {
ret[i,] <- sum(ret.int[(n*i):(n*(i+1))])
ret[((N*reps)/n),] <- sum(ret.int[(n*(i-1)):(n*i)])
}
ret2 <- ret^2
model <- arima(x = ret2, order = c(3,0,0), include.mean = TRUE)
beta_hat <- model$coef
m1 <- beta_hat[1] - b_not[1]
m2 <- beta_hat[2] - b_not[2]
m3 <- beta_hat[3] - b_not[3]
m4 <- beta_hat[4] - b_not[4]
D <- cbind(m1,m2,m3,m4)
DD <- (D)%*%t(D)/1000
DD <- as.numeric(DD)
return(DD)
}
heston.sim <- heston.II(theta)
hin <- function(theta){
h <- rep(NA, 2)
h[1] <- theta[1]
h[2] <- sqrt(2*theta[1]) - theta[3]
return(h)
}
hin(theta = theta)
.opt <- constrOptim.nl(par = theta, fn = heston.II, hin = hin)
.opt