如何定义 `f_n-chi-square 的函数并使用 `uniroot` 找到置信区间?
How to define a function of `f_n-chi-square and use `uniroot` to find Confidence Interval?
我想获得以下问题的 95% 置信区间。
我在我的 R 代码中写了函数 f_n
。我首先使用 Normal 随机抽取 100 个样本,然后为 lambda
定义函数 h
。然后我可以得到f_n
。我的问题是如何定义 f_n-chi-square and use
uniroot` 的函数来找到置信区间。
# I first get 100 samples
set.seed(201111)
x=rlnorm(100,0,2)
根据@RuiBarradas 的回答,我尝试了以下代码。
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)$root
f_n[i] = -sum(log(1 + lambda*(x - theta)))
}
j <- which.max(f_n)
max_fn <- f_n[j]
mle_theta <- theta_seq[j]
plot(theta_seq, f_n, type = "l",
main = expression(Estimated ~ theta),
xlab = expression(Theta),
ylab = expression(f[n]))
points(mle_theta, f_n[j], pch = 19, col = "red")
segments(
x0 = c(mle_theta, xmin),
y0 = c(min(f_n)*2, max_fn),
x1 = c(mle_theta, mle_theta),
y1 = c(max_fn, max_fn),
col = "red",
lty = "dashed"
)
我得到了下面的f_n
情节。
对于 95% CI,我尝试
LR <- function(theta, lambda)
{
2*sum(log(1 + lambda*(x - theta))) - qchisq(0.95, df = 1)
}
lambdamin <- (1/n-1)/(xmax - mle_theta)
lambdamax <- (1/n-1)/(xmin - mle_theta)
lambda <- uniroot(h, interval = c(lambdamin, lambdamax), n = n)$root
uniroot(LR, c(xmin, mle_theta), lambda = lambda)$root
结果是0.07198144
。那么对数就是log(0.07198144)=-2.631347
.
但是下面的代码中有NA
uniroot(LR, c(mle_theta, xmax), lambda = lambda)$root
所以 95% CI 是 theta >= -2.631347
。
但问题是95%CI应该是闭区间...
这是一个解决方案。
首先,数据生成代码错误,参数theta
在区间[1, 12]
,用rnorm(., mean = 0, .)
生成数据。我将其更改为 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)$root
#> [1] 4.774609
由 reprex package (v2.0.1)
于 2022-03-25 创建
one-sided CI95 是 theta >= 4.774609
。
我想获得以下问题的 95% 置信区间。
我在我的 R 代码中写了函数 f_n
。我首先使用 Normal 随机抽取 100 个样本,然后为 lambda
定义函数 h
。然后我可以得到f_n
。我的问题是如何定义 f_n-chi-square and use
uniroot` 的函数来找到置信区间。
# I first get 100 samples
set.seed(201111)
x=rlnorm(100,0,2)
根据@RuiBarradas 的回答,我尝试了以下代码。
set.seed(2011111)
# I define function h, and use uniroot function to find lambda
h <- function(lam, n)
{
sum((x - theta)/(1 + lam*(x - theta)))
}
# sample size
n <- 100
# the parameter of interest must be a value in [1, 12],
#true_theta<-1
#true_sd<- exp(2)
#x <- rnorm(n, mean = true_theta, sd = true_sd)
x=rlnorm(100,0,2)
xmax <- max(x)
xmin <- min(x)
theta_seq = seq(from = 1, to = 12, by = 0.01)
f_n <- rep(NA, length(theta_seq))
for (i in seq_along(theta_seq))
{
theta <- theta_seq[i]
lambdamin <- (1/n-1)/(xmax - theta)
lambdamax <- (1/n-1)/(xmin - theta)
lambda = uniroot(h, interval = c(lambdamin, lambdamax), n = n)$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"
)
我得到了下面的f_n
情节。
对于 95% CI,我尝试
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
结果是0.07198144
。那么对数就是log(0.07198144)=-2.631347
.
但是下面的代码中有NA
uniroot(LR, c(mle_theta, xmax), lambda = lambda)$root
所以 95% CI 是 theta >= -2.631347
。
但问题是95%CI应该是闭区间...
这是一个解决方案。
首先,数据生成代码错误,参数theta
在区间[1, 12]
,用rnorm(., mean = 0, .)
生成数据。我将其更改为 true_theta = 5
.
set.seed(2011111)
# I define function h, and use uniroot function to find lambda
h <- function(lam, n)
{
sum((x - theta)/(1 + lam*(x - theta)))
}
# sample size
n <- 100
# the parameter of interest must be a value in [1, 12],
true_theta <- 5
true_sd <- 2
x <- rnorm(n, mean = true_theta, sd = true_sd)
xmax <- max(x)
xmin <- min(x)
theta_seq <- seq(from = xmin + .Machine$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
由 reprex package (v2.0.1)
于 2022-03-25 创建one-sided CI95 是 theta >= 4.774609
。