R语言验证均匀分布总体参数的的矩估计是相合的consistent

作者: Cache_wood | 来源:发表于2021-03-27 00:09 被阅读0次

R语言验证均匀分布总体参数的的矩估计是相合的consistent
统计学－参数估计
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统计学第四课：参数估计
参数估计（一）矩估计法

判断估计的相合性依据的原理是6.2.1，定理6.2.2是扩充。

即需要证明两点：

（1）渐进无偏性
（2）渐进方差等于0

1.

第一段代码验证 ${X1，···，Xn}$ 是总体中值参数 $θ =µ+ φ^{-1}(0.5)\theta$ 的一致估计量.

2.

第二段代码均匀分布的最大统计量假设总体X服从均匀分布 $U(0， θ)$ ，请验证最大统计量 $X(1) = max{X1，···，Xn}$ 是θ的一致估计量。

3.

现在轮到你来验证 $Unif(a, b)$ 总体参数的矩估计是一致的。

# === medfun(): Sample median is consistent estimator of population
# median or not? Here, n: sample size; K: repeating times; mean0 and
# sd0 are mean and sd of normal distribution
medfun <- function(n = 100, K = 1000, mean0 = 0, sd0 = 1) {
  meds <- NULL
  # null vector, used to save sample medians
  for (k in 1:K) {
    sams <- rnorm(n, mean = mean0, sd = sd0)
    # samples for normal distribution
    meds <- c(meds, median(sams))
    # compute the sample median
  }
  mbias <- mean(meds) - qnorm(0.5, mean = mean0, sd = sd0)
  # empirical bias
  esd <- sd(meds)
  # empirical standard derviation
  list(n = n, mbias = mbias, esd = esd)
  # result as a list
}

# N(0,1)
n100 <- medfun(n = 100)
n200 <- medfun(n = 200)
n400 <- medfun(n = 400)
n800 <- medfun(n = 800)
n1000 <- medfun(n = 1000)
re <- rbind(n100, n200, n400, n800, n1000)
re

# N(1,2)
n100 <- medfun(n = 100, mean0 = 1, sd0 = 2)
n200 <- medfun(n = 200, mean0 = 1, sd0 = 2)
n400 <- medfun(n = 400, mean0 = 1, sd0 = 2)
n800 <- medfun(n = 800, mean0 = 1, sd0 = 2)
n1000 <- medfun(n = 1000, mean0 = 1, sd0 = 2)
re <- rbind(n100, n200, n400, n800, n1000)
re
# === consistent if: 1. empirical bias are all close to zero; and 2.
# empirical sd tends to zero as sample size increases
# === Uniform(0,theta) ===
unifun <- function(n = 100, K = 1000, theta = 1) {
  maxval <- NULL
  for (i in 1:K) {
    sams <- runif(n, 0, theta)
    maxval <- c(maxval, max(sams))
  }
  mbias <- mean(maxval) - theta
  esd <- sd(maxval)
  list(n = n, mbias = mbias, esd = esd)
}
# U(0,1)
u1 <- unifun(n = 100)
u2 <- unifun(n = 200)
u3 <- unifun(n = 400)
u4 <- unifun(n = 800)
rbind(u1, u2, u3, u4)

# U(0,2*sqrt(3))
u1 <- unifun(n = 100, theta = 2 * sqrt(3))
u2 <- unifun(n = 200, theta = 2 * sqrt(3))
u3 <- unifun(n = 400, theta = 2 * sqrt(3))
u4 <- unifun(n = 800, theta = 2 * sqrt(3))
rbind(u1, u2, u3, u4)

meanfun <- function(n = 100, K = 1000, a = 0, b = 1) {
  mea <- NULL
  # null vector, used to save sample mean number
  for (k in 1:K) {
    sams <- runif(n, a, b)
    # samples for uniform distribution
    mea <- c(mea, mean(sams))
    # compute the sample mean
  }
  mbias <- mean(mea) - mean(a:b)
  # empirical bias
  esd <- sd(mea)
  # empirical standard derviation
  list(n = n, mbias = mbias, esd = esd)
  # result as a list
}
#unif ~ (0,1)
n100 <- meanfun(n = 100, b = 2)
n200 <- meanfun(n = 200, b = 2)
n400 <- meanfun(n = 400, b = 2)
n800 <- meanfun(n = 800, b = 2)
n1000 <- meanfun(n = 1000, b = 2)
rbind(n100, n200, n400, n800, n1000)

# b = 1
   n   mbias        esd        
u1 100 -0.0352749   0.03513745 
u2 200 -0.018127    0.01760004 
u3 400 -0.008798743 0.008986948
u4 800 -0.004345918 0.004439682
# b = 2
      n    mbias         esd       
n100  100  -0.0005611266 0.05589222
n200  200  -0.000229566  0.04092171
n400  400  0.0006570319  0.02798479
n800  800  -0.0002061969 0.02009344
n1000 1000 -0.0004253045 0.01824597