如果你也在 怎样代写抽样调查Survey sampling 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。抽样调查Survey sampling是数学工程这一广泛新兴领域中的一个自然组成部分。例如，我们可以断言，数学工程之于今天的数学系，就像数学物理之于一个世纪以前的数学系一样;毫不夸张地说，数学在诸如语音和图像处理、信息理论和生物医学工程等工程学科中的基本影响。

抽样调查Survey sampling是主流统计的边缘。这里的特殊之处在于，我们有一个具有某些特征的有形物体集合，我们打算通过抓住其中一些物体并试图对那些未被触及的物体进行推断来窥探它们。这种推论传统上是基于一种概率论，这种概率论被用来探索观察到的事物与未观察到的事物之间的可能联系。这种概率不被认为是在统计学中，涵盖其他领域，以表征我们感兴趣的变量的单个值之间的相互关系。但这是由调查抽样调查人员通过任意指定的一种技术从具有预先分配概率的对象群体中选择样本而创建的。

statistics-lab™ 为您的留学生涯保驾护航 在代写抽样调查sampling theory of survey方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写抽样调查sampling theory of survey方面经验极为丰富，各种代写抽样调查sampling theory of survey相关的作业也就用不着说。

统计代写|抽样调查作业代写sampling theory of survey代考|Linear Unbiased Estimators

Let a sensitive variable $y$ be defined on a finite population $U=(1, \ldots, N)$ with values $Y_i, i=1, \ldots, N$, which are supposed to be unavailable through a DR survey. Suppose a sample $s$ of size $n$ is chosen according to a design $p$ with a selection probability $p(s)$. In order to estimate $Y=\sum_1^N Y_i$, let an RR as a value $Z_i$ be available on request from each sampled person labeled $i$ included in a sample. Before describing how a $Z_i$ may be generated, let us note the properties required of it. We will denote by $E_R\left(V_R, C_R\right)$ the operator for expectation (variance, covariance) with respect to the randomized procedure of generating RR. The basic RRs $Z_i$ should allow derivation by a simple transformation reduced RRs as $R_i$ ‘s satisfying the conditions
(a) $E_R\left(R_i\right)=Y_i$
(b) $V_R\left(R_i\right)=\alpha_i Y_i^2+\beta_i Y_i+\theta_i$ with $\alpha_i(>0), \beta_i, \theta_i$ ‘s as known constants
(c) $C_R\left(R_i, R_j\right)=0$ for $i \neq j$
(d) estimators $v_i=a_i R_i^2+b_i R_i+C_i$ exist, $a_i, b_i, c_i$ known constants, such that $E_R\left(v_i\right)=V_R\left(R_i\right)=V_i$, say, for all $i$.

We will illustrate only two possible ways of obtaining $Z_i$ ‘s from a sampled individual $i$ on request. First, let two vectors $A=$ $\left(A_1, \ldots, A_T\right)^{\prime}$ and $B=\left(B_1, \ldots, B_L\right)^{\prime}$ of suitable real numbers be chosen with means $\bar{A} \neq 0, \bar{B}$ and variances $\sigma_A^2, \sigma_B^2$. A sample person $i$ is requested to independently choose at random $a_i$ out of $A$ and $b_i$ out of $B$, and report the value $Z_i=a_i Y_i+b_i$. Then, it follows that $E_R\left(Z_i\right)=\bar{A} Y_i+\bar{B}$, giving
$$
R_i=\left(Z_i-\bar{B}\right) / \bar{A}
$$
such that
$$
\begin{aligned}
E_R\left(R_i\right) & =Y_i, \
V_R\left(R_i\right) & =\left(Y_i^2 \sigma_A^2+\sigma_B^2\right) /(\bar{A})^2=V_i, \
C_R\left(R_i, R_J\right) & =0, \quad i \neq j
\end{aligned}
$$

and
$$
v_i=\left(\sigma_A^2 R_i^2+\sigma_B^2\right) /\left(\sigma_A^2+\bar{A}^2\right)
$$
has
$$
E_R\left(v_i\right)=V_i .
$$

统计代写|抽样调查作业代写sampling theory of survey代考|A Few Specific Strategies

Let us illustrate a few familiar specific cases. Corresponding to the HTE $\bar{t}=\bar{t}(s, Y)=\sum_i \frac{Y_i}{\pi_i} I_{s i}$, we have the derived estimator $e=(s, R)=\sum_i \frac{R_i}{\pi_i} I_{s i}$ for which
$$
M=-\sum_{i<j} \sum_j\left(\pi_i \pi_j-\pi_{i j}\right)\left(Y_i / \pi_i-Y_j / \pi_j\right)^2+\sum_i \frac{V_i}{\pi_i}
$$
and
To LAHIRI’s (1951) ratio estimator $t_L=Y_i / \sum_s P_i$ based on LAHIRI-MIDZUNO-SEN (LMS, 1951, 1952, 1953) scheme corresponds the estimator
$$
e_L=\sum_s R_i / \sum_s P_i
$$
$\left(0<P_i<1, \Sigma_1^N P_i=1\right)$ for which
$$
M=\sum_{i<j} \sum_{i j}\left(1-\frac{1}{C_1} \sum_s \frac{I_{s i j}}{P_s}\right)+\sum V_i E_p\left(I_{s i} / P_s^2\right),
$$
where
$$
\begin{aligned}
C_r & =\left(\begin{array}{c}
N-r \
n-r
\end{array}\right), r=0,1,2, \ldots, P_s=\sum_s P_i, a_{i j} \
& =P_i P_j\left(Y_i / P_i-Y_j / P_j\right)^2
\end{aligned}
$$

$$
\begin{aligned}
m= & \sum \sum P_i P_j I_{s i} I_{s i j}\left(\frac{N-1}{n-1}-\frac{1}{P_s}\right) / \
& P_s\left[\left(\frac{R_i}{P_i}-\frac{R_j}{P_j}\right)^2-\left(\frac{v_i}{p_i^2}+\frac{v_j}{p_j^2}\right)\right]+\sum v_i I_{s i} / P_s^2
\end{aligned}
$$
is unbiased for $M$. If $t_L$ and $e_L$ above are based on SRSWOR in $n$ draws, then, $M$ equals
$$
\begin{aligned}
M^{\prime}= & -\frac{1}{C_0}\left[\sum_{i<j} \sum_{i j} \sum_s\left(\frac{I_{s i j}}{p_s^2}-\frac{I_{s j}}{P_s}-\frac{I_{s j}}{P_s}+1\right)\right. \
& \left.-\sum_i V_i\left(\sum_s I_{s i} / P_s^2\right)\right]
\end{aligned}
$$
and
$$
\begin{aligned}
m^{\prime}= & -\frac{N(N-1)}{n(n-1) C_0} \sum_{i<j} \sum_{\hat{a}{i j}} I{s i j} \sum_s\left(\frac{I_{s i j}}{p_s^2}-\frac{I_{s i}}{P_s}-\frac{I_{s j}}{P_s}+1\right) \
& +\frac{1}{C_0} \frac{N}{n} \sum v_i I_{s i}\left(\sum_s I_{s i} / P_s^2\right)
\end{aligned}
$$
writing
$$
\left.\widehat{a}_{i j}=\left{\left(\frac{R_i}{P_i}-\frac{R_j}{P_j}\right)^2-\frac{v_j}{P_i^2}+\frac{v_j}{P_j^2}\right)\right} P_i P_j
$$

抽样调查代考

统计代写|抽样调查作业代写sampling theory of survey代考|Linear Unbiased Estimators

假设在有限总体$U=(1, \ldots, N)$上定义一个敏感变量$y$，其值为$Y_i, i=1, \ldots, N$，该变量应该无法通过DR调查获得。假设一个大小为$n$的样本$s$是根据一个选择概率为$p(s)$的设计$p$选择的。为了估计$Y=\sum_1^N Y_i$，让RR作为一个值$Z_i$在每个样本中被标记为$i$的人的请求下可用。在描述如何生成$Z_i$之前，让我们注意一下它所需的属性。我们将用$E_R\left(V_R, C_R\right)$表示相对于生成RR的随机过程的期望(方差，协方差)的算子。基本RRs $Z_i$应允许通过简化RRs的简单变换推导出满足$R_i$条件的RRs
(a) $E_R\left(R_i\right)=Y_i$
(b) $V_R\left(R_i\right)=\alpha_i Y_i^2+\beta_i Y_i+\theta_i$，其中$\alpha_i(>0), \beta_i, \theta_i$为已知常数
(c) $i \neq j$为$C_R\left(R_i, R_j\right)=0$
(d)估计量$v_i=a_i R_i^2+b_i R_i+C_i$存在，$a_i, b_i, c_i$已知常数，使得$E_R\left(v_i\right)=V_R\left(R_i\right)=V_i$对所有$i$都成立。

我们将只说明两种可能的方法，根据要求从抽样的个人$i$获取$Z_i$。首先，取两个合适实数的向量$A=$$\left(A_1, \ldots, A_T\right)^{\prime}$和$B=\left(B_1, \ldots, B_L\right)^{\prime}$，均值$\bar{A} \neq 0, \bar{B}$，方差$\sigma_A^2, \sigma_B^2$。要求样本人员$i$从$A$和$B$中独立地随机选择$a_i$和$b_i$，并报告值$Z_i=a_i Y_i+b_i$。接下来是$E_R\left(Z_i\right)=\bar{A} Y_i+\bar{B}$，给予
$$
R_i=\left(Z_i-\bar{B}\right) / \bar{A}
$$
这样
$$
\begin{aligned}
E_R\left(R_i\right) & =Y_i, \
V_R\left(R_i\right) & =\left(Y_i^2 \sigma_A^2+\sigma_B^2\right) /(\bar{A})^2=V_i, \
C_R\left(R_i, R_J\right) & =0, \quad i \neq j
\end{aligned}
$$

和
$$
v_i=\left(\sigma_A^2 R_i^2+\sigma_B^2\right) /\left(\sigma_A^2+\bar{A}^2\right)
$$
有
$$
E_R\left(v_i\right)=V_i .
$$

统计代写|抽样调查作业代写sampling theory of survey代考|A Few Specific Strategies

让我们举例说明几个熟悉的具体案例。对应于HTE $\bar{t}=\bar{t}(s, Y)=\sum_i \frac{Y_i}{\pi_i} I_{s i}$，我们有推导出的估计量$e=(s, R)=\sum_i \frac{R_i}{\pi_i} I_{s i}$
$$
M=-\sum_{i<j} \sum_j\left(\pi_i \pi_j-\pi_{i j}\right)\left(Y_i / \pi_i-Y_j / \pi_j\right)^2+\sum_i \frac{V_i}{\pi_i}
$$
和
对LAHIRI(1951)的比率估计量$t_L=Y_i / \sum_s P_i$基于LAHIRI- midzno – sen (LMS, 1951, 1952, 1953)方案对应的估计量
$$
e_L=\sum_s R_i / \sum_s P_i
$$
$\left(0<P_i<1, \Sigma_1^N P_i=1\right)$ for which
$$
M=\sum_{i<j} \sum_{i j}\left(1-\frac{1}{C_1} \sum_s \frac{I_{s i j}}{P_s}\right)+\sum V_i E_p\left(I_{s i} / P_s^2\right),
$$
在哪里
$$
\begin{aligned}
C_r & =\left(\begin{array}{c}
N-r \
n-r
\end{array}\right), r=0,1,2, \ldots, P_s=\sum_s P_i, a_{i j} \
& =P_i P_j\left(Y_i / P_i-Y_j / P_j\right)^2
\end{aligned}
$$

$$
\begin{aligned}
m= & \sum \sum P_i P_j I_{s i} I_{s i j}\left(\frac{N-1}{n-1}-\frac{1}{P_s}\right) / \
& P_s\left[\left(\frac{R_i}{P_i}-\frac{R_j}{P_j}\right)^2-\left(\frac{v_i}{p_i^2}+\frac{v_j}{p_j^2}\right)\right]+\sum v_i I_{s i} / P_s^2
\end{aligned}
$$
对于$M$是公正的。如果上述$t_L$和$e_L$基于$n$抽签中的SRSWOR，则$M$等于
$$
\begin{aligned}
M^{\prime}= & -\frac{1}{C_0}\left[\sum_{i<j} \sum_{i j} \sum_s\left(\frac{I_{s i j}}{p_s^2}-\frac{I_{s j}}{P_s}-\frac{I_{s j}}{P_s}+1\right)\right. \
& \left.-\sum_i V_i\left(\sum_s I_{s i} / P_s^2\right)\right]
\end{aligned}
$$
和
$$
\begin{aligned}
m^{\prime}= & -\frac{N(N-1)}{n(n-1) C_0} \sum_{i<j} \sum_{\hat{a}{i j}} I{s i j} \sum_s\left(\frac{I_{s i j}}{p_s^2}-\frac{I_{s i}}{P_s}-\frac{I_{s j}}{P_s}+1\right) \
& +\frac{1}{C_0} \frac{N}{n} \sum v_i I_{s i}\left(\sum_s I_{s i} / P_s^2\right)
\end{aligned}
$$
写作
$$
\left.\widehat{a}_{i j}=\left{\left(\frac{R_i}{P_i}-\frac{R_j}{P_j}\right)^2-\frac{v_j}{P_i^2}+\frac{v_j}{P_j^2}\right)\right} P_i P_j
$$

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