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# 统计代写|抽样调查代考Survey sampling代考|STAT392 Unequal probability systematic sampling

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## 统计代写|抽样调查代考Survey sampling代考|Unequal probability systematic sampling

Let for the units $i$ of $U=(1, \cdots, i, \cdots N)$, the known positive size-measures be $x_{i}\left(0<x_{i} \leq 1, \sum_{1}^{N} x_{i}=X\right)$ and $p_{i}=\frac{x_{i}}{X}$, the normed size-measures. To draw a sample of size $n$ from $U$ using these size-measures the procedure to draw an unequal probability circular systematic sample is as follows.

Take $K=\left[\frac{X}{n}\right]$ or $K=\left[\frac{X}{n}\right]+1$; suitably reduce each $x_{i}$ to a positive integer and choose randomly a positive integer $r$ between 1 and $X$. Next calculate $C_{j}(r)=(r+j K) \bmod (X) \equiv$ the remainder on dividing $X$ by $(r+j k)$ for each of $j=0,1, \cdots,(n-1)$. Starting with $C_{0}=0$ and $C_{i}=\sum_{j=1}^{i} x_{j}, i=1, \cdots, N$, for the respective $r=1, \cdots, X$ check if
$$C_{i-1}<C_{j}(r) \leq C_{i} \text { for } i=1, \cdots N$$
and take the label $i$ into the sample; in particular, take $N$ in the sample if $C_{j}(r)$ equals zero for any $r$ and $\left.j, r=1, \cdots N, j=0,1 \cdots n-1\right)$. For the respective $r=1, \cdots, N$ we get the $r$ th sample composed of the $i$ ‘s satisfying (3.1.3) above. It is possible that the same unit may occur more than once in the same sample affecting the number of distinct sample-wise units which may not equal the desired $n$. Since the total number of samples is $X$ and each sample is chosen with the same probability $\frac{1}{X}$ counting the frequencies $f_{i}$ for $i=1, \cdots N$ and $f_{i j}$ for $i, j=1, \cdots N(i \neq j)$ of the units and the paired units the inclusion-probabilities $\pi_{i}=\frac{f_{i}}{X}$ and $\pi_{i j}=\frac{f_{i j}}{X}$ are easily calculated so the Horvitz-Thompson estimator for $Y=\sum_{1}^{N} y_{i}$ is easily obtained. But $\pi_{i j}$ ‘s may be found to be zero for many $(i, j)$ ‘s . So, unbiased estimation of the variance of the Horvitz and Thompson’s estimator may not be possible. In case $\frac{X}{n}=K$ is an integer this circular PPS systematic sampling may be revised into a linear PPS systematic sampling which may be derived easily by a mimicry of the situation when no such $x_{i}$ ‘s are available, taking $X$ equal to $N$.

How to adjust all these four procedures to render $\pi_{i j}$ positive $\forall i \neq j$ ensuring unbiased variance estimation will be now discussed.

## 统计代写|抽样调查代考Survey sampling代考|Systematic sampling modified to ensure πij > 0 ∀ i,j

Of the four alternative forms consider the most general one namely the circular PPS systematic sampling scheme. This is revised as follows to ‘modified systematic sampling’- linear, circular with equal probability and with PPS.
Let $M=X(X-1)$. Let $K=\left[\frac{M}{n}\right]$. Choose a random integer $R$ between 1 and $M$. Let $a=K$ or $K+1$. Calculate $C_{j}(R)=(R+j a) \bmod (M)$. Define $C_{i}=\sum_{j=1}^{i} x_{j}, C_{0}=0$. If $C_{j}(R)=0$, then take the label $N$ into the sample; other labels to be taken in the sample are the labels $i$ if $C_{i-1}<C_{j}(R) \leq$ $C_{i} ; j=0,1, \cdots,(n-1)$ and $R=1, \cdots, M$. So, the $M$ possible samples may easily be enumerated and written down. Then, on calculating $F_{i}$, the number of samples containing $i(=1, \cdots N)$ and $F_{i j}$, the number of samples containing $i, j(i=1, \cdots N ; j=1, \cdots N ; i \neq j)$ easily $\pi_{i j}=\frac{F_{i j}}{M}$ are obtained. Since the number of possible samples now has been enormously enhanced from $X$ to $X(X-1)$, the pairs $(i, j, i \neq j)$ are now allowed to re-appear a large number of times rendering $\pi_{i j}$ to be positive for each $i \neq j$ from 1 through $N$. Chaudhuri and Pal (2003) gave a formal proof for this. The three other special cases are easy to specify. In each case thus unbiased variance estimation becomes feasible. These results are all briefly covered in Chaudhuri (2010) but more details are reported here for the sake of an improved clarity. Before the 2003 paper, following the general practice we used to estimate variance by drawing two or more independent systematic samples. Wolter(1985) has given further alternatives which we feel we need not repeat here.

Next we consider three modifications of the PPSWR scheme of sampling and the associated estimation method of Hansen and Hurwitz (1943).

## 统计代写|抽样调查代考SURVEY SAMPLING代考|UNEQUAL PROBABILITY SYSTEMATIC SAMPLING

$$C_{i-1}<C_{j}(r) \leq C_{i} \text { for } i=1, \cdots N$$

## 统计代写|抽样调查代考SURVEY SAMPLING代考|SYSTEMATIC SAMPLING MODIFIED TO ENSURE ΠIJ > 0 ∀ I,J

$\mathrm{~ 在 四 种 萆 代 形 式 中 ， 考 虑 最 普 遍 的 一 种 ， 即 龧}$ 让 $M=X(X-1)$. 让 $K=\left[\frac{M}{n}\right]$. 选择一个随机整数 $R$ 介于 1 和 $M$. 让 $a=K$ 或者 $K+1$. 计算 $C_{j}(R)=(R+j a) \bmod (M)$. 定义 $C_{i}=\sum_{j=1}^{i} x_{j}$, $C_{0}=0$. 如果 $C_{j}(R)=0$ ，然后取标签 $N$ 进入样品；样本中要取的其他标签是标签 $i$ 如果 $C_{i-1}<C_{j}(R) \leq C_{i} ; j=0,1, \cdots,(n-1)$ 和 $R=1, \cdots, M$. 所以 $M$ 可能的样本很容易 被列举和写下来。然后，在计算 $F_{i}$, 包含的样本数 $i(=1, \cdots N)$ 和 $F_{i j}$, 包含的样本数 $i, j(i=1, \cdots N ; j=1, \cdots N ; i \neq j)$ 容易地 $\pi_{i j}=\frac{F_{i j}}{M}$ 获得。由于现在可能的 样本数量已经大大增加 $X$ 至 $X(X-1)$, 对 $(i, j, i \neq j)$ 现在允许重新出现大量渲染 $\pi_{i j}$ 对每个人都积极 $i \neq j$ 从 1 到 $N$. 夰杜里和帕尔 2003 对此给出了正式的证明。其 他三种特殊情况很容易指定。在每种情况下，因此无偏方差估计变得可行。这些结果都在 Chaudhuri 中进行了简要介绍 2010 但为了更清楚起见，这里报告了更多细 节。在 2003 年的论文之前，按照一般做法，我们通过抽取两个或多个独立的系统样本来估计方差。沃尔特 1985 已经给出了进一步的选择，我们认为我们不需要在 这里重是。

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