### 统计代写|抽样调查作业代写sampling theory of survey代考| HT Estimator t

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## 统计代写|抽样调查作业代写sampling theory of survey代考|HT Estimator t

Since $\overline$ as its variance, the following formula for which is given by HORVITZ and THOMPSON (1952)
$$V_{1}=V_{p}(\bar{t})=\sum \frac{Y_{i}^{2}}{\pi_{i}}\left(1-\pi_{i}\right)+\sum_{i \neq j} \sum_{j} \frac{Y_{i}}{\pi_{i}} \frac{Y_{j}}{\pi_{j}}\left(\pi_{i j}-\pi_{i} \pi_{j}\right) .$$
A formula for an unbiased estimator for $V_{1}$ is also given by HORVITZ and THOMPSON as
$$v_{1}=\sum \frac{Y_{i}^{2}}{\pi_{i}}\left(1-\pi_{i}\right) \frac{I_{s i}}{\pi_{i}}+\sum_{i \neq j} \sum_{j} \frac{Y_{i}}{\pi_{i}} \frac{Y_{j}}{\pi_{j}}\left(\pi_{i j}-\pi_{i} \pi_{j}\right) \frac{I_{s i j}}{\pi_{i j}}$$
assuming $\pi_{i j}>0$ for $i \neq j$.
If $Y_{i}=c \pi_{i}$ for all $i \in U$
$$\bar{t}=\sum_{i \in s} \frac{Y_{i}}{\pi_{i}}=c v(s)$$
and $Y=c \sum \pi_{i}$. If $v(s)=n$ for every $s$ with $p(s)>0$, that is, $\overline$ based on a design $p_{n}$, then, since $\sum \pi_{i}=n$ as well, the strategy $(p, \bar{t})$ is representative with respect to $\left(\pi_{1}, \pi_{2}, \ldots, \pi_{N}\right)^{\prime}$.

In this case it follows from RAO and VIJAYAN’s (1977) general result of section $2.3$ (noted earlier by SEN, 1953) that

one may write $V_{p}(\bar{t})$ alternatively as
$$V_{2}=\sum_{i0 for all i \neq j. For designs satisfying \pi_{i} \pi_{j} \geq \pi_{i j} for all i \neq j v_{2} is uniformly non-negative. If v(s) is not a constant for all s with p(s)>0 representativity of (p, \bar{t}) is violated. To cover this case, CHAUDHURI (2000a) showed that writing$$
\alpha_{i}=1+\frac{1}{\pi_{i}} \sum_{j \neq i} \pi_{i j}-\sum \pi_{j}
$$for i \in U one has a third formula for V_{p}(\overline{)}) as$$
V_{3}=V_{2}+\sum \frac{Y_{i}^{2}}{\pi_{i}} \alpha_{i}
$$and hence proposed$$
v_{3}=v_{2}+\sum \frac{Y_{i}^{2}}{\pi_{i}} \alpha_{i} \frac{I_{s i}}{\pi_{i}}
$$as an unbiased estimator for V_{p}(\bar{t}). This v_{3} is uniformly nonnegative if$$
\begin{aligned}
\pi_{i} \pi_{j} & \geq \pi_{i j} & & \text { for all } i \neq j \
\alpha_{i} &>0 & & \text { for all } i \in U .
\end{aligned}
$$CHAUDHURI and PAL (2002) illustrated a sampling scheme for which the above conditions simultaneously hold while representativity fails. ## 统计代写|抽样调查作业代写sampling theory of survey代考|Murthy’s Estimator t4 Writing$$
a_{i j}=P_{i} P_{j}\left[\frac{Y_{i}}{P_{i}}-\frac{Y_{j}}{P_{j}}\right]^{2}
$$we have$$
\begin{aligned}
M=& V_{p}\left(t_{4}\right)=-\sum_{i0}} \frac{p(s \mid i) p(s \mid j)}{p(s)}\right]
\end{aligned}
$$because$$
\begin{aligned}
E_{p}\left[\frac{p(s \mid i)}{p(s)} I_{s i}\right] &=\sum_{s} p(s \mid i) I_{s i} \
&=\sum_{s \ni i} p(s \mid i)=1 \quad \text { for } \quad i=1, \ldots, N .
\end{aligned}
$$One obvious unbiased estimator for V_{p}\left(t_{4}\right) is$$
\hat{M}=\sum_{1 \leq i<j \leq N} \sum_{i j} \frac{I_{s i j}}{p^{2}(s)}[p(s \mid i, j) p(s)-p(s \mid i) p(s \mid j)]
$$which follows from$$
\sum_{s} I_{s i j} p(s \mid i, j)=\sum_{s \ni i, j} p(s \mid i, j)=1
$$writing p(s \mid i, j) as the conditional probability of choosing s given that i and j are the first two units in s. It is assumed that the scheme of sampling is so adopted that it is meaningful to talk about the conditional probabilities p(s \mid i), p(s \mid i, j). Consider in particular the well-known sampling scheme due to LAHIRI (1951), MIDZUNO (1952), and SEN (1953) to be referred to as LMS scheme. Then on the first draw i is chosen with probability P_{i}\left(0<P_{i}<1, \Sigma_{1}^{N} P_{i}=1\right), i=1, \ldots, N and subsequently (n-1) distinct units are chosen from the remaining (N-1) units by the SRSWOR method, leaving aside the unit chosen on the first draw. For this scheme, then$$
p(s)=\sum_{i \in s} P_{i} /\left(\begin{array}{c}
N-1 \
n-1
\end{array}\right) .
$$If based on this scheme t_{4} reduces to the ratio estimator$$
t_{R}=\sum_{i \in s} Y_{i} / \sum_{i \in s} P_{i} .
$$Writing C_{r}=\left(\begin{array}{c}N-r \ n-r\end{array}\right), it follows that for this LMS scheme$$
\begin{aligned}
p(s \mid i) &=1 / C_{1}, p(s \mid i, j)=1 / C_{2} \
E_{p}\left(t_{R}\right) &=Y \
M &=E_{p}\left(t_{R}-Y\right)^{2}=V_{p}\left(t_{R}\right) \
&=\sum_{1 \leq i<j \leq N} \sum_{i j}\left[1-\frac{1}{C_{1}} \sum_{s \ni i, j} \frac{1}{\left[\sum_{i \in s} P_{i}\right]}\right] .
\end{aligned}
$$An unbiased estimator for M is$$
\hat{M}=\sum_{1 \leq i<j \leq N} \sum_{i j} \frac{I_{s i j}}{\sum_{i \in s} P_{i}}\left[\frac{N-1}{n-1}-\frac{1}{\sum_{i \in s} P_{i}}\right] .
$$It may be noted that if one takes P_{i}=X_{i} / X, then t_{R} reduces to t_{1}, which is thus unbiased for Y if based on the LMS scheme instead of SRSWOR, which is p-biased for Y in the latter case. ## 统计代写|抽样调查作业代写sampling theory of survey代考|Raj’s Estimator t5 Another popular strategy is due to RAJ (1956, 1968). The sampling scheme is called probability proportional to size without replacement (PPSWOR) with P_{i} ‘s \left(02) draw a unit i_{n}\left(\neq i_{1}, \ldots, i_{n-1}\right) is chosen with probability$$
\frac{P_{i_{n}}}{1-P_{i_{1}}-P_{i_{2}}-\ldots,-P_{i_{n-1}}}
$$out of the units of U minus i_{1}, i_{2}, \ldots, i_{n-1}. Then, e_{1}=\frac{Y_{i_{1}}}{P_{i_{1}}} e_{2}=Y_{i_{1}}+\frac{Y_{i_{2}}}{P_{i_{2}}}\left(1-P_{i_{1}}\right) e_{j}=Y_{i_{1}}+\ldots+Y_{i_{j-1}}+\frac{Y_{i_{j}}}{P_{i_{j}}}\left(1-P_{i_{1}}-\ldots-P_{i_{j-1}}\right) j=3, \ldots, n are all unbiased for Y because the conditional expectation$$
\begin{aligned}
E_{c} & {\left[e_{j} \mid\left(i_{1}, Y_{i_{1}}\right), \ldots,\left(i_{j-1}, Y_{i_{j-1}}\right)\right] } \
&=\left(Y_{i_{1}}+\ldots,+Y_{i_{j-1}}\right)+\sum_{\substack{k=1 \
\left(\neq i i_{1}, \ldots, i_{j-1}\right)}}^{N} Y_{k}=Y .
\end{aligned}
$$So, unconditionally, E_{p}\left(e_{j}\right)=Y for every j=1, \ldots, n, and t_{5}=\frac{1}{n} \sum_{j=1}^{n} e_{j}, called Raj’s (1956) estimator, is unbiased for Y. To find an elegant formula for M=V_{p}\left(t_{5}\right) is not easy, but RAJ (1956) gave a formula for an unbiased estimator for M= V_{p}\left(t_{5}\right) noting e_{j}, e_{k}(j<k) are pair-wise uncorrelated since$$
\begin{aligned}
E_{p}\left(e_{j} e_{k}\right) &=E\left[E_{c}\left(e_{j} e_{k} \mid\left(i_{1}, Y_{i_{1}}\right), \ldots,\left(i_{k-1}, Y_{i_{k-1}}\right)\right]\right.\
&=E\left[e_{j} E_{c}\left(e_{k} \mid\left(i_{1}, Y_{i_{1}}\right), \ldots,\left(i_{k-1}, Y_{i_{k-1}}\right)\right]\right.\
&=Y E\left(e_{j}\right)=Y^{2}=E_{p}\left(e_{j}\right) E_{p}\left(e_{k}\right)
\end{aligned}
$$that is, \operatorname{cov}{p}\left(e{j}, e_{k}\right)=0. So,$$
V_{p}\left(t_{5}\right)=\frac{1}{n^{2}} \sum_{j=1}^{n} V_{p}\left(e_{j}\right)
$$and$$
v_{5}=\frac{1}{n(n-1)} \sum_{j=1}^{n}\left(e_{j}-t_{5}\right)^{2}

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