# 统计代写|抽样调查作业代写sampling theory of survey代考|STAT41020

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## 统计代写|抽样调查作业代写sampling theory of survey代考|Sufficiency and Likelihood

Suppose that prior to the survey the surveyor had no idea about the values of the parameter $\mathbf{y}=\left(\gamma_{1}, \ldots, \gamma_{N}\right)$ and hence $\Omega_{\gamma}=R^{N}$, the $N$-dimensional Euclidean space was considered the parametric space. After surveying the sample $s$, the surveyor collects the ordered data $d=\left(i_{k}, \gamma_{i_{k}} ; i_{k} \in s\right)$. The data $d$ is said to be consistent with parameter vector $\mathbf{y}{0}=\left(\gamma{10}, \ldots, y_{i n}, \ldots, y_{N n}\right)$ if $y_{i_{k}}=y_{i_{k} 0}$ for $i_{k} \in s$. After collection the data $d$, the surveyor knows that the values of $y_{i}$ ‘s belong to $s$ and hence the parametric space $\Omega_{\gamma}$ reduces to $\Omega_{\gamma d}$ $\left(\subset \Omega_{\gamma}\right)$, where $\Omega_{\gamma d}$ consists of the vectors $\mathbf{y}$ with $\gamma_{j}=y_{j 0}$ for $j \in s$.

The unordered data $\widetilde{d}=\left(i_{j_{k}}, y_{j_{k}} ; j_{k} \in \widetilde{s}\right)$ obtained from $d$ will be consistent with $\mathbf{y}{0}$ if $\gamma{j k}=\gamma_{j k 0}$ for $j_{k} \in \widetilde{s}$, i.e., $\gamma_{j_{1}}=\gamma_{j_{1} 0}, \ldots, \gamma_{j_{v_{s}}}=\gamma_{j_{v_{s}},}$. Here also, given the unordered data $\widetilde{d}$, the parametric space $\Omega_{\gamma}$ reduces to $\Omega \sim{ }{y d}$, which consists of the vectors $\mathbf{y}$ with $\gamma{j}=\gamma_{j 0}$ for $j \in \widetilde{s}$. Clearly, $\Omega_{\gamma d}=\Omega \underset{\gamma d}{\sim} \underset{\sim}{\gamma d^{\prime}}$ as $s$ and $\widetilde{s}$ consist of the same set of distinct units.

## 统计代写|抽样调查作业代写sampling theory of survey代考|Minimal Sufficient Statistic

A statistic $f(d)$, a function of data $d$, partitions the sample space of $d$. Let $\mathscr{P}{f}$ be the partition associated with $f$. The statistic $f^{}(d)$ is called a minimal sufficient statistic if and only if for any statistic $f\left(\neq f^{}\right)$, each partition set of $\mathscr{P}{f}$ is a subset of a partition set $\mathscr{P}{f^{}}$ induced by $f^{}$. In other words, every set of $\mathscr{P}{f^{*}}$ can be expressed as the union of the sets of $\mathscr{P}{f}$. Theorem 2.7.2 For a noninformative sampling design $p$, the unordered data $\tilde{d}=R(d)$ derived from an ordered data $d$ is a minimal sufficient statistic for $\mathbf{y}$. Proof Let $s{1}$ and $s_{2}$ be two samples with $p\left(s_{1}\right)>0$ and $p\left(s_{2}\right)>0$ yielding data $d_{1}$ and $d_{2}$, respectively. Following Thompson and Seber (1996), we note that $\widetilde{d}$ is a minimal sufficient for $\mathbf{y}$ if for any two data points $d_{1}$ and $d_{2}$, the following holds:
$$\tilde{d}{1}=R\left(d{1}\right)=R\left(d_{2}\right)=\tilde{d}{2} \Leftrightarrow P\left(D=d{1}\right)=k P\left(D=d_{2}\right) \forall \mathbf{y} \subset \Omega_{\gamma}$$
where $k$ is a constant independent of $\mathbf{y}$.
and
$$P\left(D=d_{1}\right)=p\left(s_{1}\right) I_{d_{1}}(\mathbf{y})=\frac{p\left(s_{1}\right)}{p\left(s_{2}\right)} p\left(s_{2}\right) I_{d_{2}}(\mathbf{y})=k P\left(D=d_{2}\right)$$
where $k=\frac{p\left(s_{1}\right)}{p\left(s_{2}\right)}$ is independent of $\mathbf{y}$.
Similarly, $P\left(I=d_{1}\right)=k P\left(D=d_{2}\right) \forall \mathbf{y} \in \Omega_{\gamma}$ implies
$$p\left(s_{1}\right) I_{d_{1}}(\mathbf{y})=k p\left(s_{2}\right) I_{d_{2}}(\mathbf{y}) \quad \forall \mathbf{y} \in \boldsymbol{\Omega}{\gamma}$$ Since, $p\left(s{1}\right), p\left(s_{2}\right)>0$ and $I_{d_{1}}(\mathbf{y})$ and $I_{d_{2}}(\mathbf{y})$ can take only two values 0 or 1 , Eq. (2.7.8) implies $I_{d_{1}}(\mathbf{y})=I_{d_{2}}(\mathbf{y})$, i.e., $\widetilde{d}{1}=R\left(d{1}\right)=R\left(d_{2}\right)=\widetilde{d}_{2}$. Hence $\widetilde{d}$ is a minimal sufficient statistic for $\mathbf{y}$.

# 抽样调查代考

## 统计代写|抽样调查作业代写sampling theory of survey代考|Minimal Sufficient Statistic

(1996) 之后，我们注意到 $\tilde{d}$ 是一个足够的最小值 $\mathbf{y}$ 如果对于任何两个数据点 $d_{1}$ 和 $d_{2}$ ，以 下成立:
$$\tilde{d} 1=R(d 1)=R\left(d_{2}\right)=\tilde{d} 2 \Leftrightarrow P(D=d 1)=k P\left(D=d_{2}\right) \forall \mathbf{y} \subset \Omega_{\gamma}$$

$$P\left(D=d_{1}\right)=p\left(s_{1}\right) I_{d_{1}}(\mathbf{y})=\frac{p\left(s_{1}\right)}{p\left(s_{2}\right)} p\left(s_{2}\right) I_{d_{2}}(\mathbf{y})=k P\left(D=d_{2}\right)$$

$$p\left(s_{1}\right) I_{d_{1}}(\mathbf{y})=k p\left(s_{2}\right) I_{d_{2}}(\mathbf{y}) \quad \forall \mathbf{y} \in \boldsymbol{\Omega} \gamma$$

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