统计代写|统计推断作业代写statistical inference代考|SAMPLING ISSUES

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

统计代写|统计推断作业代写statistical inference代考|The “Mixed” Sampling Case

Another negative multinomial sampling approach stops when $r_{1}$, say, attains a given value. Here
\begin{aligned} f_{r_{1}}=& f\left(r_{2}, n_{1}, n \mid r_{1}\right)=f\left(r_{2}, n \mid n_{1}, r_{1}\right) f\left(n_{1} \mid r_{1}\right) \ =& f\left(n_{1} \mid r_{1}\right) f\left(r_{2} \mid n, n_{1}, r_{1}\right) f\left(n \mid n_{1}, r_{1}\right) \ =&\left(\begin{array}{c} n_{1}-1 \ r_{1}-1 \end{array}\right) p_{1}^{n_{1}}\left(1-p_{1}\right)^{n_{1}-r_{1}}\left(\begin{array}{c} n_{1} \ r_{2} \end{array}\right) \ & \times p_{2}^{r_{2}}\left(1-p_{2}\right)^{n_{2}-r_{2}}\left(\begin{array}{c} n-1 \ n_{1}-1 \end{array}\right) \gamma^{n_{1}}(1-\gamma)^{n-n_{1}} \end{aligned}
Again, the likelihood for $p_{1}$ and $p_{2}$ is preserved for inference that respects ELP but here we now encounter a difficulty for conditional frequentist inference regarding $p_{1}$ and $p_{2}$. What does one condition on to obtain an exact significance test on $p_{1}=p_{2}$ ? Of course, this problem would also occur when we start with one sample that is binomial, say, a control, and the other negative binomial, for say a new treatment where one would like to stop the latter trial after a given number of failures. Note that this problem would persist for $f_{r_{2}}, f_{n_{1}-r_{1}}$ and $f_{n_{2}-r_{2}}$. Hence in these four cases there is no exact conditional Fisher type test for $p_{1}=p_{2}$.

Next we examine these issues for the $2 \times 2$ table. Here we list the various ways one can sample in constructing a $2 \times 2$ table such that one of the nine values is fixed, that is, when that value appears sampling ceases. For 7 out of the 9 cases the entire likelihood is the same, where
$$L=\gamma^{n_{1}}(1-\gamma)^{n_{2}} \prod_{i-1}^{2} p_{i}^{r_{1}}\left(1-p_{i}\right)^{n_{i}-r_{i}}=L(\gamma) L\left(p_{1}, p_{2}\right)$$
with sampling probabilities
\begin{aligned} f_{n} &=\left(\begin{array}{l} n_{1} \ r_{1} \end{array}\right)\left(\begin{array}{l} n_{2} \ r_{2} \end{array}\right)\left(\begin{array}{c} n \ n_{1} \end{array}\right) L \ f_{n_{1}} &=\left(\begin{array}{l} n_{1} \ r_{1} \end{array}\right)\left(\begin{array}{l} n_{2} \ r_{2} \end{array}\right)\left(\begin{array}{c} n-1 \ n_{1}-1 \end{array}\right) L=f\left(r_{1}, r_{2}, n \mid n_{1}\right) \ &=f\left(r_{1}, r_{2} \mid n, n_{1}\right) f\left(n \mid n_{1}\right) \ f_{n_{2}} &=\left(\begin{array}{l} n_{1} \ r_{1} \end{array}\right)\left(\begin{array}{c} n_{2} \ r_{2} \end{array}\right)\left(\begin{array}{c} n-1 \ n_{2}-1 \end{array}\right) L=f\left(r_{1}, r_{2} \mid n_{2}, n\right) f\left(n \mid n_{2}\right) \end{aligned}

统计代写|统计推断作业代写statistical inference代考|OTHER PRINCIPLES

Restricted Conditionality Principle RCP: Same preliminaries as LP. $\mathcal{E}=(S, \mu, \theta, f)$ is a mixture of experiments $\mathcal{E}{i}=\left(S{i}, \mu_{i}, \theta, f_{i}\right)$ with mixture probabilities $q_{i}$ independent of $\theta$. First we randomly select $\mathcal{E}{1}$ or $\mathcal{E}{2}$ with probabilities $q_{1}$ and $q_{2}=1-q_{1}$, and then perform the chosen experiment $\mathcal{E}{i}$. Then we recognize the sample $D=\left(i, D{i}\right)$ and $f(D \mid \theta)=q_{i} f_{i}\left(D_{i} \mid \theta\right), i=1,2$. Then RCP asserts $\operatorname{Inf}(\mathcal{E}, D)=\operatorname{Inf}\left(\mathcal{E}{i}, D{i}\right)$.

Definition of Ancillary Statistic: A statistic $C=C(D)$ is ancillary with respect to $\theta$ if $f_{C}(c \mid \theta)$ is independent of $\theta$, so that an ancillary is non-informative about $\theta$. $C(D)$ maps $S \rightarrow S_{c}$ where each $c \in S_{c}$ defines $S_{c}=(D \mid C(D)=c)$. Define the conditional experiment $\mathcal{E}{D \mid C}=\left(S{c}, \mu, \theta, f_{D \mid C}(D \mid c)\right)$, and the marginal experiment $\mathcal{E}{C}=\left(S{c}, \mu, \theta, f_{c}(c)\right)$, where $\mathcal{E}{C}=$ sample from $S{c}$ or sample from $S$ and observe $c$ and $\mathcal{E}{D \mid C}=$ conditional on $C=c$, sample from $S{c}$.
Unrestricted Conditionality Principle $(U C P)$ : When $C$ is an ancillary $\operatorname{Inf}(\mathcal{E}, D)=\operatorname{Inf}\left(\mathcal{E}{D \mid C}, D\right)$ concerning $\theta$. It is as if we performed $\mathcal{E}{C}$ and then performed $\mathcal{E}_{D \mid C}$.

统计代写|统计推断作业代写statistical inference代考|ULP

Note this is just a special case of ULP.
We show that ULP $\Leftrightarrow$ (RCP, MEP). First assume ULP, so that $\operatorname{Inf}(\mathcal{E}, D)=$ $\operatorname{Inf}\left(\mathcal{E}^{\prime}, D^{\prime}\right)$. Now suppose $f(D \mid \theta)=f\left(D^{\prime} \mid \theta\right)$, then apply ULP so $\operatorname{Inf}\left(\mathcal{E}, D^{\prime}\right)=$ $\operatorname{Inf}\left(\mathcal{E}, D^{\prime}\right)$, which is MEP. Further suppose, where $C$ is an ancillary, and hence that
$$f(D \mid \theta)=f(D \mid c, \theta) h(c), \quad f\left(D^{\prime} \mid \theta\right)=f\left(D^{\prime} \mid c, \theta\right) h(c) .$$
Hence ULP implies that
$$\operatorname{Inf}(\mathcal{E}, D)=\operatorname{Inf}(\mathcal{E}, D \mid C)$$
or UCP, and UCP implies RCP.
Conversely, assume (RCP, MEP) and that $\left(\mathcal{E}{1}, D{1}\right)$ and $\left(\mathcal{E}{2}, D{2}\right)$ generate equivalent likelihoods
$$f_{1}\left(D_{1} \mid \theta\right)=h\left(D_{1}, D_{2}\right) f_{2}\left(D_{2} \mid \theta\right)$$
We will use (RCP, MEP) to show $\operatorname{Inf}\left(\mathcal{E}{1}, D{1}\right)=\operatorname{Inf}\left(\mathcal{E}{2}, D{2}\right)$. Now let $\mathcal{E}$ be the mixture experiment with probabilities $(1+h)^{-1}$ and $h(1+h)^{-1}$. Hence the sample points in $\mathcal{E}$ are $\left(1, D_{1}\right)$ and $\left(2, D_{2}\right)$ and
\begin{aligned} &f\left(1, D_{1} \mid \theta\right)=(1+h)^{-1} f_{1}\left(D_{1} \mid \theta\right)=h(1+h)^{-1} f_{2}\left(D_{2} \mid \theta\right) \ &f\left(2, D_{2} \mid \theta\right)=h(1+h)^{-1} f_{2}\left(D_{2} \mid \theta\right) \end{aligned}
Therefore $f\left(1, D_{1} \mid \theta\right)=f\left(2, D_{2} \mid \theta\right)$ so from MEP
$$\operatorname{Inf}\left(\mathcal{E},\left(1, D_{1}\right)\right)=\operatorname{Inf}\left(\mathcal{E},\left(2, D_{2}\right)\right.$$
Now apply RCP to both sides above so that
\begin{aligned} &\operatorname{Inf}\left(\mathcal{E}{1}, D{1}\right)=\operatorname{Inf}\left(\mathcal{E}, \theta,\left(1, D_{1}\right)\right)=\operatorname{Inf}\left(\mathcal{E}, \theta,\left(2, D_{2}\right)\right)=\operatorname{Inf}\left(\mathcal{E}{2}, D{2}\right) \ &\text { therefore }(\operatorname{RCP}, \mathrm{MEP}) \Longrightarrow \text { ULP } \end{aligned}

统计代写|统计推断作业代写statistical inference代考|The “Mixed” Sampling Case

Fr1=F(r2,n1,n∣r1)=F(r2,n∣n1,r1)F(n1∣r1) =F(n1∣r1)F(r2∣n,n1,r1)F(n∣n1,r1) =(n1−1 r1−1)p1n1(1−p1)n1−r1(n1 r2) ×p2r2(1−p2)n2−r2(n−1 n1−1)Cn1(1−C)n−n1

Fn=(n1 r1)(n2 r2)(n n1)大号 Fn1=(n1 r1)(n2 r2)(n−1 n1−1)大号=F(r1,r2,n∣n1) =F(r1,r2∣n,n1)F(n∣n1) Fn2=(n1 r1)(n2 r2)(n−1 n2−1)大号=F(r1,r2∣n2,n)F(n∣n2)

统计代写|统计推断作业代写statistical inference代考|ULP

F(D∣θ)=F(D∣C,θ)H(C),F(D′∣θ)=F(D′∣C,θ)H(C).

F1(D1∣θ)=H(D1,D2)F2(D2∣θ)

F(1,D1∣θ)=(1+H)−1F1(D1∣θ)=H(1+H)−1F2(D2∣θ) F(2,D2∣θ)=H(1+H)−1F2(D2∣θ)

广义线性模型代考

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MATLAB代写

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