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assignmentutor-lab™ 为您的留学生涯保驾护航 在代写统计推断Statistical inference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。我们的专家在代写统计推断Statistical inference代写方面经验极为丰富，各种代写统计推断Statistical inference相关的作业也就用不着说。

• 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

## 英国补考|统计推断代写Statistical inference代考|Detectors and their risks

Let $\Omega$ be an observation space, and $\mathcal{P}{\chi}, \chi=1,2$, be two families of probability distributions on $\Omega$. By definition, a detector associated with $\Omega$ is a real-valued function $\phi(\omega)$ of $\Omega$. We associate with a detector $\phi$ and families $\mathcal{P}{\chi}, \chi=1,2$, risks defined as follows:
(a) Risk $_{-}\left[\phi \mid \mathcal{P}{1}\right]=\sup {P \in \mathcal{P}{1}} \int{\Omega} \exp {-\phi(\omega)} P(d \omega)$
(b) $\operatorname{Risk}{+}\left[\phi \mid \mathcal{P}{2}\right]=\sup {P \in \mathcal{P}{2}} \int_{\Omega} \exp {\phi(\omega)} P(d \omega)$
(c) $\operatorname{Risk}\left[\phi \mid \mathcal{P}{1}, \mathcal{P}{2}\right]=\max \left[\operatorname{Risk}{-}\left[\phi \mid \mathcal{P}{1}\right]\right.$, Risk $\left.{+}\left[\phi \mid \mathcal{P}{2}\right]\right]$
Given a detector $\phi$, we can associate with it a simple test $\mathcal{T}{\phi}$ deciding via observation $\omega \sim P$ on the hypotheses $$H{1}: P \in \mathcal{P}{1}, H{2}: P \in \mathcal{P}{2} .$$ Namely, given observation $\omega \in \Omega$, the test $\mathcal{T}{\phi}$ accepts $H_{1}$ (and rejects $H_{2}$ ) whenever $\phi(\omega) \geq 0$, and accepts $H_{2}$ and rejects $H_{1}$ otherwise.
Let us make the following immediate observation:
Proposition 2.14. Let $\Omega$ be an observation space, $\mathcal{P}{\chi}, \chi=1,2$, be two families of probability distributions on $\Omega$, and $\phi$ be a detector. The risks of the test $\mathcal{T}{\phi}$ associated with this detector satisfy
\begin{aligned} \operatorname{Risk}{1}\left(\mathcal{T}{\phi} \mid H_{1}, H_{2}\right) & \leq \text { Risk }{-}\left[\phi \mid \mathcal{P}{1}\right] \ \operatorname{Risk}{2}\left(\mathcal{T}{\phi} \mid H_{1}, H_{2}\right) & \leq \text { Risk }{+}\left[\phi \mid \mathcal{P}{2}\right] \end{aligned}
Proof. Let $\omega \sim P \in \mathcal{P}{1}$. Then the $P$-probability of the event ${\omega: \phi(\omega)<0}$ does not exceed Risk $\left[\phi \mid \mathcal{P}{1}\right]$, since on the set ${\omega: \phi(\omega)<0}$ the integrand in $(2.45 . a)$ is $>1$, and this integrand is nonnegative everywhere, so that the integral in $(2.45 . a)$ is $\geq P{\omega: \phi(\omega)<0}$. Recalling what $\mathcal{T}{\phi}$ is, we see that the $P$-probability to reject $H_{1}$ is at most Risk $\left[\phi \mid \mathcal{P}{1}\right]$, implying the first relation in (2.47). By a similar argument, with $(2.45 . b)$ in the role of $(2.45 . a)$, when $\omega \sim P \in \mathcal{P}{2}$, the $P$-probability of the event ${\omega: \phi(\omega) \geq 0}$ is upper-bounded by Risk R $_{+}\left[\phi \mid \mathcal{P}_{2}\right]$, implying the second relation in (2.47).

## 英国补考|统计推断代写Statistical inference代考|Simple observation schemes—Motivation

A natural conclusion one can extract from the previous section is that it makes sense, to say the least, to learn how to build detector-based tests with minimal risk. Thus, we arrive at the following design problem:
Given an observation space $\Omega$ and two families, $\mathcal{P}{1}$ and $\mathcal{P}{2}$, of probability distributions on $\Omega$, solve the optimization problem
$$\text { Opt }=\min {\phi \Omega \Omega \rightarrow \mathbf{R}} \max [\underbrace{\sup {P \in \mathcal{P}{1}} \int{\Omega} \mathrm{e}^{-\phi(\omega)} P(d \omega)}{F[\phi]}, \underbrace{\sup {P \in \mathcal{P}{2}} \int{\Omega} \mathrm{e}^{\phi(\omega)} P(d \omega)}{G[\phi]}] .$$ While being convex, problem (2.53) is typically computationally intractable. First, it is infinite-dimensional – candidate solutions are multivariate functions; how do we represent them on a computer, not to mention, how do we optimize over them? Besides, the objective to be optimized is expressed in terms of suprema of infinitely many (provided $\mathcal{P}{1}$ and/or $\mathcal{P}_{2}$ are infinite) expectations, and computing just a single expectation can be a difficult task …. We are about to consider “favorable” cases – simple observation schemes – where (2.53) is efficiently solvable.

To arrive at the notion of a simple observation scheme, consider the case when all distributions from $\mathcal{P}{1}, \mathcal{P}{2}$ admit densities taken w.r.t. some reference measure $\Pi$ on $\Omega$, and these densities are parameterized by a “parameter” $\mu$ running through some parameter space $\mathcal{M}$. In other words, $\mathcal{P}{1}$ is comprised of all distributions with densities $p{\mu}(\cdot)$ and $\mu$ belonging to some subset $M_{1}$ of $\mathcal{M}$, while $\mathcal{P}{2}$ is comprised of distributions with densities $p{\mu}(\cdot)$ and $\mu$ belonging to another subset, $M_{2}$, of $\mathcal{M}$. To save words, we shall identify distributions with their densities taken w.r.t. П, so that
$$\mathcal{P}{\chi}=\left{p{\mu}: \mu \in M_{\chi}\right}, \chi=1,2,$$
where $\left{p_{\mu}(\cdot): \mu \in \mathcal{M}\right}$ is a given “parametric” family of probability densities. The quotation marks in “parametric” reflect the fact that at this point in time, the “parameter” $\mu$ can be infinite-dimensional (e.g, we can parameterize a density by itself), so that assuming “parametric” representation of the distributions from $\mathcal{P}{1}$, $\mathcal{P}{2}$ in fact does not restrict the generality.

Our first observation is that in our “parametric” setup, we can rewrite problem (2.53) equivalently as
$$\ln (\text { Opt })=\min {\phi: \Omega \rightarrow \mathrm{R}{\mu \in M_{1}, \nu \in M_{2}}} \underbrace{\frac{1}{2}\left[\ln \left(\int_{\Omega} \mathrm{e}^{-\phi(\omega)} p_{\mu}(\omega) \Pi(d \omega)\right)+\ln \left(\int_{\Omega} \mathrm{e}^{\phi(\omega)} p_{\nu}(\omega) \Pi(d \omega)\right)\right]}_{\Phi(\phi ; \mu, \nu)} .$$

# 统计推断代考

## 英国补考|统计推断代写Statistical inference代考|Detectors and their risks

(a) 风险- $[\phi \mid \mathcal{P} 1]=\sup P \in \mathcal{P} 1 \int \Omega \exp -\phi(\omega) P(d \omega)$
(二)Risk $+[\phi \mid \mathcal{P} 2]=\sup P \in \mathcal{P} 2 \int_{\Omega} \exp \phi(\omega) P(d \omega)$
(C) $\operatorname{Risk}[\phi \mid \mathcal{P} 1, \mathcal{P} 2]=\max [\operatorname{Risk}-[\phi \mid \mathcal{P} 1]$ ， 风险 $+[\phi \mid \mathcal{P} 2]]$

$$H 1: P \in \mathcal{P} 1, H 2: P \in \mathcal{P} 2 \text {. }$$

$$\operatorname{Risk} 1\left(\mathcal{T} \phi \mid H_{1}, H_{2}\right) \leq \text { Risk }-[\phi \mid \mathcal{P} 1] \operatorname{Risk} 2\left(\mathcal{T} \phi \mid H_{1}, H_{2}\right) \quad \leq \text { Risk }+[\phi \mid \mathcal{P} 2]$$

## 英国补考|统计推断代写Statistical inference代考|Simple observation schemes—Motivation

$$\text { Opt }=\min \phi \Omega \Omega \rightarrow \mathbf{R} \max [\underbrace{\sup P \in \mathcal{P} 1 \int \Omega \mathrm{e}^{-\phi(\omega)} P(d \omega)} F[\phi], \underbrace{\sup P \in \mathcal{P} 2 \int \Omega \mathrm{e}^{\phi(\omega)} P(d \omega)} G[\phi]] .$$

、eft 的分隔符缺失或无法识别

㧴们的第一个观察结果是，在我们的“参数“设置中，我们可以将问题 (2.53) 等效地重写为
$$\ln (\mathrm{Opt})=\min \phi: \Omega \rightarrow \mathrm{R} \mu \in M_{1}, \nu \in M_{2} \underbrace{\frac{1}{2}\left[\ln \left(\int_{\Omega} \mathrm{e}^{-\phi(\omega)} p_{\mu}(\omega) \Pi(d \omega)\right)+\ln \left(\int_{\Omega} \mathrm{e}^{\phi(\omega)} p_{\nu}(\omega) \Pi(d \omega)\right)\right]}_{\Phi(\phi ; \mu, \nu)}$$

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## MATLAB代写

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assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师