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    [读书笔记][更新中] Aad van der Vaart "Asymptotic Statistics"

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    • hitori_bocchiH
      hitori_bocchi
      最后由 编辑

      半参后面的理论确实有点复杂,会涉及一些泛函的东西,我不打算写的过于理论,更多还是intuition吧

      ぼっち · ざ · ろっく!

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      • hitori_bocchiH
        hitori_bocchi
        最后由 编辑

        不打算更directional derivative & pathwise derivative for functional和更多的半参理论的东西了,实在是太多了,够写一本书的。这篇文章本来也是一个偏实用的指南,还是不想太偏离主旨...

        ぼっち · ざ · ろっく!

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        • hitori_bocchiH
          hitori_bocchi
          最后由 hitori_bocchi 编辑

          It turns out that, for the purposes of constructing lower bound benchmarks for functional estimation, it often suffices to use one-dimensional parametric submodels. A common choice of submodel for nonparametric P\mathcal{P}P is, for some mean-zero function h:Z→Rh: \mathcal{Z} \rightarrow \mathbb{R}h:Z→R,

          pϵ(z)=dP(z){1+ϵh(z)} p_{\epsilon}(z)=d \mathbb{P}(z)\{1+\epsilon h(z)\} pϵ​(z)=dP(z){1+ϵh(z)}

          where ∥h∥∞≤M<∞\|h\|_{\infty} \leq M<\infty∥h∥∞​≤M<∞ and ϵ<1/M\epsilon<1 / Mϵ<1/M so that pϵ(z)≥0p_{\epsilon}(z) \geq 0pϵ​(z)≥0. Note for this submodel the score function is ∂∂ϵlog⁡pϵ(z)∣ϵ=0=∂∂ϵlog⁡{1+ϵh(z)}∣ϵ=0=h(z)\left.\frac{\partial}{\partial \epsilon} \log p_{\epsilon}(z)\right|_{\epsilon=0}=\left.\frac{\partial}{\partial \epsilon} \log \{1+\epsilon h(z)\}\right|_{\epsilon=0}=h(z)∂ϵ∂​logpϵ​(z)​ϵ=0​=∂ϵ∂​log{1+ϵh(z)}​ϵ=0​=h(z). Therefore the Cramer-Rao lower bound for some PϵP_{\epsilon}Pϵ​ in the example one-dimensional submodel Pϵ\mathcal{P}_{\epsilon}Pϵ​ above is given by

          ψ′(Pϵ)2var⁡Pϵ{sϵ(Z)}={∂∂ϵψ(Pϵ)∣ϵ=0}2EPϵ{h(Z)2}. \frac{\psi^{\prime}\left(P_{\epsilon}\right)^{2}}{\operatorname{var}_{P_{\epsilon}}\left\{s_{\epsilon}(Z)\right\}}=\frac{\left\{\left.\frac{\partial}{\partial \epsilon} \psi\left(P_{\epsilon}\right)\right|_{\epsilon=0}\right\}^{2}}{\mathbb{E}_{P_{\epsilon}}\left\{h(Z)^{2}\right\}}. varPϵ​​{sϵ​(Z)}ψ′(Pϵ​)2​=EPϵ​​{h(Z)2}{∂ϵ∂​ψ(Pϵ​)​ϵ=0​}2​.

          Comment: Why one-dimensional submodel? 详细的说明见 Michael Kosorok "Introduction to Empirical Processes and Semiparametric Inference" Chap. 18。

          还需要说明的一点是为什么我们选择了pϵ(z)=dP(z){1+ϵh(z)}p_{\epsilon}(z)=d \mathbb{P}(z)\{1+\epsilon h(z)\}pϵ​(z)=dP(z){1+ϵh(z)}作为submodel(以下内容改写自Mark van der Laan的 STAT C245B Survival Analysis and Causality 的课程材料)。

          We want to define a type of differentiability of ψ:P→Rq\psi: \mathcal{P} \rightarrow \mathbb{R}^{q}ψ:P→Rq, where ψ\psiψ is the target parameter.

          We could use the definition of a directional derivative in direction hhh :

          dψ(P)(h)=ddϵψ(P+ϵh)∣ϵ=0 d \psi(\mathbb{P})(h)=\left.\frac{d}{d \epsilon} \psi(\mathbb{P}+\epsilon h)\right|_{\epsilon=0} dψ(P)(h)=dϵd​ψ(P+ϵh)​ϵ=0​

          However, P+ϵh\mathbb{P}+\epsilon hP+ϵh might not be a path through P\mathcal{P}P, and thus ill defined. We need to define a derivative along paths that are submodels of P\mathcal{P}P.

          Let P\mathcal{P}P be nonparametric. We define a class of paths such that:

          pϵ(z)=dP(z){1+ϵh(z)} p_{\epsilon}(z)=d \mathbb{P}(z)\{1+\epsilon h(z)\} pϵ​(z)=dP(z){1+ϵh(z)}

          Two key assumptions necessary for it to be a proper submodel are as follows:

          • hhh is uniformly bounded
          • EPh(z)=0\mathbb{E}_{P} h(z)=0EP​h(z)=0

          For ϵ∈(−δ,δ)\epsilon \in(-\delta, \delta)ϵ∈(−δ,δ) with δ=1∥h∥∞\delta=\frac{1}{\|h\|_{\infty}}δ=∥h∥∞​1​, this is a submodel.

          To see why, first note that for the paths to be a proper density, we need:

          1. dP(z){1+ϵh(z)}⩾0d \mathbb{P}(z) \{1+\epsilon h(z)\} \geqslant 0dP(z){1+ϵh(z)}⩾0

          Sketch proof:

          Let h(z)h(z)h(z) be uniformly bounded and h(z)=∥h∥∞h(z)=\|h\|_{\infty}h(z)=∥h∥∞​. If ϵ⩽∣δ∣,{1+ϵh(z)}⩾0\epsilon \leqslant|\delta|, \{1+\epsilon h(z)\} \geqslant 0ϵ⩽∣δ∣,{1+ϵh(z)}⩾0. Therefore, for ϵ\epsilonϵ sufficiently small and hhh uniformly bounded, dP(z){1+ϵh(z)}⩾0d \mathbb{P}(z) \{1+\epsilon h(z)\} \geqslant 0dP(z){1+ϵh(z)}⩾0.

          1. ∫{1+ϵh(z)}dP(z)=1\int \{1+\epsilon h(z)\} d \mathbb{P}(z)=1∫{1+ϵh(z)}dP(z)=1

          Sketch proof:

          Note that ∫{1+ϵh(z)}dP(z)=∫dP(z)+ϵ∫h(z)dP(z)=1\int\{1+\epsilon h(z)\} d \mathbb{P}(z)=\int d \mathbb{P}(z)+\epsilon \int h(z) d \mathbb{P}(z)=1∫{1+ϵh(z)}dP(z)=∫dP(z)+ϵ∫h(z)dP(z)=1 since ppp is a proper density and ∫h(z)dP(z)=EPh(z)=0\int h(z) d \mathbb{P}(z)=\mathbb{E}_{P} h(z)=0∫h(z)dP(z)=EP​h(z)=0 by assumption.

          Now consider the score of this submodel.

          δδϵlog⁡dPϵdP∣ϵ=0=δδϵlog⁡{1+ϵh(z)}∣ϵ=0=h(z)1+ϵh(z)∣ϵ=0=h(z). \begin{aligned} \left.\frac{\delta}{\delta \epsilon} \log \frac{d P_{\epsilon}}{d \mathbb{P}}\right|_{\epsilon=0} & =\left.\frac{\delta}{\delta \epsilon} \log \{1+\epsilon h(z)\}\right|_{\epsilon=0} \\ & =\left.\frac{h(z)}{1+\epsilon h(z)}\right|_{\epsilon=0} \\ & =h(z). \end{aligned} δϵδ​logdPdPϵ​​​ϵ=0​​=δϵδ​log{1+ϵh(z)}​ϵ=0​=1+ϵh(z)h(z)​​ϵ=0​=h(z).​

          "Since any lower bound for the submodel Pϵ\mathcal{P}_{\epsilon}Pϵ​ is also a lower bound for P\mathcal{P}P, the best and most informative is the greatest such lower bound. Can we say anything about the best such lower bound for generic functionals and/or submodels?"

          2.2 Pathwise Differentiability

          Recall the Cramer-Rao bound

          {∂∂ϵψ(Pϵ)∣ϵ=0}2EPϵ{sϵ(Z)2} \frac{\left\{\left.\frac{\partial}{\partial \epsilon} \psi\left(P_{\epsilon}\right)\right|_{\epsilon=0}\right\}^{2}}{\mathbb{E}_{P_{\epsilon}}\left\{s_{\epsilon}(Z)^{2}\right\}} EPϵ​​{sϵ​(Z)2}{∂ϵ∂​ψ(Pϵ​)​ϵ=0​}2​

          for submodel Pϵ\mathcal{P}_{\epsilon}Pϵ​ described in the previous subsection. To find the best such lower bound, we would like to optimize the above over all PϵP_{\epsilon}Pϵ​ in some submodels. It is not a priori clear how generally this can be accomplished, since different functionals ψ\psiψ could yield very different numerators. Therefore let us first consider what we can say about the derivative in the numerator, for a large class of pathwise differentiable functionals.

          Namely, suppose the functional ψ:P↦R\psi: \mathcal{P} \mapsto \mathbb{R}ψ:P↦R is smooth, as a map from distributions to the reals, in the sense that it admits a kind of distributional Taylor expansion

          ψ(Pˉ)−ψ(P)=∫φ(z;Pˉ)d(Pˉ−P)(z)+R2(Pˉ,P) \psi(\bar{P})-\psi(P)=\int \varphi(z ; \bar{P}) d(\bar{P}-P)(z)+R_{2}(\bar{P}, P) ψ(Pˉ)−ψ(P)=∫φ(z;Pˉ)d(Pˉ−P)(z)+R2​(Pˉ,P)

          for distributions Pˉ\bar{P}Pˉ and PPP, often called a von Mises expansion, where φ(z;P)\varphi(z ; P)φ(z;P) is a mean-

          zero, finite-variance function satisfying ∫φ(z;P)dP(z)=0\int \varphi(z ; P) d P(z)=0∫φ(z;P)dP(z)=0 and ∫φ(z;P)2dP(z)<∞\int \varphi(z ; P)^{2} d P(z)<\infty∫φ(z;P)2dP(z)<∞, and R2(Pˉ,P)R_{2}(\bar{P}, P)R2​(Pˉ,P) is a second-order remainder term (which means it only depends on products or squares of differences between Pˉ\bar{P}Pˉ and P)P)P).

          Intuitively, the von Mises expansion above is just an infinite-dimensional or distributional analog of a Taylor expansion, with φ(z;Q)\varphi(z ; Q)φ(z;Q) acting as a usual derivative term; it describes how the functional ψ\psiψ changes locally when the distribution changes from PPP to Pˉ\bar{P}Pˉ. For example, when Z∈{1,…,k}Z \in\{1, \ldots, k\}Z∈{1,…,k} is discrete and so Pˉ\bar{P}Pˉ and PPP have kkk countable components, the von Mises expansion reduces to a standard multivariate Taylor expansion with

          R2(Pˉ,P)=ψ(pˉ1,…,pˉk)−ψ(p1,…,pk)−∑j∂∂tjψ(t1,…,tk)∣t=pˉ(pˉj−pj). R_{2}(\bar{P}, P)=\psi\left(\bar{p}_{1}, \ldots, \bar{p}_{k}\right)-\psi\left(p_{1}, \ldots, p_{k}\right)-\left.\sum_{j} \frac{\partial}{\partial t_{j}} \psi\left(t_{1}, \ldots, t_{k}\right)\right|_{t=\bar{p}}\left(\bar{p}_{j}-p_{j}\right). R2​(Pˉ,P)=ψ(pˉ​1​,…,pˉ​k​)−ψ(p1​,…,pk​)−j∑​∂tj​∂​ψ(t1​,…,tk​)​t=pˉ​​(pˉ​j​−pj​).

          ぼっち · ざ · ろっく!

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          • hitori_bocchiH
            hitori_bocchi
            最后由 编辑

            今天就更到这里,去看碧蓝档案3周年fes了(

            ぼっち · ざ · ろっく!

            hashhashH 1 条回复 最后回复 回复 引用 0
            • N
              nomana
              最后由 编辑

              我爱你🤟波奇酱

              hitori_bocchiH 1 条回复 最后回复 回复 引用 0
              • hitori_bocchiH
                hitori_bocchi @nomana
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                @nomana 谢谢~

                ぼっち · ざ · ろっく!

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                • N
                  nomana
                  最后由 编辑

                  波奇酱好久没更了

                  hitori_bocchiH 1 条回复 最后回复 回复 引用 0
                  • hitori_bocchiH
                    hitori_bocchi @nomana
                    最后由 编辑

                    @nomana 最近科研太忙了,等放假了再更

                    ぼっち · ざ · ろっく!

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                    • hashhashH
                      hashhash @hitori_bocchi
                      最后由 编辑

                      @hitori_bocchi 原来你也是档友!3.gif

                      hitori_bocchiH 2 条回复 最后回复 回复 引用 0
                      • hitori_bocchiH
                        hitori_bocchi @hashhash
                        最后由 编辑

                        @hashhash 阿罗娜可爱!

                        ぼっち · ざ · ろっく!

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                        • hitori_bocchiH
                          hitori_bocchi @hashhash
                          最后由 编辑

                          @hashhash 可以加贵校碧蓝档案群155199376聊天吹水

                          ぼっち · ざ · ろっく!

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