Lecture 16: Canonical Gradient and Efficient Influence Curve

Notes by Rachael Phillips for PB HLTH 290, Spring 2019

Introduction

An asymptotically linear estimator with influence curve equal to the efficient influence curve is optimal in the sense that there is no other asymptotically linear estimator with influence curve with a smaller variance. We call this estimator asymptotically efficient.

Setting

Data and Model

\(O_1, \dots, O_n \overset{iid}{\sim} P_0 \in \mathcal{M}.\) Here, \(\mathcal{M}\) denotes the statistical model, which is the collection of all possible probability distributions \(P\) that could generate the data.

Target Parameter

Estimand

Goal

Local Perturbations

Parametric submodels and scores

Parametric submodel, given h

\begin{equation*} \mathcal{M}_h(P) = \{ P^h_{\epsilon} : \epsilon \in (-\delta, \delta) \} \subset \mathcal{M} \end{equation*}

A special direction: score $S$, given $h$

\begin{equation*} S_h(O)=\left . \frac{d}{d\epsilon}\log dP_{\epsilon}^h/dP(O)\right |_{\epsilon=0} \end{equation*}

Tangent space and Hilbert space

Class of paths, Class of Scores (the Tangent Set)

The Hilbert Space $L_0^2(P)$

Scores as Random Variables

Hilbert Space defined by $P$

\begin{equation} L^2_0(P) = \{ f(O) : \mathbb{E}_P[f(O)]=0, \, \mathbb{E}_P[f(O)^2] < \infty \} \end{equation}

with inner product defined as the covariance (since they are centered):

\begin{equation} \langle f, g \rangle_P = \mathbb{E}_P[ f(O)g(O) ] = \mathrm{Cov}_P(f,g) \end{equation}

Orthogonality

Projection

Tangent Space, T(P)

Pathwise derivative

Pathwise Derivative as a Linear Operator

\begin{equation*} d\Psi(P)(S_h)=\left . \frac{d}{d\epsilon}\Psi(P_{\epsilon}^h)\right |_{\epsilon =0} \end{equation*}

\begin{equation*} \underbrace{\frac{d}{d\epsilon} \Psi(P^h_\epsilon)}_{\text{Total Change}}\bigg|_{\epsilon=0} \approx \underbrace{ \frac{d\Psi}{dP} }_{ \text{"Operator Change"} } \cdot \underbrace{ \frac{dP}{d\epsilon} }_{ \text{"Curve Change"} } \end{equation*}

Pathwise differentiability and gradient

\begin{equation*} d\Psi(P)(S_h)=E_P D(P)(O)S_h(O)= \langle D(P),S_h\rangle_P \end{equation*}

Class of gradients

Canonical gradient is projection of gradient on tangent space

Example

\begin{equation*} \left . \frac{d}{d\epsilon}\Psi(P_{\epsilon}^h)\right |_{\epsilon =0} =E_P D(P)(T)S_h(T) \end{equation*}

where gradient

\begin{equation*} D(P)(T)=I(T>5)-\Psi(P) \end{equation*}

Nonparametric model has only one gradient

\(T(P)=L_0^2(P)\) so the orthogonal complement of the tangent space is empty meaning you cannot add to the canonical gradient anything to create more gradients.

Finding canonical gradient in non-saturated models

\begin{equation*} D^*(P)=\Pi(D(P)\mid T(P)) \end{equation*}

Example

Model. \(\mathcal{M}\) is nonparametric. Here we define it as a collection of all probability distributions which have densities.

Direction \(h(o)\).

Submodel. We define \(P_\epsilon^h\) so that \(dP_{\epsilon}(o)=(1+\epsilon h(o)) dP(o)\). Defined via densities.

\begin{equation*} \int dP_\epsilon(o) = \int (1 + \epsilon h(o)) dP(o) = \underbrace{\int 1 \, dP(o)}_{=1} + \epsilon \underbrace{\int h(o) \, dP(o)}_{= E_P[h(O)] \text{ should be 0}} \end{equation*}

Intuition: To add probability mass to one area (where \(h > 0\)), we must steal it from another area (where \(h < 0\)) to keep the total mass constant.

\begin{equation*} 1 - \epsilon \|h\|_\infty \ge 0 \iff 1 \ge \epsilon \|h\|_\infty \iff \epsilon \le \frac{1}{\|h\|_\infty} \end{equation*}

Therefore, if we restrict \(\epsilon\) to be smaller than \(\delta = 1/\|h\|_\infty\), i.e. \(\epsilon\in (-\delta,\delta)\) with \(\delta=1/\|h\|_{\infty}\), this is a submodel \(\mathcal{M}_h(P)\).

Score. This construction perfectly yields the score \(h\). By the construction \(dP_{\epsilon} = (1+\epsilon h) dP\):

\begin{equation*} S(O) = \frac{d}{d\epsilon} \log \big( \frac{(1+\epsilon h(O)) dP(O)}{dP(O)} \big) \bigg|_{\epsilon=0} \end{equation*}

The derivative of \(\log(u)\) is \(u'/u\):

\begin{equation*} S(O) = \frac{h(O)}{1+\epsilon h(O)} \bigg|_{\epsilon=0} \end{equation*}

Score. This construction perfectly yields the score \(h\). By the construction \(dP_{\epsilon} = (1+\epsilon h) dP\):

\begin{equation*} S(O) = \frac{d}{d\epsilon} \log \big( \frac{(1+\epsilon h(O)) dP(O)}{dP(O)} \big) \bigg|_{\epsilon=0} \end{equation*}

The derivative of \(\log(u)\) is \(u'/u\):

\begin{equation*} S(O) = \frac{h(O)}{1+\epsilon h(O)} \bigg|_{\epsilon=0} \end{equation*}

Set \(\epsilon=0\):

\begin{equation*} S(O) = \frac{h(O)}{1} = h(O) \end{equation*}

Scores.

\begin{equation*} \mathcal{S} = \{h\in L^2_0(P) : \|h\|_{\infty}<\infty\}. \end{equation*}