Notes on ''The Asymptotic Variance of Semiparametric Estimators''
Table of Contents
0. Background: Semiparametric Model
Reference: Lecture notes from Prof. Bodhisattva Sen (Columbia University)
A semiparametric model is a statistical model that involves both parametric and nonparametric (infinite-dimensional) components. However, we are mostly interested in estimation and inference of a finite-dimensional parameter in the model.
Example 1 (population mean)
Suppose that $X_1, \ldots, X_n$ are i.i.d. $P$ belonging to the class $\mathcal{P}$ of distribution. Let $\psi(P) \equiv \mathbb{E}_P\left[X_1\right]$, the mean of the distribution, be the parameter of interest.
Question:
- Suppose that $\mathcal{P}$ is the class of all distributions that have a finite variance.
- What is the most efficient estimator of $\psi(P)$, i.e., what is the estimator with the best asymptotic performance?
Example 2 (partial linear regression model)
Suppose that we observe i.i.d. data $\left\{X_i \equiv\left(Y_i, Z_i, V_i\right): i=1, \ldots, n\right\}$ from the following partial linear regression model: $$ Y_i=Z_i^{\top} \beta+g\left(V_i\right)+\epsilon_i $$
- $Y_i$ is the scalar response variable
- $Z_i$ and $V_i$ are vectors of predictors
- $g(\cdot)$ is the unknown (nonparametric) function
- $\epsilon_i$ is the unobserved error.
For simplicity and to focus on the semiparametric nature of the problem:
- Assume that $\left(Z_i, V_i\right) \sim f(\cdot, \cdot)$, where we assume that the density $f(\cdot, \cdot)$ is known, is independent of $\epsilon_i \sim N\left(0, \sigma^2\right)$ (with $\sigma^2$ known).
- The model, under these assumptions, has a parametric component $\beta$ and a nonparametric component $g(\cdot)$
“Separated semiparametric model”: We say that the model $\mathcal{P}=\left\{P_{\nu, \eta}\right\}$ is a “separated semiparametric model”, where $\nu$ is a “Euclidean parameter” and $\eta$ runs through a nonparametric class of distributions (or some infinite-dimensional set). This gives a semiparametric model in the strict sense, in which we aim at estimating $\nu$ and consider $\eta$ as a nuisance parameter.
“Frequent questions for semiparametric model”: consider the estimation of a parameter of interest $\nu=\nu(P)$, where the data has distribution $P \in \mathcal{P}$:
- (Q1) How well can we estimate $\nu=\nu(P)$ ? What is our “gold standard”?
- (Q2) Can we compare absolute “in principle” standards for estimation of $\nu$ in a model $\mathcal{P}$ with estimation of $\nu$ in a submodel $\mathcal{P}_0 \subset \mathcal{P}$ ? What is the effect of not knowing $\eta$ on estimation of $\nu$ when $\mathcal{P}=\left\{P_\theta: \theta \equiv(\nu, \eta) \in \Theta\right\}$ ?
- (Q3) How do we construct efficient estimators of $\nu(P)$ ?
1. Introduction
- Develop a general form for the asymptotic variance of semiparametric estimators that depend on nonparametric estimators of functions.
- The formula is often straightforward to derive, requiring only some calculus.
- Although the formula is not based on primitive conditions, it should be useful for semiparametric estimators, just as analogous formulae are for parametric estimators.
- The formula gives the form of remainder terms, which facilitates specification of primitive conditions.
- It can also be used to make asymptotic efficiency comparisons and to find an efficient estimator in some class.
- - Derive the formula: **Section 2** - Propositions about semiparametric estimator - **Section 3** - **Section 4** - High-level regularity conditions: **Section 5** - Conditions for $\sqrt{n}$-consistency and asymptotic normality: **Section 6** - Primitive conditions for the examples: **Section 7**
2. The Pathwise Derivative Formula For the Asymptotic Variance
Preliminary: one-step estimators and pathwise derivatives
The formula is based on the observation that $\sqrt{n}$-consistent nonparametric estimators are often efficient.
For example, the sample mean is known to be an efficient estimator of the population mean in a nonparametric model where no restrictions, other than regularity conditions (e.g. existence of the second moment) are placed on the distribution of the data.
Idea
- Calculate the asymptotic variance of a semiparametric estimator as the variance bound for the functional that it nonparametrically estimates.
- In other words, the formula is the variance bound for the functional that is the limit of the estimator under general misspecification.
Let $z_1, \ldots, z_n$ be i.i.d. data, with (true) distribution $F_0$ of $z_i$, and let $\hat{\beta}=\beta_n\left(z_1, \ldots, z_n\right)$ denote a $q \times 1$ vector of estimators. Suppose $\hat{\beta}$ can be associated with a family of distributions and a functional as $$ \hat{\beta} \rightarrow \begin{cases}\mathscr{F}=\{F\} ; & \text { general family of distributions of } z \\ \mu: \mathscr{F} \rightarrow \mathbb{R}^q ; & \text { if } z_i \text { has distribution } F \text { then } {\color{red}\operatorname{plim}(\hat{\beta})=\mu(F)}\end{cases} $$
- The word “general” is taken to mean that $\mathscr{F}$ is unrestricted, except for regularity conditions, and allows for general misspecification.
- This equation also specifies that $\mu(F)$ is the limit of $\hat{\beta}$ when $z_i$ has distribution $F$.
- $\mu(F)$ traces out the limits of $\hat{\beta}$ as $F$ varies within the general family $\mathscr{F}$.
- The variance formula for $\hat{\beta}$ is the semiparametric bound for estimation of $\mu(F)$, calculated as in Koshevnik and Levit (1976), Pfanzagl and Wefelmeyer (1982), and others.
Let $\{F_\theta: F_\theta \in \mathscr{F}\}$ denote a one-dimensional subfamily of $\mathscr{F}$, i.e. a path in $\mathscr{F}$, that is equal to the true distribution $F_0$ when $\theta=0$.
- Suppose that $F_\theta$ has a density $d F_\theta$ and a corresponding score $${\color{red}S(z)=\frac{\partial \ln \left(d F_\theta\right)}{ \partial \theta}\Big|_{\theta=0}}.$$
- Suppose that the set of scores can approximate in mean square any mean zero, finite variance function of $z$.
- Let $E[\cdot]$ denote the expectation at the true distribution $F_0$.
The pathwise derivative of $\mu(F)$ is a $q \times 1$ vector $d(z)$ with $E[d(z)]=0$ and $E\left[|d(z)|^2\right]<\infty$ such that for every path, $$ {\color{red}\frac{\partial \mu\left(F_\theta\right)}{\partial \theta}=E[d(z) S(z)].}\qquad(\star) $$
- The variance bound for estimation of $\mu(F)$ is $\operatorname{Var}(d(z))$.
- Thus, the asymptotic variance formula suggested here is the variance of the pathwise derivative of the functional $\mu(F)$ that is estimated under general misspecification.
Example
- Consider the parameter $\beta_0=\int f_0(z)^2 d z$, where $f_0(z)$ is the density function of $z_i$.
- One estimator is $\tilde{\beta}=\sum_{i=1}^n \hat{f}\left(z_i\right) / n$, for a nonparametric density estimator $\hat{f}(z)$ of $z_i$.
- Suppose $z$ is symmetrically distributed around zero. Then one might hope to improve efficiency by using the antithetic estimate $\hat{f}(-z)$ of the density to form $$\hat{\beta}=\sum_{i=1}^n\frac{[\hat{f}(z_i)+\hat{f}(-z_i)]}{2}.$$
- The asymptotic variance can be found by calculating the limit of $\hat{\beta}$ under general misspecification, where $z$ need not be symmetric about zero, and the pathwise derivative of this limit.
- Let $E_F[\cdot]$ denote the expectation at a distribution $F$ and let $E_\theta[\cdot]=E_{F_\theta}[\cdot]$ for a path $F_\theta$.
- By an appropriate uniform law of large numbers the limit of $\hat{\beta}$ is $\mu(F)=\int[f(z)+f(-z)] f(z) d z / 2$.
- Assuming that differentiation inside the integral is allowed, $$ \begin{aligned} \frac{\partial \mu\left(F_\theta\right)}{ \partial \theta} &=\int\left[\frac{\partial f_\theta(z) }{ \partial \theta}\right] f_0(z) d z\\ &\qquad +\frac{1}{2}\left\{\int\left[\frac{\partial f_\theta(-z)}{ \partial \theta}\right] f_0(z) d z\right.\\ &\qquad\qquad\qquad\qquad\qquad +\left.\int\left[\frac{\partial f_\theta(z) }{ \partial \theta}\right] f_0(-z) d z\right\}\\ &=E\left[\left\{f_0(z)+f_0(-z)\right\} S(z)\right]\\ &=E[d(z) S(z)] \end{aligned} $$
- Thus, in this example the asymptotic variance formula is ${\color{red}\operatorname{Var}\left(2 f_0(z)\right)}$, which is the well known asymptotic variance of $\tilde{\beta}$, so no efficiency improvement results.
The pathwise derivative generalizes the Gateaux derivative formula for von-Mises estimators:
- The pathwise derivative formula works for estimators that are explicit functions of densities or expectations, where the domain of $\mu(F)$ may only include continuous distributions.
- The Gateaux derivative formula only applies when the domain of $\mu(F)$ also includes discrete distributions
A precise justification for the asymptotic variance formula is available when $\hat\beta$ is asymptotically equivalent to a sample average: define $\hat{\beta}$ to be asymptotically linear with influence function $\psi(z)$ when $z_i$ has distribution $F_0$: $$ \begin{aligned} & \sqrt{n}\left(\hat{\beta}-\beta_0\right)=\sum_{i=1}^n \psi\left(z_i\right) / \sqrt{n}+O_p(1), \\ & E[\psi(z)]=0, \quad \operatorname{Var}(\psi(z)) \text { finite. } \end{aligned} $$
Asymptotic linearity and the central limit theorem imply $\hat{\beta}$ is asymptotically normal with variance $\operatorname{Var}(\psi(z))$.
Regular path & regular estimator
-
Define the path $\left\{F_\theta: \theta \in(-\varepsilon, \varepsilon) \subset \mathbb{R}, \varepsilon>0, F_\theta \in \mathscr{F}\right\}$ to be regular if each distribution is absolutely continuous w.r.t the same dominating measure and $S(z)$ satisfies the mean-square derivative condition $$ \int\left[\theta^{-1}\left(d F_\theta^{1 / 2}-d F_0^{1 / 2}\right)-\frac{1}{2} S(z) d F_0^{1 / 2}\right]^2 d z \rightarrow 0, \text { as } \quad \theta \rightarrow 0 . $$
-
Define $\hat{\beta}$ to be a regular estimator of $\mu(F)$ if for any regular path and $\theta_n=$ $0(1 / \sqrt{n})$, when $z_i$ has distribution $F_{\theta_n}, \sqrt{n}\left(\hat{\beta}-\mu\left(F_{\theta_n}\right)\right)$ has a limiting distribution that does not depend on $\left\{\theta_n\right\}_{n=1}^{\infty}$.
The pathwise derivative asymptotic variance formula
THEOREM 2.1: Suppose that
- (i) the set of scores for regular paths is linear;
- (ii) for any $\varepsilon>0$ and measureable $s(z)$ with $E[s(z)]=0$ and $E\left[s(z)^2\right]<\infty$, there is a regular path with score $S(z)$ satisfying $E\left[|s(z)-S(z)|^2\right]<\varepsilon$;
- (iii) $\hat{\beta}$ is asymptotically linear and regular.
Then there is $d(z)$ such that equation $(\star)$ is satisfied and $\psi(z)=d(z)$
- Condition (ii), that the scores can approximate any mean zero function, is the precise version of the “generality” property of $\mathscr{F}$.
- Regularity of $\hat{\beta}$ is the precise condition that specifies that $\hat{\beta}$ is a nonparametric estimator of $\mu(F)$.
- Innovations: calculating the bound for the functional $\mu(F)$ is nonparametrically estimated by $\hat\beta$
- Asymptotic linearity and regularity imply pathwise differentiability
- Condition (ii) implies that there is only one influence function and that it equals the pathwise derivative
- Theorem 2.1 give a justification for the pathwise derivative formula, rather than an approach to showing asymptotic normality.
- A better approach:
- Solve equation $(\star)$ for the pathwise derivative, as a candidate for the influence function
- Formulate regularity conditions for the remainder $\sqrt{n}\left(\hat{\theta}-\theta_0\right)-\sum_{i=1}^n \psi\left(z_i\right) / \sqrt{n}$ to be small.
- The formula is a very important part of this approach, because it provides the form of the remainder.
- This approach, with formal calculation followed by regularity conditions, is similar to that used in parametric asymptotic theory (e.g. for Edgeworth expansions).
3. Semiparametric $M$-estimators
- Let $h$ denote a function, that can depend on the parameters $\beta$ and the data $z$.
- Let $m(z, \beta, h)$ be a vector of functions with the same dimension as $\beta$. Here $m(z, \beta, h)$ can depend on the entire function $h$, rather than just its value at particular points, so $m(z, \beta, h)$ is a vector of functionals.
- Suppose that $E\left[m\left(z, \beta_0, h_0\right)\right]=0$ for the true values $\beta_0$ and $h_0$.
- Let $\hat{h}$ denote an estimator of $h$. A semiparametric $m$-estimator is one that solves a moment equation of the form $$ {\color{red}\frac{1}{n}\sum_{i=1}^n m\left(z_i, \beta, \hat{h}\right)=0} .\qquad (\Delta) $$
- The general idea here is that $\hat{\beta}$ is obtained by a procedure that “plugs-in” an estimated function $\hat{h}$.
Example 3.1 Quasi-maximum Likelihood for a Conditional Mean Index
- The conditional mean index model: $E[y \mid x]=\tau\left(v\left(x, \beta_0\right)\right)$ for a known function $v(x, \beta)$ and an unknown function $\tau(\cdot)$.
- Let $\hat{h}(x, \beta)$ be a nonparametric estimator of $E[y \mid v(x, \beta)]$, such as a kernel estimator.
- An estimator of $\beta_0$ suggested by Ichimura (1993) minimizes $\sum_{i=1}^n\left[y_i-\hat{h}\left(x_i, \beta\right)\right]^2$.
- When $y_i$ is binary, Klein and Spady (1993) have suggested maximizing $$\sum_{i=1}^n\left\{y_i \ln [\hat{h}(x_i, \beta)]+(1-y_i) \ln [1-\hat{h}(x_i, \beta)]\right\}$$
- A generalization of these estimators is a quasi-maximum likelihood estimator (QMLE) for an exponential family.
- The estimator will be efficient when the true distribution has the exponential form.
- To describe the estimator, let $l(u, \nu)=\exp (A(\nu)+B(u)+C(\nu) u)$ be a linear exponential density, with mean $\nu$.
- Consider an estimator $\hat{\beta}$ that maximizes $\sum_{i=1}^n \ln l\left(y_i, \hat{h}\left(x_i, \beta\right)\right)$.
- The first order conditions for this estimator make it a special case of equation $(\Delta)$, with $$ m(z, \beta, h)=\left[A_\nu(h(x, \beta))+C_\nu(h(x, \beta)) y\right] \frac{\partial h(x, \beta) } {\partial \beta} $$
Example 3.2: Inverse Density Weighted Least Squares
- Let $w(x)=r((x-$ $\zeta)^{\prime} \Omega(x-\zeta)$ ) be an elliptically symmetric density function, where $\zeta$ is a vector and $\Omega$ a positive definite matrix.
- Let $\hat{h}\left(x_l\right)$ be an estimator of the density of $x$, such as a kernel estimator.
- As shown by Ruud (1986), the weighted least squares estimator $\hat{\beta}=\left[\sum_{l=1}^n w\left(x_i\right) \hat{h}\left(x_l\right)^{-1} x_l x_l^{\prime}\right]^{-1} \sum_{l=1}^n w\left(x_l\right) \hat{h}\left(x_l\right)^{-1} x_i y_i$ will be consistent up to scale, for the coefficients $\gamma_0$ of an index model $E[y \mid x]=\tau\left(x^{\prime} \gamma_0\right)$.
- This example is a special case of equation $(\Delta)$ with $$ m(z, \beta, h)=w(x) h(x)^{-1} x\left(y-x^{\prime} \beta\right) . $$
- This estimator will be used to illustrate the correction term for density estimates.
- Asymptotic normality and $\sqrt{n}$-consistency for $\hat{h}(x)$ a kernel estimator, are shown in Newey and Ruud (1991).
To use the pathwise derivative formula in this derivation, it is necessary to identify the functional that is nonparametrically estimated by $\hat{\beta}$.
- Let $h(F)$ denote the limit of $\hat{h}$ when $z$ has distribution $F$.
- The limit $\mu(F)$ of $\hat{\beta}$ for a general $F$ should be the solution to $$ E_F[m(z, \mu, h(F))]=0 .\qquad(\odot) $$
- Equation $(\Delta)$ sets $\hat{\beta}$ so that sample moments are zero, and the sample moments have a limit of $E_F[m(z, \beta, h(F))]$
- by the law of large numbers and $h(F)$ equal to the limit of $\hat{h}$
- $\Rightarrow$ $\hat{\beta}$ is consistent for the solution of $(\odot)$.
Remarks
- the estimators depend only on the limit $h(F)$, and not on the particular form of the estimator $\hat{h}$.
- Different nonparametric estimators of the same functions should result in the same asymptotic variance.
Proposition 1: The asymptotic variance of semiparametric estimators depends only on the function that is nonparametrically estimated, and not on the type of estimator.
- For a path $\left\{F_\theta\right\}$, let $h(\theta)=h\left(F_\theta\right)$. Here, $\mu\left(F_\theta\right)$ will satisfy the population moment equation $$ E_\theta[m(z, \mu, h(\theta))]=0 . $$
- Let $m(z, h)=m\left(z, \beta_0, h\right)$. Differentiation under the integral gives $$ \frac{\partial E_\theta\left[m\left(z, h_0\right)\right]}{ \partial \theta}=\int m\left(z, h_0\right)\left[\frac{\partial d F_\theta }{ \partial \theta}\right] d z=E\left[m\left(z, h_0\right) S(z)\right] . $$
- Using the chain rule, it follows that $$ \frac{\partial E_\theta[m(z, h(\theta))] }{ \partial \theta}=E\left[m\left(z, h_0\right) S(z)\right]+\frac{\partial E[m(z, h(\theta))] }{ \partial \theta} . $$
- Assuming $M \equiv \partial E\left[m\left(z, \beta, h_0\right)\right] /\left.\partial \beta\right|_{\beta_0}$ is nonsingular, by the implicit function theorem $$ \frac{\partial \mu\left(F_\theta\right) }{ \partial \theta}=-M^{-1}\left\{E\left[m\left(z, h_0\right) S(z)\right]+\frac{\partial E[m(z, h(\theta))] }{ \partial \theta}\right\} . $$
- The first term is already in an outer product form, so that the pathwise derivative can be found by putting the second term in singular form
- Suppose there is a $\alpha(z)$ such that $E[\alpha(z)]=0$ and $$\frac{\partial E[m(z, h(\theta))] }{ \partial \theta}=E[\alpha(z) S(z)]\qquad(\oplus)$$.
- Move $-M^{-1}$ inside the expectation, it follows that the pathwise derivative is $d(z)=-M^{-1}\left\{m\left(z, h_0\right)+\alpha(z)\right\}$
- By Theorem 2.1 the influence function of $\hat{\beta}$ is $$ \psi(z)=-M^{-1}\left\{m\left(z, \beta_0, h_0\right)+\alpha(z)\right\} $$ Remarks
- The leading term $-M^{-1} m\left(z, \beta_0, h_0\right)$ is the usual formula for the influence function of an $m$-estimator with moment functions $m\left(z, \beta, {\color{red}h_0}\right)$.
- The solution to equation $(\oplus)$ is an adjustment term for the estimation of $h_0$.
- Solving equation $(\oplus)$ is therefore the essential step in discovering how the estimation of $h$ affects the asymptotic variance.
- This solution can be interpreted as
- the pathwise derivative of the functional $-M^{-1} E\left[m\left(z, \beta_0, h(F)\right)\right]$
- the influence function of $-M^{-1} \int m\left(z, \beta_0, \hat{h}\right) d F_0(z)$.
When will the adjustment term be zero ?
- If the adjustment term is zero, then it should not be necessary to account for the presence of $\hat{h}$, i.e. $\hat{h}$ can be treated as if it were equal to $h_0$
- One case: equation $(\Delta)$ is the first-order condition to a maximization problem, and $\hat{h}$ has a limit that maximizes the population value of the same function.
To be specific, suppose that there is a function $q(z, \beta, h)$ and a set of functions $\mathscr{H}(\beta)$, possibly depending on $\beta$ but not on the distribution $F$ of $z$, such that $$ \begin{aligned} & m(z, \beta, h)=\partial q(z, \beta, h) / \partial \beta, \\ & h(F)=\operatorname{argmax}_{\tilde{h} \in \mathscr{H}(\beta)} E_F[q(z, \beta, \tilde{h})] .\qquad(\otimes) \end{aligned} $$
- $m(z, \beta, h)$ are the first order conditions for a maximum of the function $q$
- $h(F)$ maximizes the expected value of the same function
- i.e. $h(F)$ has been “concentrated out”.
For any parametric model $F_\theta$, since $h(\theta)=h\left(F_\theta\right)$ $\Rightarrow$ $E[q(z, \beta, h(\theta))]$ is maximized at $\theta=0$ $\Rightarrow$ $\partial E[q(z, \beta, h(\theta))] / \partial \theta=0$. Differentiating again with respect to $\beta$, $$ \begin{aligned} 0 & =\partial^2 E[q(z, \beta, h(\theta))] / \partial \theta \partial \beta\\ &=\partial E[\partial q(z, \beta, h(\theta)) / \partial \beta] / \partial \theta \\ & =\partial E[m(z, \beta, h(\theta))] / \partial \theta . \end{aligned} $$
- Evaluating this equation at $\beta_0$, it follows that $\alpha(z)=0$ will solve equation $(\oplus)$, and hence the adjustment term is zero.
Proposition 2: If equation $(\otimes)$ is satisfied, then the estimation of $h$ can be ignored in calculating the asymptotic variance, i.e. it is the same as if $\hat{h}=h_0$
A more direct condition
- Suppose that $m(z, h)$ depends on $h$ only through its value $h(v)$ at a subvector $v$ of $z$, i.e. $m(z, h)=m(z, h(v))$ where the last function depends on a real vector argument in $m(z, h)$.
- Let $h(v, \theta)$ denote the limiting value of $\hat{h}\left(v, \beta_0\right)$ for a path. For $D(z)=\partial m\left(z, \beta_0, h\right) /\left.\partial h\right|_{h=h_0(v)}$, differentiation gives $$ \frac{\partial E[m(z, h(\theta))] }{ \partial \theta}=E\left[D(z) \frac{\partial h(v, \theta) }{ \partial \theta}\right]=\frac{\partial E[D(z) h(v, \theta)] }{ \partial \theta} . $$
- If this derivative is zero for all $h(v, \theta)$, then $\alpha(z)=0$ will solve equation $(\oplus)$, and the adjustment term is zero.
- One simple condition for this is that $E[D(z) \mid v]=0$.
More generally, the adjustment term will be zero if $h(v, \theta)$ is an element of a set to which $D(z)$ is orthogonal.
Proposition 3: If $E[D(z) \mid v]=0$, or more generally, for all $F, h(v, F)$ is an element of a set $\mathscr{H}$ such that $E[D(z) \tilde{h}(v)]=0$ for all $\tilde{h} \in \mathscr{H}$, then estimation of $h$ can be ignored in calculating the asymptotic variance.
Remarks:
This condition can be checked by straightforward calculation, unlike Proposition 2, which requires finding $q(z, \beta, h)$ satisfying equation $(\otimes)$.
4. Functions of Mean-square Projection and Densities
5. Regularity Conditions
6. Series Estimation of Projection Functionals
7. Power Series Estimators For Semiparametric Individual Effects and Average Derivatives
References
- KOSHEVNIK, Y. A., AND B. Y. LEVIT (1976): “On a Non-parametric Analogue of the Information Matrix”, Theory of Probability and Applications, 21, 738-753.
- PFANZAGL, J., AND WEFELMEYER (1982): “Contributions to a General Asymptotic Statistical Theory”, New York: Springer-Verlag.