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Asymptotic Estimation Theory · Tutorial 47 of 952

Asymptotic Normality of Smooth Extremum Estimators

Learn how local criterion curvature and score variation jointly determine an extremum estimator’s asymptotic distribution.

Intermediate 12 min read

What You'll Learn

  • State the regularity conditions for asymptotic normality of a smooth extremum estimator.
  • Expand the sample first-order condition around the population parameter.
  • Combine a score central limit theorem with a Hessian probability limit.
  • Derive the asymptotic covariance matrix, including its signs and matrix order.
  • Calculate scalar and vector asymptotic distributions from hypothetical criterion moments.
  • Identify why curvature alone does not determine an estimator’s asymptotic variance.

From a Sample Optimum to a Limiting Distribution

Consistency says that an estimator approaches the target parameter as the sample grows, but it does not describe its typical sampling variation at a large, finite sample size. For a smooth extremum estimator, that variation is governed by two features: how much the sample criterion’s gradient fluctuates at the target, and how sharply the population criterion curves there. The first is summarized by a central limit theorem for the sample score; the second by the limiting Hessian.

The key step is local. At an interior optimum, the sample criterion has a zero gradient. Expanding that gradient around the population parameter relates the estimation error to the gradient evaluated at the target. If the Hessian converges to a nonsingular matrix and the score satisfies a central limit theorem, this relation yields a normal limit for the scaled estimation error.

Setup and Conditions

Let \(\theta\) be a parameter in an open subset of \(\mathbb{R}^k\). Suppose the sample criterion to be maximized is \(Q_n(\theta)=n^{-1}\sum_{i=1}^n q(W_i,\theta)\), where \(W_1,\ldots,W_n\) are identically distributed and independent observations. Let \(Q(\theta)=E[q(W_i,\theta)]\) be the population criterion, and let \(\theta_0\) be its uniquely identified maximizer. Write \(\hat{\theta}_n\) for a sample maximizer.

For the usual smooth-extremum argument, the following conditions give a useful sufficient set of assumptions. The estimator is consistent, so \(\hat{\theta}_n\to_p\theta_0\); \(\theta_0\) is interior; the sample criterion is twice continuously differentiable in a neighborhood of \(\theta_0\); and the first-order condition holds there. The sample Hessian converges to the population Hessian locally, including along the line segment between \(\theta_0\) and \(\hat{\theta}_n\). The population Hessian at \(\theta_0\) is nonsingular. Finally, the score at \(\theta_0\) has a finite covariance matrix and satisfies the multivariate central limit theorem.

Definition: Let \(s_i(\theta)=\nabla_\theta q(W_i,\theta)\) be the observation-level score, and define \(H=E[\nabla_\theta^2 q(W_i,\theta_0)]\) and \(\Omega=\operatorname{Var}(s_i(\theta_0))\). Under the stated conditions, the smooth extremum estimator is asymptotically normal if \(\sqrt{n}(\hat{\theta}_n-\theta_0)\xrightarrow{d}N(0,H^{-1}\Omega H^{-T})\), where \(H^{-T}=(H^{-1})^\top\). For a maximization problem, \(H\) is typically negative definite; for a minimization problem, it is typically positive definite. In either case, the covariance matrix is positive semidefinite.

The central limit theorem requires the score to be centered: \(E[s_i(\theta_0)]=0\). For an interior, differentiable population optimum this is the population first-order condition, provided differentiation and expectation can be interchanged. The covariance \(\Omega\) measures score variation, while the inverse Hessians translate that variation into movement in the parameter.

Why the Hessian and Score Produce This Limit

The derivation uses the sample first-order condition and a Taylor expansion with an intermediate Hessian. Define \(H_n^*=\int_0^1\nabla_\theta^2Q_n\big(\theta_0+t(\hat{\theta}_n-\theta_0)\big)\,dt\). Because \(\hat{\theta}_n\) maximizes a differentiable criterion at an interior point, \(\nabla_\theta Q_n(\hat{\theta}_n)=0\). The integral form of the expansion gives

$$ 0=\nabla_\theta Q_n(\theta_0)+H_n^*(\hat{\theta}_n-\theta_0). $$

If \(H_n^*\to_p H\) and \(H\) is nonsingular, then \(H_n^*\) is nonsingular with probability approaching one. Multiplying the equation by \(\sqrt{n}\) and solving gives

$$ \sqrt{n}(\hat{\theta}_n-\theta_0) =-(H_n^*)^{-1}\sqrt{n}\nabla_\theta Q_n(\theta_0). $$

Since \(\nabla_\theta Q_n(\theta_0)=n^{-1}\sum_{i=1}^n s_i(\theta_0)\), the multivariate central limit theorem implies \(\sqrt{n}\nabla_\theta Q_n(\theta_0)\xrightarrow{d}N(0,\Omega)\). Slutsky’s theorem then gives a normal limit with covariance \((-H^{-1})\Omega(-H^{-T})=H^{-1}\Omega H^{-T}\). The two minus signs cancel. This also explains why a maximization problem’s negative Hessian does not make the resulting covariance negative.

Justification: The local Hessian limit is what permits replacing \((H_n^*)^{-1}\) by \(H^{-1}\) in the scaled first-order condition. The score central limit theorem supplies the random limiting term. Both ingredients are needed: curvature converts score fluctuations into parameter fluctuations, but does not by itself determine their size.

Worked Scalar and Vector Calculations

Worked Example: A Scalar Extremum Estimator

Suppose a hypothetical sample consists of independent observations \(Y_i\) with \(E[Y_i]=\theta_0=12\) minutes and \(\operatorname{Var}(Y_i)=9\) minutes squared. Consider maximizing \(q(Y_i,\theta)=-\tfrac12(Y_i-\theta)^2\) over \(\theta\in\mathbb{R}\). This example illustrates the smooth-extremum calculation; it does not claim that any real data have these moments.

The population criterion is maximized at \(E[Y_i]\), and the sample maximizer is \(\hat{\theta}_n=\bar{Y}\). The observation-level score is \(s_i(\theta)=Y_i-\theta\). At \(\theta_0=12\), it has mean zero and variance \(\Omega=9\) minutes squared. The Hessian is constant: \(\partial^2 q(Y_i,\theta)/\partial\theta^2=-1\), so \(H=-1\).

The asymptotic variance is

$$ H^{-1}\Omega H^{-T} =(-1)^{-1}(9)(-1)^{-1} =9\ \text{minutes}^2. $$

Thus \(\sqrt{n}(\hat{\theta}_n-12)\xrightarrow{d}N(0,9)\), with the scaled quantity measured in minutes. Equivalently, for large \(n\), \(\hat{\theta}_n\) is approximately normal with mean 12 minutes and variance \(9/n\) minutes squared. The negative curvature affects the sign in the first-order expansion, but its square leaves the variance positive.

Worked Example: A Two-Parameter Limit

Consider a hypothetical vector criterion whose population maximizer is \(\theta_0=(1,2)^\top\), whose Hessian is \(H=\begin{pmatrix}-2&0\\0&-1\end{pmatrix}\), and whose score at \(\theta_0\) has mean zero and covariance \(\Omega=\begin{pmatrix}4&1\\1&3\end{pmatrix}\). Assume independent, identically distributed observations, finite score covariance, and the smoothness, consistency, and local Hessian conditions stated above.

The Hessian is negative definite, since its diagonal entries are negative. Its inverse is \(H^{-1}=\begin{pmatrix}-0.5&0\\0&-1\end{pmatrix}\); because it is symmetric, \(H^{-T}=H^{-1}\). Direct multiplication gives

$$ \begin{aligned} H^{-1}\Omega H^{-T} &= \begin{pmatrix}-0.5&0\\0&-1\end{pmatrix} \begin{pmatrix}4&1\\1&3\end{pmatrix} \begin{pmatrix}-0.5&0\\0&-1\end{pmatrix}\\ &= \begin{pmatrix}-2&-0.5\\-1&-3\end{pmatrix} \begin{pmatrix}-0.5&0\\0&-1\end{pmatrix}\\ &= \begin{pmatrix}1&0.5\\0.5&3\end{pmatrix}. \end{aligned} $$

For example, the upper-right entry is \((-2)(0)+(-0.5)(-1)=0.5\); the lower-left entry is \((-1)(-0.5)+(-3)(0)=0.5\). Therefore \(\sqrt{n}(\hat{\theta}_n-\theta_0)\) converges in distribution to a bivariate normal vector with mean zero and this covariance matrix. The limiting covariance of the unscaled estimator is approximately this matrix divided by \(n\). Its positive off-diagonal entry indicates that the two scaled estimation errors tend to move together, not in opposite directions.

Worked Example: Random Curvature in a Scalar Criterion

Let a hypothetical observation be \(W_i=(X_i,Y_i)\), and maximize \(q(W_i,\theta)=Y_i\theta-\tfrac12X_i\theta^2\), with \(X_i>0\). Suppose the population target is \(\theta_0=3\), \(E[X_i]=2\), and the score \(S_i=Y_i-3X_i\) has mean zero and variance \(18\). Assume the observations are independent and identically distributed, these moments are finite, and the criterion satisfies the preceding consistency and smoothness conditions.

The score is \(s_i(\theta)=Y_i-X_i\theta\), so \(s_i(\theta_0)=S_i\) and \(\Omega=18\). The Hessian is \(-X_i\), so \(H=-E[X_i]=-2\). The sample first-order condition is \(n^{-1}\sum_i(Y_i-X_i\hat{\theta}_n)=0\); since \(\sum_i X_i>0\), it has the unique solution \(\hat{\theta}_n=\sum_iY_i/\sum_iX_i\). The asymptotic variance is

$$ H^{-1}\Omega H^{-T} =(-2)^{-1}(18)(-2)^{-1} =\frac{18}{4}=4.5. $$

Consequently, \(\sqrt{n}(\hat{\theta}_n-3)\xrightarrow{d}N(0,4.5)\). The calculation uses the average curvature \(E[X_i]\), not a claim that each observation has the same curvature. A finite, positive average \(E[X_i]\) ensures nonzero population curvature in this scalar example.

Interpretation and Common Pitfalls

Asymptotic normality describes the distribution of the estimation error after scaling by \(\sqrt{n}\). In the stated result, the estimator’s large-sample covariance is approximately \(H^{-1}\Omega H^{-T}/n\), not \(H^{-1}\Omega H^{-T}\) without the division by \(n\). Units also follow from the problem: the covariance matrix has units of parameter times parameter, while the scaled error has the corresponding units multiplied by \(\sqrt{n}\).

  • Using curvature but omitting score variation. The Hessian alone cannot specify sampling uncertainty. The score covariance \(\Omega\) is also required.
  • Dropping one of the minus signs incorrectly. The expansion has a minus sign, but covariance multiplication introduces it on both sides: \((-H^{-1})\Omega(-H^{-T})=H^{-1}\Omega H^{-T}\).
  • Reversing matrix order. In general, matrix multiplication is not commutative. Use \(H^{-1}\Omega H^{-T}\), not an arbitrarily reordered product.
  • Assuming a first-order condition at a boundary. The zero-gradient condition used here requires an interior optimum. At a parameter boundary, that argument may fail.
  • Calling an approximation exact in finite samples. The normal distribution is a limiting distribution; it need not describe the exact finite-sample distribution of the estimator.

The result is useful because it separates two sources of uncertainty in a precise way: random score variation and local criterion curvature. It also gives a disciplined route from an optimization problem to a large-sample distribution: establish the first-order condition, expand it, and verify the Hessian and score limits rather than inferring normality from smoothness alone.

Check Your Understanding

Use the definitions and derivation above to answer these questions.

  1. In the first-order expansion, what role does the intermediate Hessian \(H_n^*\) play?
  2. If a maximization criterion has \(H=-3\) and score variance \(\Omega=12\) in a scalar problem, what is the asymptotic variance of \(\sqrt{n}(\hat{\theta}_n-\theta_0)\)?
  3. Why does a minus sign in the first-order expansion not make the asymptotic covariance matrix negative?
  4. In the vector example, verify the lower-right entry of the covariance matrix by multiplication.
  5. Which condition used in this derivation may fail if the optimizer is on the boundary of its parameter space?
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