Delta-Method Limits at Parameter Boundaries

Why Boundaries Change the Limit

At an interior parameter value, a smooth estimator often fluctuates on both sides of its target, and the ordinary delta method carries those fluctuations through a differentiable transformation. A parameter on the edge of its allowed set is different: the estimator cannot move outside that set. Some of the unconstrained sampling fluctuation can therefore be pushed back onto the boundary. A transformed estimator inherits this altered limit, rather than automatically having a normal limit.

The key is to separate two questions. First, what is the estimator's limit after the constraint is imposed? Second, how does the transformation act on that limit? The multivariate delta method answers the second question once the first limit and the relevant derivative are correctly specified.

Definition: A parameter value \(\theta_0\) is on the boundary of its parameter space \(\Theta\) if every neighborhood of \(\theta_0\) contains points in \(\Theta\) and points outside \(\Theta\). The tangent cone \(T\) at \(\theta_0\) describes the directions in which the parameter can move locally while remaining in \(\Theta\). For the scalar space \(\Theta=[0,\infty)\) at \(\theta_0=0\), \(T=[0,\infty)\).

The Transformation of a Boundary Limit

Suppose an estimator satisfies \(r_n(\hat\theta-\theta_0)\xrightarrow{d}W\), where \(r_n\) is its convergence rate and \(W\) may be nonnormal. If \(g\) is differentiable at \(\theta_0\), its first-order expansion gives

r_n\{g(\hat\theta)-g(\theta_0)\} = Dg(\theta_0)\,r_n(\hat\theta-\theta_0)+o_p(1) \ \xrightarrow{d}\ Dg(\theta_0)W.

Here \(Dg(\theta_0)\) is the derivative (a row vector when \(g\) is scalar-valued). This is the delta method applied to the estimator's actual limit. It does not require \(W\) to be normal. If \(W\) is a projected or positive-part normal variable, the transformed limit will generally remain nonnormal.

The formula also identifies an important exception. If \(Dg(\theta_0)W=0\) with probability one, the first-order term vanishes. A higher-order expansion may then produce a nondegenerate limit at a faster rate. For example, a quadratic transformation can turn a root-\(n\) estimator into an estimator whose nonzero limit appears at rate \(n\).

Worked Example: A Linear Transformation of a Constrained Estimate

Suppose the parameter is a nonnegative treatment-effect magnitude \(\theta\), and the true value is \(\theta_0=0\). A hypothetical unconstrained estimator \(\tilde\theta\) has \(\sqrt n\,\tilde\theta\xrightarrow{d}Z\), where \(Z\sim N(0,0.04)\). The reported estimator imposes the constraint: \(\hat\theta=\max(0,\tilde\theta)\). Consider the transformed quantity \(g(\theta)=3\theta\).

Because taking the positive part is a continuous function,

\sqrt n\,\hat\theta = \max(0,\sqrt n\,\tilde\theta) \xrightarrow{d}\max(0,Z), \qquad \sqrt n\{g(\hat\theta)-g(0)\} \xrightarrow{d}3\max(0,Z).

The limiting variable equals zero when \(Z\leq 0\), an event with probability \(0.5\). When it is positive, it equals three times a normal draw restricted to the positive side; it is not a normal variable. For example, the probability that this limit is at most \(0.6\) is

P\{3\max(0,Z)\leq 0.6\} = P(Z\leq 0.2) = \Phi(0.2/0.2) = \Phi(1) \approx 0.8413.

Thus the limit has a point mass of \(0.5\) at zero as well as a continuous positive part. The factor of 3 changes the positive scale but does not remove the boundary mass.

Why Constrained Estimators Are Projected

A common local approximation for a constrained extremum estimator is quadratic. Let \(h=\sqrt n(\theta-\theta_0)\) denote a local parameter displacement. Suppose the objective, after subtracting its value at \(\theta_0\), has the local form

-2h'Z_n+h'Hh+o_p(1), \qquad Z_n\xrightarrow{d}Z,

where \(H\) is positive definite. If there were no constraint, minimizing the quadratic gives \(h=H^{-1}Z_n\). With the constraint, allowable local displacements are restricted to the tangent cone \(T\). The limiting displacement is therefore

W=\underset{h\in T}{\operatorname{argmin}}\, \left(h-H^{-1}Z\right)'H\left(h-H^{-1}Z\right).

This is the projection of the unconstrained limit \(H^{-1}Z\) onto \(T\) using the distance defined by \(H\). When \(H\) is the identity matrix and \(T=[0,\infty)\), this reduces to the positive part \(\max(0,H^{-1}Z)\). With multiple parameters, or a nonidentity \(H\), the projection need not be coordinate-by-coordinate clipping.

Key result: Under a local quadratic approximation with a closed convex tangent cone, the root-\(n\) limit of a constrained estimator is the \(H\)-metric projection of its unconstrained Gaussian limit onto that cone. A differentiable transformation then has first-order limit \(Dg(\theta_0)W\), not necessarily a normal limit.

Worked Example: A Quadratic Transformation Changes the Rate

Continue with \(\theta_0=0\), \(\hat\theta=\max(0,\tilde\theta)\), and \(\sqrt n\,\tilde\theta\xrightarrow{d}Z\), but now let \(Z\sim N(0,0.09)\) and transform with \(g(\theta)=\theta^2\). The derivative at the boundary is \(g'(0)=0\), so the first-order delta-method limit is zero. Instead, use the exact identity

n\{g(\hat\theta)-g(0)\} = n\hat\theta^2 = \{\max(0,\sqrt n\,\tilde\theta)\}^2 \xrightarrow{d}\{\max(0,Z)\}^2.

The transformed estimator has a nondegenerate limit at rate \(n\), rather than at rate \(\sqrt n\). The limiting variable has probability \(0.5\) at zero. For a numerical threshold of \(0.09\),

P\{\max(0,Z)^2\leq 0.09\} = P(Z\leq 0.3) = \Phi(0.3/0.3) = \Phi(1) \approx 0.8413.

The first equality follows because the squared positive part is at most \(0.09\) exactly when \(Z\leq 0.3\). This example illustrates why a zero derivative is not a reason to report a degenerate answer and stop: it can signal the need for a higher-order expansion and a different rate.

A Vector Constraint Can Couple Coordinates

For a vector parameter, a boundary may restrict only some directions. Even an unconstrained normal limit can become a more complicated projected distribution. The geometry of the objective matters: its curvature determines the metric of projection. The following calculation shows why simply clipping the coordinate that violates the constraint can give the wrong answer.

Worked Example: Projection with a Nuisance Coordinate

Let \(\theta=(\beta,\gamma)'\), with the boundary restriction \(\beta\geq 0\) and \(\gamma\) unrestricted. Suppose the local quadratic criterion is \((h-z)'H(h-z)\), where

H= \begin{pmatrix} 1&0.5\\ 0.5&1 \end{pmatrix}, \qquad z= \begin{pmatrix} -0.4\\ 0.6 \end{pmatrix}.

These numbers are a hypothetical realization of the unconstrained local limit. Since \(z_1=-0.4\) violates the constraint, the constrained minimum has \(h_1=0\). Minimizing over the unrestricted coordinate \(h_2\) requires setting the derivative with respect to \(h_2\) to zero:

0.5(0-z_1)+(h_2-z_2)=0, \qquad 0.5(0.4)+(h_2-0.6)=0, \qquad h_2=0.4.

The constrained solution is \(h=(0,0.4)'\). Coordinatewise clipping would instead give \((0,0.6)'\), which does not minimize the stated criterion: the off-diagonal curvature couples the two coordinates. If the target is \(g(\beta,\gamma)=\beta+\gamma\), its local change for this realization is \(h_1+h_2=0.4\). Across repeated samples, the projected vector and its transformed value generally have nonnormal limits.

Truncation, Projection, and Interpretation

A positive-part normal limit is sometimes called a truncated normal informally, but that phrase can mislead. The variable \(\max(0,Z)\) is not the distribution of \(Z\) conditional on \(Z\geq 0\). Conditional truncation removes all negative draws and renormalizes the remaining probability to one. Positive-part projection instead maps every negative draw to zero, retaining those draws as a point mass at the boundary. In the centered scalar example, that mass is \(0.5\).

The distinction matters for inference. A normal approximation that ignores the boundary does not reproduce the point mass or the one-sided shape, so standard symmetric confidence intervals and normal critical values may not have their usual coverage. Likewise, a transformation does not make a nonnormal boundary limit normal merely because the transformation is smooth. Its derivative acts on the distribution that is actually present.

The projection result relies on the stated local quadratic structure, a suitable limiting score, and a well-behaved feasible set near the true parameter. Not every boundary problem has this form. If the constraint is irregular, the objective is not locally quadratic, or the target transformation is not differentiable, the limit must be derived from the relevant local expansion rather than assumed to be a projected normal.

Common Mistakes

Using an interior normal limit at the boundary. The constraint may place positive probability on the boundary, even when the unconstrained limit is normal. Derive the constrained limit first.
Calling the positive-part limit a conditional truncated normal. Negative draws are mapped to zero; they are not discarded. This creates an atom at zero.
Clipping each coordinate separately. That is generally correct only for particular constraint and curvature geometries. With cross-coordinate curvature, compute the metric projection.
Stopping when the first derivative is zero. A zero first-order term can mean that a higher-order term determines the limit at a faster rate, as in the quadratic example.
Assuming a smooth transformation restores normality. The delta method transforms the limit; it does not change a projected limit into a normal one by default.

Key takeaway: At a boundary, first determine the estimator's constrained limit. Then apply the derivative of the target transformation; if that derivative removes the first-order variation, examine the next nonzero term and its rate.

Check Your Understanding

Use the boundary geometry and the transformation together when answering.

If \(\sqrt n\,\hat\theta\xrightarrow{d}\max(0,Z)\) and \(Z\) is centered normal, what is the probability mass at zero?
Why is \(\max(0,Z)\) different from a normal variable conditional on being positive?
For \(g(\theta)=\theta^2\) at \(\theta_0=0\), why can the first-order delta method not give a nondegenerate limit?
In the vector example, which feature of \(H\) makes coordinatewise clipping incorrect?
State the order in which you should analyze a transformed estimator when its parameter is on a boundary.