Why Boundaries Change Likelihood Asymptotics
The usual asymptotic normality result for a maximum-likelihood estimator describes a true parameter surrounded by a full open neighborhood of allowable values. Near such an interior point, local changes in every direction are possible, and a quadratic likelihood approximation has an unconstrained normal maximizer. At a boundary, some directions are forbidden. The maximizer may then be pulled onto the boundary, even when the random score points outward.
This is not a small technical alteration. An estimator can have a point mass at the boundary and a continuous distribution elsewhere, rather than a normal limiting distribution. In other situations, such as complete separation in logistic regression, the likelihood has a supremum only as a coefficient tends to infinity, so a finite maximum-likelihood estimate does not exist.
For example, \(0\) is a boundary point of \([0,\infty)\): every neighborhood of zero contains nonnegative and negative numbers. The point \(-1\) is not a boundary point of that parameter space, because a sufficiently small neighborhood of \(-1\) does not intersect \([0,\infty)\). This distinction matters: being outside a parameter space is not the same as being on its boundary.
A Constrained Normal Limit
A simple model makes the change in asymptotic behavior exact. Suppose \(Y_1,\ldots,Y_n\) are independent with \(Y_i\sim N(\theta,1)\), but the parameter space is \(\Theta=[0,\infty)\). The unconstrained estimate of \(\theta\) is \(\bar Y\). Under the constraint, a negative sample mean is replaced by the closest allowable value, zero.
Worked Example: A Normal Mean Constrained to Be Nonnegative
The log-likelihood, omitting a constant that does not depend on \(\theta\), is \(\ell(\theta)=-\frac12\sum_{i=1}^n(Y_i-\theta)^2\). Differentiating gives \(\ell'(\theta)=\sum_i(Y_i-\theta)=n(\bar Y-\theta)\), and \(\ell''(\theta)=-n<0\). Thus the unconstrained maximum is at \(\bar Y\).
If \(\bar Y\geq0\), this maximum is allowable. If \(\bar Y<0\), then \(\ell'(\theta)<0\) for every \(\theta\geq0\), so the likelihood decreases over the allowed parameter space and is maximized at zero. Therefore \[ \hat\theta=\max(0,\bar Y). \]
When the true value is \(\theta=0\), \(\sqrt n\,\bar Y\sim N(0,1)\) exactly. It follows that \[ \sqrt n\,\hat\theta=\max(0,\sqrt n\,\bar Y)\ \sim\ \max(0,Z),\qquad Z\sim N(0,1). \] This limit has probability \(1/2\) at zero, since \(P(Z\leq0)=1/2\), and its positive part has the corresponding half of the standard normal distribution. It is not a standard normal distribution. For a fixed \(\theta>0\), by contrast, \(P(\bar Y<0)\to0\), and the constraint is asymptotically inactive.
The reason is visible in the local likelihood. At a boundary point, the local parameter \(h=\sqrt n(\theta-\theta_0)\) is restricted to \(h\geq0\), not all of \(\mathbb{R}\). Maximizing a quadratic approximation over this half-line gives a constrained, rather than an unconstrained, normal-shaped maximizer.
Proof. The limiting quadratic function on the allowed half-line is \(q(h)=hZ-\frac12 I h^2\). Completing the square gives \[ q(h)=-\frac I2(h-Z/I)^2+\frac{Z^2}{2I}. \] Its unique maximum over \(h\geq0\) is attained at \(h^*=\max(0,Z/I)\). The uniform expansion gives convergence of the local likelihood process to this quadratic on each bounded interval. Tightness of \(\hat h_n\) ensures that, with probability arbitrarily close to one, the maximizer lies in such an interval; the assumed approximate-maximization property then makes it converge in distribution to the unique maximizer of the limiting process. Since \(Z\sim N(0,I)\), this yields the stated limit. \(\square\)
The constraint in this result is essential. If the local parameter were allowed to range over the whole real line, the quadratic would instead be maximized at \(Z/I\), giving the familiar normal limit. At the boundary, the negative part is replaced by zero.
A Zero Variance Component
A nonnegative variance component is a common example of a parameter whose null value lies on the boundary. The following model isolates the boundary effect: observations are independent \(Y_i\sim N(0,\sigma^2+\tau)\), where the baseline variance \(\sigma^2>0\) is known and the additional component satisfies \(\tau\geq0\). The value \(\tau=0\) means that the additional component is absent.
Worked Example: Estimating a Variance Component at Zero
Write \(S=n^{-1}\sum_{i=1}^nY_i^2\) and \(v=\sigma^2+\tau\), so \(v\geq\sigma^2\). The log-likelihood, up to a constant, is \[ \ell(v)=-\frac n2\log v-\frac{nS}{2v}. \] Its derivative is \(\ell'(v)=\frac{n(S-v)}{2v^2}\). Thus it increases for \(v<S\) and decreases for \(v>S\); the unconstrained maximum is at \(v=S\). Restricting to \(v\geq\sigma^2\) gives \[ \hat\tau=\max(0,S-\sigma^2). \]
At the true boundary value \(\tau=0\), \(Y_i\sim N(0,\sigma^2)\). Hence \(E[Y_i^2]=\sigma^2\) and \(\operatorname{Var}(Y_i^2)=2\sigma^4\). The central limit theorem gives \[ \sqrt n(S-\sigma^2)\xrightarrow{d}N(0,2\sigma^4). \] Applying the continuous function \(x\mapsto\max(0,x)\) yields \[ \sqrt n\,\hat\tau\xrightarrow{d}\max(0,W),\qquad W\sim N(0,2\sigma^4). \] The limiting distribution has mass \(1/2\) at zero and a positive half-normal part. The mass at zero reflects samples for which the unconstrained estimate \(S-\sigma^2\) is negative and the constrained maximum is exactly the boundary value.
This derivation also shows why the usual interior-point standard error is not enough. A symmetric normal approximation puts probability on both sides of the estimate, while the estimator here cannot be negative and has a nonvanishing probability of being exactly zero after rescaling. Boundary problems can require methods that account for the constrained limiting distribution rather than treating the estimate as an ordinary normal variable.
Diverging Logistic Coefficients
Not every nonregular likelihood is a finite parameter sitting at the edge of a closed parameter space. In logistic regression, complete separation can make the likelihood increase toward its supremum as a coefficient diverges. The coefficient's parameter space is still \(\mathbb{R}\); the failure is that the supremum is not attained at any finite value.
Worked Example: Complete Separation in a Logistic Model
Consider a logistic model with no intercept, \(P(Y_i=1\mid x_i)=\operatorname{expit}(\beta x_i)\), where \(\operatorname{expit}(u)=1/(1+e^{-u})\). Suppose the observed design has \(m\) cases with \(x_i=1,Y_i=1\) and \(m\) cases with \(x_i=-1,Y_i=0\), for \(m\geq1\).
For \(x_i=1,Y_i=1\), the likelihood contribution is \(\operatorname{expit}(\beta)\). For \(x_i=-1,Y_i=0\), it is \(1-\operatorname{expit}(-\beta)=\operatorname{expit}(\beta)\). Therefore \[ L(\beta)=\operatorname{expit}(\beta)^{2m}. \] For every finite \(\beta\), \(0<\operatorname{expit}(\beta)<1\), so \(L(\beta)<1\). As \(\beta\to+\infty\), \(\operatorname{expit}(\beta)\to1\), and \(L(\beta)\to1\). Thus the likelihood has supremum \(1\), but no finite \(\beta\) attains it. The fitted probabilities approach one at \(x=1\) and zero at \(x=-1\).
This example is different from the zero-variance case: there is no finite boundary value of \(\beta\) in \(\mathbb{R}\). The coefficient diverges because increasingly extreme values continue to improve the likelihood. Consequently, a finite maximum-likelihood estimate and its usual local normal approximation are unavailable for this data set.
What to Check in Practice
Before applying an interior-point asymptotic argument, check both the parameter space and the existence of the likelihood maximum. A true value on a boundary can constrain local perturbations, while separation can prevent a finite maximizer altogether. These are distinct mechanisms, even though both undermine routine normal approximations.
Determine whether it is an interior point or a boundary point of the allowable parameter space.
At a boundary, identify which local changes are allowed; a quadratic likelihood must be maximized over those directions.
In logistic models, inspect for separation or near-separation before interpreting a coefficient and standard error.
Common mistake: “The estimator is approximately normal because the likelihood is approximately quadratic.” A quadratic approximation alone does not imply an unconstrained normal limit. At a boundary, the approximation is maximized over a restricted set, which can create a point mass at the boundary. A related definitional mistake is to call any point whose neighborhood contains points outside \(\Theta\) a boundary point: points far outside \(\Theta\) satisfy that condition too. A boundary point must have every neighborhood intersect both \(\Theta\) and \(\Theta^c\).
Check Your Understanding
Use the definitions and calculations above to answer the following questions.
- For \(\Theta=[0,\infty)\), explain why \(0\) is a boundary point but \(-1\) is not.
- In the constrained normal-mean example at \(\theta=0\), what fraction of the limiting distribution of \(\sqrt n\,\hat\theta\) is concentrated at zero, and why?
- For the variance-component model, state the constrained maximum-likelihood estimate and explain when it equals zero.
- Why does the separated logistic likelihood have no finite maximizing coefficient even though its supremum is finite?
- Distinguish a finite parameter on the boundary from a logistic coefficient that diverges under separation.