Confidence Intervals for Bounded Longitudinal Parameters

From an Influence-Curve Standard Error to an Interval

A targeted longitudinal risk estimates a probability under a specified treatment regime. Its parameter is therefore bounded between zero and one. A Wald confidence interval is easy to construct from an estimate and its standard error, but the usual symmetric interval can extend beyond those meaningful limits. This tutorial develops the ordinary interval, explains its large-sample justification, and derives a logit-scale alternative that stays within the risk range.

Let $\psi$ denote the regime-specific risk and $\hat\psi$ its targeted estimator from $n$ independent observations. Suppose $\hat D_i$ is the estimated influence-curve value for observation $i$. The standard error used here is the empirical standard deviation of those values divided by $\sqrt n$. In particular, the influence-curve standard error is not the standard deviation of the observed binary outcomes divided by $\sqrt n$: the longitudinal estimator and its estimated influence curve account for the estimation procedure and observed histories.

Definition: A two-sided, pointwise $100(1-\alpha)\%$ Wald confidence interval for $\psi$ is $$ \left[\hat\psi-z_{1-\alpha/2}\widehat{\operatorname{se}}(\hat\psi),\ \hat\psi+z_{1-\alpha/2}\widehat{\operatorname{se}}(\hat\psi)\right], $$ where $z_{1-\alpha/2}$ is the $1-\alpha/2$ quantile of the standard normal distribution. For an influence-curve standard error, $$ \widehat{\operatorname{se}}(\hat\psi) =\left\{\frac{1}{n}\left(\frac{1}{n}\sum_{i=1}^n(\hat D_i-\bar{\hat D})^2\right)\right\}^{1/2}, \qquad \bar{\hat D}=\frac{1}{n}\sum_{i=1}^n\hat D_i. $$

The interval is pointwise: its stated coverage concerns this one risk parameter, not simultaneous coverage for several treatment regimes. For a 95% interval, $z_{0.975}$ is approximately $1.96$. The choice of standard error should follow the influence-curve standard-error construction, including any centering convention used for the estimated influence curve.

Worked Example: An Ordinary Interval Away from the Boundary

Suppose a targeted estimate of a regime-specific risk is $\hat\psi=0.30$, with estimated influence-curve standard error $0.04$. Using $z_{0.975}=1.96$, the margin of error is $$ 1.96(0.04)=0.0784. $$

The lower endpoint is $0.30-0.0784=0.2216$, and the upper endpoint is $0.30+0.0784=0.3784$. The 95% Wald interval is therefore $[0.2216,0.3784]$. Both endpoints fall within the possible range for a risk, so the interval has no immediate boundary contradiction. This does not, by itself, establish that the normal approximation is adequate.

Why the Wald Interval Has Its Form

The usual justification is asymptotic. For a fixed data-generating distribution, assume the estimator is asymptotically linear: $$ \hat\psi-\psi = \frac{1}{n}\sum_{i=1}^n D(O_i)+o_p(n^{-1/2}), $$ where $O_1,\ldots,O_n$ are independent and identically distributed observations, $E[D(O)]=0$, and $0<\sigma^2=E[D(O)^2]<\infty$. Also assume the estimated influence-curve variance is consistent, so that $\widehat{\operatorname{se}}(\hat\psi)/(\sigma/\sqrt n)\xrightarrow{p}1$. These are inferential conditions, not consequences of merely having a targeted estimate.

Theorem: Under the preceding asymptotic linearity and variance conditions, $$ \frac{\hat\psi-\psi}{\widehat{\operatorname{se}}(\hat\psi)} \xrightarrow{d}N(0,1). $$ Consequently, the ordinary Wald interval has asymptotic coverage $1-\alpha$, provided the regularity conditions hold.

Justification. Since the $D(O_i)$ are independent and identically distributed, have mean zero, and have finite positive variance, the central limit theorem gives $$ \frac{n^{-1/2}\sum_{i=1}^nD(O_i)}{\sigma}\xrightarrow{d}N(0,1). $$ Multiplying the asymptotic linear representation by $\sqrt n/\sigma$, its remainder is $o_p(1)$, so Slutsky’s theorem yields $\sqrt n(\hat\psi-\psi)/\sigma\xrightarrow{d}N(0,1)$. Replacing $\sigma/\sqrt n$ by the consistent estimated standard error gives the displayed studentized limit, again by Slutsky’s theorem. Thus the probability that the studentized statistic lies between $-z_{1-\alpha/2}$ and $z_{1-\alpha/2}$ tends to $1-\alpha$, which is the asserted coverage. $\square$

For longitudinal targeted estimators, the asymptotic linearity and variance conditions require appropriate conditions on the statistical problem and estimation procedure. The influence-curve standard-error construction is useful only when its estimated values adequately represent the estimator’s sampling variability. Neither the interval formula nor targeting alone guarantees those conditions.

Logit-Scale Intervals for a Risk

When an ordinary interval extends below zero or above one, a common alternative is to transform the risk to the real line, form a Wald interval there, and transform the endpoints back. For $0<\psi<1$, define $g(\psi)=\log\{\psi/(1-\psi)\}$. Its derivative is $g'(\psi)=1/\{\psi(1-\psi)\}$. The delta method therefore estimates the standard error of $g(\hat\psi)$ by $\widehat{\operatorname{se}}(\hat\psi)/\{\hat\psi(1-\hat\psi)\}$, when $\hat\psi$ is strictly between zero and one.

Result: If $\hat\psi$ is interior to $(0,1)$, a logit-scale Wald interval, transformed back to the risk scale, is $$ \left[ \operatorname{expit}\left\{g(\hat\psi)-z_{1-\alpha/2} \frac{\widehat{\operatorname{se}}(\hat\psi)}{\hat\psi(1-\hat\psi)}\right\}, \quad \operatorname{expit}\left\{g(\hat\psi)+z_{1-\alpha/2} \frac{\widehat{\operatorname{se}}(\hat\psi)}{\hat\psi(1-\hat\psi)}\right\} \right], $$ where $g(p)=\log\{p/(1-p)\}$ and $\operatorname{expit}(x)=1/(1+e^{-x})$.

Derivation. Suppose $0<\psi<1$, the asymptotic normality and variance conditions for $\hat\psi$ hold, and the delta method applies to the differentiable function $g$ at $\psi$. Then $$ \frac{g(\hat\psi)-g(\psi)} {g'(\psi)\widehat{\operatorname{se}}(\hat\psi)} \xrightarrow{d}N(0,1), $$ using consistency of $\hat\psi$ and the estimated standard error. Replacing the unknown derivative $g'(\psi)$ by $g'(\hat\psi)$ is justified by consistency and continuity at an interior $\psi$. The normal interval on the logit scale is obtained by adding and subtracting $z_{1-\alpha/2}$ times the estimated transformed standard error. Since $\operatorname{expit}$ is strictly increasing and is the inverse of $g$, applying it to the two endpoints preserves their order and gives the displayed risk-scale interval. Its endpoints are strictly between zero and one. $\square$

Worked Example: A Logit Interval Near Zero

Suppose $\hat\psi=0.015$ and $\widehat{\operatorname{se}}(\hat\psi)=0.025$. The ordinary 95% interval is $$ 0.015\pm1.96(0.025)=0.015\pm0.049, $$ or $[-0.034,0.064]$. The lower endpoint is impossible for a risk.

For the transformed interval, $g(0.015)=\log(0.015/0.985)\approx-4.184591$, and $0.015(0.985)=0.014775$. Thus the estimated logit-scale standard error is $0.025/0.014775\approx1.692047$. The margin is $1.96(1.692047)\approx3.316412$. The logit endpoints are approximately $-7.501003$ and $-0.868179$. Applying the expit function gives approximately $[0.00055,0.2956]$. This interval respects the risk bounds, but it is wide and asymmetric on the risk scale.

Worked Example: Comparing Scales for a Larger Risk

Let $\hat\psi=0.80$ and $\widehat{\operatorname{se}}(\hat\psi)=0.05$. The ordinary 95% interval has margin $1.96(0.05)=0.098$, giving $[0.702,0.898]$.

For the logit interval, $g(0.80)=\log(4)\approx1.386294$, while $0.80(0.20)=0.16$. The transformed standard error is $0.05/0.16=0.3125$, and its margin is $1.96(0.3125)=0.6125$. The logit endpoints are $1.386294-0.6125=0.773794$ and $1.386294+0.6125=1.998794$. Transforming back gives approximately $[0.6843,0.8807]$. Both intervals are within the parameter bounds, but they differ because one is symmetric on the risk scale and the other on the logit scale.

Assumptions, Diagnostics, and Alternatives

The logit construction addresses a range problem, not every inferential problem. It relies on an interior estimate and the same underlying large-sample normal approximation as the ordinary Wald interval. Near zero or one, $1/\{\hat\psi(1-\hat\psi)\}$ can be large, so even a modest risk-scale standard error produces a large transformed standard error. If the estimate is exactly zero or one, the logit is undefined. Applying an arbitrary small offset to make it finite changes the procedure and does not supply a general coverage guarantee.

Before reporting an interval, check the estimate, the estimated standard error, and the endpoints. A negative lower endpoint or upper endpoint above one is a clear sign that the ordinary symmetric interval poorly respects the parameter space. Also investigate an unusually large standard error, highly variable influence-curve values, or a point estimate at the boundary. Such features may reflect weak information or instability; transforming the interval cannot repair them. The detailed conditions for positivity and the stability of sequential updates are separate issues that must be assessed in the longitudinal analysis.

There are several approaches, each answering a slightly different practical need:

Ordinary Wald interval: simple and locally appropriate when the estimate is well inside the bounds and the normal approximation is credible. It can yield impossible endpoints.
Logit-scale Wald interval: enforces the zero-to-one range for an interior estimate and allows asymmetry on the risk scale. It can be unstable near the boundary and is undefined at a boundary estimate.
Clipping an ordinary interval to $[0,1]$: prevents impossible reported endpoints, but clipping does not derive a new confidence procedure and does not ensure nominal coverage.

These are pointwise intervals for a single risk. If several regime-specific risks are reported, separate 95% intervals do not generally provide 95% simultaneous coverage for all of them. Finally, a contrast such as a risk difference is not itself a risk confined to $[0,1]$; the logit transformation above is not an interval method for that contrast.

Worked Example: Detecting a Boundary Problem

Suppose a targeted estimate is $\hat\psi=0.97$ with standard error $0.03$. The ordinary margin is $1.96(0.03)=0.0588$, so the interval is $[0.9112,1.0288]$. Its upper endpoint exceeds one.

The logit estimate is $\log(0.97/0.03)\approx3.476099$. Since $0.97(0.03)=0.0291$, the transformed standard error is $0.03/0.0291\approx1.030928$, and the margin is approximately $2.020619$. The transformed endpoints are about $1.455480$ and $5.496718$; applying expit gives approximately $[0.811,0.996]$. The transformed interval is bounded but very broad. It should prompt scrutiny of the normal approximation and data support rather than be taken as proof that inference is reliable.

Common Mistake

A common mistake is to compute a standard Wald interval and then simply replace a negative endpoint by zero, claiming that the result is a valid 95% interval because risks cannot be negative. The parameter bound does not preserve the original interval’s coverage after this modification. A logit-scale interval is a principled alternative for an interior risk under its stated asymptotic conditions, but it too is not a cure for boundary estimates, unstable standard errors, or a poor normal approximation.

Key takeaway: Construct the interval using the estimated influence-curve standard error, state the large-sample conditions behind the normal approximation, and inspect its relationship to the risk bounds. Use a logit-scale interval as a bounded alternative only when the estimate is interior and the approximation is credible.

Check Your Understanding

Answer each question, then compare with the solutions below.

A risk estimate is $0.40$ with standard error $0.05$. Calculate its ordinary 95% Wald interval.
Why can clipping an ordinary interval to $[0,1]$ fail to preserve its nominal coverage?
For $\hat\psi=0.20$ and standard error $0.04$, calculate the logit-scale standard error.
What happens to the logit interval formula if the estimate is exactly zero?

Solutions. (1) The margin is $1.96(0.05)=0.098$, so the interval is $[0.302,0.498]$. (2) Clipping changes the interval endpoints without deriving the coverage probability of the modified procedure; the fact that the parameter is bounded alone does not establish nominal coverage. (3) Since $0.20(0.80)=0.16$, the logit-scale standard error is $0.04/0.16=0.25$. (4) The logit $g(0)=\log\{0/(1-0)\}$ is undefined, so this formula cannot be used at that estimate.