Exact Confidence Limits for Logistic Effects

From Exact Tests to Confidence Limits

An exact conditional test answers whether a proposed logistic slope is compatible with the observed conditional data. A confidence interval collects the slope values that are not rejected. The key idea is therefore test inversion: calculate conditional tail probabilities across candidate slopes, then retain the values whose tests do not reject at the chosen level.

The resulting limits are exact in the sense that their coverage is guaranteed under the conditional model, without a large-sample approximation. Because the conditional statistic is discrete, coverage will often be greater than the nominal level, and an endpoint can be infinite when the observed statistic lies at the edge of its possible range.

The Conditional Distribution for a Slope

Consider a logistic-regression model with a scalar slope of interest $\beta$, together with nuisance parameters. After conditioning on sufficient statistics for the nuisance parameters, suppose the possible outcomes form a finite set $\Omega_s$. Let $T(y)$ be the statistic associated with $\beta$. The conditional model takes the form

P_\beta(Y=y\mid S=s) = \frac{w_s(y)\exp\{\beta T(y)\}} {\displaystyle\sum_{z\in\Omega_s}w_s(z)\exp\{\beta T(z)\}}, \qquad y\in\Omega_s,

where $w_s(y)>0$ does not depend on $\beta$, and $S$ denotes the conditioned nuisance statistics. This is a finite one-parameter exponential family. In logistic regression, conditioning on the appropriate sufficient statistics eliminates nuisance intercepts; for a slope with other nuisance slopes, those nuisance sufficient statistics must also be handled by the conditioning scheme. The conditional sample space must retain variation in $T$, or the data contain no conditional information about $\beta$.

Definition: An equal-tailed exact conditional confidence set with nominal confidence level $1-\alpha$ retains a candidate value $\beta_0$ when both conditional tail probabilities, evaluated under $\beta_0$ at the observed statistic $t_{\mathrm{obs}}$, are at least $\alpha/2$: $$ P_{\beta_0}(T\ge t_{\mathrm{obs}}\mid S=s)\ge\alpha/2 \quad\text{and}\quad P_{\beta_0}(T\le t_{\mathrm{obs}}\mid S=s)\ge\alpha/2. $$

The inclusive tails are important: the observed value is included in each tail. The confidence set is reported on the slope scale, or exponentiated to give limits for the odds ratio $\psi=\exp(\beta)$. The two scales tell the same inferential story, but their endpoints are different numerical quantities.

Why Tail Probabilities Give Ordered Limits

The upper-tail probability increases as the candidate slope increases, while the lower-tail probability decreases. Consequently, the retained values of $\beta_0$ form an interval, possibly with one or both endpoints infinite.

Result: In a finite conditional exponential family with positive weights, $P_\beta(T\ge t_{\mathrm{obs}}\mid S=s)$ is nondecreasing in $\beta$, and $P_\beta(T\le t_{\mathrm{obs}}\mid S=s)$ is nonincreasing in $\beta$.

Proof. Write $p_\beta(y)=P_\beta(Y=y\mid S=s)$. Since the support is finite, differentiating the normalizing sum term by term is valid. For any function $g$ of the conditional outcome,

\frac{d}{d\beta}E_\beta[g(Y)\mid S=s] = E_\beta[g(Y)T(Y)\mid S=s] - E_\beta[g(Y)\mid S=s]E_\beta[T(Y)\mid S=s] = \operatorname{Cov}_\beta(g(Y),T(Y)\mid S=s).

For the upper tail take $g(Y)=\mathbf{1}\{T(Y)\ge t_{\mathrm{obs}}\}$. This indicator is a nondecreasing function of $T$. If $T'$ is an independent copy of $T$, then

2\operatorname{Cov}(g(T),T) = E[(g(T)-g(T'))(T-T')]\ge 0,

because the two differences always have the same sign or one is zero. Thus the derivative of the upper-tail probability is nonnegative. For the lower tail, $g(T)=\mathbf{1}\{T\le t_{\mathrm{obs}}\}$ is nonincreasing, so the product in the covariance identity is nonpositive and the derivative is nonpositive. This proves the result. $\square$

Define the lower confidence limit by solving the upper-tail equation at $\alpha/2$, and the upper confidence limit by solving the lower-tail equation at $\alpha/2$, whenever the equations have finite solutions. Monotonicity ensures that all retained values lie between these limits. If a tail never reaches $\alpha/2$ at a finite slope, the corresponding confidence limit is infinite.

Conditional Coverage

The confidence set has at least its stated coverage conditional on the statistics used to define the conditional sample space. The inequality may be strict because the test statistic is discrete.

Theorem: Suppose the conditional sample space is finite and the true slope is $\beta$. Inverting the two inclusive one-sided tests at tail level $\alpha/2$ gives $$ P_\beta\{\beta\text{ is in the confidence set}\mid S=s\}\ge 1-\alpha. $$

Proof. At the true $\beta$, define the upper-tail probability at the observed statistic by $U=P_\beta(T\ge T_{\mathrm{obs}}\mid S=s)$. The event $U<\alpha/2$ is an upper-tail rejection event. If it is nonempty, the finite ordered support has a smallest value $t_*$ for which $P_\beta(T\ge t_*\mid S=s)<\alpha/2$. The event $U<\alpha/2$ is then contained in $\{T_{\mathrm{obs}}\ge t_*\}$, whose probability is $P_\beta(T\ge t_*\mid S=s)<\alpha/2$. If the event is empty, its probability is zero. Hence $P_\beta(U<\alpha/2\mid S=s)\le\alpha/2$. The same argument, ordering the support in the opposite direction, gives probability at most $\alpha/2$ for the lower-tail rejection event.

The true value is excluded only if at least one of these two tests rejects. By the union bound, the conditional probability of exclusion is at most $\alpha/2+\alpha/2=\alpha$. Therefore conditional coverage is at least $1-\alpha$. $\square$

Worked Conditional Limits

Worked Example: A Boundary Table and an Infinite Limit

In a two-by-two table, condition on both row and column totals, each equal to 2, with grand total 4. Let $X$ be the upper-left cell count. Its support is $0,1,2$, and the conditional probabilities are proportional to

\binom{2}{x}\binom{2}{2-x}\psi^x, \qquad \psi=\exp(\beta).

Thus the probabilities have weights $1,4\psi,\psi^2$, with normalizing constant $D=1+4\psi+\psi^2$. Suppose the observed value is $X=2$, and construct a 95% equal-tailed interval, so $\alpha/2=0.025$.

The lower limit solves $P_\psi(X\ge2)=0.025$:

\frac{\psi^2}{1+4\psi+\psi^2}=0.025 \quad\Longrightarrow\quad 0.975\psi^2-0.1\psi-0.025=0.

The positive root is $\psi=(0.1+\sqrt{0.1075})/1.95\approx0.2194$, giving a lower slope limit $\log(0.2194)\approx-1.52$. For the upper limit, the required lower tail is $P_\psi(X\le2)$, which equals 1 for every $\psi>0$. It never falls to 0.025, so the upper odds-ratio limit is infinite. The interval is approximately $[0.2194,\infty)$ on the odds-ratio scale, or $[-1.52,\infty)$ on the slope scale.

Worked Example: Interior Observation in the Same Table

Use the same conditional distribution, but suppose the observed count is $X=1$. The lower odds-ratio limit solves $P_\psi(X\ge1)=0.025$:

\frac{4\psi+\psi^2}{1+4\psi+\psi^2}=0.025 \quad\Longrightarrow\quad 0.975\psi^2+3.9\psi-0.025=0.

The positive root is $\psi=(-3.9+\sqrt{15.3075})/1.95\approx0.00640$.

The upper limit solves $P_\psi(X\le1)=0.025$:

\frac{1+4\psi}{1+4\psi+\psi^2}=0.025 \quad\Longrightarrow\quad 0.025\psi^2-3.9\psi-0.975=0.

Its positive root is $\psi=(3.9+\sqrt{15.3075})/0.05\approx156.25$. The 95% conditional interval is therefore approximately $[0.00640,156.25]$, or $[-5.05,5.05]$ for $\beta=\log(\psi)$. The very wide interval is a consequence of the small conditional sample space: observing one count provides limited precision even though the limits are calculated exactly.

Worked Example: Matched Pairs with No Exposed Case

Suppose there are two matched pairs, each with one discordant pair member exposed and the other unexposed, and exactly one case in each pair. Let $T$ count pairs in which the exposed member is the case. Conditional on the pair totals, each pair contributes a factor $1+\psi$, so

P_\psi(T=t)=\frac{\binom{2}{t}\psi^t}{(1+\psi)^2}, \qquad t=0,1,2.

If the observed value is $T=0$, the lower odds-ratio limit is infinite toward zero: $P_\psi(T\ge0)=1$ for every $\psi>0$, so there is no finite lower limit. The upper limit solves

P_\psi(T\le0)=\frac{1}{(1+\psi)^2}=0.025.

Therefore $1+\psi=1/\sqrt{0.025}=\sqrt{40}$, and $\psi=\sqrt{40}-1\approx5.325$. The interval is $(0,5.325]$ on the odds-ratio scale, equivalently $(-\infty,1.672]$ on the slope scale. The endpoint at zero is a limiting boundary: a finite logistic slope has $\psi>0$, but the confidence set has no positive lower bound.

Interpreting Discreteness and Limits

An exact interval is obtained from the conditional distribution actually specified by the model and conditioning scheme. Since a discrete tail probability changes in jumps as the observed statistic changes, the coverage guarantee is generally “at least” the nominal level, not exactly equal to it. The resulting interval can also look much wider than a Wald interval; that is not a computational defect but a reflection of the information in the conditional sample space.

A finite endpoint is usually found by numerical root-finding on a monotone tail probability. It need not be a tail probability that equals the target exactly at every attainable observation if a different exact-test convention is chosen; the interval must be reported with its tail convention made clear. Here the specified convention uses inclusive tails and equal tail allocation. Exponentiating slope limits preserves their ordering and produces odds-ratio limits, but an infinite slope endpoint becomes an odds-ratio endpoint of zero or infinity.

A more fundamental limitation occurs if conditioning leaves only one possible value of $T$. Then the conditional distribution does not depend on $\beta$, both inclusive tail probabilities at the observed value equal 1, and inversion retains every finite $\beta$. The conditional data provide no information about that slope. This is distinct from a boundary observation: in the examples above there is still variation in the conditional sample space, even though one confidence limit is unbounded.

Common Mistake

Wrong reasoning: “The conditional test is exact, so the 95% interval must have exactly 95% coverage and two finite endpoints.” Exactness guarantees coverage of at least the nominal level under the stated conditional model; discreteness can make it conservative, and boundary observations can produce an infinite endpoint.
Correction: Inspect the support and calculate both inclusive tails. A tail that cannot reach $\alpha/2$ has no finite root, so report the corresponding limit as unbounded rather than inventing a finite estimate or substituting a Wald limit.

Key takeaway: Exact conditional confidence limits for a logistic slope are obtained by inverting its conditional tail tests. Monotone tails give an interval, finite-sample discreteness can make it conservative, and boundary or uninformative conditional samples can yield unbounded limits.

Check Your Understanding

Use the conditional distribution and tail convention stated above.

Why does the conditional distribution need variation in $T$ to contain information about the slope?
For a candidate slope, which inclusive tail probability increases as the slope increases, and which decreases?
Why is the conditional coverage at least, rather than necessarily exactly, 95%?
In the two-by-two table example with $X=2$, why is the upper odds-ratio limit infinite?
What does an interval $(-\infty,b]$ for a slope become on the odds-ratio scale?