Discrete Pivots and Conservative Coverage

Why Discrete Pivots Need Careful Quantiles

For a continuous pivot, a chosen tail probability can often be matched exactly by a quantile. A discrete pivot has probability concentrated at individual values, so its cumulative distribution function jumps. A requested tail probability may fall between two attainable cumulative probabilities. The resulting confidence procedure can therefore cover more often than its nominal level. That is not a failure: the essential requirement for a conservative confidence procedure is coverage of at least the stated level.

The distinction matters when deriving intervals by pivot inversion. A quantile convention determines which value is selected at a jump, and that choice determines the interval and its repeated-sampling coverage. We will also distinguish a parameter-free discrete pivot from an exact interval obtained by inverting binomial tail probabilities; both use discreteness, but their constructions are not identical.

Definition: A discrete pivot is a statistic $Q(X,\theta)$ whose probability distribution is known and does not depend on the unknown parameter $\theta$, under the stated model. Its cumulative distribution function is $F_Q(q)=P(Q\le q)$. For $0<\gamma<1$, define its lower quantile by $$ q_\gamma=\inf\{q:F_Q(q)\ge \gamma\}. $$ For an integer-valued pivot, this is the smallest support value $q$ such that $F_Q(q)\ge\gamma$. Write $F_Q(q-)=P(Q<q)$ for the probability strictly below $q$.

Coverage From Two Discrete Quantiles

Suppose a model supplies a discrete pivot $Q(X,\theta)$, and the parameter values are found by inverting the event $q_L\le Q(X,\theta)\le q_U$. Choose positive tail allocations $\alpha_L$ and $\alpha_U$ with $\alpha_L+\alpha_U=\alpha<1$, and set $q_L=q_{\alpha_L}$ and $q_U=q_{1-\alpha_U}$. These quantile levels are strictly between zero and one, so the quantiles are finite for a proper discrete distribution on the real line.

C(X)=\{\theta:q_L\le Q(X,\theta)\le q_U\}, \qquad P_\theta(\theta\in C(X))=F_Q(q_U)-F_Q(q_L-).

The equality follows because the event $q_L\le Q\le q_U$ is the event $Q\le q_U$ with the disjoint event $Q<q_L$ removed. By the quantile definition, $F_Q(q_U)\ge1-\alpha_U$ and $F_Q(q_L-)<\alpha_L$. Consequently,

P_\theta(\theta\in C(X)) =F_Q(q_U)-F_Q(q_L-) \ge 1-\alpha_U-\alpha_L =1-\alpha.

This is the conservative-coverage result: the probability of covering the fixed true parameter is at least the target $1-\alpha$. In a discrete distribution, the inequality will generally be strict because a selected quantile can carry positive probability mass. The statement does not mean that the parameter itself is random; the probability is over repeated samples, as in the definition of coverage probability.

Important condition: For finite quantiles under the definition above, require $0<\alpha_L$, $0<\alpha_U$, and $\alpha_L+\alpha_U=\alpha<1$. A zero allocation would call for $q_0$ or $q_1$; these are not defined by this quantile convention, and $q_1$ need not be finite for an unbounded pivot. One-sided bounds use an allocation $\alpha$ satisfying $0<\alpha<1$.

Worked Example: A Two-Sided Interval From a Binomial-Noise Pivot

A sensor reports $X=\theta+Z$, where $\theta$ is an unknown integer and $Z\sim\operatorname{Binomial}(4,0.5)$ is independent measurement error. The observed report is $X=27$. Construct the equal-tailed interval using $\alpha_L=\alpha_U=0.10$, and calculate its actual coverage. This is a hypothetical measurement model; no interest rate or mortality assumption applies.

Method. Since $Q=X-\theta=Z$ has a distribution independent of $\theta$, it is a discrete pivot. The binomial probabilities for $Z=0,1,2,3,4$ are respectively $1/16,4/16,6/16,4/16,1/16$. Thus the cumulative probabilities are

F_Z(0)=0.0625,\quad F_Z(1)=0.3125,\quad F_Z(2)=0.6875,\quad F_Z(3)=0.9375,\quad F_Z(4)=1.

The lower quantile is $q_{0.10}=1$: the CDF at 0 is below 0.10, while the CDF at 1 is at least 0.10. The upper quantile is $q_{0.90}=3$: the CDF at 2 is below 0.90, while the CDF at 3 is at least 0.90.

To invert $1\le X-\theta\le3$, subtract $X$ and solve for $\theta$, reversing the inequalities when multiplying by $-1$:

1\le27-\theta\le3 \quad\Longleftrightarrow\quad 24\le\theta\le26.

The actual coverage is $P(1\le Z\le3)=(4+6+4)/16=14/16=0.875$. The target is $1-\alpha=0.80$, so coverage exceeds the target by $0.075$. The interval is $[24,26]$, and its coverage is 0.875. Probabilities are reported to four decimals.

Check and trap avoided. The excluded pivot values are $Z=0$ and $Z=4$, each with probability $1/16$, so coverage is $1-2/16=14/16$, agreeing with the direct calculation. A common mistake is to assume that choosing 10% in each tail makes each actual tail probability exactly 0.10. Here each actual tail is only 0.0625 because the CDF jumps over 0.10.

Why Coverage Has Gaps

As the requested confidence level changes, discrete quantiles do not necessarily change at every step. For example, for the binomial pivot in the previous calculation, the CDF moves from 0.0625 at 0 to 0.3125 at 1. Every probability level strictly between these values selects the same lower quantile, namely 1. That creates a range of nominal tail allocations with the same endpoint. Actual coverage is consequently a stepwise function of the requested level, not a smooth function that always equals it.

The tail calculation makes the overcoverage visible. With $q_L=q_{\alpha_L}$, the actual lower excluded probability is $F_Q(q_L-)$, which is strictly less than $\alpha_L$. With $q_U=q_{1-\alpha_U}$, the upper excluded probability is $1-F_Q(q_U)$, which is at most $\alpha_U$. The actual total probability outside the interval is therefore no more than $\alpha$. Probability mass at the selected boundary values is included in the interval, which is why coverage can exceed the target.

Specify the pivot and its distribution.
Verify that the distribution does not depend on the unknown parameter. If it does depend on the parameter, do not call it a parameter-free pivot; a tail-inversion method may still be available.

Choose valid tail allocations.
For a two-sided target $1-\alpha$, use $0<\alpha_L$, $0<\alpha_U$, and $\alpha_L+\alpha_U=\alpha<1$.

Find the discrete quantiles.
For each requested level, select the smallest support value whose CDF is at least that level. Do not interpolate between support values.

Invert the event and check coverage.
Pay attention to whether solving for the parameter reverses an inequality. Then calculate the probability mass included by the resulting pivot bounds.

Worked Example: A One-Sided Bound With an Unbounded Pivot

Suppose a hypothetical counter records $X=\theta+Z$, where $\theta$ is an unknown integer and $Z$ is the number of failures before a success in independent Bernoulli trials with success probability $0.5$. Thus $P(Z=k)=0.5^{k+1}$ for $k=0,1,\ldots$. If the observed count is $X=20$, find a one-sided lower confidence bound for $\theta$ with coverage at least 95%. No interest rate or mortality assumption applies.

Method. The pivot is $Q=X-\theta=Z$. Its CDF at a nonnegative integer $k$ is the sum of its probabilities from 0 through $k$:

F_Z(k)=\sum_{j=0}^{k}0.5^{j+1}=1-0.5^{k+1}.

For a 95% lower bound, find $q_{0.95}$. At $k=3$, $F_Z(3)=1-1/16=0.9375$, which is below 0.95. At $k=4$, $F_Z(4)=1-1/32=0.96875$, which is at least 0.95. Hence $q_{0.95}=4$.

The event $Z\le4$ is equivalent to $X-\theta\le4$, or $\theta\ge X-4$. The observed lower confidence bound is therefore $20-4=16$, giving $[16,\infty)$. Its coverage is $P(Z\le4)=0.96875$, which is at least 0.95. The answer is $\theta\ge16$, with coverage 0.9688 to four decimals.

Check and trap avoided. The next lower quantile, $q_{0.95}$, is finite because the requested level is below 1, even though the pivot has unbounded support. Do not substitute a level of 1 for a zero tail allocation: this distribution has no finite support value where its CDF equals 1. The selected quantile also gives conservative rather than exactly 95% coverage.

Exact Binomial Bounds by Tail Inversion

A binomial count $X\sim\operatorname{Binomial}(n,p)$ is not itself a parameter-free pivot: its distribution depends on $p$. Nevertheless, one can obtain a conservative lower confidence bound by inverting an upper-tail test. Assume $n$ independent Bernoulli trials with a common, fixed success probability $p$. For an observed count $x\ge1$, define $p_L(x)$ as the solution to

P_{p_L(x)}(X\ge x)=\alpha.

The binomial upper-tail probability is increasing in $p$, so this equation identifies a unique bound between 0 and 1 for $1\le x\le n$. Set $p_L(0)=0$. The resulting procedure has coverage at least $1-\alpha$. To see why, fix the true $p$. The event that the bound exceeds $p$ is $p_L(X)>p$. By the definition of the bound and the increasing binomial tail, this can occur only for counts $X$ whose upper-tail probability under the true $p$ is less than $\alpha$. Those counts form an upper tail of the count distribution, and that tail has probability less than $\alpha$. Thus the probability of failing to cover is at most $\alpha$.

Worked Example: A Lower Bound After Three Successes

A hypothetical quality audit samples three components independently, with each component passing with probability $p$. All three pass. Find the one-sided lower confidence bound obtained by tail inversion when $\alpha=0.125$. This is a binomial model with $n=3$; no interest rate or mortality assumption applies.

Method. With $X\sim\operatorname{Binomial}(3,p)$ and observed $x=3$, the required bound solves the probability of three successes equal to $\alpha$:

P_p(X\ge3)=P_p(X=3)=p^3=0.125.

Taking cube roots gives $p_L=0.5$. The resulting confidence set is $[0.5,1]$, with coverage at least $1-\alpha=0.875$ for every true $p$. The reported lower endpoint is $0.5000$ to four decimals.

Check and trap avoided. Substitution verifies $0.5^3=0.125$, as required. The equation uses the probability of $X\ge x$, not $P(X\le x)$: the observed result is an unusually high count when $p$ is small, so it is the upper tail that determines this lower bound.

Common Mistakes and Practical Interpretation

Assuming nominal coverage is exact. For a discrete pivot, calculate $F_Q(q_U)-F_Q(q_L-)$ if actual coverage is requested. Quantile levels alone do not give the exact included probability.
Using the wrong boundary probability. The probability strictly below the lower endpoint is $F_Q(q_L-)$, not $F_Q(q_L)$. The mass at $q_L$ is included in the interval.
Interpolating a discrete quantile. Use the smallest support value at which the CDF reaches the requested level. A value between support points is not a pivot quantile under this convention.
Allowing a zero tail allocation without defining endpoints. For finite quantiles, keep both two-sided allocations strictly positive. A zero allocation can require an undefined or infinite endpoint.
Calling every parameter-dependent statistic a pivot. A binomial count depends on $p$; its confidence bound above comes from tail inversion, not from a parameter-free pivot distribution.

Key takeaway: Discrete quantiles select attainable support values, not arbitrary probability levels. Invert the selected event carefully, then verify that its probability is at least the nominal coverage. The resulting interval or bound may be conservative because jumps in the CDF leave gaps between attainable coverage levels.

Check Your Understanding

For each problem, use the stated quantile convention and check the relevant tail probability.

Let $X=\theta+Z$, where $\theta$ is an integer and $Z\sim\operatorname{Binomial}(3,0.5)$. If $X=12$, find the equal-tailed interval using $\alpha_L=\alpha_U=0.10$, and calculate its actual coverage. (Answer: $q_L=0,\ q_U=3$, interval $[9,12]$, actual coverage $1.0000$.)
Let $X=\theta+Z$, where $Z$ is geometric with success probability $0.5$, counting failures before the first success. For observed $X=14$, find a one-sided lower bound with at least 90% coverage. (Answer: $q_{0.90}=3$, bound $\theta\ge11$, actual coverage $0.9375$.)
For $X\sim\operatorname{Binomial}(2,p)$, two successes are observed. Find the tail-inversion lower bound when $\alpha=0.04$. (Answer: solve $p_L^2=0.04$, giving $p_L=0.2000$ and confidence set $[0.2000,1]$.)
A pivot $Q$ has support $-1,0,2,5$ with respective probabilities $0.15,0.35,0.30,0.20$. Use $\alpha_L=0.20$ and $\alpha_U=0.10$. Find the quantile endpoints and actual coverage of the pivot event. (Answer: $q_L=0,\ q_U=5$, actual coverage $0.8500$, at least the nominal $0.7000$.)
For a two-sided discrete-pivot interval with total error probability $\alpha<1$, is it valid under the finite-quantile definition to set $\alpha_L=0$ and use $q_{\alpha_L}$? Explain the correction. (Answer: No; $q_0$ is undefined under the stated convention. Require $0<\alpha_L$, $0<\alpha_U$, and $\alpha_L+\alpha_U=\alpha$, or explicitly define an extended endpoint.)

What You'll Learn

Why Discrete Pivots Need Careful Quantiles

Coverage From Two Discrete Quantiles

Worked Example: A Two-Sided Interval From a Binomial-Noise Pivot

Worked Example: A Two-Sided Interval From a Binomial-Noise Pivot

Why Coverage Has Gaps

Worked Example: A One-Sided Bound With an Unbounded Pivot

Worked Example: A One-Sided Bound With an Unbounded Pivot

Exact Binomial Bounds by Tail Inversion

Worked Example: A Lower Bound After Three Successes

Common Mistakes and Practical Interpretation

Check Your Understanding

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