Confidence Sets From Accepted Pivot Values

From Accepted Pivot Values to a Confidence Set

A confidence interval is one possible result of inverting a probability statement about a pivot. The inversion does not automatically produce an interval: its shape depends on the accepted pivot values and on how the pivot changes with the parameter. A careful solution therefore has two parts: establish the probability of the accepted region, then solve the resulting inequalities for the parameter.

This viewpoint is useful whenever the sampling distribution provides a pivot but the desired confidence set is not already in a familiar formula. It also prevents a common error: treating a confidence level as though it were a probability statement about the parameter after the data have been observed.

Definition: A pivot is a statistic $Q(X,\theta)$, involving the data $X$ and an unknown parameter $\theta$, whose probability distribution does not depend on $\theta$. If a set $A$ of pivot values satisfies $P_\theta(Q(X,\theta)\in A)=1-\alpha$ for every allowed $\theta$, the inverted confidence set is $$ C(X)=\{\theta:Q(X,\theta)\in A\}. $$ It has coverage $1-\alpha$, because the event $\theta\in C(X)$ is exactly the event $Q(X,\theta)\in A$.

The probability in this definition is calculated under repeated sampling with $\theta$ fixed. Once the data are observed, $C(X)$ is a set of parameter values; the coverage statement does not say that the fixed parameter has probability $1-\alpha$ of lying in that realized set.

Why the Inversion Has Its Stated Coverage

Let $X$ have a sampling distribution indexed by $\theta$, and suppose the pivot $Q(X,\theta)$ has the same distribution for every allowed value of $\theta$. Choose an accepted-value region $A$ whose probability under that distribution is $1-\alpha$. The set of parameter values consistent with the observed pivot being accepted is $C(X)$. For any fixed true value $\theta_0$, membership of $\theta_0$ in $C(X)$ is equivalent to acceptance of the pivot calculated at $\theta_0$.

\begin{aligned} P_{\theta_0}\{\theta_0\in C(X)\} &=P_{\theta_0}\{Q(X,\theta_0)\in A\}\\ &=1-\alpha. \end{aligned}

The first equality follows from the definition of $C(X)$; the second follows from the pivot distribution and the probability assigned to $A$. This establishes exact coverage when the probability is exactly $1-\alpha$. If the pivot is discrete, an available accepted region may have probability greater than the target, giving conservative coverage; the set inversion is the same, but its coverage is then at least the target rather than necessarily equal to it.

Worked Example: A Connected Accepted Region Gives an Interval

Suppose one observation $X$ has the model $X\sim\operatorname{Uniform}(\theta-1,\theta+1)$, with $\theta\in\mathbb{R}$. Thus $Q(X,\theta)=X-\theta$ is uniform on $[-1,1]$, regardless of $\theta$. Use the accepted pivot region $A=[-0.9,0.9]$, and suppose the observed value is $x=4.2$. Find and check the confidence set.

Method and acceptance probability. The interval $A$ has length $1.8$, while the full pivot range has length $2$. Since the pivot is uniform, $$ P_\theta\{Q(X,\theta)\in A\}=\frac{1.8}{2}=0.9. $$ The target coverage is therefore 90%.

Invert the region. Substitute the observed value into the pivot and solve both inequalities: $$ -0.9\leq 4.2-\theta\leq 0.9. $$ The left inequality gives $\theta\leq 5.1$, and the right inequality gives $\theta\geq 3.3$. Hence the confidence set is $[3.3,5.1]$.

Coverage check. For any true $\theta$, this set contains $\theta$ exactly when $-0.9\leq X-\theta\leq0.9$. That event has probability $0.9$ under the stated model. The result is a single interval because the accepted pivot values form an interval and $X-\theta$ is monotone in $\theta$. The common mistake avoided here is reversing an inequality without reversing its direction: subtracting $\theta$ requires solving each bound explicitly.

When Is the Inverted Set an Interval?

A confidence set is an interval when the accepted pivot values pull back to a connected set of parameter values. A useful sufficient condition is that the allowed parameter space is an interval, the accepted pivot region is an interval, and, for each fixed observed data value, $Q(x,\theta)$ is continuous and monotone in $\theta$. The inverse image of an interval under such a function is an interval, possibly empty or with an endpoint excluded depending on the region and parameter space.

These conditions are sufficient, not necessary. A non-monotone pivot can still yield an interval for a particular accepted region or data value. Conversely, a disconnected accepted region can produce a disconnected confidence set even when the pivot is monotone. The right procedure is to solve the inequalities and inspect the resulting set rather than assume its shape from the word “confidence interval.”

Key distinction: A confidence set is the general term for the parameter values retained by inversion. It is a confidence interval only when those values form one interval. Disconnected accepted pivot values, non-monotone dependence on the parameter, or parameter constraints can affect the shape.

Worked Example: A Disconnected Accepted Region

Keep the model $X\sim\operatorname{Uniform}(\theta-1,\theta+1)$, with $\theta\in\mathbb{R}$, so $Q(X,\theta)=X-\theta\sim\operatorname{Uniform}(-1,1)$. Now accept pivot values in $A=[-1,-0.1]\cup[0.1,1]$, and suppose $x=0.4$. Determine the confidence set and its coverage.

Method and acceptance probability. Each of the two pieces has length $0.9$, so their combined length is $1.8$. The pivot’s full range has length $2$, giving $$ P_\theta\{Q(X,\theta)\in A\}=\frac{0.9+0.9}{2}=0.9. $$ Thus inversion gives a 90% confidence set.

Invert the negative piece. Solve $-1\leq0.4-\theta\leq-0.1$. The first inequality implies $\theta\leq1.4$; the second implies $\theta\geq0.5$. This piece gives $[0.5,1.4]$.

Invert the positive piece. Solve $0.1\leq0.4-\theta\leq1$. The first inequality implies $\theta\leq0.3$; the second implies $\theta\geq-0.6$. This piece gives $[-0.6,0.3]$.

The full confidence set is therefore $[-0.6,0.3]\cup[0.5,1.4]$. It is not one interval: the values between $0.3$ and $0.5$ are excluded. For any fixed true $\theta$, containment is equivalent to the pivot falling in the two accepted pieces, an event with probability $0.9$. The common mistake avoided is replacing a union of accepted regions by the single interval between its outermost endpoints; doing so would accept pivot values in the gap and change the coverage.

Solving with a Ratio Pivot and a Parameter Constraint

The direction of an inequality can change when the unknown parameter appears in a denominator. In such cases, first state the parameter’s allowable range and positivity assumptions, then solve the pivot inequalities. Intersect the result with the parameter space, even if the resulting bounds appear to imply the constraint automatically.

Worked Example: Inverting a Sample-Maximum Pivot

Let $X_1,X_2$ be independent observations from $\operatorname{Uniform}(0,\theta)$, where $\theta>0$, and let $M=\max(X_1,X_2)$. The pivot $R=M/\theta$ has distribution $P_\theta(R\leq r)=r^2$ for $0\leq r\leq1$. Use $A=[0.25,1]$. If the observed maximum is $m=1.6$, find the confidence set and its coverage.

Method and acceptance probability. The pivot is always between 0 and 1. Its probability of falling in $A$ is $$ P_\theta(0.25\leq R\leq1)=1-P_\theta(R<0.25) =1-(0.25)^2=1-0.0625=0.9375. $$ The set will have 93.75% coverage.

Invert the accepted values. Because $\theta>0$, the inequality $0.25\leq1.6/\theta\leq1$ can be solved without changing inequality direction when multiplying by $\theta$: $$ 0.25\theta\leq1.6\leq\theta. $$ The right inequality gives $\theta\geq1.6$; the left gives $\theta\leq1.6/0.25=6.4$. The confidence set is $[1.6,6.4]$, which also respects the model constraint $\theta>0$ and the observed-data requirement $\theta\geq M$.

Coverage check. For any true $\theta$, the event that the set contains $\theta$ is $0.25\leq M/\theta\leq1$. The upper inequality always holds because the sample maximum cannot exceed $\theta$; the lower inequality has probability $1-(0.25)^2=0.9375$. The result agrees with the pivot calculation. A common trap is multiplying an inequality by a negative parameter or forgetting its sign; here $\theta>0$ is essential to preserving the directions.

A Practical Inversion Checklist

For an unfamiliar pivot, separate the probability calculation from the algebra. That makes both the intended coverage and the source of any interval endpoints transparent.

State the model and pivot.
Specify the sampling assumptions, parameter space, and the parameter-free distribution of $Q(X,\theta)$.

Choose accepted values.
Write the region $A$ and calculate its probability under the pivot distribution. Do not infer its probability just from its appearance unless the pivot is uniform.

Substitute the observed data.
Set $Q(x,\theta)\in A$, split unions into separate inequalities, and solve with the sign and monotonicity of each expression in view.

Apply the parameter space and check coverage.
Intersect with the allowed parameter values, inspect whether the answer is connected, and verify that containment of the true parameter is the original acceptance event.

This method clarifies what the chosen accepted region controls. Two regions can have the same probability, and hence the same coverage, but produce sets with different shapes and widths. Choosing among procedures according to width or other criteria is a separate question; first ensure the inversion itself is correct.

Do not assume every confidence set is an interval. Keep unions and gaps when the inequalities produce them.
Do not lose the parameter constraint. Check positivity, endpoint restrictions, or other stated limits after solving.
Do not confuse coverage with a probability for the realized parameter. Coverage refers to repeated samples with the parameter fixed.
Do not skip the coverage check. Re-expressing containment as pivot acceptance is a direct way to catch an algebraic error.

Key takeaway: Start with a pivot region of known probability, then invert its inequalities exactly. The probability of acceptance gives the coverage; the algebra and the parameter space determine whether the resulting confidence set is a single interval.

Check Your Understanding

For each problem, identify the accepted pivot probability, invert the region, and note whether the result is one interval.

Suppose $X\sim\operatorname{Uniform}(\theta-2,\theta+2)$, $\theta\in\mathbb{R}$, so $Q=X-\theta\sim\operatorname{Uniform}(-2,2)$. Accept $[-1.5,1.5]$. For observed $x=7$, find the confidence set. (Answer: $[5.5,8.5]$, coverage $0.75$.)
Suppose $X\sim\operatorname{Uniform}(\theta-1,\theta+1)$, $\theta\in\mathbb{R}$, and accept $Q=X-\theta$ in $[-1,-0.2]\cup[0.2,1]$. For $x=-0.3$, find the set and coverage. (Answer: $[-1.3,-0.5]\cup[-0.1,0.7]$, coverage $0.8$.)
Let $X_1,X_2,X_3$ be independent $\operatorname{Uniform}(0,\theta)$ observations, $\theta>0$, and $M=\max(X_1,X_2,X_3)$. Accept $M/\theta\in[0.5,1]$. If $M=2$, find the confidence set and coverage. (Answer: $[2,4]$, coverage $0.875$.)
Suppose $X\sim\operatorname{Uniform}(\theta-1,\theta+1)$, $\theta\in\mathbb{R}$, and use the pivot $Q=|X-\theta|$, which is uniform on $[0,1]$. Accept $Q\in[0,0.8]$. For $x=3$, find the set and coverage. (Answer: $[2.2,3.8]$, coverage $0.8$.)
For two independent observations from $\operatorname{Uniform}(0,\theta)$, let $M$ be their maximum and accept $M/\theta\in[0.2,0.8]$. If $M=3$, find the confidence set and coverage. (Answer: $[3.75,15]$, coverage $0.8^2-0.2^2=0.60$.)

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