Computation of exact confidence limits from discrete data

Abstract

Suppose that the data have a discrete distribution determined by (∞, ψ) where θ is the scalar parameter of interest and ψ is a nuisance parameter vector. The Buehler 1 - α upper confidence limit for θ is as small as possible, subject to the constraints that (a) its coverage probability is at least 1 - α and (b) it is a nondecreasing function of a pre-specified statisticT. This confidence limit has important biostatistical and reliability applications. The main result of the paper is that for a wide class of models (including binomial and Poisson), parameters of interest 9 and statisticsT (which possess what we call the “logical ordering” property) there is a dramatic increase in the ease with which this upper confidence limit can be computed. This result is illustrated numerically for θ a difference of binomial probabilities. Kabaila & Lloyd (2002) also show that ifT is poorly chosen then an assumption required for the validity of the formula for this confidence limit may not be satisfied. We show that for binomial data this assumption must be satisfied whenT possesses the “logical ordering” property.

Appendices

Appendix A: Proof of Theorem 3.1

Proof of Theorem 3.1 (a).

If j is such that b(λ) is a nonincreasing and nonconstant function of λ_j (where λ = (λ₁,…, λ_m)) then define $\tilde{Y}_{j}=-Y_{j}$, $\tilde{y}_{j}=-y_{j}$ and $\tilde{\lambda}_{j}=-\lambda_{j}$; otherwise define $\tilde{Y}_{j}=Y_{j}$, $\tilde{y}_{j}=y_{j}$ and $\tilde{\lambda}_{j}=\lambda_{j}(j=1, \ldots, m)$. Let $\tilde{F}_{i}\left(\cdot | \tilde{\lambda}_{i}\right)$ denote the distribution function of $\tilde{Y}_{i}(i=1, \ldots, m)$. Also let $\tilde{\lambda}=\left(\tilde{\lambda}_{1}, \ldots, \tilde{\lambda}_{m}\right)$ and $\tilde{y}=\left(\tilde{y}_{1}, \ldots, \tilde{y}_{m}\right)$. Define $\tilde{b}(\tilde{\lambda})$ to be that function of $\tilde{\lambda}$ such that $\tilde{b}(\tilde{\lambda})=b(\lambda)$ for all λ ∈ A. Also define $\tilde{t}(\tilde{y})$ to be that function of $\tilde{y}$ such that $\tilde{b}(\tilde{y})=b(y)$ for all $y \in \mathcal{Y}_{1} \times \cdots \times \mathcal{Y}_{m}$.

$\tilde{Y}_{1}, \ldots, \tilde{Y}_{m}$ are independent. For each y ∈ R, $\tilde{F}_{i}\left(y | \tilde{\lambda}_{i}\right)$ is a nonincreasing function of $\tilde{\lambda}_{i}(i=1, \ldots, m)$. Also, $\tilde{b}(\tilde{\lambda})$ and $\tilde{t}(\tilde{y})$ are nondecreasing in each of the components of $\tilde{\lambda}=\left(\tilde{\lambda}_{1}, \ldots, \tilde{\lambda}_{m}\right)$ and $\tilde{y}=\left(\tilde{y}_{1}, \ldots, \tilde{y}_{m}\right)$ respectively.

Proof of Theorem 3.1 (b).

By condition B2, t(y₁,…, y_m) is a nondecreasing function of y₁,…, y_m. Fix $y \in \mathcal{Y}$ and i ∈ {1,…,m}. Let $\mathcal{C}_{i}$ denote the set of (y₁,…, y_{i‒ 1}, y_{i +1}, …, y_m)’s, where $y_{j} \in \mathcal{Y}_{j}$ (j = 1,…,m), such that $\left(y_{1}, \ldots, y_{i-1}, y_{i}^{*}, y_{i+1}, \ldots, y_{m}\right) \in D(y)$ for some $y_{i}^{*} \in \mathcal{Y}_{i}$. Fix $\left(y_{1}, \ldots, y_{i-1}, y_{i+1}, \dots, y_{m}\right) \in \mathcal{C}_{i}$. Observe that by condition B2 the set of all $z \in \mathcal{Y}_{i}$ such that $\left(y_{1}, \dots, y_{i-1}, z, y_{i+1}, \dots, y_{m}\right) \in D(y)$ is the set of all $z \in \mathcal{Y}_{i}$ less than or equal to some $k \in \mathcal{Y}_{i}$. Define G to be the set $\left\{\left(y_{1}, \ldots, y_{i-1}, z, y_{i+1}, \ldots, y_{m}\right) : z \in \mathcal{Y}_{i}, z \leq k\right\}$, It now follows from Assumption A that P(Y ∈ G ∣ λ) is a nonincreasing function of λ_i for each fixed (λ₁,…, λ_{i ‒1}, y_{i +1}, …, y_m) This argument applies for each fixed i and each fixed element of $\mathcal{C}_{i}$. Hence P(Y ∈ D(y) ∣ λ) is a nonincreasing function of λ_i (i = 1,…, m).

Fix θ₁, θ₂ ∈ B satisfying θ₁ > θ₂. By condition C, there exists a function d such that for every λ = (λ₁,…, λ_m) ∈ A satisfying b(λ) = θ₁, $\lambda^{*}=\left(\lambda_{1}^{*}, \ldots, \lambda_{m}^{*}\right)=d(\lambda)$ satisfies $\lambda_{i}^{*} \leq \lambda_{i}(i=1, \ldots, m)$ and b(λ*) = θ₂. For each λ ∈ A satisfying b(λ) = θ₁,

$$P(Y \in D(y) | \lambda) \leq P(Y \in D(y)) | d(\lambda) ).$$

Hence

$$\begin{array}{l}{\sup \left\{P(Y \in D(y) | \lambda) : b(\lambda)=\theta_{1}, \lambda \in A\right\}} \\ { \leq \sup \left\{P(Y \in D(y) | d(\lambda)) : b(\lambda)=\theta_{1}, \lambda \in A\right\}} \\ { \leq \sup \left\{P(Y \in D(y) | \lambda) : b(\lambda)=\theta_{2}, \lambda \in A\right\}}.\end{array}$$

In other words, sup_ψ P(T ≤ t(y) ∣ θ, ψ) is a nonincreasing function of θ. This argument applies for every $y \in \mathcal{Y}$.

Appendix B: Proof of Theorem 3.2

We may suppose, without loss of generality, that b(λ) is nondecreasing in each of the components of λ = (λ₁,…, λ_m). This is because if b(λ) is nonincreasing in λ_j then n_j ‒ ∼ Binomial(n_j, 1 ‒ λ_j) and b(λ) is nondecreasing in 1 ‒ λ_j. Define y* = (0,…, 0). Since T = t(Y) satisfies the “logical ordering” property, t(y) achieves its minimum value at y = y*. Thus

$$P(T \leq t(y) | \lambda) \geq P\left(Y=y^{*} | \lambda\right)=\prod_{i=1}^{m}\left(1-\lambda_{i}\right)^{n_{i}}$$

In some cases, such as θ = λ₁λ₂, we need to disallow some boundary values of λ_i so that (θ, ψ) is a one-one function of λ = (λ₁,…, λ_m). We therefore restrict attention to λ = (λ₁,…, λ_m) ∈ (0, 1)^m. Clearly, we may choose λ = (λ₁,…, λ_m) ∈ (0, 1)^m such that $\prod\nolimits_{i=1}^{m}\left(1-\lambda_{i}\right)^{n_{i}}>\alpha$. Hence there exists (θ, ψ) such that P(Y = y*∣θ, ψ) > α. Therefore there exists θ such that sup_ψ P(T ≤ t(y) ∣θ, ψ) > α. This argument holds for every

$$y \in \mathcal{Y}$$