Bahadur intercept with applications to one-sided testing

Abstract

In this paper we propose Bahadur intercept for measuring the Bahadur deficiency of two test procedures that have the same Bahadur slope. We illustrate our results in an example of testing a one-sided hypothesis in which the comparison of Bahadur intercepts of one-sided and two-sided tests reveals an important explanation for the test deficiency. We also show that the Bahadur deficiency coincides with the Pitman deficiency under additional conditions.

Appendix: Proofs

Proof of Theorem 1

For the sequence $r=1,2,\ldots$ let $\alpha _{r}\rightarrow 0$ as $r\rightarrow \infty$. Let $\theta \in \Theta -\Theta _{0}$. For a given $\varepsilon >0$ it follows from Assumptions 1 and 3 that there exist $r_{0}$ such that for $r\ge r_{0}$

$$\begin{aligned} \gamma (\alpha _{r})-\varepsilon \le K_{n_{r}}\le \gamma (\alpha _{r})+\varepsilon , \end{aligned}$$

(A. 1)

holds with the probability $1-\delta (\varepsilon )$, where $\delta (\varepsilon )\rightarrow 0$ as $\varepsilon \rightarrow 0$. By Eq. (3) it follows that there exists $r_{1}$ corresponding to the given $\varepsilon$ and $\delta (\varepsilon )$ such that for all $r\ge r_{1}\ge r_{0}$

$$\begin{aligned} d(\theta )-\varepsilon \le K_{n_{r}}-n_{r}c(\theta )\le d(\theta )+\varepsilon \end{aligned}$$

(A. 2)

holds with the probability $1-\delta (\varepsilon )$. This in turn means that the probability $1-\delta (\varepsilon )$ is guaranteed if the sample size suggested by Inequality (A. 2) is bounded as

$$\begin{aligned} \frac{K_{n_{r}}-d(\theta )-\varepsilon }{c(\theta )}\le n_{r}\le \frac{ K_{n_{r}}-d(\theta )+\varepsilon }{c(\theta )}. \end{aligned}$$

(A. 3)

It then follows from Inequalities (A. 1) and (A. 3) that

$$\begin{aligned} \frac{\gamma (\alpha _{r})-d(\theta )-2\varepsilon }{c(\theta )}\le n_{r}\le \frac{\gamma (\alpha _{r})-d(\theta )+2\varepsilon }{c(\theta )}. \end{aligned}$$

For the two tests indexed by i and j that have $c_{i}(\theta )=c_{j}(\theta )=c(\theta )$ let

$$\begin{aligned} D_{j,i}^{B+}= & {} \sup \{n_{r,j}-n_{r,i}\} \\= & {} \frac{\gamma _{j}(\alpha _{r})-d_{j}(\theta )+2\varepsilon }{c(\theta )} -\frac{\gamma _{i}(\alpha _{r})-d_{i}(\theta )-2\varepsilon }{c(\theta )}. \end{aligned}$$

For the two tests to be equally successful in rejecting $H_{0}$ they require $\gamma _{i}(\alpha _{r})=\gamma _{j}(\alpha _{r})$, hence

$$\begin{aligned} D_{j,i}^{B+}=\frac{\Delta d_{i,j}(\theta )+4\varepsilon }{c(\theta )}. \end{aligned}$$

Similarly, we have

$$\begin{aligned} D_{j,i}^{B-}= & {} \inf \{n_{r,j}-n_{r,i}\} \\= & {} \frac{\gamma _{j}(\alpha _{r})-d_{j}(\theta )-2\varepsilon }{ c_{j}(\theta )}-\frac{\gamma _{i}(\alpha _{r})-d_{i}(\theta )+2\varepsilon }{c_{i}(\theta )} \\= & {} \frac{\Delta d_{i,j}(\theta )-4\varepsilon }{c(\theta )}. \end{aligned}$$

As $\varepsilon$ can be arbitrarily small it follows that

$$\begin{aligned} D_{j,i}^{B}=\lim _{r\rightarrow \infty ,\alpha _{r}\rightarrow 0}D_{j,i}^{B+}=\lim _{r\rightarrow \infty ,\alpha _{r}\rightarrow 0}D_{j,i}^{B-}=\frac{\Delta d_{i,j}(\theta )}{c(\theta )}. \end{aligned}$$

This completes the proof. $\square$

Proof of Theorem 2

For the sequence $r=1,2,\ldots$ let $\theta _{r}\rightarrow \theta _{0}$ as $r\rightarrow \infty$. From Assumption 5 we can choose $\delta >0$ and an integer $r_{0}$ such that for $r\ge r_{0}$ we have $\delta \le \beta _{r}\le 1-\delta$, $b(\theta _{r})\le 1$ and $\theta _{r}\in \mathcal {N}(\theta _{0},\theta ^{*})$. Let $x_{0}>2(A+1)^{1/2}$ such that F is strictly increasing on $(x_{0},\infty )$. Choose an integer $r_{1}\ge r_{0}$ such that for $r\ge r_{1}$ and $n_{r}\in (A/b^{2}(\theta _{r}),A/b^{2}(\theta _{r})+1]$, the rejection region of the level $\alpha$ test based $T_{n}$ is contained in the interval $(2(A+1)^{1/2},\infty )$ which is a subset of $(2b(\theta _{r})n_{r}^{1/2},\infty )$ because

$$\begin{aligned} 2b(\theta _{r})n_{r}^{\frac{1}{2}}\le 2(A+b^{2}(\theta _{r}))^{\frac{1}{2} }\le 2(A+1)^{\frac{1}{2}}. \end{aligned}$$

That is, the power $\beta _{r}\le \beta _{r_{1}}$ for $r\ge r_{1}$. As $K_{n}$ is the monotonic transformation of $T_{n}$ we have $K_{n_{r}}\le K_{n_{r_{1}}}$ for $r\ge r_{1}$. Because the inequalities (A. 1) and (A. 2) hold as $n_{r}\rightarrow \infty$ ($r\rightarrow \infty$). Then following the proof of Theorem 1 it easily leads to

$$\begin{aligned} D_{j,i}^{P}= & {} \lim _{r\rightarrow \infty ,\theta _{r}\rightarrow \theta _{0}}\inf \{n_{r,j}-n_{r,i}\} \\= & {} \lim _{r\rightarrow \infty ,\theta _{r}\rightarrow \theta _{0}}\sup \{n_{r,j}-n_{r,i}\} \\= & {} \frac{\Delta d_{i,j}(\theta )}{c(\theta )}. \end{aligned}$$

$\square$

Proof of Proposition 1

For the tests based on $T_{n,3}$ and $T_{n,4}$ Assumption 5 can be verified by following the discussions in Example 3 of Wieand (1976) while the other remaining assumptions can be verified easily.

For notational convenience let $\pi (Q,m,\Sigma )=\pi _{m}$. Let the random variable appearing in the subscript of F be the cumulative distribution of that random variable; for example, $F_{\chi _{m}^{2}}$ denotes the cumulative distribution of $\chi _{m}^{2}$.

Because for $x>0$ and $0\le m<m^{\prime }$,

$$\begin{aligned} \Pr (\chi _{m}^{2}>x)<\Pr (\chi _{m^{\prime }}^{2}>x), \end{aligned}$$

we denote by $\varrho (m)$ a strictly decreasing function of m such that for $x\rightarrow \infty$

$$\begin{aligned} 1-F_{\chi _{m}^{2}}(x)=\exp \left\{ -\frac{x}{2}(1+o(1))-\varrho (m)\right\} . \end{aligned}$$

It then follows that for $x>0$, we have

$$\begin{aligned}&1-F_{\overline{\chi }^{2}}\left( x\right) \\&\quad =\sum _{m=1}^{k}\pi _{m}\{1-F_{\chi _{m}^{2}}\left( x\right) \} \\&\quad =\sum _{m=1}^{k}\pi _{m}\exp \left\{ -\frac{x}{2}\left( 1+o\left( 1\right) \right) -\varrho \left( m\right) \right\} \\&\quad =\sum _{m=1}^{k}\exp \left\{ -\frac{x}{2}\left( 1+o\left( 1\right) \right) -\varrho \left( m\right) +\log \pi _{m}\right\} \\&\quad =\exp \left( -\frac{x}{2}\right) \sum _{m=1}^{k}\exp \{-\varrho \left( m\right) +\log \pi _{m}\}+o\left( \exp \left( -\frac{x}{2}\right) \right) . \end{aligned}$$

Therefore,

$$\begin{aligned} -\log \{1-F_{\overline{\chi }^{2}}(x)\}=\frac{x}{2}-\log \left[ \sum _{m=1}^{k}\exp \{-\varrho (m)+\log \pi _{m}\}\right] +o(x). \end{aligned}$$

It follows that QLR tests based on $T_{n,3}$ have the approximate Bahadur slope as

$$\begin{aligned} c_{3}(\theta )= & {} p\lim _{n\rightarrow \infty }-\frac{2}{n}\log \left[ 1-F_{ \overline{\chi }_{k}^{2}}\{T_{n,3}\}\right] \\= & {} p\lim _{n\rightarrow \infty }\frac{T_{n,3}}{n} \\= & {} \theta ^{\prime }\Sigma ^{-1}\theta . \end{aligned}$$

Similarly, for two-sided tests we have

$$\begin{aligned} -\log \{1-F_{\chi _{k}^{2}}(x)\}=\frac{x}{2}+\varrho (k)+o(x). \end{aligned}$$

and the approximate Bahadur slope based on $T_{n,4}$ is

$$\begin{aligned} c_{4}(\theta )= & {} p\lim _{n\rightarrow \infty }-\frac{2}{n}\log [1-F_{\chi _{k}^{2}}\{T_{n,4}\}] \\= & {} p\lim _{n\rightarrow \infty }\frac{T_{n,4}}{n} \\= & {} \theta ^{\prime }\Sigma ^{-1}\theta . \end{aligned}$$

Having shown $c_{3}(\theta )=c_{4}(\theta )$ we now show that

$$\begin{aligned} \Delta d_{3,4}(\theta )= & {} d_{3}(\theta )-d_{4}(\theta ) \\= & {} p\lim _{n\rightarrow \infty }\left\{ -2\log \left[ 1-F_{\overline{\chi } _{k}^{2}}\{T_{n,3}\}\right] +2\log \left[ 1-F_{\chi _{k}^{2}}\{T_{n,4}\}\right] \right\} \\= & {} p\lim _{n\rightarrow \infty }\left\langle T_{n,3}-2\log \left[ \sum _{m=1}^{k}\exp \{-\varrho (m)+\log \pi _{m}\}\right] -T_{n,4}-2\varrho (k)\right\rangle \\= & {} -2\log \left[ \sum _{m=1}^{k}\exp \{-\varrho (m)+\log \pi _{m}\}\right] -2\varrho (k), \end{aligned}$$

where the last line follows because of $T_{n,3}=T_{n,4}+o_{p}(1)$. Because $\sum _{m=0}^{k}\pi _{m}=1$ and $0<\pi _{m}<1$, $m\in \{1,\ldots ,k\}$, we have $\sum _{m=1}^{k}\pi _{m}<1$. Together with that $\varrho (m)$ is a strictly decreasing function of m it follows

$$\begin{aligned} \sum _{m=1}^{k}\exp \{-\varrho (m)+\log \pi _{m}\}< & {} \sum _{m=1}^{k}\exp \{-\varrho (k)+\log \pi _{m}\} \\= & {} \exp \{-\varrho (k)\}\sum _{m=1}^{k}\pi _{m} \\< & {} \exp \{-\varrho (k)\}. \end{aligned}$$

Therefore, it follows that $0<\Delta d_{3,4}(\theta )<\infty$, so the deficiency result follows by applying Theorem 1. $\square$