Three-stage confidence intervals for a linear combination of locations of two negative exponential distributions

Abstract

Mukhopadhyay and Padmanabhan (Metrika 40:121–128, 1993) considered the construction of fixed-width confidence intervals for the difference of location parameters of two negative exponential distributions via triple sampling when the scale parameters are unknown and unequal. Under the same setting, this paper deals with the problem of fixed-width confidence interval estimation for a linear combination of location parameters, using the above mentioned three-stage procedure.

Appendix

In this appendix we will give the uniform integrability of $\{\tilde{S}_i^{\,p},\,0<d\le d_0\}$ for each $p\ge 1$ in Lemma 3. Let $Y_2,\,Y_3,\ldots$ be a sequence of independent and identically distributed positive continuous random variables having a finite mean $\theta =E(Y_2).$ We consider the following three-stage procedure defined by Mukhopadhyay (1990):

$$\begin{aligned} R=R(d)={\max }\left\{ m,\ N_1\right\} \quad \text{ and }\quad S=S(d)={\max }\left\{ R,\ N_2\right\} , \end{aligned}$$

where $N_1=\langle \rho \lambda \overline{Y}_m \rangle +1$, $N_2=\langle \lambda \overline{Y}_R \rangle +1$, $0<\rho <1$, $0<\lambda <\infty$, $\overline{Y}_n=(n-1)^{-1}\sum _{i=2}^n Y_i$ for $n\ge 2$ and $m=m(d)\, (\ge 2)$ is the starting sample size such that $m\rightarrow \infty$ as $d\rightarrow 0$. Let $n^*=\lambda \theta$ and we suppose the following conditions

$$\begin{aligned} \lambda =\lambda (m)\rightarrow \infty \ \text{ as } m\rightarrow \infty ,\quad \limsup _{d\rightarrow 0}{m}/n^* <\rho ^2 \end{aligned}$$

(21)

and for some $r>1$, as $m\rightarrow \infty$

$$\begin{aligned} n^*=O(m^r) . \end{aligned}$$

(22)

In the following we assume that $E(Y_2^p)<\infty$ for some $p\ge 2$ and let M denote a generic positive constant, not depending on d. Let $V_j=Y_j/\theta$ for $j=2,3,\cdots$ and $\overline{V}_n=\sum _{j=2}^{n}V_j/(n-1)$. Then $N_1=\langle \rho n^* \overline{V}_m \rangle +1$ and $N_2=\langle n^* \overline{V}_R \rangle +1$. For $\varepsilon \in (0,1)$, define a set $B_{m,\varepsilon }$ by $B_{m,\varepsilon }=\left\{ \overline{V}_m<1-\varepsilon \right\}$.

Lemma 8

As $d\rightarrow 0$, we have $P(B_{m,\varepsilon })=O(m^{-p/2})$.

Proof

Since $\{\overline{V}_n -1,\ n\ge m\}$ is a reversed martingale, we have from the submartingale inequality,

$$\begin{aligned} P(B_{m,\varepsilon })\le P\left\{ \sup _{n\ge m}\left| \overline{V}_n -1\right| >\varepsilon \right\} \le \varepsilon ^{-p}E\left| \overline{V}_m -1\right| ^{p}=O(m^{-p/2}) . \end{aligned}$$

$\square$

Lemma 9

As $d\rightarrow 0$, we have

$$\begin{aligned} P(R\ne N_1)=O(m^{-p/2})\quad \text{ and }\quad P(S\ne N_2)= & {} O(m^{-p/2}). \end{aligned}$$

(23)

Proof

Fix $\varepsilon _0\in (0,1-\rho )$. By (21) and Lemma 8, for sufficiently small d,

$$\begin{aligned} P(R\ne N_1)\le P(\overline{V}_m<m/(\rho n^*))\le P(\overline{V}_m<1-\varepsilon _0)=O(m^{-p/2}), \end{aligned}$$

which implies the left side of (23). Next,

$$\begin{aligned}&P(S\ne N_2)\le P(\langle \rho \lambda \overline{Y}_m \rangle +1>\langle \lambda \overline{Y}_R \rangle +1, R= N_1)+P(R\ne N_1)\\&\le P(\rho \lambda \overline{Y}_m > \lambda \overline{Y}_R ) +O(m^{-p/2}) \end{aligned}$$

from the left side of (23). The first term is evaluated as follows.

$$\begin{aligned}&P(\rho \lambda \overline{Y}_m> \lambda \overline{Y}_R )\\&= P(\rho n^* \overline{V}_m> n^* \overline{V}_R, \overline{V}_R<1-\varepsilon _0)+P(\rho n^* \overline{V}_m> n^* \overline{V}_R, \overline{V}_R\ge 1-\varepsilon _0)\\&\le P\left( \left| \overline{V}_R-1\right|>\varepsilon _0\right) +P(\rho \overline{V}_m>1-\varepsilon _0). \end{aligned}$$

As in the proof of Lemma 8, we have that $P\left( \left| \overline{V}_R-1\right| >\varepsilon _0\right) =O(m^{-p/2})$ and $P(\rho \overline{V}_m>1-\varepsilon _0)=P\left( \left| \overline{V}_m-1\right| >(1-\varepsilon _0-\rho )/\rho \right) =O(m^{-p/2}).$ Hence, the right side of (23) holds.$\square$

Lemma 10

If $0<q<p/(2r)$, where r is as in (22), then $\left\{ (n^*/R)^{q}, 0<d\le d_0\right\}$ and $\left\{ (n^*/S)^{q}, 0<d\le d_0\right\}$ are uniformly integrable for some $d_0>0$.

Proof

Note that $(n^*/S)^{q}\le (n^*/R)^{q}$. From Lemma 1 of Chow and Yu (1981), it suffices to show that $P(R<\varepsilon _1 n^*)=o(n^{*\,-q})$ for some $\varepsilon _1 \in (0,1)$. By choosing $\varepsilon _1 \in (0,\rho )$, we have from (22)

$$\begin{aligned} P(R<\varepsilon _1 n^*)\le P(\rho \overline{V}_m<\varepsilon _1 )\le P\left( \left| \overline{V}_m-1\right| >1-\varepsilon _1/\rho \right) =o({n^*}^{-q}).\square \end{aligned}$$

Lemma 11

For $0<q\le p, \left\{ (R/n^*)^{q}, 0<d\le d_0\right\}$ and $\left\{ (S/n^*)^{q}, 0<d\le d_0\right\}$ are uniformly integrable for some $d_0>0$.

Proof

From Corollary 4.1 of Gut (2005), if $E\left\{ \sup _{0<d\le d_0}(R/n^*)^q\right\} <\infty$, then $\left\{ (R/n^*)^{q}, 0<d\le d_0\right\}$ is uniformly integrable. By the definition of R, Doob’s maximal inequality for the reversed martingale and (21),

$$\begin{aligned}&E\left\{ \sup _{0<d\le d_0}(R/n^*)^{q}\right\} \le M\cdot E\left[ \sup _{0<d\le d_0}\{(m/n^*)^{q}+(\rho \overline{V}_m+(1/n^*))^{q}\}\right] \\&\quad \le M +M \rho ^{q}E\left( \sup _{0<d\le d_0}\overline{V}_m^q\right) \le M+M E\left( \sup _{n\ge 2}\overline{V}_n^{q}\right) \le M\quad \text{ for } 1<q\le p, \end{aligned}$$

which yields the uniform integrability of $\left\{ (R/n^*)^{q}, 0<d\le d_0\right\}$ for $1<q\le p$. When $0<q\le 1$, we have that $\sup _{0<d\le d_0}E(R/n^*)^{q\zeta }= \sup _{0<d\le d_0}E(R/n^*)^{p}<\infty$ for $\zeta =p/q>1$. Therefore, $\left\{ (R/n^*)^{q}, 0<d\le d_0\right\}$ is uniformly integrable for $0<q\le p$. Next, we shall show the uniform integrability of $\left\{ (S/n^*)^{q}, 0<d\le d_0\right\}$. Since $S\le N_2+R$, it suffices to show that $E\left\{ \sup _{0<d\le d_0}(N_2/n^*)^{q}\right\} <\infty$ which can be proved similarly. $\square$

Lemma 12

For $0<q\le p$,

$$\begin{aligned} \left\{ \left| {n^*}^{-\frac{1}{2}}\sum _{j=2}^{R}(V_j-1)\right| ^{q}\!, 0<d\le d_0\right\} \ \text{ and }\ \left\{ \left| {n^*}^{-\frac{1}{2}}\sum _{j=2}^{S}(V_j-1)\right| ^{q}\!, 0<d\le d_0\right\} \end{aligned}$$

are uniformly integrable for some $d_0>0$.

Proof

Follows from Lemma 5 of Chow and Yu (1981) and Lemma 11. $\square$

Proposition 2

We assume that $E(Y_2^p)<\infty$ for some $p\ge 2$. Let $\tilde{S}={n^*}^{-\frac{1}{2}}(S-n^*)$. Under the conditions (21) and (22), if $0<q< p/(2r+1)$, then $\left\{ \tilde{S}^{q}, 0<d\le d_0\right\}$ is uniformly integrable for some $d_0>0$.

Proof

Now,

$$\begin{aligned} |\tilde{S}^q|= & {} |n^{*\,-1/2}(S-n^*)|^{q}\\= & {} |n^{*\,-1/2}(\langle n^* \overline{V}_R \rangle +1-n^*)|^{q}I(S=N_2) +|n^{*\,-1/2}(R-n^*)|^{q}I(S\ne N_2)\\\equiv & {} K_1+K_2,\quad \text{ say }. \end{aligned}$$

Since $K_3\equiv n^{*\,-1/2}(\langle n^* \overline{V}_R \rangle +1-n^* \overline{V}_R)\le n^{*\,-1/2}\le 1$ and $0<R/(R-1)\le 2$, we have for some $\zeta >1$, $u=2r+1$ and $v=\frac{1}{2r}+1$,

$$\begin{aligned}&E(K_1 ^{\zeta }) \le \ E|n^{*\,-1/2}(n^* \overline{V}_R -n^*)+ K_3|^{q\zeta } \\&\le \ M E\left| {n^*}^{-\frac{1}{2}}\sum _{j=2}^{R}(V_j-1)\cdot (R/(R-1))\cdot (n^*/R)\right| ^{q\zeta }+M \\&\le \ M \left\{ E\left| {n^*}^{-\frac{1}{2}}\sum _{j=2}^{R}(V_j-1)\right| ^{u q \zeta }\right\} ^{\frac{1}{u}} \left\{ E(n^*/R)^{v q \zeta }\right\} ^{\frac{1}{v}}+M=O(1) \end{aligned}$$

by Lemmas 10 and 12. Finally, for some $\zeta >1$, $u_0=r+1$ and $v_0=\frac{1}{r}+1$, we have from (23) and Lemma 11

$$\begin{aligned} E(K_2^{\zeta })\le & {} M\, {n^*}^{\frac{1}{2} q \zeta }\{E(R/n^*)^{u_0 q\zeta }+1\}^{\frac{1}{u_0}}\left\{ P(S\ne N_2)\right\} ^{\frac{1}{v_0}} = O(m^{q\zeta r/2-p/(2v_0)})\\= & {} O(1). \end{aligned}$$

Hence, the proposition is proved.$\square$

Proof of the uniform integrability

We will show the uniform integrability of $\{\tilde{S}_i^{\,p},\ 0<d\le d_0\}$ for each $p\ge 1$. Let $Y_{i j}^{\prime }=Y_{i j}/\sigma _i$ and $C_i=\lambda _i=a_{*}\sigma _i d^{-1}$, where $Y_{ij}$ has the exponential distribution $\mathrm{{E_{XP}}}(0,\sigma _i)$. Then $Y_{i2}^{\prime },\,Y_{i3}^{\prime },\ldots$ are i.i.d random variables according to $\mathrm{{E_{XP}}}(0,1)$, and $R_i$ and $S_i$ defined by (10) can be written as $R_i={\max }\left\{ m,\ N_{1i}\right\}$ and $S_i={\max }\left\{ R_i,\ N_{2i}\right\} ,$ where $N_{1i}=\langle \rho _i \lambda _i \overline{Y^{\prime }}_{i m} \rangle +1,\ N_{2i}=\langle \lambda _i \overline{Y^{\prime }}_{i R_i} \rangle +1$ and $0<\rho _i<1$. Put $n^*=C_i,\, \lambda =\lambda _i,\, \rho =\rho _i,\,R=R_i,\,S=S_i$, $Y_j=Y_{ij}$ and $V_j=Y_{ij}^{\prime }$ for $i=1,2$. Since $E(Y_{i 2}^p)<\infty$ for all $p>0$ and $m=O(d^{-1/r})$ for some $r>1$, the conditions (21) and (22) are satisfied. Therefore from Proposition 2, $\{\tilde{S_i}^{p}, 0<d\le d_0\}$ is uniformly integrable for some $d_0>0$.

$\square$