Shrinkage estimation for the mean of the inverse Gaussian population

Abstract

We consider improved estimation strategies for a two-parameter inverse Gaussian distribution and use a shrinkage technique for the estimation of the mean parameter. In this context, two new shrinkage estimators are suggested and demonstrated to dominate the classical estimator under the quadratic risk with realistic conditions. Furthermore, based on our shrinkage strategy, a new estimator is proposed for the common mean of several inverse Gaussian distributions, which uniformly dominates the Graybill–Deal type unbiased estimator. The performance of the suggested estimators is examined by using simulated data and our shrinkage strategies are shown to work well. The estimation methods and results are illustrated by two empirical examples.

Appendices

Appendix A: Proofs

1.1 A.1 Proof for Theorem 1

Denote $\Delta = R(\mu ,\tilde{\mu }_{\lambda }) - R(\mu ,\bar{X})$. It is easy to see that

$$\begin{aligned} R(\mu ,\tilde{\mu }_{\lambda })&= \frac{1}{\mu ^{3}}E\left[ \left( 1-\frac{c}{n\lambda +4a}\text { exp }\{{-a/\bar{X}}\}\right) \bar{X}-\mu \right] ^{2} \nonumber \\&= R(\mu ,\bar{X})-\frac{2c}{(n\lambda +4a)\mu ^{3}}E[(\bar{X}-\mu )\bar{X}\text { exp }\{{-a/\bar{X}}\}] \nonumber \\&\quad +\frac{c^{2}}{(n\lambda +4a)^{2}\mu ^{3}}E(\bar{X}^{2}\text { exp }\{{-2a/\bar{X}}\}) \nonumber \\&= R(\mu ,\bar{X})+\Delta . \end{aligned}$$

(8.1)

Next, we rewrite $\Delta$ and provide a simple condition for $\Delta <0$. Using $\bar{X}\sim IG(\mu ,n\lambda )$ and Lemma 2, we get

$$\begin{aligned} \Delta&= -\frac{2c}{(n\lambda +4a)\mu }\left( \frac{\mu }{n\lambda }+\frac{\sqrt{n\lambda +2a}}{\sqrt{n\lambda }}-1\right) \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +2a)}}{\mu }\right\} \nonumber \\&\quad +\frac{c^{2}}{(n\lambda +4a)^{2}\mu } \left( \frac{\mu }{n\lambda }+\frac{\sqrt{n\lambda +4a}}{\sqrt{n\lambda }}\right) \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +4a)}}{\mu }\right\} \nonumber \\&= -\frac{2c}{\lambda (n\lambda +4a)n} \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +2a)}}{\mu }\right\} \nonumber \\&\quad +\frac{c^{2}}{\lambda (n\lambda +4a)^{2}n} \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +4a)}}{\mu }\right\} \nonumber \\&\quad -\frac{2c}{(n\lambda +4a)\mu }\frac{\sqrt{n\lambda +2a}-\sqrt{n\lambda }}{\sqrt{n\lambda }} \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +2a)}}{\mu }\right\} \nonumber \\&\quad +\frac{c^{2}}{(n\lambda +4a)^{2}\mu } \frac{\sqrt{n\lambda +4a}}{\sqrt{n\lambda }} \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +4a)}}{\mu }\right\} \nonumber \\&= J_{1}+J_{2}, \end{aligned}$$

(8.2)

where

$$\begin{aligned} J_{1}&= \frac{c^{2}}{\lambda (n\lambda +4a)^{2}n} \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +4a)}}{\mu }\right\} \\&\quad - \frac{2c}{\lambda (n\lambda +4a)n} \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +2a)}}{\mu }\right\} \!,\\ J_{2}&= -\frac{2c}{(n\lambda +4a)\mu }\frac{\sqrt{n\lambda +2a}-\sqrt{n\lambda }}{\sqrt{n\lambda }} \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +2a)}}{\mu }\right\} \\&\quad +\frac{c^{2}}{(n\lambda +4a)^{2}\mu } \frac{\sqrt{n\lambda +4a}}{\sqrt{n\lambda }}\text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +4a)}}{\mu }\right\} \!. \end{aligned}$$

Therefore, it follows from $\text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +4a)}}{\mu }\right\} <\text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +2a)}}{\mu }\right\}$ that a sufficient condition for $\Delta <0$ is

$$\begin{aligned}&2(n\lambda +4a)\ge c,\end{aligned}$$

(8.3a)

$$\begin{aligned}&2(n\lambda +4a)(\sqrt{n\lambda +2a}-\sqrt{n\lambda })\ge c\sqrt{n\lambda +4a}. \end{aligned}$$

(8.3b)

This is a relaxed condition, and it obviously holds when $2a \ge c > 0$.

This completes the proof.

1.2 A.2 Proof for Theorem 2

Similar to the proof of Theorem 1, we get

$$\begin{aligned} \Delta&= R(\mu ,\tilde{\mu }_{U})-R(\mu ,\bar{X}) \nonumber \\&= -\frac{1}{\mu ^{3}}E\left[ \frac{2c}{n/S+4a}\right] E[(\bar{X}-\mu )\bar{X}e^{-a/\bar{X}}] +\frac{1}{\mu ^{3}}E\left[ \frac{c^{2}}{(n/S+4a)^{2}}\right] E[\bar{X}^{2}e^{-2a/\bar{X}}] \nonumber \\&= -E\left[ \frac{2c\lambda }{(n/S+4a)\mu }\right] \left( \frac{\mu }{n\lambda }+\frac{\sqrt{n\lambda +2a}}{\sqrt{n\lambda }}-1\right) \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +2a)}}{\mu }\right\} \nonumber \\&\quad +E\left[ \frac{c^{2}}{(n/S+4a)^{2}\mu }\right] \left( \frac{\mu }{n\lambda }+\frac{\sqrt{n\lambda +4a}}{\sqrt{n\lambda }}\right) \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +4a)}}{\mu }\right\} \nonumber \\&= -E\left[ \frac{2c}{\lambda (n/S+4a)n}\right] \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +2a)}}{\mu }\right\} \nonumber \\&\quad +E\left[ \frac{c^{2}}{\lambda (n/S+4a)^{2}n}\right] \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +4a)}}{\mu }\right\} \nonumber \\&\quad -E\left[ \frac{2c}{(n/S+4a)\mu }\right] \frac{\sqrt{n\lambda +2a}-\sqrt{n\lambda }}{\sqrt{n\lambda }} \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +2a)}}{\mu }\right\} \nonumber \\&\quad +E\left[ \frac{c^{2}}{(n/S+4a)^{2}\mu }\right] \frac{\sqrt{n\lambda +4a}}{\sqrt{n\lambda }} \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +4a)}}{\mu }\right\} \nonumber \\&= L_{1}+L_{2}, \end{aligned}$$

(8.4)

where

$$\begin{aligned} L_{1}&= E\left[ \frac{c^{2}}{\lambda (n/S+4a)^{2}n}\right] \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +4a)}}{\mu }\right\} \\&\quad -E\left[ \frac{2c}{\lambda (n/S+4a)n}\right] \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +2a)}}{\mu }\right\} \!,\\ L_{2}&= E\left[ \frac{c^{2}}{(n/S+4a)^{2}\mu }\right] \frac{\sqrt{n\lambda +4a}}{\sqrt{n\lambda }}\text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +4a)}}{\mu }\right\} \\&\quad -E\left[ \frac{2c}{(n/S+4a)\mu }\right] \frac{\sqrt{n\lambda +2a}-\sqrt{n\lambda }}{\sqrt{n\lambda }} \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +2a)}}{\mu }\right\} \!. \end{aligned}$$

Obviously, $\Delta <0$ when $L_{1}<0$ and $L_{2}<0$. Note that

$$\begin{aligned} \text { exp }\left\{ \frac{n\lambda -\sqrt{n\lambda (n\lambda +4a)}}{\mu }\right\} <\text { exp }\left\{ \frac{n\lambda - \sqrt{n\lambda (n\lambda +2a)}}{\mu }\right\} \!. \end{aligned}$$

Then $\Delta <0$ is true if the following two inequalities hold

$$\begin{aligned}&E\left[ \frac{2c}{(n/S+4a)n}\right] \ge E\left[ \frac{c^{2}}{(n/S+4a)^{2}n}\right] \!, \end{aligned}$$

(8.5a)

$$\begin{aligned}&(\sqrt{n\lambda +2a}-\sqrt{n\lambda })E\left[ \frac{2c\lambda }{(n/S+4a)\mu }\right] \ge \sqrt{n\lambda +4a}E\left[ \frac{c^{2}\lambda }{(n/S+4a)^{2}\mu }\right] \!.\quad \quad \end{aligned}$$

(8.5b)

It is easy to see that (8.5b) is a sufficient condition for (8.5a). Then we get that $\Delta <0$ if (8.5b) is true. Next, we discuss (8.5b).

Note that $\sqrt{n\lambda +2a}-\sqrt{n\lambda }=\frac{2a}{\sqrt{n\lambda +2a}+\sqrt{n\lambda }}>\frac{a}{\sqrt{n\lambda +4a}}$. After some manipulation, a sufficient condition for (8.5b) is found to be

$$\begin{aligned} E\left[ \frac{2ac(n/S+4a)-c^{2}(n\lambda +4a)}{(n/S+4a)^{2}}\right] \ge 0. \end{aligned}$$

(8.6)

Obviously, (8.6) is equivalent to

$$\begin{aligned} E\left[ \frac{2acnS-(c^{2}n\lambda +4ac(c-2a))S^{2}}{(n+4aS)^{2}}\right] \ge 0. \end{aligned}$$

(8.7)

Using $c\le 2a$, from (8.5) to (8.6), we get a sufficient condition for $\Delta <0$

$$\begin{aligned} E\left[ \frac{2aS-c\lambda S^{2}}{(n+4aS)^{2}}\right] \ge 0. \end{aligned}$$

(8.8)

It follows from Lemma 1 that

$$\begin{aligned} E\left[ \frac{2aS-c\lambda S^{2}}{(n+4aS)^{2}}\right] \ge E\left[ \frac{2aS-c\lambda S^{2}}{(n+8a^{2}/c\lambda )^{2}}\right] =\frac{\frac{2a(n-1)}{n\lambda }-\frac{c(n^{2}-1)}{n^{2}\lambda }}{(n+8a^{2}/c\lambda )^{2}}, \end{aligned}$$

(8.9)

where the equality holds when $n\lambda S$ is distributed as $\chi _{n-1}^{2}$.

From (8.8) and (8.9), it is therefore enough to see that $\Delta <0$ holds for $0<c\le \frac{2an}{n+1}$.

This completes the proof.

1.3 A.3 Proof for Theorem 3

Note that

$$\begin{aligned} R(\tilde{\mu }_{S},\mu )&= E(\tilde{\mu }_{S}-\mu )^{2}/\mu ^{3}\nonumber \\&= \frac{1}{\mu ^{3}}E\left[ \frac{1}{\sum _{i=1}^{k}\frac{n_{i}-1}{S_{i}}} \sum _{i=1}^{k}\frac{n_{i}-1}{S_{i}}\left( \left( 1-\frac{c_{i}}{n_{i}/S_{i}+4a_{i}}\text { exp }\{-a_{i}/\bar{X}_{i}\}\right) \bar{X}_{i}-\mu \right) \right] ^{2}\nonumber \\&= \frac{1}{\mu ^{3}}E\left[ \frac{1}{\sum _{i=1}^{k}\frac{n_{i}-1}{S_{i}}} \sum _{i=1}^{k}\frac{n_{i}-1}{S_{i}}\left( \bar{X}_{i}-\mu -\frac{c_{i}}{n_{i}/S_{i}+4a_{i}}\text { exp }\{-a_{i}/\bar{X}_{i}\}\bar{X}_{i}\right) \right] ^{2}\nonumber \\&= \frac{1}{\mu ^{3}}E\left[ \frac{1}{\sum _{i=1}^{k}\frac{n_{i}-1}{S_{i}}} \left( \sum _{i=1}^{k}\frac{n_{i}-1}{S_{i}}(\bar{X}_{i}-\mu ) -\sum _{i=1}^{k}\frac{c_{i}(n_{i}-1)}{n_{i}+4a_{i}S_{i}}\text { exp }\{-a_{i}/\bar{X}_{i}\}\bar{X}_{i}\right) \right] ^{2}\nonumber \\&= \frac{1}{\mu ^{3}}E\left[ \frac{1}{\sum _{i=1}^{k}\frac{n_{i}-1}{S_{i}}}\sum _{i=1}^{k}\frac{n_{i}-1}{S_{i}}(\bar{X}_{i}-\mu )\right] ^{2}\nonumber \\&-\frac{2}{\mu ^{3}}E\left[ \frac{1}{(\sum _{i=1}^{k}\frac{n_{i}-1}{S_{i}})^{2}}\sum _{i=1}^{k}\frac{n_{i}-1}{S_{i}}(\bar{X}_{i}-\mu )\sum _{i=1}^{k}\frac{c_{i}(n_{i}-1)}{n_{i}+4a_{i}S_{i}}\text { exp }\{-a_{i}/\bar{X}_{i}\}\bar{X}_{i}\right] \nonumber \\&+\frac{1}{\mu ^{3}}E\left[ \frac{1}{\sum _{i=1}^{k}\frac{n_{i}-1}{S_{i}}} \sum _{i=1}^{k}\frac{c_{i}(n_{i}-1)}{n_{i}+4a_{i}S_{i}}\text { exp }\{-a_{i}/\bar{X}_{i}\}\bar{X}_{i}\right] ^{2}\nonumber \\&= R_{3}(\tilde{\mu }_\text {GD},\mu )-2I_{1}+I_{2}. \end{aligned}$$

(8.10)

Then it is only required to prove $I_{2}-2I_{1}\le 0$ if we want to get the dominance result of $\tilde{\mu }_\text {S}$.

As $\bar{X}_{i}'s$ are mutually independent and $E(\bar{X}_{i})=\mu$, using Lemma 2 we get

$$\begin{aligned} I_{1}&= \frac{1}{\mu ^{3}}E\left[ \frac{1}{(\sum _{j=1}^{k}\frac{n_{j}-1}{S_{j}})^{2}} \sum _{i=1}^{k}\frac{c_{i}(n_{i}-1)^{2}}{(n_{i}+4a_{i}S_{i})S_{i}} \text { exp }\{-a_{i}/\bar{X}_{i}\}(\bar{X}_{i}-\mu )\bar{X}_{i}\right] \nonumber \\&= \frac{1}{\mu ^{3}}E\left[ E\left( \sum _{i=1}^{k}\frac{c_{i}(n_{i}-1)^{2}\text { exp }\{-a_{i}/\bar{X}_{i}\}(\bar{X}_{i}-\mu )\bar{X}_{i}}{(n_{i}+4a_{i}S_{i})S_{i}(\sum _{j=1}^{k}\frac{n_{j}-1}{S_{j}})^{2}} | S_{i} \right) \right] \nonumber \\&= \frac{1}{\mu ^{3}}E\left( \frac{1}{(\sum _{j=1}^{k}\frac{n_{j}-1}{S_{j}})^{2}} \sum _{i=1}^{k}\frac{c_{i}(n_{i}-1)^{2}\mu ^{2}}{(n_{i}+4a_{i}S_{i})S_{i}} \left( \frac{\mu }{n_{i}\lambda _{i}}+\frac{\sqrt{n_{i}\lambda _{i}+2a_{i}}}{\sqrt{n_{i}\lambda _{i}}}\right) \right. \nonumber \\&\qquad \qquad \,\,\times \left. \text { exp }\left\{ \frac{n_{i}\lambda _{i}-\sqrt{n_{i}\lambda _{i}(n_{i}\lambda _{i}+2a_{i})}}{\mu }\right\} \right) \nonumber \\&-\frac{1}{\mu ^{3}}E\left( \frac{1}{\left( \sum _{j=1}^{k}\frac{n_{j}-1}{S_{j}}\right) ^{2}} \sum _{i=1}^{k}\frac{c_{i}(n_{i}-1)^{2}\mu ^{2}}{(n_{i}+4a_{i}S_{i})S_{i}} \text { exp }\left\{ \frac{n_{i}\lambda _{i}-\sqrt{n_{i}\lambda _{i}(n_{i}\lambda _{i}+2a_{i})}}{\mu }\right\} \right) \!.\nonumber \\ \end{aligned}$$

(8.11)

By Cauchy inequality and Lemma 2, we have

$$\begin{aligned} I_{2}&\le \frac{k}{\mu ^{3}}E\left[ \left( \frac{1}{\sum _{j=1}^{k}\frac{n_{j}-1}{S_{j}}}\right) ^{2} \sum _{i=1}^{k}\frac{c_{i}^{2}(n_{i}-1)^{2}}{(n_{i}+4a_{i}S_{i})^{2}}\text { exp }\{-2a_{i}/\bar{X}_{i}\}\bar{X}_{i}^{2}\right] \nonumber \\&= \frac{k}{\mu ^{3}}E\left[ E\left( \sum _{i=1}^{k}\frac{c_{i}^{2}(n_{i}-1)^{2}\text { exp }\{-2a_{i}/\bar{X}_{i}\}\bar{X}_{i}^{2}}{(n_{i}+4a_{i}S_{i})^{2}(\sum _{j=1}^{k}\frac{n_{j}-1}{S_{j}})^2}\mid S_{i} \right) \right] \nonumber \\&= \frac{k}{\mu ^{3}}E\left[ \left( \frac{1}{\sum _{j=1}^{k}\frac{n_{j}-1}{S_{j}}}\right) ^{2} \sum _{i=1}^{k}\frac{c_{i}^{2}(n_{i}-1)^{2}}{(n_{i}+4a_{i}S_{i})^{2}}\mu ^{2} \left( \frac{\mu }{n_{i}\lambda _{i}}+\frac{\sqrt{n_{i}\lambda _{i}+4a_{i}}}{\sqrt{n_{i}\lambda _{i}}}\right) \right. \nonumber \\&\qquad \qquad \times \,\,\left. \text { exp }\left\{ \frac{n_{i}\lambda _{i}-\sqrt{n_{i}\lambda _{i}(n_{i}\lambda _{i}+4a_{i})}}{\mu }\right\} \right] \!. \end{aligned}$$

(8.12)

So we get $I_{2}-2I_{1}\le \frac{1}{\mu ^{3}}E\left[ \left( \frac{\mu }{\sum _{j=1}^{k}\frac{n_{j}-1}{S_{j}}}\right) ^{2}\right.$

$$\begin{aligned}&\sum _{i=1}^{k}\left( \frac{kc_{i}^{2}(n_{i}-1)^{2}}{(n_{i}+4a_{i}S_{i})^{2}} \frac{\sqrt{n_{i}\lambda _{i}+4a_{i}}}{\sqrt{n_{i}\lambda _{i}}} \text { exp }\left\{ \frac{n_{i}\lambda _{i}-\sqrt{n_{i}\lambda _{i}(n_{i}\lambda _{i}+4a_{i})}}{\mu }\right\} \right. \nonumber \\&\quad \left. -\frac{2c_{i}(n_{i}-1)^{2}}{(n_{i}+4a_{i}S_{i})S_{i}} \frac{\sqrt{n_{i}\lambda _{i}+2a_{i}}-\sqrt{n_{i}\lambda _{i}}}{\sqrt{n_{i}\lambda _{i}}} \text { exp }\left\{ \frac{n_{i}\lambda _{i}-\sqrt{n_{i}\lambda _{i}(n_{i}\lambda _{i}+2a_{i})}}{\mu }\right\} \right) \nonumber \\&\quad +\sum _{i=1}^{k}\left( \frac{kc_{i}^{2}(n_{i}-1)^{2}}{(n_{i}+4a_{i}S_{i})^{2}}\frac{\mu }{n_{i}\lambda _{i}} \text { exp }\left\{ \frac{n_{i}\lambda _{i}-\sqrt{n_{i}\lambda _{i}(n_{i}\lambda _{i}+4a_{i})}}{\mu }\right\} \right. \nonumber \\&\quad \left. \left. -\frac{2c_{i}(n_{i}-1)^{2}}{(n_{i}+4a_{i}S_{i})S_{i}}\frac{\mu }{n_{i}\lambda _{i}} \text { exp }\left\{ \frac{n_{i}\lambda _{i}-\sqrt{n_{i}\lambda _{i}(n_{i}\lambda _{i}+2a_{i})}}{\mu }\right\} \right) \right] \!. \end{aligned}$$

(8.13)

Note that $\text { exp }\left\{ \frac{n_{i}\lambda _{i}-\sqrt{n_{i}\lambda _{i}(n_{i}\lambda _{i}+4a_{i})}}{\mu }\right\} <\text { exp }\left\{ \frac{n_{i}\lambda _{i}-\sqrt{n_{i}\lambda _{i}(n_{i}\lambda _{i}+2a_{i})}}{\mu }\right\}$. Then the following two inequalities is a sufficient condition for $I_{2}-2I_{1}\le 0$

$$\begin{aligned}&E\left[ \left( \frac{\mu }{\sum _{j=1}^{k}\frac{n_{j}-1}{S_{j}}}\right) ^{2}\left( \frac{kc_{i}^{2}(n_{i}-1)^{2}}{(n_{i}+4a_{i}S_{i})^{2}} \frac{\sqrt{n_{i}\lambda _{i}+4a_{i}}}{\sqrt{n_{i}\lambda _{i}}}\right. \right. \nonumber \\&\qquad \quad \left. \left. -\frac{2c_{i}(n_{i}-1)^{2}}{(n_{i}+4a_{i}S_{i})S_{i}}\frac{\sqrt{n_{i}\lambda _{i}+2a_{i}}-\sqrt{n_{i}\lambda _{i}}}{\sqrt{n_{i}\lambda _{i}}}\right) \right] \le 0, \end{aligned}$$

(8.14)

$$\begin{aligned}&E\left[ \left( \frac{\mu }{\sum _{j=1}^{k}\frac{n_{j}-1}{S_{j}}}\right) ^{2}\left( \frac{kc_{i}^{2}(n_{i}-1)^{2}}{(n_{i}+4a_{i}S_{i})^{2}}- \frac{2c_{i}(n_{i}-1)^{2}}{(n_{i}+4a_{i}S_{i})S_{i}}\right) \right] \le 0,\quad \quad \end{aligned}$$

(8.15)

$i=1,\ldots ,k.$

Note that (8.14) is a sufficient condition for (8.15) and the fact that $\sqrt{n_{i}\lambda _{i}+2a_{i}}-\sqrt{n_{i}\lambda _{i}}>\frac{a_{i}}{\sqrt{n_{i}\lambda _{i}+4a_{i}}}$. So, a sufficient condition for $I_{2}-2I_{1}\le 0$ is given for $i=1,\ldots ,k$ by

$$\begin{aligned} E\left[ \left( \frac{\mu }{\sum _{j=1}^{k}\frac{n_{j}-1}{S_{j}}}\right) ^{\!2}\!\left( \frac{kc_{i}^{2}(n_{i}-1)^{2}}{(n_{i}+4a_{i}S_{i})^{2}} \frac{n_{i}\lambda _{i}+4a_{i}}{a_{i}} -\frac{2c_{i}(n_{i}-1)^{2}}{(n_{i}+4a_{i}S_{i})S_{i}}\right) \right] \!\le \! 0.\quad \quad \quad \end{aligned}$$

(8.16)

After some manipulation, for each $i=1,\ldots ,k$, (8.16) is equivalent to

$$\begin{aligned} E\left[ \left( \frac{\mu }{S_{i}\sum _{j=1}^{k}\frac{n_{j}-1}{S_{j}}}\right) ^{2}\frac{c_{i}(n_{i}-1)^{2}}{(n_{i}+4a_{i}S_{i})} \left( \frac{kc_{i}(n_{i}\lambda _{i}+4a_{i})S_{i}^{2}}{a_{i}(n_{i}+4a_{i}S_{i})} -2S_{i}\right) \right] \le 0.\quad \quad \quad \quad \end{aligned}$$

(8.17)

As ${\small \left( \frac{\mu }{S_{i}\sum _{j=1}^{k}\frac{n_{j}-1}{S_{j}}}\right) ^{2}\frac{c_{i}(n_{i}-1)^{2}}{(n_{i}+4a_{i}S_{i})^{2}}}$ is a continuous nonnegative decreasing function of $S_{i}$, using Lemma 1 we get for $kc_{i}\le 2a_{i}, i=1,\ldots ,k,$

$$\begin{aligned}&E\left[ \left( \frac{\mu }{S_{i}\sum _{j=1}^{k}\frac{n_{j}-1}{S_{j}}}\right) ^{2}\frac{c_{i}(n_{i}-1)^{2}}{(n_{i}+4a_{i}S_{i})} \left( \frac{kc_{i}(n_{i}\lambda _{i}+4a_{i})S_{i}^{2}}{a_{i}(n_{i}+4a_{i}S_{i})} -2S_{i}\right) \right] \nonumber \\&\qquad \le E\left[ \left( \frac{\mu }{S_{i}\sum _{j=1}^{k}\frac{n_{j}-1}{S_{j}}}\right) ^{2}\frac{c_{i}n_{i}(n_{i}-1)^{2}}{(n_{i}+4a_{i}S_{i})^{2}} \left( \frac{kc_{i}\lambda _{i}}{a_{i}}S_{i}^{2}-2S_{i}\right) \right] \nonumber \\&\qquad \le \left( \frac{\mu }{n_{i}-1+\sum _{j\ne i}\frac{a_{i}(n_{j}-1)}{c_{i}\lambda _{i}S_{j}}}\right) ^{2}\frac{c_{i}n_{i}(n_{i}-1)^{2}}{(n_{i}+8a_{i}^{2}/kc_{i}\lambda _{i})^{2}} \left( \frac{kc_{i}(n_{i}^{2}-1)}{a_{i}n_{i}^{2}\lambda _{i}}-\frac{2(n_{i}-1)}{n_{i}\lambda _{i}}\right) \!.\nonumber \\ \end{aligned}$$

(8.18)

Then, from (8.18) we get a sufficient condition for (8.17)

$$\begin{aligned} 0\le c_{i}\le \frac{2a_{i}n_{i}}{k(n_{i}+1)},\quad for\quad i=1,\ldots ,k. \end{aligned}$$

(8.19)

This completes the proof.

Appendix B: Derivation of (3.5)

The derivation of (3.5) is as follows:

$$\begin{aligned} \tilde{\mu }_{B}(\bar{X})&=\frac{\int _{0}^{\infty }\mu ^{-2}\text { exp }\{-n\lambda (\bar{X}-\mu )^{2}/(2\mu ^{2}\bar{X})\}d\mu }{\int _{0}^{\infty }\mu ^{-3}\text { exp }\{-n\lambda (\bar{X}-\mu )^{2}/(2\mu ^{2}\bar{X})\}d\mu } \nonumber \\&\overset{\theta =1/\mu }{=}\frac{\int _{0}^{\infty }\text { exp }\{-n\lambda \bar{X}(\theta -1/\bar{X})^{2}/2\}d\theta }{\int _{0}^{\infty }\theta \text { exp }\{-n\lambda \bar{X}(\theta -1/\bar{X})^{2}/2\}d\theta } \nonumber \\&\overset{y=\sqrt{n\lambda \bar{X}}(\theta -1/\bar{X})}{=}\frac{\int _{-\sqrt{n\lambda /\bar{X}}}^{\infty }\text { exp }\{-y^{2}/2\}dy}{\int _{-\sqrt{n\lambda /\bar{X}}}^{\infty }(y/\sqrt{n\lambda \bar{X}}+1/\bar{X}) \text { exp }\{-y^{2}/2\}dy} \nonumber \\&=\frac{\varPhi (\sqrt{n\lambda /\bar{X}})}{\int _{-\sqrt{n\lambda /\bar{X}}}^{\infty }\frac{y/\sqrt{n\lambda \bar{X}}+1/\bar{X}}{\sqrt{2\pi }} \text { exp }\{-y^{2}/2\}dy} \nonumber \\&=\frac{\varPhi (\sqrt{n\lambda /\bar{X}})}{\varPhi (\sqrt{n\lambda /\bar{X}})/\bar{X}+\text { exp }\{-n\lambda /2\bar{X}\}/\sqrt{2\pi n\lambda \bar{X}}} \nonumber \\&=\bar{X}\frac{\varPhi (\sqrt{n\lambda /\bar{X}})}{\varPhi (\sqrt{n\lambda /\bar{X}})+\sqrt{\bar{X}}\text { exp }\{-n\lambda /2\bar{X}\}/\sqrt{2\pi n\lambda }} \nonumber \\&=\bar{X}\frac{\sqrt{n\lambda /\bar{X}}\sqrt{2\pi }\varPhi (\sqrt{n\lambda /\bar{X}})}{\sqrt{n\lambda /\bar{X}}\sqrt{2\pi }\varPhi (\sqrt{n\lambda /\bar{X}})+\text { exp }\{-n\lambda /2\bar{X}\}} \nonumber \\&=\bar{X}\left[ 1-\frac{\sqrt{\bar{X}}/\sqrt{2\pi n\lambda }}{\varPhi (\sqrt{n\lambda /\bar{X}})+\sqrt{\bar{X}}\text { exp }\{-n\lambda /2\bar{X}\}/\sqrt{2\pi n\lambda }}\text { exp }\{-n\lambda /2\bar{X}\}\right] \nonumber \\&=\bar{X}\left[ 1-\frac{1}{\sqrt{n\lambda /\bar{X}}\sqrt{2\pi }\varPhi (\sqrt{n\lambda /\bar{X}})+\text { exp }\{-n\lambda /2\bar{X}\}}\text { exp }\{-n\lambda /2\bar{X}\}\right] \!. \nonumber \end{aligned}$$