Empirical phi-divergence test statistics for the difference of means of two populations

Abstract

Empirical phi-divergence test statistics have demostrated to be a useful technique for the simple null hypothesis to improve the finite sample behavior of the classical likelihood ratio test statistic, as well as for model misspecification problems, in both cases for the one population problem. This paper introduces this methodology for two-sample problems. A simulation study illustrates situations in which the new test statistics become a competitive tool with respect to the classical z test and the likelihood ratio test statistic.

Appendix

1.1 Proof of Lemma 1

In a similar way as in Hall and Scala (1990), we can establish

$$\begin{aligned} \lambda _{1}= & {} \lambda _{1}\left( \mu \right) =-\sigma _{1}^{-2}\left( \overline{X}-\mu \right) +O_{p}(m^{-1})\text { and }\\&\quad \lambda _{2}=\lambda _{2}\left( \mu \right) = -\sigma _{2}^{-2}\left( \overline{Y}-\mu -\delta _{0}\right) +O_{p}(n^{-1}). \end{aligned}$$

Now applying that

$$\begin{aligned} m\lambda _{1}\left( \mu \right) +n\lambda _{2}\left( \mu \right) =0, \end{aligned}$$

we have

$$\begin{aligned} m\sigma _{1}^{-2}\left( \overline{X}-\mu \right) +O_{p}(1)+n\sigma _{2} ^{-2}\left( \overline{Y}-\mu -\delta _{0}\right) +O_{p}(1)=0. \end{aligned}$$

(35)

Solving the equation for $\mu$, we have the enunciated result.

1.2 Proof of Theorem 2

First, we are going to establish

$$\begin{aligned}&\frac{2}{\phi ^{\prime \prime }\left( 1\right) }\sum \limits _{i=1}^{m}mp_{i} \phi \left( \frac{1}{mp_{i}}\right) =m\left( \frac{\overline{X}-\mu }{\sigma _{1}}\right) ^{2}+o_{p}(1) \end{aligned}$$

(36)

$$\begin{aligned}&\frac{2}{\phi ^{\prime \prime }\left( 1\right) } {\textstyle \sum \limits _{j=1}^{n}} nq_{j}\phi \left( \frac{1}{nq_{j}}\right) =n\left( \frac{\overline{Y} -\mu -\delta _{0}}{\sigma _{2}}\right) ^{2}+o_{p}(1). \end{aligned}$$

(37)

If we denote $W_{i}=\lambda _{1}\left( \mu \right) \left( X_{i}-\mu \right)$ we have $\phi \left( \frac{1}{mp_{i}}\right) =\phi \left( 1+W_{i}\right)$. A Taylor expansion gives

$$\begin{aligned} \phi \left( 1+W_{i}\right) =\phi \left( 1\right) +\phi ^{\prime }\left( 1\right) W_{i}+\frac{1}{2}\phi ^{\prime \prime }\left( 1\right) W_{i} ^{2}+o\left( W_{i}^{2}\right) . \end{aligned}$$

On the other hand

$$\begin{aligned} mp_{i}=\frac{1}{1+\lambda _{1}\left( \mu \right) \left( X_{i}-\mu \right) }=\frac{1}{1+W_{i}}=1-W_{i}+W_{i}^{2}+o\left( W_{i}^{2}\right) . \end{aligned}$$

Then

$$\begin{aligned}&\frac{2}{\phi ^{\prime \prime }\left( 1\right) }\sum \limits _{i=1}^{m}mp_{i} \phi \left( \frac{1}{mp_{i}}\right) =\frac{2}{\phi ^{\prime \prime }\left( 1\right) }\sum \limits _{i=1}^{m}\left( 1-W_{i}+W_{i}^{2}+o\left( W_{i} ^{2}\right) \right) \left( \frac{1}{2}\phi ^{\prime \prime }\left( 1\right) W_{i}^{2}\right. \\&\quad \quad \left. +\,o\left( W_{i}^{2}\right) \right) =\frac{2}{\phi ^{\prime \prime }\left( 1\right) }\left( \frac{1}{2} \phi ^{\prime \prime }\left( 1\right) \sum \limits _{i=1}^{m}W_{i}^{2} +\sum \limits _{i=1}^{m}o\left( W_{i}^{2}\right) -\frac{1}{2}\phi ^{\prime \prime }\left( 1\right) \sum \limits _{i=1}^{m}W_{i}^{3}\right. \\&\quad \quad \left. -\,\sum \limits _{i=1}^{m}o\left( W_{i}^{2}\right) W_{i}+\frac{1}{2}\phi ^{\prime \prime }\left( 1\right) \sum \limits _{i=1}^{m}W_{i}^{4} +\sum \limits _{i=1}^{m}o\left( W_{i}^{2}\right) W_{i}^{2}\right. \\&\quad \quad \left. +\,\frac{1}{2}\phi ^{\prime \prime }\left( 1\right) \sum \limits _{i=1}^{m}o\left( W_{i}^{2}\right) W_{i}^{2}+\sum \limits _{i=1} ^{m}o\left( W_{i}^{2}\right) o\left( W_{i}^{2}\right) \right) \\&\quad =\sum \limits _{i=1}^{m}W_{i}^{2}+\frac{2}{\phi ^{\prime \prime }\left( 1\right) }\sum \limits _{i=1}^{m}o\left( W_{i}^{2}\right) -\sum \limits _{i=1}^{m}W_{i}^{3}-\sum \limits _{i=1}^{m}o\left( W_{i}^{2}\right) W_{i}+\sum \limits _{i=1}^{m}W_{i}^{4}\\&\quad \quad +\,\frac{2}{\phi ^{\prime \prime }\left( 1\right) }\sum \limits _{i=1} ^{m}o\left( W_{i}^{2}\right) +\sum \limits _{i=1}^{m}o\left( W_{i} ^{2}\right) W_{i}^{2}+\frac{2}{\phi ^{\prime \prime }\left( 1\right) } \sum \limits _{i=1}^{m}o\left( W_{i}^{2}\right) o\left( W_{i}^{2}\right) . \end{aligned}$$

But

$\bullet$ $\begin{array}{ll}\frac{2}{\phi ^{\prime \prime }\left( 1\right) } {\textstyle \sum \limits _{i=1}^{m}} o\left( W_{i}^{2}\right) &{}=\frac{2}{\phi ^{\prime \prime }\left( 1\right) } {\textstyle \sum \limits _{i=1}^{m}} o\left( \lambda _{1}^{2}\left( \mu \right) \left( X_{i}-\mu \right) ^{2}\right) =\frac{2}{\phi ^{\prime \prime }\left( 1\right) }m\frac{1}{m} {\textstyle \sum \limits _{i=1}^{m}} \left( X_{i}-\mu \right) ^{2}o\left( \lambda _{1}^{2}\left( \mu \right) \right) \\ &{}=mo_{p}(1)o\left( O_{p}\left( m^{-1}\right) \right) =o_{p}(1)\end{array}$, because

$$\begin{aligned} \lambda _{1}\left( \mu \right) =O_{p}(m^{-1/2}) \end{aligned}$$

(see page 220 in Owen (2001)), and

$$\begin{aligned} \frac{1}{m}\sum \limits _{i=1}^{m}\left( X_{i}-\mu \right) ^{2}=o_{p}(1) \end{aligned}$$

applying the strong law of large numbers.

$\bullet$ $\left| {\textstyle \sum \limits _{i=1}^{m}} W_{i}^{3}\right| \le \left| \lambda _{1}\left( \mu \right) ^{3}\right| m\frac{1}{m} {\textstyle \sum \limits _{i=1}^{m}} \left| X_{i}-\mu \right| ^{3}=O_{p}(m^{-3/2})mo\left( m^{1/2}\right) =o_{p}(1)$, because

$$\begin{aligned} \frac{1}{m}\sum \limits _{i=1}^{m}\left| X_{i}-\mu \right| ^{3}=o\left( m^{1/2}\right) , \end{aligned}$$

by Lemma 11.3 in page 218 in Owen (2001).

$\bullet$ $\left| {\textstyle \sum \limits _{i=1}^{m}} o\left( W_{i}^{2}\right) W_{i}\right| \le o\left( \lambda _{1} ^{2}\left( \mu \right) \right) \lambda _{1}\left( \mu \right) m\frac{1}{m} {\textstyle \sum \limits _{i=1}^{m}} \left| X_{i}-\mu \right| ^{3}=o\left( O_{p}(m^{-1}\right) O_{p}(m^{-1/2})o(m^{3/2})=o_{p}(1)$.

$\bullet$ $\left| {\textstyle \sum \limits _{i=1}^{m}} W_{i}^{4}\right| \le \left| \lambda _{1}\left( \mu \right) ^{4}\right| {\textstyle \sum \limits _{i=1}^{m}} \left| X_{i}-\mu \right| ^{4}\le O_{p}\left( m^{-2}\right) mZ_{m}\frac{1}{m} {\textstyle \sum \limits _{i=1}^{m}} \left| X_{i}-\mu \right| ^{3}=O_{p}\left( m^{-2}\right) mo(m^{1/2})O(m^{1/2})=o_{p}(1),$ because

$$\begin{aligned} Z_{m}=\max _{1\le i\le m}\left| X_{i}-\mu \right| =O(m^{1/2}) \end{aligned}$$

applying Lemma 11.2 in page 218 in Owen (2001).

$\bullet$ $\left| \frac{2}{\phi ^{\prime \prime }\left( 1\right) } {\textstyle \sum \limits _{i=1}^{m}} o\left( W_{i}^{2}\right) o\left( W_{i}^{2}\right) \right| \le \frac{2}{\phi ^{\prime \prime }\left( 1\right) }\left| \lambda _{1}\left( \mu \right) ^{4}\right| {\textstyle \sum \limits _{i=1}^{m}} \left| X_{i}-\mu \right| ^{4}=o_{p}(1)$. Therefore

$$\begin{aligned} \frac{2}{\phi ^{\prime \prime }\left( 1\right) }\sum \limits _{i=1}^{m}mp_{i} \phi \left( \frac{1}{mp_{i}}\right)&=\sum \limits _{i=1}^{m}W_{i}^{2} +o_{p}(1)\\&=\sum \limits _{i=1}^{m}\lambda _{1}^{2}\left( \mu \right) (X_{i}-\mu )^{2}+o_{p}(1)\\&=\sigma _{1}^{-4}\left( \overline{X}-\mu \right) ^{2}m\frac{1}{m} \sum \limits _{i=1}^{m}(X_{i}-\mu )^{2}+o_{p}(1)\\&=m\left( \frac{\overline{X}-\mu }{\sigma _{1}}\right) ^{2}+o_{p}(1). \end{aligned}$$

In a similar way, we can get

$$\begin{aligned} \frac{2}{\phi ^{\prime \prime }\left( 1\right) }\sum \limits _{j=1}^{n}nq_{j} \phi \left( \frac{1}{nq_{j}}\right) =n\left( \frac{\overline{Y}-\mu -\delta _{0}}{\sigma _{2}}\right) ^{2}+o_{p}(1). \end{aligned}$$

Therefore,

$$\begin{aligned} T_{\phi }\left( \delta _{0}\right)&=\frac{2}{\phi ^{\prime \prime } (1)}\left\{ \sum \limits _{i=1}^{m}m\widetilde{p}_{i}\phi \left( \frac{1}{m\widetilde{p}_{i}}\right) +\sum \limits _{j=1}^{n}n\widetilde{q}_{j} \phi \left( \frac{1}{n\widetilde{q}_{j}}\right) \right\} \\&=m\left( \frac{\overline{X}-\widetilde{\mu }}{\sigma _{1}}\right) ^{2}+n\left( \frac{\overline{Y}-\widetilde{\mu }-\delta _{0}}{\sigma _{2} }\right) ^{2}+o_{p}(1). \end{aligned}$$

Applying (35),

$$\begin{aligned} n\sigma _{2}^{-2}\left( \overline{Y}-\widetilde{\mu }-\delta _{0}\right) =-m\sigma _{1}^{-2}\left( \overline{X}-\widetilde{\mu }\right) +O_{p}(1) \end{aligned}$$

and

$$\begin{aligned} T_{\phi }\left( \delta _{0}\right)&=m\left( \frac{\overline{X}-\widetilde{\mu }}{\sigma _{1}}\right) ^{2}-m\sigma _{1}^{-2}\left( \overline{X}-\widetilde{\mu }\right) \left( \overline{Y}-\widetilde{\mu }-\delta _{0}\right) +o_{p}(1)\\&=m\left( \frac{\overline{X}-\widetilde{\mu }}{\sigma _{1}^{2}}\right) \left( \overline{X}-\widetilde{\mu }-\overline{Y}+\widetilde{\mu }+\delta _{0}\right) +o_{p}(1)\\&=m\left( \frac{\overline{X}-\widetilde{\mu }}{\sigma _{1}^{2}}\right) \left( \overline{X}-\overline{Y}+\delta _{0}\right) +o_{p}(1)\\&=-n\sigma _{2}^{-2}\left( \overline{Y}-\widetilde{\mu }-\delta _{0}\right) \left( \overline{X}-\overline{Y}+\delta _{0}\right) +o_{p}(1). \end{aligned}$$

From (19) we have

$$\begin{aligned} \overline{Y}-\mu -\delta _{0}&=\overline{Y}-\dfrac{\dfrac{m\overline{X} }{\sigma _{1}^{2}}+n\dfrac{\left( \overline{Y}-\delta _{0}\right) }{\sigma _{2}^{2}}}{\dfrac{m}{\sigma _{1}^{2}}+\dfrac{n}{\sigma _{2}^{2}}}-\delta _{0}\\&=\frac{m}{\sigma _{1}^{2}}\frac{\left( -\overline{X}+\overline{Y} -\delta _{0}\right) }{\frac{\sigma _{1}^{2}}{m}+\frac{\sigma _{2}^{2}}{n}}. \end{aligned}$$

Therefore,

$$\begin{aligned} T_{\phi }\left( \delta _{0}\right)&=\frac{nm\sigma _{1}^{-2}\left( \overline{Y}-\overline{X}-\delta _{0}\right) }{\frac{\sigma _{1}^{2}}{m} +\frac{\sigma _{2}^{2}}{n}}(-1)\left( \overline{X}-\overline{Y}+\delta _{0}\right) \sigma _{2}^{-2}\nonumber \\&=\frac{mn}{m+n}\left( \overline{X}-\left( \overline{Y}-\delta _{0}\right) \right) ^{2}\sigma _{1}^{-2}\sigma _{2}^{-2}\left( \frac{m\sigma _{1} ^{-2}+n\sigma _{2}^{-2}}{m+n}\right) ^{-1}+o_{p}(1)\nonumber \\&=\frac{1}{\frac{m}{m+n}\sigma _{2}^{2}+\frac{n}{m+n}\sigma _{1}^{2}}\frac{mn}{m+n}\left( \overline{X}-\left( \overline{Y}-\delta _{0}\right) \right) ^{2}+o_{p}(1). \end{aligned}$$

(38)

Now we have,

$$\begin{aligned}&\sqrt{m}\left( \overline{X}-\mu \right) \underset{m\rightarrow \infty }{\overset{\mathcal {L}}{\rightarrow }}\mathcal {N}(0,\sigma _{1}^{2}),\\&\sqrt{n}\left( \overline{Y}-\left( \mu -\delta _{0}\right) \right) \underset{n\rightarrow \infty }{\overset{\mathcal {L}}{\rightarrow }} \mathcal {N}(0,\sigma _{2}^{2}). \end{aligned}$$

and

$$\begin{aligned}&\sqrt{\frac{mn}{m+n}}\left( \overline{X}-\mu \right) \underset{m,n\rightarrow \infty }{\overset{\mathcal {L}}{\rightarrow }} \mathcal {N}(0,\left( 1-\nu \right) \sigma _{1}^{2}).\\&\sqrt{\frac{mn}{m+n}}\left( \overline{Y}-\left( \mu +\delta _{0}\right) \right) \underset{m,n\rightarrow \infty }{\overset{\mathcal {L}}{\rightarrow } }\mathcal {N}(0,\nu \sigma _{2}^{2}), \end{aligned}$$

where is such that (2). Hence

$$\begin{aligned} \sqrt{\frac{mn}{m+n}}\left( \overline{X}-\overline{Y}+\delta _{0}\right) \underset{m,n\rightarrow \infty }{\overset{\mathcal {L}}{\rightarrow }} \mathcal {N}\left( 0,\left( 1-\nu \right) \sigma _{1}^{2}+\nu \sigma _{2}^{2}\right) , \end{aligned}$$

from which is obtained

$$\begin{aligned} \sqrt{\frac{1}{\frac{m}{m+n}\sigma _{2}^{2}+\frac{n}{m+n}\sigma _{1}^{2}}} \sqrt{\frac{mn}{m+n}}\left( \overline{X}-\overline{Y}+\delta _{0}\right) \underset{m,n\rightarrow \infty }{\overset{\mathcal {L}}{\rightarrow }} \mathcal {N}(0,1) \end{aligned}$$

and now the result follows.