Robust tests for the equality of two normal means based on the density power divergence

Abstract

Statistical techniques are used in all branches of science to determine the feasibility of quantitative hypotheses. One of the most basic applications of statistical techniques in comparative analysis is the test of equality of two population means, generally performed under the assumption of normality. In medical studies, for example, we often need to compare the effects of two different drugs, treatments or preconditions on the resulting outcome. The most commonly used test in this connection is the two sample $t$ test for the equality of means, performed under the assumption of equality of variances. It is a very useful tool, which is widely used by practitioners of all disciplines and has many optimality properties under the model. However, the test has one major drawback; it is highly sensitive to deviations from the ideal conditions, and may perform miserably under model misspecification and the presence of outliers. In this paper we present a robust test for the two sample hypothesis based on the density power divergence measure (Basu et al. in Biometrika 85(3):549–559, 1998), and show that it can be a great alternative to the ordinary two sample $t$ test. The asymptotic properties of the proposed tests are rigorously established in the paper, and their performances are explored through simulations and real data analysis.

Appendix

Proof of Theorem 1

As $\widehat{\mu }_{i\beta }$ is the solution of the estimating equation $_{1}h_{n_{i},\beta }^{\prime }\left( \mu _{i},\sigma \right) =0$, we get from Eq. (7)

$$\begin{aligned} \sqrt{n_{i}}(\widehat{\mu }_{i\beta }-\mu _{i0})=\sqrt{n_{i}} \varvec{J}_{11,\beta }^{-1}( \sigma _{0})\,_{1}h_{n_{i},\beta }^{\prime }\left( \mu _{i0},\sigma _0\right) +o_{p}(1),\quad i=1,2. \end{aligned}$$

Hence, using (9) we get

$$\begin{aligned} \sqrt{n_{i}}(\widehat{\mu }_{i\beta }-\mu _{i0})\underset{n_{i}\rightarrow \infty }{\overset{\mathcal {L}}{\longrightarrow }}{\mathcal {N}}\left( 0,K_{11,\beta }\varvec{(}\sigma _0)J_{11,\beta }^{-2}( \sigma _0) \right) ,\quad i=1,2, \end{aligned}$$

(27)

where

$$\begin{aligned} K_{11,\beta }\varvec{(}\sigma _0)J_{11,\beta }^{-2}(\sigma _{0})=\sigma _0^{2}\left( \beta +1\right) ^{3}\left( 2\beta +1\right) ^{- \frac{3}{2}}. \end{aligned}$$

(28)

It is clear that $\widehat{\mu }_{1\beta }$ and $\widehat{\mu }_{2\beta }$ are based on two independent set of observations, hence, $Cov(\widehat{\mu }_{1\beta },\widehat{\mu }_{2\beta })=0$. As $_{2}h_{n_1,n_2,\beta }^{\prime }(\widehat{\varvec{\eta }}_\beta )=0$, taking a Taylor series expansion around $\varvec{\eta }_0$ we get

$$\begin{aligned} _{2}h_{n_1,n_2,\beta }^{\prime }(\widehat{\varvec{\eta }}_\beta ) =&_{2}h_{n_1,n_2,\beta }^{\prime }( \varvec{\eta }_0) + \left. \frac{\partial }{\partial \mu _1}\,_{2}h_{n_1,n_2,\beta }^{\prime }( \varvec{\eta })\right| _{\varvec{\eta }=\varvec{\eta }_0} (\widehat{\mu }_{1\beta }-\mu _{10}) \nonumber \\&+ \left. \frac{\partial }{\partial \mu _2}\,_{2} h_{n_1,n_2,\beta }^{\prime \prime }( \varvec{\eta }) \right| _{\varvec{\eta }=\varvec{\eta }_0} (\widehat{ \mu }_{2\beta }-\mu _{20}) \nonumber \\&+ \left. \frac{\partial }{\partial \sigma }\,_{2}h_{n_1,n_2,\beta }^{\prime \prime }( \varvec{\eta })\right| _{\varvec{\eta }=\varvec{\eta }_0} \left( \widehat{\sigma }_\beta -\sigma _0\right) +o_{p}\left( (n_1+n_2)^{-1/2}\right) \nonumber \\ =&\ 0. \end{aligned}$$

(29)

Notice that

$$\begin{aligned}&\lim _{n_1,n_2\rightarrow \infty } \left. \frac{\partial }{\partial \mu _1}\,_{2}h_{n_1,n_2,\beta }^{\prime }( \varvec{\eta }) \right| _{\varvec{\eta }=\varvec{\eta }_0} \nonumber \\&\quad = \lim _{n_1,n_2\rightarrow \infty } \frac{\partial }{ \partial \mu _1}\left( \frac{n_1}{n_1+n_2}\,_{2}h_{n_1,\beta }^{\prime }\left( \mu _{10},\sigma _0\right) +\frac{ n_2}{n_1+n_2}\,_{2}h_{n_2}^{\prime }(\mu _{10},\sigma _{0})\right) \nonumber \\&\quad =\lim _{n_1,n_2\rightarrow \infty } \frac{n_1}{n_1+n_2} \lim _{n_1,n_2\rightarrow \infty } \left. \frac{\partial }{\partial \mu _1}\,_{2}h_{n_1,\beta }^{\prime }\left( \mu _{1},\sigma _0\right) \right| _{\mu _1 = \mu _{10} }\nonumber \\&\quad = w \varvec{J}_{12,\beta }\left( \sigma _0\right) = 0. \end{aligned}$$

(30)

Similarly we get

$$\begin{aligned} \lim _{n_1,n_2\rightarrow \infty } \left. \frac{\partial }{\partial \mu _2}\,_{2}h_{n_1,n_2,\beta }^{\prime }( \varvec{\eta })\right| _{\varvec{\eta }=\varvec{\eta }_0} =0. \end{aligned}$$

(31)

Moreover,

$$\begin{aligned} \lim _{n_1,n_2\rightarrow \infty } \left. \frac{\partial }{\partial \sigma } _{2}h_{n_1,n_2,\beta }^{\prime }\left( \varvec{\eta } \right) \right| _{\varvec{\eta }=\varvec{\eta }_0}&= \lim _{n_1,n_2\rightarrow \infty }\tfrac{ n_1}{n_1+n_2}\,_{22}h_{n_1,\beta }^{\prime \prime }(\mu _{10},\sigma _0)\nonumber \\&+\lim _{n_1,n_2\rightarrow \infty }\tfrac{n_2}{ n_1+n_2}\,_{22}h_{n_2,\beta }^{\prime \prime }\left( \mu _{20},\sigma _0\right) \nonumber \\&= w\varvec{J}_{22,\beta }\left( \sigma _0\right) +(1-w)\varvec{J}_{22,\beta }( \sigma _0)\nonumber \\&= \varvec{J}_{22,\beta } ( \sigma _0). \end{aligned}$$

(32)

Therefore, using Eqs. (30), (31) and (32) we get from Eq. (29)

$$\begin{aligned} \sqrt{n_1+n_2}\left( \widehat{\sigma }_\beta -\sigma _0\right) =-\varvec{J}_{22,\beta }^{-1}\left( \sigma _0\right) \sqrt{n_1+n_2} \,_{2}h_{n_1,n_2,\beta }^{\prime }( \varvec{\eta }_0) +o_{p}(1). \end{aligned}$$

(33)

Applying (9) and (14) we get

$$\begin{aligned}&\lim _{n_1,n_2\rightarrow \infty } \text {E}\left[ \sqrt{n_1+n_2}\,_{2}h_{n_1,n_2,\beta }^{\prime }(\varvec{\eta }_0) \right] \\&\quad = \lim _{n_1,n_2\rightarrow \infty } \frac{\sqrt{n_1+n_2}}{n_1+n_2}E\left[ n_1\,_{2}h_{n_1,\beta }^{\prime }\left( \mu _{10},\sigma _0\right) +n_2\,_{2}h_{n_2,\beta }^{\prime }\left( \mu _{20},\sigma _{0}\right) \right] \\&\quad = \lim _{n_1,n_2\rightarrow \infty }\sqrt{\frac{n_1}{n_1+n_2}} \lim _{n_1,n_2\rightarrow \infty } E\left[ \sqrt{n_1}\,_{2}h_{n_1,\beta }^{\prime }(\mu _{10},\sigma _0)\right] \\&\qquad + \lim _{n_1,n_2\rightarrow \infty } \sqrt{\frac{ n_2}{n_1+n_2}} \lim _{n_1,n_2\rightarrow \infty } E\left[ \sqrt{n_2}\,_{2}h_{n_2,\beta }^{\prime }\left( \mu _{20},\sigma _0\right) \right] \\&\quad = 0. \end{aligned}$$

Similarly we also have

$$\begin{aligned}&\lim _{n_1,n_2\rightarrow \infty } Var\left[ \sqrt{n_1+n_2}\,_{2}h_{n_1,n_2,\beta }^{\prime }(\varvec{\eta }_0) \right] \\&\quad =\lim _{n_1,n_2\rightarrow \infty } (n_1+n_2)Var\left[ \frac{1}{n_1+n_2}\left( n_1\,_{2}h_{n_1,\beta }^{\prime }(\mu _{10},\sigma _{0})+n_2\,_{2}h_{n_2,\beta }^{\prime }(\mu _{20},\sigma _{0}\right) \right] \\&\quad = \lim _{n_1,n_2\rightarrow \infty } \frac{n_1}{n_1+n_2} \lim _{n_1,n_2\rightarrow \infty } \text {Var}\left[ \sqrt{n_1}\,_{2}h_{n_1,\beta }^{\prime }(\mu _{10},\sigma _0)\right] \\&\qquad + \lim _{n_1,n_2\rightarrow \infty } \frac{n_2}{ n_1+n_2} \lim _{n_1,n_2\rightarrow \infty } Var\left[ \sqrt{n_2}\,_{2}h_{n_2,\beta }^{\prime }(\mu _{20},\sigma _0)\right] \\&\quad = w \varvec{K}_{22,\beta }(\sigma _0) + (1-w) \varvec{K}_{22,\beta }(\sigma _0) \\&\quad = \varvec{K}_{22,\beta }(\sigma _0). \end{aligned}$$

Hence,

$$\begin{aligned} \sqrt{n_1+n_2}\,_{2}h_{n_1,n_2,\beta }^{\prime }\left( \varvec{\eta }_0\right) \underset{n_1,n_2\rightarrow \infty }{\overset{\mathcal {L}}{\longrightarrow }}{\mathcal {N}}\left( 0,\varvec{K}_{22,\beta }(\sigma _0)\right) . \end{aligned}$$

Now, from Eq. (33) we get

$$\begin{aligned} \sqrt{n_1+n_2}\left( \widehat{\sigma }_\beta -\sigma _0\right) \underset{n_1,n_2\rightarrow \infty }{\overset{\mathcal {L}}{ \longrightarrow }}{\mathcal {N}}\left( 0,\varvec{K}_{22,\beta }(\sigma _0)\varvec{J}_{22,\beta }^{-2}(\sigma _0)\right) , \end{aligned}$$

(34)

where

$$\begin{aligned} \varvec{K}_{22,\beta }(\sigma _0)\varvec{J}_{22,\beta }^{-2}(\sigma _0)=\sigma _0^{2} \frac{\left( \beta +1\right) ^{5}}{\left( \beta ^{2}+2\right) ^{2}}\left( \frac{4\beta ^{2}+2}{(1+2\beta )^{5/2}}-\frac{\beta ^{2}}{(1+\beta )^{3}} \right) . \end{aligned}$$

(35)

As $\varvec{J}_{12,\beta }(\sigma _0)=\varvec{J}_{21,\beta }(\sigma _0)=0$, it is clear that

$$\begin{aligned} \lim _{n_1,n_2\rightarrow \infty } \left. \frac{\partial ^2 }{\partial \mu _1 \partial \sigma }\,h_{n_1,n_2,\beta }( \varvec{\eta })\right| _{\varvec{\eta } = \varvec{\eta }_0} = \lim _{n_1,n_2\rightarrow \infty } \left. \frac{\partial ^2 }{\partial \mu _2 \partial \sigma }\,h_{n_1,n_2,\beta }( \varvec{\eta }) \right| _{\varvec{\eta } = \varvec{\eta }_0} =0. \end{aligned}$$

Therefore, $Cov(\widehat{\mu }_{1\beta },\widehat{\sigma }_{\beta })=Cov(\widehat{\mu }_{2\beta },\widehat{\sigma }_{\beta })=0$. Moreover, $Cov(\widehat{\mu }_{1\beta },\widehat{\mu }_{2\beta })=0$. Combining the results in (27) and (34) we get the variance-covariance matrix of $\sqrt{\frac{n_1n_2}{n_1+n_2}}\widehat{\varvec{\eta }}_{\beta }$ as follows

$$\begin{aligned} \varvec{\Sigma }_{w,\beta }(\sigma _0)=\left( \begin{array}{c@{\quad }c@{\quad }c} \left( 1-w\right) \varvec{K}_{11,\beta }\varvec{(}\sigma _0) \varvec{J}_{11,\beta }^{-2}\left( \sigma _0\right) &{} 0 &{} 0 \\ 0 &{} w \varvec{K}_{11,\beta }\varvec{(}\sigma _0) \varvec{J}_{11,\beta }^{-2}(\sigma _0) &{} 0 \\ 0 &{} 0 &{} w\left( 1-w\right) \varvec{K}_{22,\beta }(\sigma _0)\varvec{J}_{22,\beta }^{-2}\left( \sigma _0\right) \end{array} \right) , \end{aligned}$$

where the values of the diagonal elements are given in (28) and (35). Hence, the theorem is proved.$\square$

Proof of Theorem 2

A Taylor expansion of $d_{\gamma }(f_{\widehat{\mu }_{1\beta },\widehat{\sigma }_\beta },f_{ \widehat{\mu }_{2\beta },\widehat{\sigma }_\beta })$ around $\varvec{\eta }_{0}$ gives

$$\begin{aligned} d_{\gamma }(f_{\widehat{\mu }_{1\beta },\widehat{\sigma }_\beta },f_{ \widehat{\mu }_{2\beta },\widehat{\sigma }_\beta })=d_{\gamma }(f_{\mu _{10},\sigma _0},f_{\mu _{20},\sigma _0})+\varvec{t}_{\gamma }^{T}\left( \varvec{\eta }_{0}\right) (\widehat{\varvec{\eta }}_{\beta }-\varvec{\eta }_{0})+o_{p}\left( \left\| \widehat{ \varvec{\eta }}_{\beta }-\varvec{\eta }_{0}\right\| \right) , \end{aligned}$$

where $\varvec{t}_{\gamma }\left( \varvec{\eta }_{0}\right) =\frac{ \partial }{\partial \varvec{\eta }}\left. d_{\gamma }(f_{\mu _1,\sigma },f_{\mu _2,\sigma })\right| _{\varvec{\eta }=\varvec{\eta }_{0}}$; the expressions of the components $t_{\gamma , i}\left( \varvec{\eta }_{0}\right)$, $i=1,2,3$, are given in (18)–(20). Hence, the result directly follows from Theorem 1. $\square$

Proof of Theorem 3

If $\mu _{10}=\mu _{20}$, it is obvious that $d_{\gamma }(f_{\mu _{10},\sigma _0},f_{\mu _{20},\sigma _0})=0$, and $\varvec{t}_{\gamma } ( \varvec{\eta }_{0})=0$. Hence, a second order Taylor expansion of $d_{\gamma }(f_{\widehat{\mu }_{1\beta },\widehat{\sigma }_\beta },f_{ \widehat{\mu }_{2\beta },\widehat{\sigma }_\beta })$ around $\varvec{\eta }_{0}$ gives

$$\begin{aligned} 2d_{\gamma }(f_{\widehat{\mu }_{1\beta },\widehat{\sigma }_\beta },f_{ \widehat{\mu }_{2\beta },\widehat{\sigma }_\beta })=(\widehat{\varvec{ \eta }}_{\beta }-\varvec{\eta }_{0})^{T}\varvec{A}_{\gamma }\left( \sigma _0\right) (\widehat{\varvec{\eta }}_{\beta }-\varvec{\eta } _{0})+o_p(\left\| \widehat{\varvec{\eta }}_{\beta }-\varvec{\eta }_{0}\right\| ^{2}), \qquad \end{aligned}$$

(36)

where $\varvec{A}_{\gamma }(\sigma _0)$ is the matrix containing the second derivatives of $d_{\gamma }(f_{\mu _1,\sigma },f_{\mu _2,\sigma })\$ evaluated at $\mu _{10}=\mu _{20}$. It can be shown that

$$\begin{aligned} \varvec{A}_{\gamma }\left( \sigma _0\right) \varvec{=}\ell _{\gamma }(\sigma _0)\left( \begin{array}{ccc} 1 &{}\quad -1 &{}\quad 0 \\ -1 &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 \end{array} \right) , \end{aligned}$$

where

$$\begin{aligned} \ell _{\gamma }(\sigma _0)=\sigma ^{-(\gamma +2)}\left( 2\pi \right) ^{- \frac{\gamma }{2}}\left( \gamma +1\right) ^{-\frac{1}{2}}. \end{aligned}$$

Therefore, Eq. (36) simplifies to

$$\begin{aligned} 2d_{\gamma }(f_{\widehat{\mu }_{1\beta },\widehat{\sigma }_\beta },f_{ \widehat{\mu }_{2\beta },\widehat{\sigma }_\beta })&= \left( \begin{pmatrix} \widehat{\mu }_{1\beta } \\ \widehat{\mu }_{2\beta } \end{pmatrix} - \begin{pmatrix} \mu _{10} \\ \mu _{20} \end{pmatrix} \right) ^{T}\varvec{A}_{\gamma }^{*}\left( \sigma _0\right) \left( \begin{pmatrix} \widehat{\mu }_{1\beta } \\ \widehat{\mu }_{2\beta } \end{pmatrix} - \begin{pmatrix} \mu _{10} \\ \mu _{20} \end{pmatrix} \right) \nonumber \\&\quad +\, o_{p}\left( \left\| \widehat{\varvec{\eta }}_{\beta }- \varvec{\eta }_{0}\right\| ^{2}\right) , \end{aligned}$$

where

$$\begin{aligned} \varvec{A}_{\gamma }^{*}\left( \sigma _0\right) =\ell _{\gamma }(\sigma _0)\left( \begin{array}{cc} 1 &{}\quad -1 \\ -1 &{}\quad 1 \end{array} \right) . \end{aligned}$$

From Theorem 1 we know that

$$\begin{aligned} \sqrt{\frac{n_1n_2}{n_1+n_2}}\left( \begin{pmatrix} \widehat{\mu }_{1\beta } \\ \widehat{\mu }_{2\beta } \end{pmatrix} - \begin{pmatrix} \mu _{10} \\ \mu _{20} \end{pmatrix} \right) ^{T}\underset{}{\overset{\mathcal {L}}{\longrightarrow }}{\mathcal {N}} \left( \varvec{0}_{2},\varvec{\Sigma }_{w,\beta }^{*}(\sigma _{0})\right) \!, \end{aligned}$$

where

$$\begin{aligned} \varvec{\Sigma }_{w,\beta }^{*}(\sigma _0)=K_{11,\beta } \varvec{(}\sigma _0)J_{11,\beta }^{-2}\left( \sigma _0\right) \left( \begin{array}{cc} 1-w &{}\quad 0 \\ 0 &{}\quad w \end{array} \right) \!. \end{aligned}$$

Therefore, $\frac{2 n_1n_2}{n_1+n_2} d_{\gamma }(f_{\widehat{\mu }_{1\beta },\widehat{\sigma }_{\beta }},f_{\widehat{\mu }_{2\beta },\widehat{\sigma }_\beta })$ has the same asymptotic distribution (see Dik and de Gunst 1985) as the random variable

$$\begin{aligned} \sum \limits _{i=1}^{2}\lambda _{i,\beta ,\gamma }(\sigma _0)Z_{i}^{2}, \end{aligned}$$

where $Z_{1}$ and $Z_{2}$ are independent standard normal variables, and

$$\begin{aligned} \lambda _{1,\beta ,\gamma }(\sigma _0)=0\text {,\quad and \quad }\lambda _{2,\beta ,\gamma }(\sigma _0)=K_{11,\beta }\varvec{(}\sigma _0)J_{11,\beta }^{-2}\left( \sigma _0\right) \ell _{\gamma }(\sigma _0)=\lambda _{\beta ,\gamma }(\sigma _0) \end{aligned}$$

are the eigenvalues of the matrix $\varvec{\Sigma }_{w,\beta }^{*}(\sigma _0)\varvec{A}_{\gamma }^{*}\left( \sigma _0\right)$. Hence,

$$\begin{aligned} \frac{2n_1n_2}{n_1+n_2}\frac{d_{\gamma }(f_{\widehat{\mu }_{1\beta },\widehat{\sigma }_\beta },f_{\widehat{\mu }_{2\beta },\widehat{\sigma }_\beta })}{\lambda _{\beta ,\gamma }\,(\sigma _{0} )} \underset{ n_1,n_2\rightarrow \infty }{\overset{\mathcal {L}}{\longrightarrow }}\chi ^{2}(1). \end{aligned}$$

Finally, since $\widehat{\sigma }_\beta$ is a consistent estimator of $\sigma$, replacing $\lambda _{\beta ,\gamma }(\sigma _0)$ by $\lambda _{\beta ,\gamma }(\widehat{\sigma }_\beta )$ and by following Slutsky’s theorem we obtain the desired result. $\square$