The Inference on the Location Parameters Under Multivariate Skew Normal Settings

Abstract

In this paper, the sampling distributions of multivariate skew normal distribution are studied. Confidence regions of the location parameter, $\varvec{\mu }$, with known scale parameter and shape parameter are obtained by the pivotal method, Inferential Models (IMs), and robust method, respectively. The hypothesis test is proceeded based on the pivotal method and the power of the test is studied using non-central skew Chi-square distribution. For illustration of these results, the graphs of confidence regions and the power of the test are presented for combinations of various values of parameters. A group of Monte Carlo simulation studies is proceeded to verify the performance of the coverage probabilities at last.

Appendix

1.1 Inferential Models (IMs) for Location Parameter $\varvec{\mu }$ When $\varSigma$ Is Known

In general, IMs consist three steps, association step, predict step and combination step. We will follow this three steps to set up an IM for the location parameter $\varvec{\mu }$.

Association Step. Based on the sample matrix Y which follows the distribution (7), we use the sample mean $\overline{\varvec{Y}}$ defined by (8) following the distribution (10). Thus we obtain the potential association

$$\begin{aligned} \mathbf {\overline{Y}}=a(\varvec{\mu },\mathbf {W})=\varvec{\mu }+\mathbf {W}, \end{aligned}$$

where the auxiliary random vector $\mathbf {W}\sim SN_p(0,\varSigma /n,\sqrt{n}\varvec{\lambda })$ but the components of $\varvec{W}$ are not independent. So we use transformed IMs as follow, (see Martin and Liu [20] Sect. 4.4 for more detail on validity of transformed IMs). By Lemmas 1 and 3, we use linear transformations

$$\varvec{V}=A' \varSigma ^{-1/2}\varvec{W}$$

where A is an orthogonal matrix with the first column is $\varvec{\lambda }/||\varvec{\lambda }||$, then

$$ \varvec{V}\sim SN_p(0,I_p,\varvec{\lambda ^{*}}) $$

where $\varvec{\lambda ^{*}}=(\lambda ^{*},0,\ldots ,0)'$ with $\lambda ^*=||\varvec{\lambda }||$. Thus each component of $\varvec{V}$ are independent. To be concrete, let $\varvec{V}=(V_1,\ldots ,V_p)'$, $V_1\sim SN(0,1,\lambda ^*)$ and $V_i\sim N(0,1)$ for $i=2,\ldots ,p$. Therefore, we obtain a new association

$$\begin{aligned} A'\varSigma ^{-1/2}\overline{\varvec{Y}}= & {} A'\varSigma ^{-1/2}\varvec{\mu }+\varvec{V}\\= & {} A'\varSigma ^{-1/2}\varvec{\mu }+G^{-1}\left( \varvec{U}\right) \end{aligned}$$

where $\varvec{U}=\left( U_{1},U_{2},\ldots ,U_{p}\right) '$, $G^{-1}\left( \varvec{U}\right) =\left( G_{1}^{-1}\left( U_{1}\right) ,G_{2}^{-1}\left( U_{2}\right) ,\ldots ,G_{p}^{-1}\left( U_{p}\right) \right) '$ with $G_{1}\left( \cdot \right)$ is the cdf of $SN\left( 0,1,\lambda ^{*}\right)$, $G_{i}\left( \cdot \right)$ is the cdf of $N\left( 0,1\right)$ for $i=2,\ldots ,p$, and $U_{i}$’s follow $U\left( 0,1\right)$ independently for $i=1,\ldots ,p$.

To make the association to be clearly presented, we write down the component wise associations as follows

where $\left( A'\varSigma ^{-1/2}\overline{\varvec{Y}}\right) _{i}$ and $\left( A'\varSigma ^{-1/2}\varvec{\mu }\right) _{i}$ represents the ith component of $A'\varSigma ^{-1/2}\overline{\varvec{Y}}$ and $A'\varSigma ^{-1/2}\varvec{\mu }$, respectively. $G_{1}\left( \cdot \right)$ represents the cdf of $SN\left( 0,1,\lambda ^{*}\right)$ and $G_{i}\left( \cdot \right)$ represents the cdf of $N\left( 0,1\right)$ for $i=2,\ldots ,p$, and $U_{i}\sim U\left( 0,1\right)$ are independently distributed for $i=1,\ldots ,p$.

Thus for any observation $\overline{\varvec{y}}$, and $u_i \in (0,1)$ for $i=1,\ldots ,p$, we have the solution set

$$\begin{aligned} \varTheta _{\overline{\varvec{y}}}\left( \varvec{\mu }\right)&=\left\{ \varvec{\mu }:A'\varSigma ^{-1/2}\overline{\varvec{y}}=A'\varSigma ^{-1/2}\varvec{\mu }+G^{-1}\left( \varvec{U}\right) \right\} \\&=\left\{ \varvec{\mu }:G\left( A'\varSigma ^{-1/2}\left( \overline{\varvec{y}}-\varvec{\mu }\right) \right) =\varvec{U}\right\} \end{aligned}$$

Predict Step. To predict the auxiliary vector $\varvec{U}$, we use the default predictive random set for each components

Combine Step. By the above two steps, we have the combined set

$$ \varTheta _{\overline{\mathbf{Y}}}\left( S\right) =\left\{ \varvec{\mu }:\max \left\{ |G\left( A'\varSigma ^{-1/2}\left( \overline{\varvec{y}}-\varvec{\mu }\right) \right) -0.5|\right\} \le \max \left\{ |\mathbf{U}-0.5|\right\} \right\} . $$

where

$$\begin{aligned} \max \left\{ \left| G\left( A'\varSigma ^{-1/2}\left( \overline{\varvec{y}}-\varvec{\mu }\right) \right) -0.5\right| \right\} =\max _{i=1,\ldots ,p}\left\{ \left| \left( G\left( A'\varSigma ^{-1/2}\left( \overline{\varvec{y}}-\varvec{\mu }\right) \right) \right) _{i}-0.5\right| \right\} \end{aligned}$$

and

$$ max\left\{ \left| \varvec{U}-0.5\right| \right\} =\max _{i=1,\ldots ,p}\left\{ \left| U_{i}-0.5\right| \right\} . $$

Thus, apply above IM, for any singleton assertion $A=\left\{ \varvec{\mu }\right\}$, by definition of believe function and plausibility function, we obtain

$$ \mathsf {bel}_{\overline{\varvec{Y}}}\left( A;\mathscr {S}\right) =P\left( \varTheta _{\overline{\varvec{Y}}}\left( \mathscr {S}\right) \subseteq A\right) =0 $$

since $\left\{ \varTheta _{\overline{\varvec{Y}}}\left( \mathscr {S}\right) \subseteq A\right\} =\emptyset$, and

$$\begin{aligned} \mathsf {pl}_{\overline{\varvec{Y}}}\left( A;\mathscr {S}\right)&=1-\mathsf {bel}_{\overline{\varvec{Y}}}\left( A^{C};\mathscr {S}\right) =1-P_{\mathscr {S}}\left( \varTheta _{\overline{\varvec{Y}}}\left( \mathscr {S}\right) \subseteq A^{C}\right) \\&=1-\left( \max \left\{ |2 G\left( A'\varSigma ^{-1/2}\left( \overline{\varvec{y}}-\varvec{\mu }\right) \right) -1|\right\} \right) ^{p}. \end{aligned}$$

Then the Theorem 3 follows by above computations.

1.2 Robust Method for Location Parameter $\varvec{\mu }$ When $\varSigma$ and $\varvec{\lambda }$ Are Known

Based on the distribution of $\varvec{\overline{Y}}\sim SN_p(\varvec{\mu },\frac{\varSigma }{n},\sqrt{n}\varvec{\lambda })$, we obtain the confidence distribution of $\varvec{\mu }$ given $\overline{\varvec{y}}$ has pdf

$$ f(\varvec{\mu }|\overline{\varvec{Y}}=\overline{\varvec{y}})=2\phi (\varvec{\mu };\varvec{\overline{y}},\frac{\varSigma }{n})\varPhi (n{\varvec{\lambda } \varSigma ^{-1/2}}(\varvec{\overline{y}}-\varvec{\mu })). $$

At confidence level $1-\alpha$, it is natural to construct the confidence set $\mathscr {S}$, i.e. a set $\mathscr {S}$ such that

$$\begin{aligned} P(\varvec{\mu } \in \mathscr {S})=1-\alpha . \end{aligned}$$

(20)

To choose one set out of infinity many possible sets satisfying condition (20), we follow the idea of the most robust confidence set discussed by Kreinovich [4], for any connected set $\mathscr {S}$, defines the measure of robustness of the set $\mathscr {S}$

$$ r\left( \mathscr {S}\right) \equiv \max _{\varvec{y}\in \partial \mathscr {S}}f_{\overline{\varvec{Y}}}\left( \varvec{y}\right) . $$

Then at confidence level $1-\alpha$, we obtain the most robust confidence set

$$ \mathscr {S}=\left\{ \varvec{y}:f_{\overline{Y}}\left( \varvec{y}\right) \ge c_{0}\right\} , $$

where $c_0$ is uniquely determined by the conditions $f_{\overline{\varvec{Y}}}(\varvec{y})\equiv c_0$ and $\int _{\mathscr {S}}f_{\overline{\varvec{Y}}}\left( \varvec{y}\right) \text {d}\varvec{y}=1-\alpha$.

Remark 2

As mentioned by Kreinovich in [4], for Gaussian distribution, such an ellipsoid is indeed selected as a confidence set.