A stochastic ordering based on the canonical transformation of skew-normal vectors

Abstract

In this paper, we define a new skewness ordering that enables stochastic comparisons for vectors that follow a multivariate skew-normal distribution. The new ordering is based on the canonical transformation associated with the multivariate skew-normal distribution and on the well-known convex transform order applied to the only skewed component of such canonical transformation. We examine the connection between the proposed ordering and the multivariate convex transform order studied by Belzunce et al. (TEST 24(4):813–834, 2015). Several standard skewness measures like Mardia’s and Malkovich–Afifi’s indices are revisited and interpreted in connection with the new ordering; we also study its relationship with the J-divergence between skew-normal and normal random vectors and with the Negentropy. Some artificial data are used in simulation experiments to illustrate the theoretical discussion; a real data application is provided as well.

Appendix

Lemma 1

(Sign patterns) Let us consider $\alpha _1$ and $\alpha _2$ nonnegative parameters such that $0< \alpha _1< \alpha _2$. The function m defined by

$$\begin{aligned} m(x)=-\frac{x^2}{2}+\frac{(a+bx)^2}{2}+\log \varPhi (\alpha _1 x)-\log \varPhi (\alpha _2(a+bx))-\log b\text{: } x\in {\mathbb {R}}, \end{aligned}$$

with $b>0$, has one, two or three zeros.

Proof

We know that $\hbox {e}^{m(x)}$ is the quotient of the densities of two scalar random variables S and Z such that $S\sim \hbox {SN}_1(0,1,\alpha _1)$ and $\displaystyle Z=\frac{T-a}{b}$ with $T\sim \hbox {SN}_1(0,1,\alpha _2)$. Therefore, we can assert that the number of zeros of the function m is at least one. Since m can be rewritten as $m(x)=m_1(x)-m_2(x)$, with $\displaystyle m_1(x)= -\frac{x^2}{2}+\log \varPhi (\alpha _1 x)$ and $\displaystyle m_2(x)=-\frac{(a+bx)^2}{2}+\log \varPhi (\alpha _2(a+bx))+\log b$, the number of intersections between $m_1$ and $m_2$ gives the number of zeros of m. We now examine the behavior of both functions.

$$\begin{aligned} \lim _{x\rightarrow -\infty }m_1(x)=\lim _{x\rightarrow \infty }m_1(x)=-\infty \quad \text{ and } \quad m_1(x)<0\text{, } x\in {\mathbb {R}}. \end{aligned}$$

(11)

The first derivative of $m_1$ is given by

$$\begin{aligned} m_1^\prime (x)=-x+\frac{\alpha _1\phi (\alpha _1 x)}{\varPhi (\alpha _1 x)}=-x+\frac{\alpha _1}{\frac{1}{\phi (\alpha _1 x)}-r(\alpha _1 x)} \text{, } \text{ with } r(t)=\frac{1-\varPhi (t)}{\phi (t)} \end{aligned}$$

for $t>0$ the well-known Mills’ ratio, which is a convex and strictly decreasing function Baricz (2008). Therefore, when $x>0$ we get

$$\begin{aligned} \displaystyle m_1^{\prime \prime }(x)=-1-\frac{\alpha _1}{(\frac{1}{\phi (\alpha _1 x)}-r(\alpha _1 x))^2}\left( \frac{\alpha _1^2 x}{\phi (\alpha _1 x)}-\alpha _1 r^\prime (\alpha _1 x)\right) <0. \end{aligned}$$

When $x<0$, taking into account that $\displaystyle m_1^\prime (x)=-x+\frac{\alpha _1\phi (\alpha _1 x)}{1-\varPhi (-\alpha _1 x)}=-x+\frac{\alpha _1}{r(-\alpha _1 x)}$ we obtain once again that

$$\begin{aligned} m_1^{\prime \prime }(x)=-1+\frac{\alpha _1^2}{r(-\alpha _1 x)^2}r^\prime (-\alpha _1 x)<0. \end{aligned}$$

Therefore, $m_1$ is a negative concave function having a unique maximum.

On the other hand, for $m_2$ it is straightforward to show that, as long $b\le 1$, we have

$$\begin{aligned} \lim _{x\rightarrow -\infty }m_2(x)=\lim _{x\rightarrow \infty }m_2(x)=-\infty \text{ with } m_2(x)<0 \text{, } x\in {\mathbb {R}}. \end{aligned}$$

(12)

Now, using an analogous argument as previously for $m_1$ we conclude that $m_2$ is a negative concave function with a unique maximum provided that $b\le 1$. Similarly, when $b>1$ we can see that $m_2$ is a concave function with two zeros, being nonnegative between the zeros.

Next, we study the asymptotic relative positions of $m_1$ and $m_2$, as well as the limit behavior of the function m.

Firstly, we calculate the limit

$$\begin{aligned} \displaystyle \lim _{x\rightarrow \infty } m(x)= & {} \frac{a^2}{2}-\log b+\lim _{x\rightarrow \infty }\left( \frac{(b^2-1)x^2}{2}+abx\right) \nonumber \\= & {} {\left\{ \begin{array}{ll} -\infty &{} \quad \text{ if } b<1 \, \text{ or } \text{ if } \,b=1 \, \text{ and } \,a<0 \\ \infty &{} \quad \text{ if } b>1 \text{ or } \text{ if } b=1 \, \text{ and } \,a>0 \\ \end{array}\right. }. \end{aligned}$$

(13)

Secondly, we need the limit

$$\begin{aligned} \displaystyle \lim _{x\rightarrow -\infty } m(x)=\frac{a^2}{2}-\log b+\lim _{x\rightarrow -\infty } x^2\left[ \frac{b^2-1}{2}+\frac{ab}{x}+\frac{1}{x^2}\log \left( \frac{\varPhi (\alpha _1 x)}{\varPhi (\alpha _2(a+bx))}\right) \right] . \end{aligned}$$

The calculus of this limit requires looking into some details about the asymptotic behavior of the last summand in the expression above. We sketch the computations as follows: Taking into account that

$$\begin{aligned}&\displaystyle \lim _{x\rightarrow -\infty }\frac{\varPhi (\alpha _1 x)}{\varPhi (\alpha _2(a+bx))}=\frac{\alpha _1}{\alpha _2 b}\lim _{x\rightarrow -\infty }\frac{\phi (\alpha _1 x)}{\phi (\alpha _2(a+bx))}\\&\quad =\frac{\alpha _1}{\alpha _2 b}\exp \left\{ \lim _{x\rightarrow -\infty }\left( -\frac{\alpha _1^2x^2}{2}+\frac{\alpha _2^2(a+bx)^2}{2}\right) \right\} =0 \, \text{ or } \,\infty , \end{aligned}$$

depending on whether the limit of the exponential equals $-\infty$ or $\infty$, we would obtain that $\displaystyle \lim \nolimits _{x\rightarrow -\infty }\log \left( \frac{\varPhi (\alpha _1 x)}{\varPhi (\alpha _2(a+bx))}\right) =\pm \,\infty$. It then stands to reason the application of L’hopital’s rule in order to get

$$\begin{aligned} \lim _{x\rightarrow -\infty }\frac{1}{x^2}\log \left( \frac{\varPhi (\alpha _1 x)}{\varPhi (\alpha _2(a+bx))}\right)= & {} \lim _{x\rightarrow -\infty }\frac{1}{2x}\left[ \frac{\alpha _1\phi (\alpha _1 x)}{\varPhi (\alpha _1 x)}- \frac{b\alpha _2\phi (\alpha _2(a+bx))}{\varPhi (\alpha _2(a+bx))}\right] \\= & {} \lim _{x\rightarrow -\infty }\frac{1}{2x}\left[ b\alpha _2^2(a+bx)-\alpha _1^2x\right] =\frac{b^2\alpha _2^2-\alpha _1^2}{2}, \end{aligned}$$

where the steps in the previous expressions have been carried out using the relation between the quotients in the square brackets and the Mills’ ratio together with the asymptotic expansion of the Mills’ ratio (Abramowitz and Stegun 1965, (26.2.12)). Consequently,

$$\begin{aligned} \displaystyle \lim _{x\rightarrow -\infty } m(x)=\frac{a^2}{2}-\log b+\lim _{x\rightarrow -\infty } x^2\left( \frac{b^2-1}{2}+\frac{ab}{x}+\frac{b^2\alpha _2^2-\alpha _1^2}{2}\right) \end{aligned}$$

whose result, that depends on the signs of the difference $b^2(\alpha _2^2+1)-(\alpha _1^2+1)$ and a, can be established as follows:

$$\begin{aligned} \lim _{x\rightarrow -\infty }m(x)= {\left\{ \begin{array}{ll} -\infty &{} \quad \text{ if } b<\sqrt{\frac{\alpha _1^2+1}{\alpha _2^2+1}}\, \text{ or } \text{ if } \,b=\sqrt{\frac{\alpha _1^2+1}{\alpha _2^2+1}} \, \text{ and } \,a>0 \\ \infty &{} \quad \text{ if } b>\sqrt{\frac{\alpha _1^2+1}{\alpha _2^2+1}} \, \text{ or } \text{ if } \, b=\sqrt{\frac{\alpha _1^2+1}{\alpha _2^2+1}} \, \text{ and } \,a<0 \\ \end{array}\right. } \end{aligned}$$

(14)

The limits (13) and (14) along with the fact $\sqrt{\frac{\alpha _1^2+1}{\alpha _2^2+1}}<1$, resulting from the condition of the statement, are suggesting the cases to be considered in items (i)–(vii). For all of them, we will take into account that both $m_1$ and $m_2$ are concave functions with a unique maximum.

Before studying the sign pattern of the function m in all these cases, let us introduce some notation: The symbol − is used to denote negativity, and the symbol $+$ is for positivity; for example, with the pattern $-+$ we denote that a function is changing from negative to positive.

• (i) $0<b<\sqrt{\frac{\alpha _1^2+1}{\alpha _2^2+1}}$. In this case, taking into account the conclusions derived in (13) and (14), we obtain: $\displaystyle \lim \nolimits _{x\rightarrow -\infty }m(x)=\lim \nolimits _{x\rightarrow \infty }m(x)=-\infty$, which implies that $m_1$ and $m_2$ have two intersection points and the function m has two zeros with sign pattern $-+-$.

• (ii) $\sqrt{\frac{\alpha _1^2+1}{\alpha _2^2+1}}<b<1$. Now, the results obtained in (13) and (14) lead to the following limits: $\displaystyle \lim \nolimits _{x\rightarrow -\infty }m(x)=\infty$ and $\displaystyle \lim \nolimits _{x\rightarrow \infty }m(x)=-\infty$, from which we can deduce that $m_1$ and $m_2$ may intersect once or three times, depending on the values of a. Hence, m may have one or three zeros with respective sign patterns $+-$ and $+-+-$.

• (iii) $b>1$. In this case, from (13) and (14) we obtain: $\lim \nolimits _{x\rightarrow -\infty }m(x)=\lim _{x\rightarrow \infty }m(x)=\infty$, which implies that $m_1$ and $m_2$ will intersect one each other twice and m has two zeros with sign pattern $+-+$.

• (iv) $b=1, a>0$. For this case, once again $\lim \nolimits _{x\rightarrow -\infty }m(x)=\lim _{x\rightarrow \infty }m(x)=\infty$ and m will have two zeros with the same sign pattern as in (iii).

• (v) $b=1, a<0$. This case leads to the same limits as those in item (ii): $\displaystyle \lim \nolimits _{x\rightarrow -\infty }m(x)=\infty$ and $\displaystyle \lim \nolimits _{x\rightarrow \infty }m(x)=-\infty$. However, in this case m has only one zero, with $+-$ sign pattern, due to the stochastic dominance of the distribution functions of variables S and $Z=T-a$.

• (vi) $b=\sqrt{\frac{\alpha _1^2+1}{\alpha _2^2+1}}, a>0$. In this case, m has the same limit behavior as in item (i), so that it has two zeros with sign pattern $-+-$.

• (vii) $b=\sqrt{\frac{\alpha _1^2+1}{\alpha _2^2+1}}, a<0$. This case leads to the same limits for m as in item (ii) so the function m may have one or three zeros which in turn yield the sign patterns $+-$ and $+-+-$, respectively.

The conclusions derived in all these previous cases prove the statement of the lemma.$\square$