Shape measures based on the convex transform order

Abstract

Three functional measures of the shape of univariate distributions are proposed which are consistent with respect to the convex transform order. The first two are weighted tail indices that characterize location-scale families of distributions, whilst the third is a skewness measure. Properties of the new measures are established for various classes of symmetric and asymmetric distributions, and the generalized Pareto distribution characterized in terms of them. Kernel density based estimation of the measures is also considered, and the use of the estimated functionals is illustrated in the analysis of two real data sets.

Appendix A: Proofs

Proof of Proposition 2.1

 From (2) we observe that

$$\begin{aligned} G^{-1}F(\cdot ) \hbox { is convex }\Leftrightarrow & {} G^{-1}F(t)-G^{-1}F(s) \ge \frac{d}{dx}G^{-1}F(s)(t-s), \, \, \forall \, s,\,t \in D_X\\\Leftrightarrow & {} G^{-1}F(t)-G^{-1}F(s)\ge \frac{f(s)}{g(G^{-1}(F(s)))}(t-s), \,\, \forall \, s,t \in D_X\\\Leftrightarrow & {} y_u-y_{p} \ge \frac{f(x_{p})}{g(y_{p})}(x_u-x_{p}), \, \, \forall \, u,\,p \in (0,1)\\\Leftrightarrow & {} (x_p-x_u)f(x_p)\ge (y_p-y_u)g(y_p),\quad \forall u,p\in (0,1), \end{aligned}$$

where the third equivalence follows by choosing $t=x_u$, $s=x_{p}$ with $u,\, p\in (0,1)$ and using the fact that $G^{-1}F(\cdot )$ maps the quantiles of X to the corresponding quantiles of Y. $\square$

Proof of Corollary 2.2

 Using $F^{-1}(U)=_{st} X$, where U denotes a uniform random variable on (0, 1), we see that $(x_p-x_u)^+ f(x_p)$ and $(x_p-x_u)^- f(x_p)$ can be thought of as realizations of the random variables $(F^{-1}(U)-x_u)^+f(F^{-1}(U))$ and $(F^{-1}(U)-x_u)^-f(F^{-1}(U))$, respectively. Analogously for Y. The proof then follows directly from Corollary 2.1 and Theorem 1.A.1. in Shaked and Shanthikumar (2007). $\square$

Proof of Lemma 2.2

 The proof follows directly from Definition 2.1 and the facts that $-\,x_u$ is the $(1-u)$th quantile of $-\,X$, $a^+=(-\,a)^-$ for all $a\in \mathbb R$ and $f_{-X}(-X)=f_X (X)$. $\square$

Proof of Lemma 2.3

 From (3) we can write

$$\begin{aligned} R_X(u)= \displaystyle \int _{u}^{1}x_pf(x_p)dp-x_u \displaystyle \int _{u}^{1}f(x_p)dp, \quad \forall u\in (0,1). \end{aligned}$$

The result follows on taking derivatives and using $x'_u=1/f(x_u)$. $\square$

Proof of Theorem 2.1

 The sufficient condition of (i) for $R_X(u)$ follows from (3) and the fact that $(x_p-x_u)^+ f(x_p)$, with $u,p\in (0,1)$, is not affected by translations and positive changes of scale. To prove the necessary condition, assume that $R_X(u)=R_Y(u)$, for all $u\in (0,1)$. Taking derivatives and using Lemma 2.3 we obtain that

$$\begin{aligned} \left[ \log \left( \displaystyle \int _{u}^{1}f(x_p)dp\right) \right] '=\left[ \log \left( \displaystyle \int _{u}^{1}g(y_p)dp\right) \right] ', \quad \forall u\in (0,1), \end{aligned}$$

which trivially implies that there exists an $a>0$ such that

$$\begin{aligned} \displaystyle \int _{u}^{1}f(x_p)dp=a \displaystyle \int _{u}^{1}g(y_p)dp, \quad \forall u\in (0,1). \end{aligned}$$

By taking derivatives again, we have $f(x_u)=ag(y_u)$, for all $u\in (0,1)$ or, equivalently, $y'_u=ax'_u$, for all $u\in (0,1)$. Therefore, there exists a $b\in \mathbb {R}$ such that $y_u=ax_u+b$, for all $u\in (0,1)$, which means that $Y=_{st}aX+b$, i.e., $G\in [F]$. The proof for $L_X(u)$ follows straightforwardly using Lemma 2.2.

Part (ii) holds because of the constructions of $R_X(u)$ and $L_X(u)$.

In order to prove (iii), assume that X and Y are symmetric and $X\le _{k}Y$, which means that $G^{-1}F(x)$ is convex for all $x>m_X$, where $m_X$ denotes the median of X. This implies, by restricting Proposition 2.1 to $D=(m_X,+\,\infty )$, that $R_X(u) \ge R_Y(u)$, for all $u \in (1/2,1)$. Now, from Lemma 2.2, the location invariance property and the symmetry of X we have

$$\begin{aligned} L_{X}(u)= & {} R_{-X}(1-u)\\= & {} R_{-X+m_X}(1-u) =R_{X-m_X}(1-u)=R_{X}(1-u),\ u\in (0,1), \end{aligned}$$

and it follows that $L_X(u) \ge L_Y(u)$, for all $u\in (0,1/2)$. $\square$

Proof of Theorem 2.2

 From the assumptions, and using L’Hôpital’s rule and Lemma 2.3, we obtain

$$\begin{aligned} \lim _{u \rightarrow 1^{-}} \frac{R_X(u)}{ R_Y(u)}= & {} \lim _{u \rightarrow 1^{-}} \frac{\log \left( \displaystyle \int _{u}^{1}g(y_p)dp\right) }{\log \left( \displaystyle \int _{u}^{1}f(x_p)dp\right) } \\= & {} \lim _{u \rightarrow 1^{-}} \frac{-\log \left( \displaystyle \int _{y_u}^{+\,\infty } g^2(t)dt\right) }{-\log \left( \displaystyle \int _{x_u}^{+\,\infty }f^2(t)dt\right) } > 1, \end{aligned}$$

where the second equality follows by changing the variables in the integrals. It follows that there exists an $u_0 \in (0,1)$ such that

$$\begin{aligned} H_2(Y;y_u) = -\log \left( \displaystyle \int _{y_u}^{+\,\infty } \frac{g^2(t)}{ (1-u)^2}dt\right) \ge -\log \left( \displaystyle \int _{x_u}^{+\,\infty }\frac{ f^2(t)}{(1-u)^2}dt\right) =H_2(X;x_u), \end{aligned}$$

for all $u\ge u_0$. $\square$

Proof of Proposition 3.1

 The necessary condition follows from Theorem 2.1(i). In order to prove the sufficient condition, let X be a GP(c, k) random variable with $c < 2$ and $k > 0$. Observe that, if $c=0,$, X follows an exponential distribution and the result follows from Table 1. Suppose, therefore, that $c\ne 0$. From (5) the density and quantile functions of X are given by

$$\begin{aligned}f_{X}(x)=\frac{1}{k }\left( 1-\frac{c}{k}x\right) ^{\frac{1-c}{c}}, ~~~~~~ x_p=\frac{k}{c}[1-(1-p)^{c}] \end{aligned}$$

and therefore

$$\begin{aligned} R_{X}(u)= & {} \displaystyle \int _{u}^{1}(x_p-x_u)f_{X}(x_p)dp\\= & {} \frac{1}{c}\left[ (1-u)^{c}\displaystyle \int _{u}^{1}(1-p)^{1-c}dp -\displaystyle \int _{u}^{1}(1-p)dp \right] \\= & {} \frac{1}{4-2c}(1-u)^{2},\quad \hbox {for all }u\in (0,1). \end{aligned}$$

$\square$

Proof of Proposition 3.2

 Let f denote the density of X. From Lemma 2.3 we have

$$\begin{aligned} \left[ \log \displaystyle \int _{u}^{1}f(x_p)dp \right] ^{\prime }=\frac{1}{R^{\prime }_X(u)}. \end{aligned}$$

Then, there exists a $c>0$ such that

$$\begin{aligned} \displaystyle \int _{u}^{1}f(x_p)dp = c\,\exp \left( \displaystyle \int \frac{1}{R^{\prime }_X(u)}du \right) . \end{aligned}$$

Using $f(x_u)=1/x_u^{\prime }$, we have

$$\begin{aligned} x^{\prime }_u = -\,a\,R^{\prime }_X(u)\,\exp \left( -\displaystyle \int \frac{1}{R^{\prime }_X(u)}du \right) ,\quad \hbox {with}\,\, a=\frac{1}{c}>0, \end{aligned}$$

and by integration we obtain

$$\begin{aligned} x_u = - a \displaystyle \int \,R^{\prime }_X(u)\,\exp \left( -\displaystyle \int \frac{1}{R^{\prime }_X(u)}du \right) du+b,\quad \hbox {with}\,\, a>0,\, b\in \mathbb {R}, \end{aligned}$$

which concludes the proof. $\square$

Proof of Proposition 4.1

 Part (i) follows from Theorem 2.1(i). Similarly, (iii) and (iv) follow straightforwardly from Lemma 2.2 and Theorem 2.1(ii). To prove (ii), first suppose that X is symmetric about the median $m_X$, which implies that $X-m_X=_{st} -X+m_X$. Hence, using Lemma 2.2 and Theorem 2.1(i), we have

$$\begin{aligned} L_{X}(u)=R_{-X}(1-u)=R_{-X+m_X}(1-u) =R_{X-m_X}(1-u)=R_{X}(1-u), \end{aligned}$$

which implies that $\textit{SK}_X(u) = 0$, for all $u\in (0,1)$.

Conversely, assume $\textit{SK}_X(u) = 0$, for all $u\in (0,1)$. From the definition of $\textit{SK}_X(u)$, this means that $L_{X}(u)=R_{X}(1-u)$ holds for $u\in (0,1)$. It follows from Lemma 2.2 that $R_{-X}(1-u)=R_{X}(1-u)$ for $u\in (0,1)$. Applying Theorem 2.1(i), there exists an $a>0$ and a $b\in \mathbb {R}$ such that $-\,X=_{st}aX+b$. By taking variances and expectations it follows that $a=1$ and $b=-2 \mu _X$, where $\mu _X$ represents the expectation of X. Therefore $-\,X+\mu _X=_{st} X -\mu _X$ which implies that X is symmetrically distributed. $\square$

Proof of Proposition 4.2

 Suppose that $-\,X\le _c X$. The proof for when $X\le _c -X$ is analogous. It follows from Proposition 4.1(iv) that $\textit{SK}_{-X}(u)\le \textit{SK}_X(u)$, for all $u\in (0,1)$, which implies, by Proposition 4.1(iii), that $-\,\textit{SK}_{X}(u)\le \textit{SK}_X(u)$ holds for all $u\in (0,1)$. This concludes the proof. $\square$

Proof of Corollary 4.1

 Assume that f is increasing on $D_X$. The proof for when f is decreasing is similar. From Proposition 4.2 it is sufficient to verify that $X \le _{c} -X$. Denoting the distribution function of $-\,X$ by $G(x)=1-F(-x)$, we have

$$\begin{aligned} G^{-1}F(x) \hbox { is convex on } D_X\Longleftrightarrow & {} -F^{-1}(1-F(x)) \hbox { is convex on } D_X, \\\Longleftrightarrow & {} \frac{f(x)}{f(F^{-1}(1-F(x)))} \hbox { is increasing on } D_X, \\\Longleftrightarrow & {} \frac{f(x_u)}{f(x_{1-u})} \hbox { is increasing on } u\in (0,1), \end{aligned}$$

where the second equality follows by taking derivatives and the third one by making a change of variable. Since f is increasing, we see that $f(x_u)$ and $f(x_{1-u})$ are increasing and decreasing functions of u, respectively, in the interval (0, 1). This concludes the proof. $\square$