On the robustness of an epsilon skew extension for Burr III distribution on the real line

Abstract

Burr III (BIII) distribution is used in a wide variety of fields, such as lifetime data analysis, reliability theory, and financial literature, and suchlike. It is defined on the positive axis and has two shape parameters, say c and k. These shape parameters make the distribution quite flexible. They also control the tail behaviour of the distribution. In this study, we extend BIII distribution to the real line and also add a skewness parameter, say $\varepsilon$, with an epsilon skew extension approach. When the parameters c and k have a relationship such that $ck \le 1$, it is skew unimodal. Otherwise, it is skew bimodal with the same level of peaks on the negative and positive sides of the real line. Thus, the epsilon skew extension of Burr III (ESBIII) distribution with only three parameters can provide adequate fits for data sets that may have heavy-tailedness, skewness, unimodality or bimodality. A location-scale form of this distribution is also given. Distributional properties are investigated. The maximum likelihood (ML) estimation method for the parameters of ESBIII is considered. The robustness properties of the ML estimators are studied in terms of the boundedness of the influence function. Further, tail behaviour of ESBIII distribution is also examined to explore the robustness of ESBIII distribution against the outliers. The modelling capacity of this distribution is illustrated using two real data examples.

Appendix

Proof of Theorem 2.4

$$\begin{aligned} \int _{{\mathbb {R}}}f^{\alpha }(x)dx =\int _{-\infty }^{\mu } f_{-}^{\alpha }(x)dx+\int _{\mu }^{\infty } f_{+}^{\alpha }(x)dx, \end{aligned}$$

where

$$\begin{aligned} f_{-}(x)= & {} \frac{ck}{2\sigma } \bigg (\frac{-(x-\mu )}{\sigma (1-\varepsilon )}\bigg )^{-(c+1)}\Big \{1+\bigg (\frac{-(x-\mu )}{\sigma (1-\varepsilon )}\bigg )^{-c} \Big \}^{-(k+1)}, \\ f_{+}(x)= & {} \frac{ck}{2\sigma } \bigg (\frac{x-\mu }{\sigma (1+\varepsilon )}\bigg )^{-(c+1)}\Big \{1+\bigg (\frac{x-\mu }{\sigma (1+\varepsilon )}\bigg )^{-c} \Big \}^{-(k+1)}. \end{aligned}$$

Now consider the integral $\int _{\mu }^{\infty } f_{+}^{\alpha }(x)dx$. This integral is analytically obtained by applying the transformations, $y=\frac{x-\mu }{\sigma (1+\varepsilon )}$, $u=y^{-c}$, $v=1+u$ and $z=1/v$, consecutively. These steps are given as follows.

$$\begin{aligned} \int _{\mu }^{\infty } f_{+}^{\alpha }(x)dx= & {} \int _{\mu }^{\infty } \bigg [\frac{ck}{2\sigma } \bigg (\frac{x-\mu }{\sigma (1+\varepsilon )}\bigg )^{-(c+1)}\Big \{1+\bigg (\frac{x-\mu }{\sigma (1+\varepsilon )}\bigg )^{-c} \Big \}^{-(k+1)}\bigg ]^{\alpha }dx \\= & {} \bigg (\frac{ck}{2\sigma }\bigg )^{\alpha }\sigma (1+\varepsilon )\int _{0}^{\infty } y^{-\alpha (c+1)}\Big (1+y^{-c}\Big )^{-\alpha (k+1)}dy \\= & {} \bigg (\frac{ck}{2\sigma }\bigg )^{\alpha }\frac{\sigma (1+\varepsilon )}{c}\int _{0}^{\infty } u^{(\alpha -1)(1+1/c)}(1+u)^{-\alpha (k+1)}du \\= & {} \bigg (\frac{ck}{2\sigma }\bigg )^{\alpha }\frac{\sigma (1+\varepsilon )}{c}\int _{1}^{\infty } (v-1)^{(\alpha -1)(1+1/c)}v^{-\alpha (k+1)}dv \\= & {} \bigg (\frac{ck}{2\sigma }\bigg )^{\alpha }\frac{\sigma (1+\varepsilon )}{c}\int _{0}^{1} z^{\alpha (k-1/c)+1/c-1}(1-z)^{(\alpha -1)(1+1/c)}dz \\= & {} \bigg (\frac{ck}{2\sigma }\bigg )^{\alpha }\frac{\sigma (1+\varepsilon )}{c}B\Big (\alpha (k-1/c)+1/c , \alpha (1+1/c)-1/c\Big ), \end{aligned}$$

where $\alpha (k-1/c)+1/c>0$, $\alpha (1+1/c)-1/c>0$. Similarly, the other integral is obtained as

$$\begin{aligned} \int _{-\infty }^{\mu } f_{-}^{\alpha }(x)dx=\bigg (\frac{ck}{2\sigma }\bigg )^{\alpha }\frac{\sigma (1-\varepsilon )}{c}B\Big (\alpha (k-1/c)+1/c , \alpha (1+1/c)-1/c\Big ), \end{aligned}$$

where $\alpha (k-1/c)+1/c>0$, $\alpha (1+1/c)-1/c>0$.

Then, the Rényi entropy is

$$\begin{aligned} J_{\mathbb {R}}(\alpha )=\frac{1}{1-\alpha }\log \bigg [\Big (\frac{ck}{2\sigma }\Big )^{\alpha }\frac{2\sigma }{c}B\Big (\alpha (k-1/c)+1/c , \alpha (1+1/c)-1/c\Big ) \bigg ], \end{aligned}$$

where $\alpha (k-1/c)+1/c>0$,  $\alpha (1+1/c)-1/c>0$,  $\alpha >0$ and $\alpha \ne 1$.

The Shannon entropy is obtained by finding $\underset{\alpha \rightarrow 1}{\lim } J_{\mathbb {R}}(\alpha )$. One can apply L’Hôpital rule to find this limit. Thus,

$$\begin{aligned} \lim _{\alpha \rightarrow 1}J_{\mathbb {R}}(\alpha )=(k+1)\psi (k+1)+(1+1/c)\gamma +(1/c-k)\psi (k)+\log \Big (\frac{2\sigma }{ck}\Big ), \end{aligned}$$

which is the Shannon entropy. Here $B(\cdot )$ is beta function, $\psi (\cdot )$ is digamma function and $\gamma$ is Euler constant. $\square$