Shape mixtures of skew-t-normal distributions: characterizations and estimation

Abstract

This paper introduces the shape mixtures of the skew-t-normal distribution which is a flexible extension of the skew-t-normal distribution as it contains one additional shape parameter to regulate skewness and kurtosis. We study some of its main characterizations, showing in particular that it is generated through a mixture on the shape parameter of the skew-t-normal distribution when the mixing distribution is normal. We develop an Expectation Conditional Maximization Either algorithm for carrying out maximum likelihood estimation. The asymptotic standard errors of estimators are obtained via the information-based approximation. The numerical performance of the proposed methodology is illustrated through simulated and real data examples.

Appendices

Appendix A: The score function and Hessian matrix

From (14), the log-likelihood function corresponding to the jth observation is

$$\begin{aligned} \log f(y_{j};\varvec{\theta })= & {} -\log \sigma -\frac{1}{2}\log \nu +\log \varGamma \left( \frac{\nu +1}{2}\right) -\log \varGamma \left( \frac{\nu }{2}\right) \nonumber \\- & {} \frac{\nu +1}{2} \log \left( 1+\frac{u_{j}^{2}}{\nu }\right) +\log \varPhi \left( \frac{\lambda u_{j}}{\sqrt{1+\alpha u_{j}^{2}}}\right) , \end{aligned}$$

(A.1)

where $u_{j}=(y_{j}-\xi )/\sigma$. Let $s_{j}(\varvec{\theta })=(s_{j,\xi },s_{j,\sigma },s_{j,\lambda },s_{j,\alpha },s_{j,\nu })$ be a $5\times 1$ vector. Based on the definition of $s_{j}(\varvec{\theta })$ in Sect. 3.2, explicit expressions for the components of $s_{j}(\varvec{\theta })$ are obtained by differentiation from (A.1) with respect to each parameter. They are given by

$$\begin{aligned} s_{j,\xi }= & {} \frac{1}{\sigma }\left( \zeta _{j}u_{j}-\frac{\lambda R_{j} }{\omega _{j}}\right) ,~\quad s_{j,\sigma }=\frac{1}{\sigma }\left( \zeta _{j}u_{j}^{2}-1-\frac{\lambda R_{j}u_{j}}{\omega _{j}}\right) ,\nonumber \\ s_{j,\lambda }= & {} \frac{R_{j}u_{j}}{\omega _{j}^{1/3}},~\quad s_{j,\alpha }=-\frac{\lambda R_{j}u_{j}^{3}}{2\omega _{j}},\nonumber \\ s_{j,\nu }= & {} \frac{1}{2}\left\{ \mathrm{DG}\left( \frac{\nu +1}{2}\right) -\mathrm{DG}\left( \frac{\nu }{2}\right) -\frac{1}{\nu }-\log \left( 1+\frac{u_{j}^{2} }{\nu }\right) +\zeta _{j}\frac{u_{j}^{2}}{\nu }\right\} . \end{aligned}$$

(A.2)

where $\zeta _{j}=(\nu +1)/(\nu +u_{j}^{2})$, $\omega _{j}=(1+\alpha u_{j}^{2})^{3/2}$ and $R_{j}=\phi (\lambda u_{j}\omega _{j}^{-1/3})/\varPhi (\lambda u_{j}\omega _{j}^{-1/3})$. The Hessian matrix consisting of the second partial derivatives of the SMSTN log-likelihood takes the form of

$$\begin{aligned} H_{j}(\varvec{\theta })=\left[ \begin{array} [c]{ccccc} H_{j}^{\xi \xi } &{} H_{j}^{\xi \sigma } &{} H_{j}^{\xi \lambda } &{} H_{j}^{\xi \alpha } &{} H_{j}^{\xi \nu }\\ H_{j}^{\sigma \xi } &{} H_{j}^{\sigma \sigma } &{} H_{j}^{\sigma \lambda } &{} H_{j}^{\sigma \alpha } &{} H_{j}^{\sigma \nu }\\ H_{j}^{\lambda \xi } &{} H_{j}^{\lambda \sigma } &{} H_{j}^{\lambda \lambda } &{} H_{j}^{\lambda \alpha } &{} H_{j}^{\lambda \nu }\\ H_{j}^{\alpha \xi } &{} H_{j}^{\alpha \sigma } &{} H_{j}^{\alpha \lambda } &{} H_{j}^{\alpha \alpha } &{} H_{j}^{\alpha \nu }\\ H_{j}^{\nu \xi } &{} H_{j}^{\nu \sigma } &{} H_{j}^{\nu \lambda } &{} H_{j}^{\nu \alpha } &{} H_{j}^{\nu \nu } \end{array} \right] . \end{aligned}$$

The detailed expressions for the components of $H_{j}(\varvec{\theta })$ are shown below.

$$\begin{aligned} H_{j}^{\xi \xi }= & {} -\frac{1}{\sigma ^{2}}\left\{ \frac{\zeta _{j}(\nu -u_{j}^{2})}{(\nu +u_{j}^{2})}+\frac{\lambda R_{j}}{\omega _{j}^{2}}(A_j+3\alpha u_{j} \omega _{j}^{1/3}) \right\} , \nonumber \\ H_{j}^{\xi \sigma }= & {} -\frac{1}{\sigma ^{2}}\left\{ \frac{2\zeta _{j}\nu u_{j}}{(\nu +u_{j}^{2})}-\frac{\lambda R_{j}}{\omega _{j} }+\frac{\lambda R_{j}u_{j}}{\omega _{j}^{2}}(A_j+3\alpha u_{j}\omega _{j}^{1/3}) \right\} ,\nonumber \\ H_{j}^{\sigma \sigma }= & {} -\frac{1}{\sigma ^{2}}\left\{ \frac{\zeta _{j}u_{j}^{2} }{(\nu +u_{j}^{2})}\left( 3\nu +u_{j}^{2}\right) -1-\frac{2\lambda R_{j}u_{j} }{\omega _{j}}+\frac{\lambda R_{j}u_{j}^{2}}{\omega _{j}^{2}}(A_j+3\alpha u_{j}\omega _{j} ^{1/3})\right\} ,\nonumber \\ H_{j}^{\xi \lambda }= & {} -\frac{R_{j}}{\sigma \omega _{j} }(1-u_{j}\omega _{j}^{-1/3}A_j), \quad H_{j}^{\sigma \lambda }=-\frac{R_{j}u_{j}}{\sigma \omega _{j}}(1-u_{j}\omega _{j}^{-1/3}A_j), \nonumber \\ H_{j}^{\lambda \lambda }= & {} -\frac{R_{j}u_{j}^{2}}{\omega _{j}^{2/3}}(R_{j}+\lambda u_{j}\omega _{j}^{-1/3}), \quad H_{j}^{\alpha \lambda } =-\frac{R_{j}u_{j}^{3} }{2\omega _{j}}(1-u_{j}\omega _{j}^{-1/3}A_j), \nonumber \\ H_{j}^{\xi \alpha }= & {} -\frac{\lambda R_{j}u_{j}^{2} }{2\sigma \omega _{j}^{2}}(u_jA_j-3\omega _{j}^{1/3}),\quad H_{j}^{\sigma \alpha } = -\frac{\lambda R_{j}u_{j}^{3} }{2\sigma \omega _{j}^{2}}(u_jA_j -3\omega _{j}^{1/3}), \nonumber \\ H_{j}^{\alpha \alpha }= & {} -\frac{\lambda R_{j}u_{j}^{5}}{4\omega _{j}^{2} }(u_jA_j-3\omega _{j}^{1/3}),\quad H_{j}^{\xi \nu } = -\frac{1}{\sigma }\frac{u_{j}(1-u_{j}^{2} )}{(\nu +u_{j}^{2})^{2}}, \nonumber \\ H_{j}^{\nu \nu }= & {} \frac{1}{2}\left\{ \frac{1}{2}TG\left( \frac{\nu +1}{2}\right) -\frac{1}{2}TG\left( \frac{\nu }{2}\right) +\frac{\nu +u_{j}^{4} }{\nu \left( \nu +u_{j}^{2}\right) ^{2}}\right\} ,\nonumber \\ H_{j}^{\sigma \nu }= & {} -\frac{1}{\sigma }\frac{u_{j}^{2}(1-u_{j}^{2})}{(\nu +u_{j}^{2})^{2}},\quad H_{j}^{\lambda \nu }=H_{j}^{\alpha \nu }=0, \end{aligned}$$

(A.3)

where $H_j^{\theta _r\theta _s}=H_j^{\theta _s\theta _r}$, $A_j=\lambda R_{j}+\lambda ^{2}u_{j}\omega _{j}^{-1/3}$ and $\mathrm{TG}(x)=d^{2}\log \varGamma (x)/dx^{2}$ is the trigamma function.

Appendix B: The procedure of the Kolmogorov–Smirnov test for continuous data

• 1. Sort data values into ascending order $y_{(1)}\le y_{(2)}\le \cdots \le y_{(n)}$.

• 2. Compute the KS test statistic

  $$\begin{aligned} D=\mathop {\max _{j=1,\ldots ,n}}\left\{ \frac{j}{n}-\hat{F}(y_{(j)}),~\hat{F}(y_{(j)})-\frac{j-1}{n}\right\} , \end{aligned}$$

  where $\hat{F}(\cdot )$ is the fitted cdf under a specific distribution.

• 3. Generate n random number from U(0, 1) and sort them into ascending order, we have $u^{(i)}_{(1)}\le u^{(i)}_{(2)} \le \cdots \le u^{(i)}_{(n)}$ for $i=1,\ldots ,n$.

• 4. Compute

  $$\begin{aligned} d_i=\mathop {\max _{j=1,\ldots ,n}}\left\{ \frac{j}{n}-u_{(j)}^{(i)},~u_{(j)}^{(i)}-\frac{j-1}{n}\right\} . \end{aligned}$$

• 5. Set $I_i=1$ if $d_i\ge D$ and 0 otherwise. Repeat Steps 3 and 4 M times, we get $I_1,...,I_M$. The p-value is estimated by $\sum ^{M}_{i=1} I_i/M$.