Semiparametric mixture of linear regressions with nonparametric Gaussian scale mixture errors

Abstract

In finite mixture of regression models, normal assumption for the errors of each regression component is typically adopted. Though this common assumption is theoretically and computationally convenient, it often produces inefficient and undesirable estimates which undermine the applicability of the model particularly in the presence of outliers. To reduce these defects, we propose to use nonparametric Gaussian scale mixture distributions for component error distributions. By this means, we can lessen the risk of misspecification and obtain robust estimators. In this paper, we study the identifiability of the proposed model and develop a feasible estimating algorithm. Numerical studies including simulation studies and real data analysis to demonstrate the performance of the proposed method are also presented.

Appendix: Proof of Theorem 1

In (3), suppose that there exists $\tilde{{\varvec{\theta }}}= (\tilde{\pi }_1, \tilde{{\varvec{\beta }}}_1, \ldots , \tilde{\pi }_J, \tilde{{\varvec{\beta }}}_J)$ and $\tilde{{\varvec{Q}}}= (\tilde{Q}_1, \ldots , \tilde{Q}_J)$ satisfying

$$\begin{aligned} \sum _{k=1}^{K} \pi _k \int \frac{1}{\sigma } \phi \left( \frac{y - \eta ({{\varvec{x}}};{{\varvec{\beta }}}_k)}{\sigma }\right) dQ_k(\sigma ) = \sum _{j=1}^{J} \tilde{\pi }_j \int \frac{1}{\sigma } \phi \left( \frac{y - \eta ({{\varvec{x}}};\tilde{{\varvec{\beta }}}_j)}{\sigma }\right) d\tilde{Q}_j(\sigma ). \end{aligned}$$

(11)

The characteristic function of the left hand side of (11) is

$$\begin{aligned} \psi _{Y \mid {{\varvec{X}}}} (t)&= E_{Y \mid {{\varvec{X}}}} [\exp (iyt)] \\&= \sum _{k=1}^{K} \pi _k \int \int \exp (iyt) \frac{1}{\sigma } \phi \left( \frac{y - \eta ({{\varvec{x}}};{{\varvec{\beta }}}_k)}{\sigma }\right) dy dQ_k(\sigma ) \\&=\sum _{k=1}^{K} \pi _k \exp (i\eta ({{\varvec{x}}};{{\varvec{\beta }}}_k)t) \psi _{Q_k}(t), \end{aligned}$$

where $i=\sqrt{-1}$ and $\psi _{Q_k}(t)=\int \exp (-t^2\sigma ^2/2)dQ_k(\sigma )$, $k=1,\ldots ,K$. The characteristic function of the right hand side of (11) can be similarly represented as

$$\begin{aligned} \tilde{\psi }_{Y \mid {{\varvec{X}}}} (t) = \sum _{k=1}^{J} \tilde{\pi }_j \exp (i\eta ({{\varvec{x}}};\tilde{{{\varvec{\beta }}}}_j)t) \psi _{\tilde{Q}_j}(t) \end{aligned}$$

where $\psi _{G_j}(t)=\int \exp (-t^2\sigma ^2/2)d\tilde{Q}_j(\sigma )$, $j=1,2,\cdots ,J$.

Then the following equality must hold:

$$\begin{aligned} \sum _{k=1}^{K} \pi _k \exp (i\eta ({{\varvec{x}}};{{\varvec{\beta }}}_k)t) \psi _{Q_k}(t) = \sum _{j=1}^{J} \tilde{\pi }_j \exp (i\eta ({{\varvec{x}}};\tilde{{\varvec{\beta }}}_j)t) \psi _{\tilde{Q}_j}(t). \end{aligned}$$

(12)

Because ${{\varvec{\beta }}}_{k_1}\ne {{\varvec{\beta }}}_{k_2}$ if $k_1\ne k_2$, there exists an open set $U =\{{{\varvec{u}}}\in \mathbf {\mathcal {X}} |\eta ({\varvec{u}};{{\varvec{\beta }}}_{k_1}) \ne \eta ({\varvec{u}};{{\varvec{\beta }}}_{k_2})\}$, where $\mathbf {\mathcal {X}}$ is the support of ${{\varvec{X}}}$. For any fixed $k^{*} \in \{1, 2, \cdots , K \}$, multiplying $\exp (-i\eta ({{\varvec{u}}};{{\varvec{\beta }}}_{k^*})t)$ on both sides of equation (12), we have

$$\begin{aligned}&\pi _{k^{*}} \psi _{Q_{k^{*}}}(t) + \sum _{k \ne k^{*}}^{} \pi _k \exp (i(\eta ({{\varvec{u}}};{{\varvec{\beta }}}_k) - \eta ({{\varvec{u}}};{{\varvec{\beta }}}_{k^*}))t) \psi _{Q_k}(t) \nonumber \\&\quad = \sum _{j=1}^{J} \tilde{\pi }_j \exp (i({\eta }({{\varvec{u}}};\tilde{{\varvec{\beta }}}_j) - \eta ({{\varvec{u}}};{{\varvec{\beta }}}_{k^*}))t) \psi _{\tilde{Q}_j}(t) \end{aligned}$$

(13)

for all ${{\varvec{u}}}\in U$.

Assume that $\eta ({{\varvec{u}}};\tilde{{\varvec{\beta }}}_j) - \eta ({{\varvec{u}}};{{\varvec{\beta }}}_{k^*}) \ne 0$ for all $j = 1, \dots , J$, and let H be an open subset of $\{{{\varvec{u}}}\in U| \eta ({{\varvec{u}}};\tilde{{\varvec{\beta }}}_j) \ne \eta ({{\varvec{u}}};{{\varvec{\beta }}}_{k^*}), j=1,\ldots ,J \}$. Then, for ${{\varvec{u}}}\in H$, (13) should hold. Because $\eta (\cdot )$ is a continuous function, there exists open sets V and $\tilde{V}$ in $\mathbb {R}$ satisfying

$$\begin{aligned} \pi _{k^{*}} \psi _{Q_{k^{*}}}(t) + \sum _{k \ne k^{*}}^{} \pi _k \exp (iv_kt) \psi _{Q_k}(t) = \sum _{j=1}^{J} \tilde{\pi }_j \exp (i\tilde{v}_jt) \psi _{\tilde{Q}_j}(t), \end{aligned}$$

(14)

for all $v_k \in V$ for $k \in \{1,2,\ldots , K \} {\setminus } \{k^*\}$ and $\tilde{v}_j \in \tilde{V}$ for $j \in \{1,2,\ldots , J \}$. Because V and $\tilde{V}$ are open sets, there also exists $\xi >0$ such that

$$\begin{aligned} \pi _{k^{*}} \psi _{Q_{k^{*}}}(t) + \sum _{k \ne k^{*}}^{} \pi _k \exp (icv_kt) \psi _{Q_k}(t) = \sum _{j=1}^{J} \tilde{\pi }_j \exp (ic\tilde{v}_jt) \psi _{\tilde{Q}_j}(t) \end{aligned}$$

(15)

holds for all $c\in (1-\xi ,1+\xi )$. Because the exponential function is analytic on the whole complex plane, if (15) holds for $c\in (1-\xi ,1+\xi )$, (15) should also hold for all $-\infty<c<\infty$. Hence, the following equality should also hold:

$$\begin{aligned}&\pi _{k^{*}} \psi _{Q_{k^{*}}}(t) + \sum _{k \ne k^{*}}^{} \pi _k \Bigg \{ \frac{1}{2T}\int _{-T}^T\exp (icv_kt)dc\Bigg \} \psi _{Q_k}(t) \nonumber \\&\quad = \sum _{j=1}^{J} \tilde{\pi }_j \Bigg \{ \frac{1}{2T}\int _{-T}^T \exp (ic\tilde{v}_jt)dc \Bigg \} \psi _{\tilde{Q}_j}(t), \end{aligned}$$

(16)

for all $T>0$. Letting $T\rightarrow \infty$, we can conclude $\pi _{k^{*}} \psi _{Q_{k^{*}}}(t)=0$ from $\lim _{T\rightarrow \infty }\frac{1}{2T}\int _{-T}^T\exp (ic\lambda )dc=0$ for any $\lambda$. Further, letting $t \rightarrow 0$ implies $\pi _{k^*}=0$ which contradicts the assumption that $\pi _{k^*}>0$. Therefore, there must exist $j^{*} \in \{1, 2, \ldots, J \}$ such that $\eta ({\varvec{u}};\tilde{{\varvec{\beta }}}_{j^{*}}) = \eta ({\varvec{u}};{{\varvec{\beta }}}_{k^{*}})$, implying $\tilde{{\varvec{\beta }}}_{j^{*}}={{\varvec{\beta }}}_{k^{*}}$. Accordingly, (16) follows $\pi _{k^{*}} \psi _{Q_{k^{*}}}(t) = \tilde{\pi }_{j^{*}} \psi _{\tilde{Q}_{j^{*}}}(t)$. Letting $t \rightarrow 0$, we have $\pi _{k^{*}} = \tilde{\pi }_{j^{*}}$. Thus, $\psi _{Q_{k^{*}}}(t) = \psi _{\tilde{Q}_{j^{*}}}(t)$ is obtained, which in turns implies that $Q_{k^{*}} = \tilde{Q}_{j^{*}}$ from the uniqueness of the Laplace transform.

Repeating the prior argument, we obtain $\pi _{k^*} = \tilde{\pi }_{j^*}$, ${{\varvec{\beta }}}_{k^*} = \tilde{{{\varvec{\beta }}}}_{j^*}$, $Q_{k^*} = \tilde{Q}_{j^*}$ for $k^*,j^* \in \{ 1,\ldots ,K \}$ and $K \le J$. However, if $K < J$ is hold, there exist some $\tilde{\pi }_j$’s, $j = 1,\ldots ,J$, are 0 in contradiction to $\pi _j > 0, j = 1,\ldots ,J$. Therefore, K and J must be the same.