Multilevel cluster-weighted models for the evaluation of hospitals

Abstract

In recent years, increasing attention has been directed toward problems inherent to quality control in healthcare services. In particular, it is necessary to measure effectiveness with respect to improving healthcare outcomes of diagnostic procedures or specific treatment episodes. The performance of hospitals is usually evaluated by multilevel models and other methods for risk adjustment. However, these approaches are not suitable for data with large unobserved heterogeneity. A potentially large source of unobserved heterogeneity comes from the variation of the regression coefficients between groups of individuals sharing similar but unobserved characteristics. To overcome such drawbacks, we propose the multilevel cluster-weighted model, a new mixture model approach for handling hierarchical data.

Appendix: Proof of Proposition 1

Assume $\sigma ^2_{X|c}=\sigma ^2_{X}$ and $\pi _c=\pi =1/K$, $c=1,\ldots ,K$; the density in (8) yields

$$\begin{aligned} p(x_{ij}, y_{ij}; \varvec{\vartheta })&= \sum ^{K}_{c=1} p(y_{ij}|x_{ij}; \varvec{\xi }_c) \phi (x_{ij};\mu _{X|c},\sigma ^2_{X}) \pi \\&=p(x_{ij}; \varvec{\psi }) \sum ^{K}_{c=1} p(y_{ij}|x_{ij}; \varvec{\xi }_c) \frac{\phi (x_{ij};\mu _{X|c},\sigma ^2_{X}) \pi }{p(x_{ij}; \varvec{\psi }) } \\&= p(x_{ij}; \varvec{\psi }) \sum ^{K}_{c=1} p(y_{ij}|x_{ij}; \varvec{\xi }_c) \frac{\exp \left( \displaystyle {-\frac{1}{2 \sigma ^2_{X}}(x_{ij}-\mu _{X|c})^2 } \right) }{\sum _{k=1}^{K}\exp \left( \displaystyle {-\frac{1}{2 \sigma ^2_{X}}(x_{ij}-\mu _{X|k})^2 } \right) } , \end{aligned}$$

where

$$\begin{aligned} p(x_{ij}; \varvec{\psi }) = \sum _{c=1}^K \phi (x_{ij};\mu _{X|c},\sigma ^2_{X}) \pi = \sum _{c=1}^K \frac{1}{\sqrt{2 \pi \sigma ^2_{X}}} \exp \left( \displaystyle {-\frac{1}{2 \sigma ^2_{X}}(x_{ij}-\mu _{X|c})^2 } \right) \pi . \end{aligned}$$

The proposition is proven once we show that the term

$$\begin{aligned} \frac{\exp \left( \displaystyle {-\frac{1}{2 \sigma ^2_{X}}(x_{ij}-\mu _{X|c})^2 } \right) }{\sum _{k=1}^{K}\exp \left( \displaystyle {-\frac{1}{2 \sigma ^2_{X}}(x_{ij}-\mu _{X|k})^2 } \right) } \end{aligned}$$

is equivalent to the probability of latent class membership (4) for suitable parameters $a_c$ and $b_c$. To this end, consider that

$$\begin{aligned}&\frac{\exp \left( \displaystyle {-\frac{1}{2 \sigma ^2_{X}}(x_{ij}-\mu _{X|c})^2 } \right) }{\sum _{k=1}^{K}\exp \left( \displaystyle {-\frac{1}{2 \sigma ^2_{X}}(x_{ij}-\mu _{X|k})^2 } \right) } = \frac{\exp \left( \displaystyle {-\frac{1}{2 \sigma ^2_{X}}(x_{ij}^2-2\mu _{X|c}x_{ij}+\mu _{X|c}^2) } \right) }{\sum _{k=1}^{K}\exp \left( \displaystyle {-\frac{1}{2 \sigma ^2_{X}}(x_{ij}^2-2\mu _{X|k}x_{ij}+\mu _{X|k}^2) } \right) } \nonumber \\&\qquad = \frac{\exp \left( \displaystyle {-\frac{x_{ij}^2}{2 \sigma ^2_{X}} } \right) \exp \left( \displaystyle {\frac{2\mu _{X|c}x_{ij}-\mu _{X|c}^2}{2 \sigma ^2_{X}} } \right) }{\sum _{k=1}^{K}\exp \left( \displaystyle {-\frac{x_{ij}^2}{2 \sigma ^2_{X}} } \right) \exp \left( \displaystyle {\frac{2\mu _{X|k}x_{ij}-\mu _{X|k}^2}{2 \sigma ^2_{X}} } \right) } \nonumber \\&\qquad = \frac{ \exp \left( \displaystyle {\frac{2\mu _{X|c}x_{ij}-\mu _{X|c}^2}{2 \sigma ^2_{X}} } \right) }{\sum _{k=1}^{K} \exp \left( \displaystyle {\frac{2\mu _{X|k}x_{ij}-\mu _{X|k}^2}{2 \sigma ^2_{X}} } \right) } = \frac{ \exp \left( \displaystyle {\frac{\mu _{X|c}}{\sigma ^2_{X}} x_{ij}- \frac{\mu _{X|c}^2}{2 \sigma ^2_{X}} } \right) }{\sum _{k=1}^{K} \exp \left( \displaystyle {\frac{\mu _{X|k}}{\sigma ^2_{X}} x_{ij}- \frac{\mu _{X|k}^2}{2 \sigma ^2_{X}} } \right) }, \end{aligned}$$

which assumes the form (4) when $b_c=\mu _{X|c}/\sigma ^2_{X}$ and $a_c=\mu _{X|c}^2 / 2 \sigma ^2_X$. This completes the proof.