Probit Model

Modelo Probito (Questão em português)

Mostrar a equivalência dos seguintes modelos probitos e obter as distribuições condicionais completas para \(\beta\) e \(u_{i}\) assumindo uma priori normal para \(\beta\).

Solução

Dado o MLG Probito para variáveis binárias, \[\begin{equation*} \begin{array}{rclll} Y_{i} & \sim & \textrm{Ber}(\Phi(\textbf{X}_{i}\boldsymbol{\beta})) \\ \end{array} \end{equation*}\]

Em que para tornar o MCMC para este modelo mais eficiente, temos o modelo ‘aumentado’ com a inserção de uma variável auxiliar \(u\), dado por: \[\begin{equation*} \begin{array}{rclll} u_{i} & \sim & \mathcal{N}(\textbf{X}_{i}\boldsymbol{\beta}, 1) \\ Y_{i} & = & \begin{cases} 1, & \text{se } u_{i} > 0 \\ 0, & \text{se } u_{i} \leq 0. \end{cases} \end{array} \end{equation*}\]

Iremos demonstrar a equivalência dos modelos com a variável auxiliar \(u_{i}\) e do modelo sem a variável auxiliar \(u_{i}\). Observe que, a probabilidade marginal do modelo com a variável auxiliar, é dada por: \[\begin{equation*} \begin{array}{rclll} \mathbb{P}(Y_{i} = 1) & = & \mathbb{E}(Y_{i}) \; = \; \mathbb{E}_{u_{i}}[\mathbb{E}(Y_{i}\mid u_{i})] \; = \; \mathbb{E}_{u_{i}}[\mathbb{1}(u_{i} > 0)] \\ & = & \mathbb{P}(u_{i} > 0) \\ & = & 1 - \mathbb{P}(u_{i} \leq 0) \\ & = & 1 - \mathbb{P}\left( Z_{i} \leq \dfrac{0 - \textbf{X}_{i}\boldsymbol{\beta}}{1} \right)\\ & = & 1 - \Phi(- \textbf{X}_{i}\boldsymbol{\beta}) \\ & = & 1 - \left[ 1 - \Phi(\textbf{X}_{i}\boldsymbol{\beta})\right] \\ & = & \Phi(\textbf{X}_{i}\boldsymbol{\beta}) \\ \end{array} \end{equation*}\]

O que é equivalente ao modelo probito na estrutura padrão. Portanto, fica demonstrado que os dois modelos são equivalentes.

Na segunda parte desta questão, a fim de encontrar as condicionais completas referentes os parâmetros desconhecido, precisamos completar o modelo. Assim, iremos atribuir uma priori normal para os \(\beta\)’s, ou seja, \[\begin{equation*} \begin{array}{rclll} \pi(\boldsymbol{\beta}) & \sim & \mathcal{N}_{p}(\textbf{b}, \textbf{V} ) \\ \end{array} \end{equation*}\]

Portanto, o modelo ‘aumentado’ terá a seguinte estrutura hierárquica: \[\begin{equation*} \begin{array}{lclll} Y_{i} & = & \begin{cases} 1, & \text{se } u_{i} > 0 \\ 0, & \text{se } u_{i} \leq 0. \\ \end{cases} \\ u_{i}\mid \boldsymbol{\beta} & \sim & \mathcal{N}(\textbf{X}_{i}\boldsymbol{\beta}, 1) \\ \boldsymbol{\beta} & \sim & \mathcal{N}_{p}(\textbf{b}, \textbf{V} ) \end{array} \end{equation*}\]

Assim, podemos encontrar a distribuição conjunta a posteriori para os parâmetros, a qual é dada por: \[\begin{equation*} \begin{array}{rclll} \pi(\boldsymbol{\beta}, \textbf{u}\mid\textbf{Y}) & \propto & \pi(\boldsymbol{\beta}, \textbf{u},\textbf{Y})\\ & \propto & \pi(\textbf{Y}\mid\boldsymbol{\beta}, \textbf{u})\pi( \textbf{u}\mid\boldsymbol{\beta})\pi(\boldsymbol{\beta})\\ & \propto & \pi(\textbf{Y}\mid\textbf{u})\pi( \textbf{u}\mid\boldsymbol{\beta})\pi(\boldsymbol{\beta})\\ \end{array} \end{equation*}\]

Começando pela análise do produto das distribuições: \(\pi(\textbf{Y}\mid\textbf{u})\pi( \textbf{u}\mid\boldsymbol{\beta})\), temos: \[\begin{equation*} \begin{array}{rclll} \pi(\textbf{Y}\mid\textbf{u})\pi( \textbf{u}\mid\boldsymbol{\beta}) & = & \displaystyle\prod_{i=1}^{n}[\pi(Y_{i}\mid u_{i})\pi(u_{i}\mid\boldsymbol{\beta})]\\[12pt] & = & \displaystyle\prod_{i=1}^{n} [ \mathbb{1}(Y_{i} = 1)\mathbb{1}(u_{i} > 0) + \mathbb{1}(Y_{i} = 0)\mathbb{1}(u_{i} \leq 0) ] \times \\[10pt] & & \times \dfrac{1}{\sqrt{2\pi}}\exp{\left\{-\dfrac{1}{2}(u_{i} - \textbf{X}_{i}\boldsymbol{\beta})^2\right\}}\\[10pt] & = & \displaystyle\prod_{i=1}^{n} \left[ \dfrac{1}{\sqrt{2\pi}}\exp{\left\{-\dfrac{1}{2}(u_{i} - \textbf{X}_{i}\boldsymbol{\beta})^2\right\}}\mathbb{1}(Y_{i} = 1)\mathbb{1}(u_{i} > 0) \right. + \\[10pt] & & + \left. \dfrac{1}{\sqrt{2\pi}}\exp{\left\{-\dfrac{1}{2}(u_{i} - \textbf{X}_{i}\boldsymbol{\beta})^2\right\}}\mathbb{1}(Y_{i} = 0)\mathbb{1}(u_{i} \leq 0) \right] \\ \end{array} \end{equation*}\]

Logo, a distribuição conjunta a posteriori para os parâmetros, é dada por: \[\begin{equation*} \begin{array}{rclll} \pi(\boldsymbol{\beta}, \textbf{u}\mid\textbf{Y}) & \propto & \pi(\textbf{Y}\mid\textbf{u})\pi( \textbf{u}\mid\boldsymbol{\beta})\pi(\boldsymbol{\beta})\\[10pt] & \propto & \displaystyle\prod_{i=1}^{n} \left[ \dfrac{1}{\sqrt{2\pi}}\exp{\left\{-\dfrac{1}{2}(u_{i} - \textbf{X}_{i}\boldsymbol{\beta})^2\right\}}\mathbb{1}(Y_{i} = 1)\mathbb{1}(u_{i} > 0) \right. + \\[12pt] & & + \left. \dfrac{1}{\sqrt{2\pi}}\exp{\left\{-\dfrac{1}{2}(u_{i} - \textbf{X}_{i}\boldsymbol{\beta})^2\right\}}\mathbb{1}(Y_{i} = 0)\mathbb{1}(u_{i} \leq 0) \right] \times\\[12pt] & & \times \exp{\left\{-\dfrac{1}{2}(\boldsymbol{\beta} - \textbf{b})^{\top}\textbf{V}^{-1}(\boldsymbol{\beta} - \textbf{b})\right\}} \end{array} \end{equation*}\]

Agora, podemos encontrar as distribuições condicionais completas para \(\boldsymbol{\beta}\) e \(u_{i}\), para \(i = \{1, \cdots, n\}\), a partir da distribuição conjunta a posteriori. Tais distribuições são dadas por:

Para \(u_{i}\): \[\begin{equation*} \begin{array}{rclll} \pi(u_{i}\mid\boldsymbol{\beta}, Y_{i},\textbf{X}_{i}) & \propto & [\pi(Y_{i}\mid u_{i})\pi(u_{i}\mid\boldsymbol{\beta})]\pi(\boldsymbol{\beta})\\[8pt] & \propto & \left[ \dfrac{1}{\sqrt{2\pi}}\exp{\left\{-\dfrac{1}{2}(u_{i} - \textbf{X}_{i}\boldsymbol{\beta})^2\right\}}\mathbb{1}(Y_{i} = 1)\mathbb{1}(u_{i} > 0) \right. + \\[8pt] & & + \left. \dfrac{1}{\sqrt{2\pi}}\exp{\left\{-\dfrac{1}{2}(u_{i} - \textbf{X}_{i}\boldsymbol{\beta})^2\right\}}\mathbb{1}(Y_{i} = 0)\mathbb{1}(u_{i} \leq 0) \right] \\[8pt] & \propto & \begin{cases} \mathcal{N}(\textbf{X}_{i}\boldsymbol{\beta}, 1)\mathbb{1}(u_{i} > 0) , & \text{se } Y_{i} = 1 \\ \mathcal{N}(\textbf{X}_{i}\boldsymbol{\beta}, 1)\mathbb{1}(u_{i} \leq 0), & \text{se } Y_{i} = 0. \\ \end{cases} \\[5pt] \end{array} \end{equation*}\]

Para \(\boldsymbol{\beta}\): \[\begin{equation*} \begin{array}{rclll} \pi(\boldsymbol{\beta}\mid\textbf{u}, \textbf{Y},\textbf{X}) & \propto & \pi(\textbf{Y}\mid\textbf{u})\pi( \textbf{u}\mid\boldsymbol{\beta})\pi(\boldsymbol{\beta})\\[5pt] & \propto & \exp{\left\{-\dfrac{1}{2}(\textbf{u} - \textbf{X}\boldsymbol{\beta})^{\top}(\textbf{u} - \textbf{X}\boldsymbol{\beta})\right\}} \cdot \exp{\left\{-\dfrac{1}{2}(\boldsymbol{\beta} - \textbf{b})^{\top}\textbf{V}^{-1}(\boldsymbol{\beta} - \textbf{b})\right\}} \\[8pt] & \propto & \exp{\left\{-\dfrac{1}{2}\left(\textbf{u}^{\top}\textbf{u} - \textbf{u}^{\top}\textbf{X}\boldsymbol{\beta} - \boldsymbol{\beta}^{\top}\textbf{X}^{\top}\textbf{u} + \boldsymbol{\beta}^{\top}\textbf{X}^{\top}\textbf{X}\boldsymbol{\beta} + \right. \right.}\\[8pt] & & \left. \left. + \boldsymbol{\beta}^{\top}\textbf{V}^{-1}\boldsymbol{\beta} - \boldsymbol{\beta}^{\top}\textbf{V}^{-1}\textbf{b} - \textbf{b}^{\top}\textbf{V}^{-1}\boldsymbol{\beta} + \textbf{b}^{\top}\textbf{V}^{-1}\textbf{b}\right) \right\} \\[8pt] \end{array} \end{equation*}\]

\[\begin{equation*} \begin{array}{rclll} & \propto & \exp{\left\{-\dfrac{1}{2}\left(\boldsymbol{\beta}^{\top}\textbf{X}^{\top}\textbf{X}\boldsymbol{\beta} + \boldsymbol{\beta}^{\top}\textbf{V}^{-1}\boldsymbol{\beta} - \boldsymbol{\beta}^{\top}\textbf{X}^{\top}\textbf{u} - \boldsymbol{\beta}^{\top}\textbf{V}^{-1}\textbf{b}\right)\right\}}\\[8pt] & \propto & \exp{\left\{-\dfrac{1}{2}\left[\boldsymbol{\beta}^{\top} \left( \textbf{X}^{\top}\textbf{X} + \textbf{V}^{-1} \right) \boldsymbol{\beta} - \boldsymbol{\beta}^{\top} \left( \textbf{X}^{\top}\textbf{u} + \textbf{V}^{-1}\textbf{b} \right)\right]\right.}\\[9pt] & & \left. \left. - (\textbf{u}^{\top}\textbf{X} + \textbf{b}^{\top}\textbf{V}^{-1})\boldsymbol{\beta} \right] \right\} \\[8pt] & \propto & \mathcal{N}_{p}\left[\left( \textbf{X}^{\top}\textbf{X} + \textbf{V}^{-1} \right)^{-1}\left( \textbf{X}^{\top}\textbf{u} + \textbf{V}^{-1}\textbf{b}\right), \left( \textbf{X}^{\top}\textbf{X} + \textbf{V}^{-1} \right)^{-1} \right]\\[9pt] \end{array} \end{equation*}\]

Portanto, as distribuições condicionais completas para \(\beta\) e \(u_{i}\), para \(i = \{1, \cdots, n\}\) são dadas por: \[\begin{equation*} \begin{array}{rclll} \pi(u_{i}\mid\boldsymbol{\beta}, Y_{i},\textbf{X}_{i}) & \sim & \begin{cases} \mathcal{N}(\textbf{X}_{i}\boldsymbol{\beta}, 1)\mathbb{1}(u_{i} > 0) , & \text{se } Y_{i} = 1 \\ \mathcal{N}(\textbf{X}_{i}\boldsymbol{\beta}, 1)\mathbb{1}(u_{i} \leq 0), & \text{se } Y_{i} = 0. \\ \end{cases} \\[16pt] \pi(\boldsymbol{\beta}\mid\textbf{u}, \textbf{Y},\textbf{X}) & \sim & \mathcal{N}_{p}\left[\left( \textbf{X}^{\top}\textbf{X} + \textbf{V}^{-1} \right)^{-1}\left( \textbf{X}^{\top}\textbf{u} + \textbf{V}^{-1}\textbf{b}\right), \left( \textbf{X}^{\top}\textbf{X} + \textbf{V}^{-1} \right)^{-1} \right] \end{array} \end{equation*}\]