Estimación de máxima verosimilitud EM para la distribución de Weibull

Nota: estoy publicando una pregunta de un ex alumno mío que no puede publicar por su cuenta por razones técnicas.

Dada una muestra de iid de una distribución de Weibull con pdf ¿hay una representación variable faltante útil y, por lo tanto, un algoritmo EM (expectativa-maximización) asociado que podría usarse para encontrar el MLE de , en lugar de usarlo directamente optimización numérica?$x_1,\ldots,x_n$

$$f_k(x) = k x^{k-1} e^{-x^k} \quad x>0$$ $$f_k(x) = \int_\mathcal{Z} g_k(x,z)\,\text{d}z$$ $k$

Creo que la respuesta es sí, si he entendido la pregunta correctamente.

Escribe . A continuación, un tipo de algoritmo de EM de la iteración, a partir de, por ejemplo, k = 1 , es$z_i = x_i^k$$\hat k = 1$

• E ${\hat z}_i = x_i^{\hat k}$

• M $\hat k = \frac{n}{\left[\sum({\hat z}_i - 1)\log x_i\right]}$

Este es un caso especial (el caso sin censura ni covariables) de la iteración sugerida para los modelos de riesgos proporcionales de Weibull por Aitkin y Clayton (1980). También se puede encontrar en la Sección 6.11 de Aitkin et al (1989).

• Aitkin, M. y Clayton, D., 1980. El ajuste de distribuciones exponenciales, de Weibull y de valores extremos a datos de supervivencia censurados complejos utilizando GLIM. Estadística Aplicada , pp.156-163.

• Aitkin, M., Anderson, D., Francis, B. e Hinde, J., 1989. Modelización estadística en GLIM . Prensa de la Universidad de Oxford. Nueva York.

El Weibull MLE solo tiene solución numérica:

Deje conβ

$$f_{\lambda,\beta}(x) = \begin{cases} \frac{\beta}{\lambda}\left(\frac{x}{\lambda}\right)^{\beta-1}e^{-\left(\frac{x}{\lambda}\right)^{\beta}} & ,\,x\geq0 \\ 0 &,\, x<0 \end{cases}$$ .$\beta,\,\lambda>0$

1) Likelihoodfunction :

$$\mathcal{L}_{\hat{x}}(\lambda, \beta) =\prod_{i=1}^N f_{\lambda,\beta}(x_i) =\prod_{i=1}^N \frac{\beta}{\lambda}\left(\frac{x_i}{\lambda}\right)^{\beta-1}e^{-\left(\frac{x_i}{\lambda}\right)^{\beta}} = \frac{\beta^N}{\lambda^{N \beta}} e^{-\sum_{i=1}^N\left(\frac{x_i}{\lambda}\right)^{\beta}} \prod_{i=1}^N x_i^{\beta-1}$$

log-Likelihoodfunction :

$$\ell_{\hat{x}}(\lambda, \beta):= \ln \mathcal{L}_{\hat{x}}(\lambda, \beta)=N\ln \beta-N\beta\ln \lambda-\sum_{i=1}^N \left(\frac{x_i}{\lambda}\right)^\beta+(\beta-1)\sum_{i=1}^N \ln x_i$$

2) Problema MLE : 3) Maximizaciónpor0-gradientes: ∂

$$\begin{equation*} \begin{aligned} & & \underset{(\lambda,\beta) \in \mathbb{R}^2}{\text{max}}\,\,\,\,\,\, & \ell_{\hat{x}}(\lambda, \beta) \\ & & \text{s.t.} \,\,\, \lambda>0\\ & & \beta > 0 \end{aligned} \end{equation*}$$ $0$ Sigue: -Nβ $$\begin{align*} \frac{\partial l}{\partial \lambda}&=-N\beta\frac{1}{\lambda}+\beta\sum_{i=1}^N x_i^\beta\frac{1}{\lambda^{\beta+1}}&\stackrel{!}{=} 0\\ \frac{\partial l}{\partial \beta}&=\frac{N}{\beta}-N\ln\lambda-\sum_{i=1}^N \ln\left(\frac{x_i}{\lambda}\right)e^{\beta \ln\left(\frac{x_i}{\lambda}\right)}+\sum_{i=1}^N \ln x_i&\stackrel{!}{=}0 \end{align*}$$ $$\begin{align*} -N\beta\frac{1}{\lambda}+\beta\sum_{i=1}^N x_i^\beta\frac{1}{\lambda^{\beta+1}} &= 0\\\\ -\beta\frac{1}{\lambda}N +\beta\frac{1}{\lambda}\sum_{i=1}^N x_i^\beta\frac{1}{\lambda^{\beta}} &= 0\\\\ -1+\frac{1}{N}\sum_{i=1}^N x_i^\beta\frac{1}{\lambda^{\beta}}&=0\\\\ \frac{1}{N}\sum_{i=1}^N x_i^\beta&=\lambda^\beta \end{align*}$$ $$\Rightarrow\lambda^*=\left(\frac{1}{N}\sum_{i=1}^N x_i^{\beta^*}\right)^\frac{1}{\beta^*}$$

$\lambda^*$ into the second 0-gradient condition:

$$\begin{align*} \Rightarrow \beta^*=\left[\frac{\sum_{i=1}^N x_i^{\beta^*}\ln x_i}{\sum_{i=1}^N x_i^{\beta^*}}-\overline{\ln x}\right]^{-1} \end{align*}$$

This equation is only numerically solvable, e.g. Newton-Raphson algorithm. $\hat{\beta}^*$ can then be placed into $\lambda^*$ to complete the ML estimator for the Weibull distribution.

Though this is an old question, it looks like there is an answer in a paper published here: http://home.iitk.ac.in/~kundu/interval-censoring-REVISED-2.pdf

In this work the analysis of interval-censored data, with Weibull distribution as the underlying lifetime distribution has been considered. It is assumed that censoring mechanism is independent and non-informative. As expected, the maximum likelihood estimators cannot be obtained in closed form. In our simulation experiments it is observed that the Newton-Raphson method may not converge many times. An expectation maximization algorithm has been suggested to compute the maximum likelihood estimators, and it converges almost all the times.

In this case the MLE and EM estimators are equivalent, since the MLE estimator is actually just a special case of the EM estimator. (I am assuming a frequentist framework in my answer; this isn't true for EM in a Bayesian context in which we're talking about MAP's). Since there is no missing data (just an unknown parameter), the E step simply returns the log likelihood, regardless of your choice of $k^{(t)}$. The M step then maximizes the log likelihood, yielding the MLE.

EM would be applicable, for example, if you had observed data from a mixture of two Weibull distributions with parameters $k_1$ and $k_2$, but you didn't know which of these two distributions each observation came from.