EM-Maximum-Likelihood-Schätzung für die Weibull-Verteilung

Hinweis: Ich stelle eine Frage eines ehemaligen Studenten, der aus technischen Gründen nicht in der Lage ist, selbst zu posten.

Ausgehend von einer iid-Stichprobe $x_1,\ldots,x_n$ aus einer Weibull-Verteilung mit pdf

$$f_k(x) = k x^{k-1} e^{-x^k} \quad x>0$$ gibt es eine nützliche fehlende Variablendarstellung und daher einen zugehörigen EM-Algorithmus (Erwartungsmaximierung), der sein könnte verwendet, um die MLE vonzu finden, anstatt eine einfache numerische Optimierung zu verwenden? $$f_k(x) = \int_\mathcal{Z} g_k(x,z)\,\text{d}z$$ $k$

Ich denke die Antwort ist ja, wenn ich die Frage richtig verstanden habe.

Schreibe . Dann wird eine EM - Algorithmus Art von Iteration, beginnend mit beispielsweise k = 1 , ist$z_i = x_i^k$$\hat k = 1$

• E - Schritt: z i = x k i${\hat z}_i = x_i^{\hat k}$

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

Dies ist ein Sonderfall (der Fall ohne Zensur und ohne Kovariaten) der von Aitkin und Clayton (1980) für Weibull-Proportional-Hazards-Modelle vorgeschlagenen Iteration. Es ist auch in Abschnitt 6.11 von Aitkin et al. (1989) zu finden.

• Aitkin, M. und Clayton, D., 1980. Die Anpassung von Exponential-, Weibull- und Extremwertverteilungen an komplexe zensierte Überlebensdaten unter Verwendung von GLIM. Applied Statistics , S. 156-163.

• Aitkin, M., Anderson, D., Francis, B. und Hinde, J., 1989. Statistical Modeling in GLIM . Oxford University Press. New York.

Das Weibull MLE ist nur numerisch lösbar:

Sei mitβ

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

1) Likelihoodfunction :

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

lügt-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) MLE-Problem : 3) Maximierungum0-Gradienten: ∂

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

Plugging $\lambda^*$ 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.