Оценка максимального правдоподобия EM для распределения Вейбулла

24

Примечание: я отправляю вопрос от моего бывшего студента, который не может публиковать сообщения самостоятельно по техническим причинам.

Для данного iid образца $x_1,\ldots,x_n$ из распределения Вейбулла pdf

е_{К} (Икс) знак равно К {Икс}^{К - 1} е^{- {Икс}^{К}} Икс > 0

$f_k(x) = k x^{k-1} e^{-x^k} \quad x>0$ есть ли полезное представление отсутствующей переменной

е_{К} (Икс) знак равно \int_{Z} г_{К} (Икс, Z) d Z

$f_k(x) = \int_\mathcal{Z} g_k(x,z)\,\text{d}z$ и, следовательно, связанный с ним алгоритм EM (ожидание-максимизация), который можно использовать для нахождения MLE из

k

$k$ вместо простой численной оптимизации?

— Сиань
источник

2

Есть ли цензура?

— Октябрь

2

Что не так с Ньютоном Рапсоном?

— вероятностная

2

@probabilityislogic: все в порядке! Мой ученик хотел бы знать, есть ли версия EM, вот и все ...

— Сиань

1

Не могли бы вы привести пример того, что вы ищете в другом, более простом контексте, например, возможно, с наблюдениями гауссовой или равномерной случайной величины? Когда все данные соблюдаются, я (и некоторые другие постеры, основываясь на их комментариях) не вижу, как EM относится к вашему вопросу.

— Ахфосс

1

@probabilityislogic Я думаю, вы должны были сказать: «О, вы хотите сказать, что хотите использовать Ньютона Рафсона?». Вейбуллы - это обычные семьи ... Я думаю, поэтому решения ML уникальны. Таким образом, EM не имеет ничего против «E», так что вы просто «M»… и поиск корней уравнений для оценки - лучший способ сделать это!

— AdamO

7

Я думаю, что ответ - да, если я правильно понял вопрос.

Напишите . Тогда ЭМ типа алгоритм итерации, начиная с, например , , является $z_i = x_i^k$ $\hat k = 1$

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

Это особый случай (случай без цензуры и ковариат) итерации, предложенной для моделей пропорционального риска Вейбулла Эйткином и Клейтоном (1980). Это также можно найти в разделе 6.11 Aitkin et al (1989).

Aitkin, M. and Clayton, D., 1980. Подгонка экспоненциального распределения, распределения Вейбулла и экстремальных значений к сложным цензурированным данным о выживании с использованием GLIM. Прикладная статистика , с.156-163.
Айткин М., Андерсон Д., Фрэнсис Б. и Хинде Дж., 1989. Статистическое моделирование в GLIM . Издательство Оксфордского университета. Нью-Йорк.

— DavidF
источник

Большое спасибо, Дэвид! Относиться к

как к отсутствующему варианту никогда не приходило мне в голову ...!

x_{i}^{k}

$x_i^k$

— Сиань

7

Вейбулла MLE только численно разрешима:

Пусть с

f_{λ, β} (x) = {\begin{cases} \frac{β}{λ} {(\frac{x}{λ})}^{β - 1} e^{- {(\frac{x}{λ})}^{β}} & , x \geq 0 \\ 0 & , x < 0 \end{cases}

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

.

β, λ > 0

$\beta,\,\lambda>0$

1) Likelihoodfunction :

L_{\hat{x}} (λ, β) = \prod_{i = 1}^{N} f_{λ, β} (x_{i}) = \prod_{i = 1}^{N} \frac{β}{λ} {(\frac{x_{i}}{λ})}^{β - 1} e^{- {(\frac{x_{i}}{λ})}^{β}} = \frac{β^{N}}{λ^{N β}} e^{- \sum_{i = 1}^{N} {(\frac{x_{i}}{λ})}^{β}} \prod_{i = 1}^{N} x_{i}^{β - 1}

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

лог-Likelihoodfunction :

ℓ_{\hat{x}} (λ, β) := \ln L_{\hat{x}} (λ, β) = N \ln β - N β \ln λ - \sum_{i = 1}^{N} {(\frac{x_{i}}{λ})}^{β} + (β - 1) \sum_{i = 1}^{N} \ln x_{i}

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

2) MLE-проблема :

\begin{aligned} max_{(λ, β) \in R^{2}} & ℓ_{\hat{x}} (λ, β) \\ s.t. λ > 0 \\ β > 0 \end{aligned}

$\begin{equation*} \begin{aligned} & & \underset{(\lambda,\beta) \in \mathbb{R}^2}{\text{max}}\,\,\,\,\,\, & \ell_{\hat{x}}(\lambda, \beta) \\ & & \text{s.t.} \,\,\, \lambda>0\\ & & \beta > 0 \end{aligned} \end{equation*}$ 3) Maximization by

0

$0$ -gradients:

\begin{aligned} \frac{\partial l}{\partial λ} & = - N β \frac{1}{λ} + β \sum_{i = 1}^{N} x_{i}^{β} \frac{1}{λ^{β + 1}} & \overset{!}{=} 0 \\ \frac{\partial l}{\partial β} & = \frac{N}{β} - N \ln λ - \sum_{i = 1}^{N} \ln (\frac{x_{i}}{λ}) e^{β \ln (\frac{x_{i}}{λ})} + \sum_{i = 1}^{N} \ln x_{i} & \overset{!}{=} 0 \end{aligned}

$\begin{align*} \frac{\partial l}{\partial \lambda}&=-N\beta\frac{1}{\lambda}+\beta\sum_{i=1}^N x_i^\beta\frac{1}{\lambda^{\beta+1}}&\stackrel{!}{=} 0\\ \frac{\partial l}{\partial \beta}&=\frac{N}{\beta}-N\ln\lambda-\sum_{i=1}^N \ln\left(\frac{x_i}{\lambda}\right)e^{\beta \ln\left(\frac{x_i}{\lambda}\right)}+\sum_{i=1}^N \ln x_i&\stackrel{!}{=}0 \end{align*}$ It follows:

\begin{aligned} - N β \frac{1}{λ} + β \sum_{i = 1}^{N} x_{i}^{β} \frac{1}{λ^{β + 1}} & = 0 \\ - β \frac{1}{λ} N + β \frac{1}{λ} \sum_{i = 1}^{N} x_{i}^{β} \frac{1}{λ^{β}} & = 0 \\ - 1 + \frac{1}{N} \sum_{i = 1}^{N} x_{i}^{β} \frac{1}{λ^{β}} & = 0 \\ \frac{1}{N} \sum_{i = 1}^{N} x_{i}^{β} & = λ^{β} \end{aligned}

$\begin{align*} -N\beta\frac{1}{\lambda}+\beta\sum_{i=1}^N x_i^\beta\frac{1}{\lambda^{\beta+1}} &= 0\\\\ -\beta\frac{1}{\lambda}N +\beta\frac{1}{\lambda}\sum_{i=1}^N x_i^\beta\frac{1}{\lambda^{\beta}} &= 0\\\\ -1+\frac{1}{N}\sum_{i=1}^N x_i^\beta\frac{1}{\lambda^{\beta}}&=0\\\\ \frac{1}{N}\sum_{i=1}^N x_i^\beta&=\lambda^\beta \end{align*}$

\Rightarrow λ^{*} = {(\frac{1}{N} \sum_{i = 1}^{N} x_{i}^{β^{*}})}^{\frac{1}{β^{*}}}

$\Rightarrow\lambda^*=\left(\frac{1}{N}\sum_{i=1}^N x_i^{\beta^*}\right)^\frac{1}{\beta^*}$

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

\begin{aligned} \Rightarrow β^{*} = {[\frac{\sum_{i = 1}^{N} x_{i}^{β^{*}} \ln x_{i}}{\sum_{i = 1}^{N} x_{i}^{β^{*}}} - \bar{\ln x}]}^{- 1} \end{aligned}

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

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.

— emcor
источник

11

Unfortunately, this does not appear to answer the question in any discernible way. The OP is very clearly aware of Newton-Raphson and related approaches. The feasibility of N-R in no way precludes the existence of a missing-variable representation or associated EM algorithm. In my estimation, the question is not concerned at all with numerical solutions, but rather is probing for insight that might become apparent if an interesting missing-variable approach were demonstrated.

— cardinal

@cardinal It is one thing to say there was only numerical solution, and it is another thing to show there is only numerical solution.

— emcor

5

Dear @emcor, I think you may be misunderstanding what the question is asking. Perhaps reviewing the other answer and associated comment stream would be helpful. Cheers.

— cardinal

@cardinal I agree it is not direct answer, but it is the exact expressions for the MLE's e.g. can be used to verify the EM.

— emcor

4

Хотя это старый вопрос, похоже, что в опубликованной здесь статье есть ответ: 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.

— user3204720
источник

1

Can you post a full citation for the paper at the link, in case it goes dead?

— gung - Reinstate Monica

1

This is an EM algorithm, but does not do what I believe the OP wants. Rather, the E-step imputes the censored data, after which the M-step uses a fixed point algorithm with the complete data set. So the M-step is not in closed form (which I think is what the OP is looking for).

— Cliff AB

1

@CliffAB: thank you for the link (+1) but indeed the EM is naturally induced in this paper by the censoring part. My former student was looking for a plain uncensored iid Weibull likelihood optimisation via EM.

— Xi'an

-1

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.

— ahfoss
источник

6

I think you may have misinterpreted the point of the question, which is: Does there exist some missing-variable interpretation from which one would obtain the given Weibull likelihood (and which would allow an EM-like algorithm to be applied)?

— cardinal

4

The question statement in @Xi'an's post is quite clear. I think the reason it hasn't been answered is because any answer is likely nontrivial. (It's interesting, so I wish I had more time to think about it.) At any rate, your comment appears to betray a misunderstanding of the EM algorithm. Perhaps the following will serve as an antidote:

— cardinal

6

Let

f (x) = π φ (x - μ_{1}) + (1 - π) φ (x - μ_{2})

$f(x) = \pi \varphi(x-\mu_1) + (1-\pi) \varphi(x-\mu_2)$ where

φ

$\varphi$ is the standard normal density function. Let

F (x) = \int_{- \infty}^{x} f (u) d u

$F(x) = \int_{-\infty}^x f(u)\,\mathrm{d}u$ . With

U_{1}, \dots, U_{n}

$U_1,\ldots,U_n$ iid standard uniform, take

X_{i} = F^{- 1} (U_{i})

$X_i = F^{-1}(U_i)$ . Then,

X_{1}, \dots, X_{n}

$X_1,\ldots,X_n$ is a sample from a Gaussian mixture model. We can estimate the parameters by (brute-force) maximum likelihood. Is there any missing data in our data-generation process? No. Does it have a latent-variable representation allowing for the use of an EM algorithm? Yes, absolutely.

— cardinal

4

My apologies @cardinal; I think I have misunderstood two things about your latest post. Yes, in the GMM problem you could search

R^{2} \times [0, 1]

$\mathbb{R}^2 \times [0,1]$ via a brute force ML approach. Also, I now see that the original problem looks for a solution that involves introducing a latent variable that allows for an EM approach to estimating the parameter

k

$k$ in the given density

k x^{k - 1} e^{- x^{k}}

$k x^{k-1}e^{-x^k}$ . An interesting problem. Are there any examples of using EM like this in such a simple context? Most of my exposure to EM has been in the context of mixture problems and data imputation.

— ahfoss

3

@ahfoss: (+1) to your latest comment. Yes! You got it. As for examples: (i) it shows up in censored data problems, (ii) classical applications like hidden Markov models, (iii) simple threshold models like probit models (e.g., imagine observing the latent

Z_{i}

$Z_i$ instead of Bernoulli

X_{i} = 1_{(Z_{i} > μ)}

$X_i = \mathbf{1}_{(Z_i > \mu)}$ ), (iv) estimating variance components in one-way random effects models (and much more complex mixed models), and (v) finding the posterior mode in a Bayesian hierarchical model. The simplest is probably (i) followed by (iii).

— cardinal