Estimasi kemungkinan maksimum EM untuk distribusi Weibull

24

Catatan: Saya memposting pertanyaan dari mantan mahasiswa saya yang tidak dapat memposting sendiri karena alasan teknis.

Diberikan sampel iid $x_1,\ldots,x_n$ dari distribusi Weibull dengan pdf

f_{k} (x) = k x^{k - 1} e^{- x^{k}} x > 0

$f_k(x) = k x^{k-1} e^{-x^k} \quad x>0$ apakah ada variabel yang hilang representasi berguna

f_{k} (x) = \int_{Z} g_{k} (x, z) d z

$f_k(x) = \int_\mathcal{Z} g_k(x,z)\,\text{d}z$ dan karenanya sebuah terkait EM (harapan-maksimisasi) algoritma yang dapat digunakan untuk mencari MLE dari

k

$k$ , daripada menggunakan optimasi numerik langsung?

— Xi'an
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2

Apakah ada sensor?

— ocram

2

Apa yang salah dengan newton rhapson?

— probabilityislogic

2

@probabilityislogic: tidak ada yang salah dengan apa pun! Murid saya ingin tahu apakah ada versi EM, itu saja ...

— Xi'an

1

Bisakah Anda memberikan contoh apa yang Anda cari dalam konteks yang berbeda dan lebih sederhana, misalnya dengan pengamatan terhadap variabel acak Gaussian atau seragam? Ketika semua data diamati, saya (dan beberapa poster lainnya, berdasarkan komentar mereka) tidak melihat bagaimana EM relevan dengan pertanyaan Anda.

— ahfoss

1

@probabilityislogic Saya pikir Anda seharusnya mengatakan, "Oh, maksud Anda Anda ingin MENGGUNAKAN Newton Raphson?". Weibulls adalah keluarga biasa ... Saya pikir, jadi solusi ML adalah unik. Karenanya, EM tidak memiliki apa pun untuk "E" selesai, jadi Anda hanya "M" ing ... dan menemukan akar persamaan skor adalah cara terbaik untuk melakukannya!

— AdamO

7

Saya pikir jawabannya adalah ya, jika saya telah memahami pertanyaan dengan benar.

Tulis . Kemudian EM jenis algoritma iterasi, dimulai dengan misalnya , adalah $z_i = x_i^k$ $\hat k = 1$

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

Ini adalah kasus khusus (kasus tanpa sensor dan tanpa kovariat) dari iterasi yang disarankan untuk model bahaya proporsional Weibull oleh Aitkin dan Clayton (1980). Itu juga dapat ditemukan di Bagian 6.11 dari Aitkin et al (1989).

Aitkin, M. dan Clayton, D., 1980. Pemasangan distribusi nilai eksponensial, Weibull dan ekstrim untuk data survival yang disensor kompleks menggunakan GLIM. Statistik Terapan , hal.156-163.
Aitkin, M., Anderson, D., Francis, B. dan Hinde, J., 1989. Pemodelan Statistik dalam GLIM . Oxford University Press. New York.

— DavidF
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Terima kasih banyak, David! Memperlakukan

sebagai varian yang hilang tidak pernah terlintas di benak saya ...!

x_{i}^{k}

$x_i^k$

— Xi'an

7

The Weibull MLE hanya numerik dipecahkan:

Biarkan dengan

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}

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

log-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-Masalah : 3) Maksimalisasidengangradien:

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

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

0

$0$

\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
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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

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.

— user3204720
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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
sumber

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