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As I was doing some research about survival analysis, I found the following article stating that, if we have right censored data, we can estimate the parameters of a distribution with the following modified MLE :

$L(\theta, X) = \prod_{i=1}^{n}f(x_i;\theta)^{\delta_i} [1-F(x_i;\theta)]^{1-\delta_i}$

$\delta_i = \begin{cases} 1 \quad \text{if $x_i$ is censored} \\ 0 \quad \text{if $x_i$ is not censored} \end{cases}$

My question would be the following : If, in some cases, we only have access to right censored data, would it still possible to have a estimation of the parameters using :

$L(\theta,X) = \prod_{i=1}^{n}(1-F(x_i; \theta))$

If you have additional content on the topic, I would be very happy to have a look on it.

kjetil b halvorsen
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Adem A.
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    No, it is not a problem! For an example see https://stats.stackexchange.com/questions/133347/ml-estimate-of-exponential-distribution-with-censored-data/133360#133360 – kjetil b halvorsen May 31 '22 at 12:54
  • @kjetilbhalvorsen Thank you very much ! I saw that you've written the post, do you have any resources (books, articles) I can refer to? – Adem A. May 31 '22 at 13:06
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    Beware: if $\theta$ has two or more components and all data are censored at the same value, you usually cannot estimate $\theta.$ In general, when all data are censored then it is difficult to estimate $\theta$ with much precision unless you have a lot of data and a wide range of censoring thresholds. – whuber May 31 '22 at 13:38
  • @whuber It is exactly my problem.. I have to estimate a Weibull distribution with two parameters with only right-censored parameters.. Look like I have to find an other solution. Do you have any recommendation? – Adem A. May 31 '22 at 14:22
  • I am saying that unless you are in one of the situations I described, what you are looking for is mathematically impossible. It can be made statistically possible by adopting stronger assumptions, such as specifying a property of the distribution or assuming a Bayesian prior distribution for the parameters. – whuber May 31 '22 at 14:27
  • what do you mean by right-censored parameters? usually, the data is censored ... Maybe ask a question with your real problem, all details? – kjetil b halvorsen May 31 '22 at 20:45
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    @kjetilbhalvorsen I have access to data of maintenance of equipment which have never broken down over their lifetime. The only information I have is the time between each maintenance/revision (we did not let the machine break down, but changed the components before). This type of data is said to be right-censored as the failure has never occurred. And with all that info I'm trying to estimate the lifetime distribution of the considered equipment. – Adem A. Jun 01 '22 at 05:17
  • This raises interesting questions regarding the censoring process. For instance, was maintenance performed strictly according to a prescribed schedule, or was it perhaps conditioned on observations of imminent failure? "Which have never broken down" implies some equipment has broken down, but you don't have information about it: why not, exactly? Another issue concerns the reference to "components," suggesting an analysis of this equipment as systems rather than individual items. – whuber Jun 01 '22 at 15:07
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    @whuber Indeed, the maintenance is based on prescribed schedule, which is based on experts' judgment. So far, it has worked successfully (i.e., they've never raised any failure). Concerning the last issue, the equipment is indeed composed of several components. But as a first approach, I only consider the equipment as an individual item. – Adem A. Jun 03 '22 at 05:41

2 Answers2

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This is an extended comment on the answer from @kjetil b halvorsen and on other comments.

I would put it more like "it is not a problem" provided that you understand the limitations.

First, as comments from @whuber note, you can't approach this problem reliably with anything beyond a 1-parameter model. If you use the location-scale modeling of the R survreg() function, there could be an infinite set of location-scale combinations equally compatible with having no events in your data. Even if you managed to have a solution that converged numerically the estimates might be unreliable.

Second, you aren't going to get a reliable point estimate even under the assumption of an exponential distribution. The confidence interval in the linked answer is OK if the exponential distribution holds. You might think of that as an extension of the rule of three.

Third, Bayesian survival approaches have limits here. With an exponential distribution, Ibrahim et al. show a closed-form solution for the posterior expected value of the parameter $\lambda$, with the conjugate gamma prior:

$$ E(\lambda |D) = \frac{\alpha_0 + d}{\lambda_0 + \sum_{i=1}^n y_i}.$$

Here, $D$ represents the data, $\alpha_0$ and $\lambda_0$ are the parameter-value choices for the gamma prior, $d$ is the number of events/deaths in the data, and $y_i$ is the event/censoring time for observation $i$ out of $n$ total observations. If $d = 0$ then the numerator is completely determined by the prior choice of $\alpha_0$.

In your situation the best working solution is to assume the exponential distribution (Weibull with scale parameter in location-scale form fixed at 1) and estimate the confidence interval as @kjetil b halvorsen suggests. I recommend playing with simple simulated data sets to see the dangers of trying to go farther.

EdM
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No, it is not a problem! For a post with an example see

Note that for the exponential distribution example there, when all observations are censored, it is reduced to a binomial likelihood where the probability parameter is a function of the rate $\lambda$.

For a good reference treating this, see In All Likelihood: Statistical Modelling and Inference Using Likelihood by Yudi Pawitan (Chapter 11).

kjetil b halvorsen
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