Parameters estimations with right censored data 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.
 A: 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.
A: No, it is not a problem!  For a post with an example see

*

*ML estimate of exponential distribution (with censored data)
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).
