I am reading "Monte Carlo Statistical Methods" by Robert and Cassella, and problem 1.3 asks

In example 1.1, the distribution of the random variable $Z=\min(X,Y)$ was of interest. Derive the distribution of $Z$ in the following case of informative censoring, where $Y\sim N(\theta,\sigma^2)$ and $X\sim N(\theta,\theta^2\sigma^2)$. Pay attention to the identifiability issues.

Now I am almost certainly missing something, but my understanding is that informative censoring happens when $X$ and $Y$ are not independent. However, just knowing that they are not independent is not enough information to get the joint distribution, but if they are independent, I do not see any identifiability issues.

Added: If $X$ and $Y$ are independent it is straightforward but tedious to write down the distribution of $Z$, the tedium exacerbated by the fact that the distribution of $X$ is a delta function when $\theta$ is $0$. For a given distribution however, we can find $\theta$ as the (obviously unique) third quartile of the distribution, and given $\theta$, $\sigma^2$ is just a scaling parameter, so there are no identifiability issues that I can see.

So, in summary, my questions are:

  • What precisely is the definition of informative censoring and why is the censoring in this exercise informative?

  • If we are meant to take $X$ and $Y$ as independent, what are the identifiability issues that need to be paid attention to?

Further addition

With Ocram's explanation of informative censoring it is now clear that the identifiability issues that neede to be paid attention to were that there were not any. If the parameters for the failure and censoring distributions were separate then there would be identifiability issues as we could swap the two distributions and get the same result.

If someone more knowledgeable than I is feeling particularly Quixotic, please consider clarifying the Censoring wikipedia page.

  • $\begingroup$ I don't quite follow: given $\theta$ and $\sigma$, you know the distributions of $X$ and $Y$. Assuming independence gives you the joint distribution of $(X,Y)$, whence you have all you need to determine the distribution of any measurable function of $X$ and $Y$, such as $\min$. What is the "more information" you seek, then? $\endgroup$
    – whuber
    Commented Feb 9, 2012 at 18:36
  • $\begingroup$ @whuber Yes, but assuming independence, there are no identifiability issues to deal with. Also, as I understand it independence means that the censoring is non-informative. So the "more information" is : What exactly is "informative censoring" and what are the "identifiability issues" that I need to pay attention to. If the question was: Given $X$ and $Y$ what is the distribution of $\min(X,Y)$ I'd have no problems. Unfortunately, there are terms like "informative censoring" that I think I understand, but my understanding does not coincide with the use in the question. $\endgroup$
    – deinst
    Commented Feb 9, 2012 at 18:54
  • $\begingroup$ @deinst: To my point of view, independence does not mean "non-informative censoring". If you want, I can distinguish between them a little bit later on... $\endgroup$
    – ocram
    Commented Feb 10, 2012 at 13:46
  • $\begingroup$ @Ocram Yes, please do. Understanding what informative censoring is is the gist of my question. In particular, how it relates to the question at hand. I'll try to update the question once I extinguish the morning fires at work. $\endgroup$
    – deinst
    Commented Feb 10, 2012 at 13:59

1 Answer 1


This is an attempt to answer the request you made in the comments.

Independence between $T$ and $C$ vs non-informative censoring

In the following, I assume random right censoring.

Take a sample of i.i.d. survival times $$(y_1, \delta_1), \ldots{}, (y_n, \delta_n),$$ where $y_i = \min(t_i, c_i)$ is the minimum between the survival time and the censoring time, and where $\delta_i = I(t_i \leq c_i)$ is the event indicator. So, using my notation, $T$ is the event time random variable with density $f(\cdot)$ and survival time $S(\cdot)$, while $C$ is the censoring time random variable with density $g(\cdot)$ and survival $G(\cdot)$.

Under independence between $T$ and $C$, the likelihood function's contribution to an event time $(y_i, 1)$ is easily seen to be $$"\Pr[T=y_i, C > y_i]" = G(y_i) f(y_i).$$ Similarly, the likelihood function's contribution to censored data $(y_i, 0)$ is $$"\Pr[C=y_i, T > y_i]" = S(y_i) g(y_i). $$

The likelihood function for the complete data can therefore be written as $$L = \prod_{i=1}^{n} \left[G(y_i) f(y_i)\right]^{\delta_i} \left[S(y_i) g(y_i)\right]^{1- \delta_i}.$$

Now, assume that the distribution of $C$ does not depend on the parameters of the distribution of $T$. Then the factors $G(y_i)^{\delta_i} g(y_i)^{1-\delta_i}$ are non-informative and can be factored out: $$L \propto \prod_{i=1}^{n} f(y_i)^{\delta_i} S(y_i)^{1- \delta_i}.$$

This is the usual likelihood when dealing with survival data. Loosely speaking, independence between $T$ and $C$ allows you to split the joint contribution of $T$ and $C$ into their marginal contributions whereas the non-informative censoring assumption allows you to get rid of $g(\cdot)$ and $G(\cdot)$.

  • $\begingroup$ Ok, so for the problem at hand the "informativeness" of the censoring comes from the fact that censoring time and survival time distributions depend on the same parameters. Thanks. $\endgroup$
    – deinst
    Commented Feb 10, 2012 at 20:23
  • $\begingroup$ So it might be better to say "noninformative on $\theta$". Aren't there different attempts to define noninformative censoring in survival analysis ? $\endgroup$ Commented Feb 11, 2012 at 8:29
  • $\begingroup$ I do not know whether there were any other attempts. Anyway, I think this is what is classicaly meant by non-informative since this is the classical likelihood... $\endgroup$
    – ocram
    Commented Feb 11, 2012 at 10:57
  • $\begingroup$ @ocram could you recommend a text or online source that goes into reasonable depth on these issues? Ideally roughly at the level of your post, but failing that, more detail is better than less. $\endgroup$
    – Paul
    Commented Jun 12, 2017 at 22:12
  • $\begingroup$ It's worth noting that the assumptions in your derivation can be relaxed; we don't need complete independence but only conditional independence on observable covariates. See slide 5 of myweb.uiowa.edu/pbreheny/7210/f15/notes/9-3.pdf $\endgroup$
    – Paul
    Commented Jun 12, 2017 at 23:00

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