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I have seen several ways to write (and calculate and interpret) a survivor function in discrete time survival analysis and I wonder which is correct or if they both are, but the interpretation and/or setup of the problem is different and I am missing it.

Here is an example.

  • Customers open an account at a bank.
  • Record the month of opening and month of close and the discrete random variable T is in the set {0,1..,23}. After 23, there is right censoring. T denotes the number of month boundaries crossed between open and close. For example:
  • A customer that opens in January 2012 and closes in January 2012 is labeled as 0.
  • A customer that opens in January 2012 and closes in Feb 2012 is labeled as 1

An interval is of the form [a,a+1) where a=0,..,A.

The following hazards, h(t), are calculated.

For example, a customer opens and closes in the same month with probability 0.019194.

Question: What is the proper way to construct the survivor function?

  1. I am seeing that some sources set S(t) as the probability that the event will occur AFTER the period t: S(t) = Pr(T>t). This is the yellow column below. It is calculated as S(t) = $\prod_{t=0}^{t}(1-h(t))$
  2. Others will set the first period S(0) = 1 and then continue. This seems to be saying S(t) = Pr(T>=t). It is calculated as S(t) = $\prod_{t=0}^{t}(1-h(t-1))$

In the continuous case, I guess it doesn't matter between Pr(T>t) and Pr(T>=t), but in the discrete case it does.

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  • $\begingroup$ This is how actuaries calculate their p's and q's. $\endgroup$ – DWin May 29 '13 at 3:36
  • $\begingroup$ "This" meaning method 1 or 2? $\endgroup$ – B_Miner May 29 '13 at 13:47
  • $\begingroup$ I didn't see that there was any substantive difference, since one is just a shifted version of the other: en.wikipedia.org/wiki/Life_table Generally S(0) is by definition 1, but if you labeled the yellow column P(t) you would get no complaints. $\endgroup$ – DWin May 29 '13 at 14:16
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The answer is that both are used, unfortunately. In the continuous case, you are right the distinction is unimportant. In the discrete case, the interpretation would be slightly different and therefore clarity is important.

In my experience, the most common definition of the survival function is $S(t) = Pr(T>t)$ and so would match your yellow column. This is the one used in the derivation of the Kaplan-Meier estimator: $\hat{S}(t) = \frac{\text{individuals with } T>t}{\text{total individuals}} = \prod_{j=1}^k{(1-\frac{d_j}{r_j})} $ where $d_j$ is the number of events in interval $j$, $r_j$ is the number of individuals at risk in interval $j$, and $\frac{d_j}{r_j} = h(t)$

An important note is that the survival function should start with $S(0) = 1$ if 0 is the first time point, in the absence of left censoring (i.e. assuming no one starts follow-up already having had the event). In the case of your example, I presume that someone who opens and closes a bank account in January 2012 opens the account before they close it; so if time intervals were shortened (for example using weeks or days as the time scale) then S(0) would equal 1 in both cases.

How much the distinction between the two definitions matters may depend on the specific application. The degree of divergence between the two calculations will likely depend on the length of follow-up, the frequency of the event, and the number of ties and how these are considered.

In addition, in many applications we are interested in comparing hazards or survival between two groups rather than the absolute survival or hazard in a specific group. In this case, I think the distinction should be even less important, but I would have to check into that to be sure.

For more detail on survival analysis where $S(t)$ is clearly defined as $Pr(T>t)$, see: Allison: Survival Analysis Using the SAS System

For more detail, with $S(t) = Pr(T>=t)$, see: Collett: Modelling Survival Data in Medical Research, 2nd ed. (note, most of the analytic details will be the same as in Allison, but the interpretations may differ)

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  • $\begingroup$ Thank you. I am finding the same thing (that both definitions are used) and that was confusing. I had found examples of both in different works on the same subject by the same author in fact. $\endgroup$ – B_Miner May 29 '13 at 13:07
  • $\begingroup$ What do you mean by "An important note is that the survival function should start with S(0)=1 if 0 is the first time point"? Are you saying S(0) =1 is correct if your time interval is granular enough (days versus months)? You still agree the most common definition is the yellow column for my application (monthly data), correct? $\endgroup$ – B_Miner May 29 '13 at 13:08
  • $\begingroup$ Yes, the yellow column is the most common definition, and is perfectly correct for monthly data. I just mean that in most data sets, everyone starts out without the event and then some proportion of people have the event over follow-up. So, if the time intervals are granular enough, then there will exist some scale on which S(t=0)=1. For example, perhaps if you were using days then S(t=Jan 1,2012) = 1. The exception to this rule is the case with left-censoring, where some individuals start out already having had the event. In this case, special methods are needed to ensure unbiased estimates. $\endgroup$ – Ellie May 29 '13 at 16:47
  • $\begingroup$ Cool. I have a text that deals with left truncation in a discrete time hazard model, simply by setting the customer's time (or whatever is the unit being measure) t= their tenure at the truncation date. All previous dates are excluded. Then, they only contribute to the likelihood (assuming fitting with a logistic regression) where observed. $\endgroup$ – B_Miner May 29 '13 at 18:24
  • $\begingroup$ Yeah, you can deal with left censoring by using interval censoring methods, where you explicitly give the time of left (and/or right) censoring and individuals only contribute to the likelihood when observed as you say. However, I think that option is not available in all software packages. I know SAS has the option (proc lifereg), though. There are a variety of non-parametric approaches as well, but that probably deserves its own question. $\endgroup$ – Ellie May 29 '13 at 18:52

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