# Understanding survival at time function

This question is related to a few others (Here,Here) on the topic as I have been searching for information. Hopefully this one is sufficient.

1) I am seeing differences in the relationship between survival at time t, S(t) and the hazard at time t, h(t).

In Singer and Willet page 337 (10.5) they define the estimated survival $\hat{S}(t_{j})$ = $[1-\hat{h}(t_{j})][1-\hat{h}(t_{j-1})][1-\hat{h}(t_{j-2})]....[1-\hat{h}(t_{1})]$. They also define a life table as such:

In other sources Will Potts and others, they define $\hat{S}(t)$=$\prod_{t_{j}< t}(1-h(t_{j}))$ implying that the value of the hazard at time t is not included in the calculation for Survival at time t.

Which is correct? Is the difference in how the life table is set up?

2) How would you calculate the mean time time (restricted since there is censoring at the end of the collection period). Would you sum all $\hat{S}(t)$ (including the 1 at time zero) to arrive at 7.7? You always her it is the "area under the survival curve".

3) How would you calculate the residual mean lifetime for those that survived to the end of, say year 4? Is it summing all $\hat{S}(t)$ from 0.6209 through 0.41123 divided by 0.6209

Using the fact that the formula for the mean residual survival time mrl(u) = E(T -u | T>u) is given by $\frac{1}{S(u)}\int_u^{\infty} S(t)\, dt$ and our "u" is 4 here (since we are saying T>4 in the conditional expectation)?

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 The numbers and presentation are slightly suspicious: nobody seems to have left in the first year at all, and in educational terms, many teachers leave exactly at the end of an academic year, so the number employed at the end of one year is usually not equal to the number employed at the beginning of the next. – Henry Dec 11 '11 at 18:36

Let us take your example of those who lasted at least four years, and then look at the residual mean lifetime.

$295$ of those left in the time interval $[4,5)$, which means that these lasted strictly less than five years, i.e. they did not last a full further year.

By including $0.6209$ in the sum in the numerator of your calculation, you are treating them as all lasting the full fifth year, when in fact they all left earlier in that year. Similarly with all the other years. So your method is biased upwards.

If you wanted to adjust for this bias and you thought that teachers left on average a quarter of the way through the year, you could take your method and subtract $0.75$ from the final result, or you could leave out the initial term in the sum in the numerator of your calculation and add $0.25$ to the final result. Both methods come to the same answer.

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I think I am finally seeing the point here. But, what is the intuition for dividing by S(u) in the MRL? That one escapes me so far. – B_Miner Dec 11 '11 at 23:49
If you were working out the mean residual lifetime for the 2742 teachers still there at $t=4$, you might add up their remaining lifetimes and divide by 2742. The survival function $S(t)$ is essentially the number of teachers remaining at time $t$, divided by the 3941 teachers at the start. So using the survivor function, you need to adjust the results by dividing by a factor of $\frac{2742}{3941} \approx 0.6958$ which is $S(4)$ though shown at the end of the row that starts "3 [3,4)" – Henry Dec 12 '11 at 0:01

I give some elements for 1 and 2 and I let the residual survival time for you ;-) You can get more details by googling "grouped survival data" or something like that... (Take care of distinguishing between "grouped" and "clustered"!)

1

Let $h(t_j)$ be the probability that a teacher employed at the beginning of the year $t_{j - 1}$ left during the year:

$h(t_j) = \Pr[T < t_j \mid T \geq t_{j - 1}]; \hspace{0.75cm} j = 1, 2, \ldots$

As for the survival function at $t_j$, we have

$S(t_j) = \Pr(T \geq t_1) \Pr(T \geq t_2 \mid T \geq t_1) \ldots \Pr(T \geq t_j \mid T \geq t_{j - 1})$ $\phantom{S(t_j) } = (1 - h(t_1)) (1 - h(t_2)) \ldots (1 - h(t_j))$

For example

$\hat{S}(t_3) = (1 - \hat{h}(t_1)) (1 - \hat{h}(t_2)) (1 - \hat{h}(t_3))$ $\phantom{S(t_3)} = (1 - 0.1157) (1 - 0.1102) (1 - 0.1158) = 0.6957$

Now, $S(.)$ is constant between two $t_j$'s and therefore

$S(t) = \prod_{t_j \leq t} (1 - h(t_j))$

which should be your second formula.

For example, $\hat{S}(t) = 0.6957$ for all $t \in [t_3, t_4)$.

2

$\hat{S}(t_6)=0.5189$ and $\hat{S}(t_7)=0.4877$. Therefore, the median survival time is $t_7$, i.e., the year at which $50 \%$ of the teachers have already left.

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 Thank you. I found the same thing for #1 - I believe that Will Potts paper had a typo and that the calculation of the survivorship function is as shown in Singer and Willett. For #2) I was asking about the mean, not the median. I later found explicit calculation of this in Hosmer and Lemeshow where they did exactly as I show - i.e. including the "1" of S(0). This is at odds with answers I received in the linked questions. Which leads me to believe that my #3 question above (MRL) is as I show here and not in the linked answer..... – B_Miner Dec 11 '11 at 17:19

I believe you are correct in questioning the Potts' citation (which I took the liberty of using as my template for editing the expression in your question). Looking at both Therneau and Gramsch "Modeling Survival Data" and Kalbfleisch and Prentice "Statistical Analysis of Failure Time Data", the definition of S(t) is Pr(T>t) and all of their interval survival estimates are closed on the right ... so it should be $\hat{S}(t)$=$\prod_{t_{j} \leq t}(1-\hat{h}(t_{j}))$. Any other interpretation would either leave vacant all of the interval between time_0 and the first event or would assign it a lower probability than unity. Either choice seems counter-factual.

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