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First, the expected survival time when time is time to event is treated as a continuous r.v. is given as the following:

$$\int_0^{\infty} S(t)\, dt$$

The formula for the mean residual survival time $\text{mrl}(u) = E(T -u | T>u)$ is given by

$$\frac{1}{S(u)}\int_u^{\infty} S(t)\, dt.$$

Can anyone explain the logic of dividing by $S(u)$?

For discrete time, this step does not appear needed (cf related question). Why?

Thanks!

ADD after Henry's explanation:

Lets say I have discrete time (12 months with the following hazards and Survival).

month<-seq(0,12)
h_x<-c(0,0.115706673,0.110186514,0.115769107,0.108296623,0.08908868,0.082548228,0.060146699,0.048112058,0.042197452,0.036919831,0.024691358,0.012787724)
S_x<-c(1,0.8843,0.7869,0.6958,0.6204,0.5651,0.5185,0.4873,0.4639,0.4443,0.4279,0.4173,0.4120)

plot(month,S_x,type="b")

1) Is the mean (truncated) lifetime sum(S_x[2:12]) #6.3117?

2) Lets say that we are know that a person survives up to 5th month (start of 5) i.e. they get through 4 complete months and we now want to estimated their mean (truncated) residual life. So, $\gamma$ =7 Is this computed as sum(S_x[5:12])/S_x[5] #6.358317 since 6.358 < $\gamma$+1 ?

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    $\begingroup$ I'm a little unsure of the notation, but I think $\prod_{k=t}^{j}\left(1-h(k|\mathbf{x})\right)$ corresponds to $S(t)/S(u)\ dt$. Notice how the former is automatically normalized to $1$ for $j\lt t$ and, similarly, that $S(t)/S(u)\ dt = dt$ at $u=t$. Therefore the answer to your question must be lurking in the definition of $h$. Could you provide that definition so that this pair of questions can stand on their own? $\endgroup$ – whuber Oct 18 '11 at 21:42
  • $\begingroup$ Not sure I follow @whuber. Can you explain more? $\endgroup$ – B_Miner Oct 18 '11 at 23:03
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$S(t)$ is the probability of surviving to time $t$ or beyond.

So the probability of surviving to time $t$ or beyond given that you have survived to time $u$ (with $u \le t$) is $\frac{S(t)}{S(u)}$, and so this is also the probability of a residual survival time of $t-u$ or beyond after time $u$.

This is implicit in your earlier discrete question, which could have looked like $$\frac{\prod_{k=0}^{j}\left(1-h(k|\mathbf{x})\right)}{\prod_{k=0}^{t-1}\left(1-h(k|\mathbf{x})\right)}$$ but this is more easily written as ${\prod_{k=t}^{j}\left(1-h(k|\mathbf{x})\right)}.$

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  • $\begingroup$ Henry: I posted an addition to my question to help clarify. Many thanks! $\endgroup$ – B_Miner Oct 19 '11 at 1:10
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    $\begingroup$ Your extended question is slightly spoilt by R starting counting at 1 not 0. I would given sum(S_x[2:13]) (6.724) and sum(S_x[6:13])/S_x[5] (6.022). $\endgroup$ – Henry Oct 19 '11 at 1:24
  • $\begingroup$ should the mrl be sum(S_x[6:13])/S_x[6] as well? $\endgroup$ – B_Miner Oct 19 '11 at 1:33
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    $\begingroup$ @B_Miner: To expand briefly on Henry's nice answer, the formulas you provided in the question are completely general in the sense that they hold for all distributions of nonnegative random variables. $\endgroup$ – cardinal Oct 19 '11 at 9:34
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    $\begingroup$ @B_Miner: As events $[T>x+t] ∩ [T>t] = [T>x+t]$ because $[T>x+t] \subset [T>t]$ $\endgroup$ – Henry Oct 19 '11 at 15:18

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