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Restricted standard deviation of survival time

I’m working through a paper on restricted mean survival time and I cannot figure out how this paper jumps from one step to the next.

The restricted mean survival time $mu$ is the mean survival time of $X = min(T, t^*)$ where T is a random variable (time-to-event) and $t^{*}$ is a specified time horizon. Then,

$ mu = E(X) = E[min(T, t^*)] = int_{0}^{t^*} S(t) dt $

where $S(t) = Pr(T > t)$ is the survival function. To find the variance of $mu$,

$E(X^2) = E(T^2|T leq t^{*}) Pr(T leq t^{*}) + {t^{*}}^2 Pr(T > t^{*})$

Then, since $Pr(T leq t^{*}) = 1 – S(t^{*})$,

$E(T^2|T leq t^{*}) Pr(T leq t^{*}) = int_{0}^{t^*} t^2 f(t) dt$

I am able to follow up to this part, but they then go on to state,

$E(T^2|T leq t^{*}) Pr(T leq t^{*}) = {t^{*}}^2 [1 – S(t^{*})] – int_{0}^{t^*} 2t[1-S(t)] dt$

I’m not sure where they got this relationship from, would anyone be able to provide some intuition / a point in the right direction?

Cross Validated Asked by Emma Jean on November 12, 2021

1 Answers

One Answer

They used integration by parts.

If $h$ and $g$ are functions then since $(hg)' = h'g + hg'$, $$ int h'g = [hg] - int hg' $$

Here in the integral $$ int_{0}^{t^*} t^2 f(t) dt $$ they used $h'(t)=f(t)$ (thus $h(t) = F(t)$) and $g(t)=t^2$ (and $g'(t) = 2t$) which gives, begin{align*} int_{0}^{t^*} t^2 f(t) dt &= left[t^2 F(t) right]_0^{t^*} - int_0^{t^*}2tF(t)dt \ &= left[t^2 left (1-S(t) right ) right]_0^{t^*} - int_0^{t^*}2tleft(1-S(t) right)dt end{align*} Since $S(0) = 1$,

$$ int_{0}^{t^*} t^2 f(t) dt = t^{*2}left(1-S(t^*) right) - int_0^{t^*}2tleft(1-S(t) right)dt $$

Answered by periwinkle on November 12, 2021

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