# 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

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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