Let me explain my question. Consider, for example, a Weibull mortality curve with covariates included:

$$\mu_{x}=\bigg(\frac{a}{b}\bigg)\bigg(\frac{x}{b}\bigg)^{a-1}\text{exp}(\beta Z)\quad\quad a,b,x>0$$

where $\beta$ is a $(1\times n)$ matrix and $Z$ is an $(n\times 1)$ matrix. We can derive the cumulative hazard function as follows:

$$\begin{align} H_{x}(t)&=\int_{0}^{t}\mu_{x+t}\,dt\\ &=\bigg(\frac{a}{b}\bigg)\text{exp}(\beta Z)\int_{0}^{t}\bigg(\frac{x+t}{b}\bigg)^{a-1}\,dt\\ &=a\,\text{exp}(\beta Z)\bigg[\bigg(\frac{x+t}{b}\bigg)^{a}\bigg]^{t}_{0}\\ &=\frac{a\,\text{exp}(\beta Z)}{b^{a}}\big((x+t)^{a}-x^{a}\big) \end{align}$$

From here the survival curve is:

$$\begin{align} S_{x}(t)&=\text{exp}(-H_{x}(t))\\ &=\frac{\text{exp}\Big(-\frac{a\,\text{exp}(\beta Z)}{b^{a}}(x+t)^{a}\Big)}{\text{exp}\Big(-\frac{a\,\text{exp}(\beta Z)}{b^{a}}x^{a}\Big)}\\ &=\frac{\text{exp}\Big(-H_{0}(x+t)\Big)}{\text{exp}\Big(-H_{0}(x)\Big)}\\ &=\frac{S_{0}(x+t)}{S_{0}(x)} \end{align}$$

So in general is it true that:


i.e. surviving from $0$ to $(x+t)$ is equivalent to surviving from $0$ to $x$ and then surviving $t$ from $x$? It seems intuitive and is probably very obvious but I just wanted to make sure my understanding is sound.


1 Answer 1


You don't need any distribution assumption.

With your notation, $$H_x(t) = \int_0^t \mu_{x+s}ds = \int_x^{t+x}\mu_sds$$ is the cumulative hazard between $x$ and $t+x$ and then $$ S_x(t) = e^{-H_x(t)} $$ is the probability of surviving up to time $t + x$ given that the individual is alive at $x$, i.e. $$ S_x(t) = P(T>x+t|T>x) $$ and by simple probability theory, this is $$ P(T>x+t|T>x) = \frac{P(T>x+t)}{P(T>x)}. $$

I think it's more clear to see why this is true if you keep in mind that, with your notation, $S_x(t)$ refers to a conditional probability (which I made explicit here).


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